There are quite a few errors in this book. Read with your own risk. Errors are Printing errors, Calculation Errors, Conceptual Errors etc

There are quite a few errors in this book. Read with your own risk. Errors are Printing errors, Calculation Errors, Conceptual Errors etc

PART A

Chapter 0 Basic Mathematics for Physics

Chapter 1 Measurement

chapter 2 Motion in a Straight Line 30

chapter 3 Motion in a Plane $77-106$

chapter 4 Force and Laws of Motion 107.156

chapter 5 Work, Energy and Power

CHAPTER 0

$\int x^{n}dx=\frac{x^{n+1}}{n+1}+c$

$\frac{d{x}^n}{dx}=n{x}^{n-1}$

Basic Mathematics For Physics

If you want to excel in science field, it is very important to have good mathematical aptitude. This chapter is presenting Basic Mathematics that is very useful for developing problem solving strategies whenever required. It includes application in physics section that will develop your scientific temper. Go through it twice or thrice before start learning physics. Take your teacher help wherever you feel any doubt or require more illustrations. “Mathematics is the language of physics”.

TRIGONOMETRIC IDENTITIES

1. $\sin (-\theta )=-\sin \theta$
2. $\cos (\mathrm{\neg }\theta )=\cos \theta$
3. ${\sin }^2\theta +{\cos }^2\theta =1$
4. $\tan (-\theta )=-\tan \theta$
5. $(\sin \theta )/(\cos \theta )=\tan \theta$
6. $\sin (\theta \pm \beta )=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta$; If $\alpha ={90}^{\circ },\sin ({90}^{\circ }\pm \beta )=\cos \beta$; If $\alpha =\beta ,\sin 2\beta =2\sin \beta \cos \beta$
7. $\cos (\alpha \pm \beta )=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta$; If $\alpha ={90}^{\circ },\cos ({90}^{\circ }\pm \beta )=\mp \sin$; If $\alpha =\beta ,\cos 2\beta ={\cos }^2\beta -{\sin }^2\beta =1-2{\sin }^2\beta$.
8. $\sin ({90}^{\circ }-\theta )=\cos \theta ;\cos ({90}^{\circ }-\theta )=\sin \theta ;\tan ({90}^{\circ }-\theta )=\cot \theta$
9. $\sin ({90}^{\circ }+\theta )=\cos \theta ;\cos ({90}^{\circ }+\theta )=-\sin \theta ;\tan ({90}^{\circ }+\theta )=-\cot \theta$
10. $\sin ({180}^{\circ }-\theta )=\sin \theta ;\cos ({180}^{\circ }-\theta )=-\cos \theta ;\tan ({180}^{\circ }-\theta )=-\tan \theta$
11. $\sin ({180}^{\circ }+\theta )=-\sin \theta ;\cos ({180}^{\circ }+\theta )=-\cos \theta ;\tan ({180}^{\circ }+\theta )=\tan \theta$
12. $\sin ({270}^{\circ }-\theta )=-\cos \theta ;\cos ({270}^{\circ }-\theta )=-\sin \theta ;\tan ({270}^{\circ }-\theta )=\cot \theta$
13. $\sin ({270}^{\circ }+\theta )=-\cos \theta ;\cos ({270}^{\circ }+\theta )=\sin \theta ;\tan ({270}^{\circ }+\theta )=-\cot \theta$
14. $\sin ({360}^{\circ }-\theta )=-\sin \theta ;\cos ({360}^{\circ }-\theta )=\cos \theta ;\tan ({360}^{\circ }-\theta )=-\tan \theta$
15. ${\sin }^2\theta +{\cos }^2\theta =1;{\sec }^2\theta -{\tan }^2\theta =1;cose{c}^2\theta -{\cot }^2\theta =1$
16. $\cos (A+B)\cos (A-B)={\cos }^{2}A-{\sin }^{2}B$
17. $\sin A+\sin B=2\sin \frac{(A+B)}{2}\cos \frac{(A-B)}{2}$
18. $\cos A-\cos B=2\cos \frac{(A+B)}{2}\cos \frac{(A-B)}{2}$
19. $\sin (A+B)\sin (A-B)={\sin }^{2}A-{\sin }^{2}B$
20. $\tan (A+B)=\frac{\tan A+\tan B}{1+\tan A\tan B}$
21. $\sin A-\sin B=2\cos \frac{(A+B)}{2}\cos \frac{(A-B)}{2}$
22. $\cos A-\cos B=-2\sin \frac{(A+B)}{2}\sin \frac{(A-B)}{2}$
23. $1-\cos \theta =2{\sin }^2\frac{\theta }{2}$

DIFFERENTIAL CALCULUS

The derivative $y$ with respect to $x$ is defined as the limit of the slopes of chords drawn between two points on the $y$ verses $x$ curve as $\mathrm{\Delta }x$ approaches zero. Mathematically, we write this defintion as

$\frac{dy}{dx}=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{y(x+\mathrm{\Delta }x)-y(x)}{\mathrm{\Delta }x}$

Where $\mathrm{\Delta }y$ and $\mathrm{\Delta }x$ are defined as $\mathrm{\Delta }x=x_2-x_1$ and $\mathrm{\Delta }y=y_2-y_1$ (See Figure above)

A useful expression to remember when $y(x)=a{x}^n$, where $a$ is a constant and $n$ is any positive or negative number (integer or fraction), is

$\frac{dy}{dx}=na{x}^{n-1}$

If $y(x)$ is a polynomial or algebraic function of $x$, we apply above equation to each term in the polynomial and take $\frac{da}{dx}=0$. It is important to note that $\frac{dy}{dx}$ does not mean $dy$ divided by $dx$, but is simply a notation of the limiting process of the derivative. In

Examples, we evaluate the derivatives of several well-behaved functions.

Illustration :

$y(x)=8x^5+4x^3+2x+7$

SOLUTION: Applying equation 2 to each term independently, and remembring that $\frac{d}{dx}$ (constant) $=0$,

We have

$\begin{array}{rl}& \frac{dy}{dx}=8(5)x^4+4(3)x^2+2(1)x^0+0\\ & \frac{dy}{dx}=40x^4+12x^2+2\end{array}$

SPECIAL PROPERTIES OF THE DERIVATIVE

(a) Derivative of the Product of Two Functions. If a function $y$ is given by the product of two functions, say, $g(x)$ and $h(x)$, then the derivative of $y$ is defined as

$\frac{d}{dx}[f(x)]=\frac{d}{dx}[g(x)h(x)]=g\frac{dh}{dx}+h\frac{dg}{dx}$

(b) Derivative of the Sum of Two Functions. If a function $y$ is equal to the sum of two functions, then the derivative of the sum is equal to the sum of the derivatives :

$\frac{d}{dx}[f(x)]=\frac{d}{dx}[g(x)+h(x)]=\frac{dg}{dx}+\frac{dh}{dx}$

(c) Chain Rule of Differential Calculus. If $y=f(x)$ and $x$ is a function of some other variable $z$.

Then $\frac{dy}{dx}$ can be written as the product of two derivatives :

$\frac{dy}{dx}=\frac{dy}{dz}\frac{dz}{dx}$

(d) The Second Derivative. The second derivative of $y$ with respect to $x$ is defined as the derivative of the function $\frac{dy}{dx}$ (or the derivative of the derivative). It is usually written

$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}(\frac{dy}{dx})$

TABLE 1 : DERIVATIVES FOR SEVERAL FUNCTIONS

$\frac{d}{dx}(a)=0$	$\frac{d}{dx}(\tan ax)=a{\sec }^{2}ax$
$\frac{d}{dx}(a{x}^n)=na{x}^{n-1}$	$\frac{d}{dx}(\cot ax)=-acose{c}^{2}ax$
$\frac{d}{dx}(e^{ax})=a{e}^{ax}$	$\frac{d}{dx}(\sec x)=\tan x\sec x$
$\frac{d}{dx}(\sin ax)=a\cos ax$	$\frac{d}{dx}(cosecx)=-\cot xcosecx$
$\frac{d}{dx}(\cos ax)=-a\sin ax$	$\frac{d}{dx}(\ln ax)=\frac{1}{x}$

MAXIMA AND MINIMA

To find the maxima and minima for a function or extreme values of the function $f(x)$ we proceed as follows

STEP-I : Find $f^{\prime }(x)=\frac{d}{dx}[f(x)]$

STEP-II : Put $f^{\prime }(x)=0$ and solve the obtained equations for values of $x$ or simply saying, find the zero of the function $f^{\prime }(x)$. Let $x=a_1,a_2,\dots a_n$ be the zeros of the function $f^{\prime }(x)$ also called the stationary values of $f^{\prime }(x)$.

STEP-III : Find $f^{\prime \prime }(x)=\frac{d^2}{d{x}^2}[f(x)]=\frac{d}{dx}[f^{\prime }(x)]$

STEP-IV : Find the values of $f^{\prime \prime }(x)$ at $x=a_1,a_2,\dots a_n$.

STEP-V : If $f^{\prime \prime }(x)>0$ then we get MINIMA

If $f^{\prime \prime }(x)<0$ then we get MAXIMA

STEP-VI : If $f^{\prime \prime }(x)=0$ then we get the point of INFLEXION

i.e. a point which is neither MINIMA nor MAXIMA.

INTEGRAL CALCULUS

Integration is the inverse of differentiation.

TABLE 2 : SOME INDEFINITE INTEGRALS

$\int x^{n}dx=\frac{x^{n+1}}{n+1}($ provided $n\ne -1)$	$\int \frac{dx}{a^2+x^2}=\frac{1}{a}{\tan }^{-1}\frac{x}{a}$
$\int \frac{dx}{x}=\int x^{-1}dx=\ln x$	$\int e^{ax}dx=\frac{1}{a}{e}^{ax}$
$\int \frac{dx}{a+bx}=\frac{1}{b}\ln (a+bx)$	$\int \sin axdx=-\frac{1}{a}\cos ax$
$\int \frac{dx}{(a+bx{)}^2}=\frac{1}{b(a+bx)}$	$\int \cos axdx=\frac{1}{a}\sin ax$

VECTOR

Physical Quantities (Anything which can be measured)

Scalar	Vector
1. Physical quantities which need only magnitude to express them.	Physical quantities which need both magnitude and direction to express them & they should obey vector rules.
2. They may have direction but not needed for their expression.
Example : Mass, energy, work, distance, speed, pressure, current, Time, $Are{a}^{\ast }$ etc.	Examples: Velocity, Displacement, Accelaration, Force, Momentum, Current density, $Are{a}^{\ast }$ etc.

VECTOR REPRESENTATION

Vectors are represented either graphically or by symbols as following :

Graphical Representation of Vectors

Vectors are represented by directed line segments. Their terminology is as shown in the figure.

The length of the line segment represents the magnitude of the vector and its arrow-head depicts its direction. The \to head can be put on the extreme end of the line or at any other point of the line.

represent a vector whose direction is perpendicular to the plane of this paper, coming out of the paper i.e. towards you. Similarly, $\otimes$ represents a vector which is perpendicular to the plane of this paper but going inside it.

Symbolic representation of vector

1. Vectors are represented in either of the following two ways.

Either as $\overrightarrow{r},\overrightarrow{PQ}$, etc. (arrow on the head) Or as $r,PQ$, etc. (bold letters)

2. Magnitude (also called modulus or mod values) of vectors are represented by normal alphabets, e.g.

Magnitude of $\overrightarrow{r}$ or $\overline{PQ}=r$ or $|\overrightarrow{r}|,PQ$ or $\overrightarrow{PQ}$

3. Unit vectors (detail study in types of vector) (i.e. whose magnitude is one unit) are represented as $\overrightarrow{r},\widehat{PQ}$ etc. (read as $r$-cap, $PQ$-cap, $r$-hat, $PQ$-hat or as $r$-caret, $PQ$-caret etc.)

TYPES OF VECTOR

1. Equal vectors : Two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ are said to be equal when they have equal magnitudes and same direction.

2. Parallel vector: Two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ are said to be parallel when

(i) Both have same direction.

(ii) One vector is scalar (positive non-zero) multiple of another vector.

Use : If a vector is shifted parallel to itself their will be no change in it.

3. Anti-parallel vectors : Two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ are said to be anti-parallel when

(i) Both have opposite direction.

(ii) One vector is scalar (non-zero negative) multiple of another vector. Use : In subtraction of vector.

4. Collinear vectors: When the vectors under consideration can share the same support or have a common support then the considered vectors are colinear.

5. Zero vector $(\overrightarrow{0})$ : A vector having zero magnitude and arbitrary direction (not known to us) is a zero vector.

Properties of zero vector:

(i) The sum of a finite vector $\overrightarrow{A}$ and the zero vector is equal to the finite vector : $\overrightarrow{A}+\overrightarrow{0}=\ddot{A}$

(ii) The multiplication of a zero vector by a finite number $n$ is equal to the zero vector : $n\overrightarrow{0}=\overrightarrow{0}$

(iii) The multiplication of a finite vector $\overrightarrow{A}$ by zero is equal to zero vector : $\overrightarrow{0}\overrightarrow{A}=\overrightarrow{0}$

Examples : The position vector of the origin in a system of coordinates is a zero vector. If an object is stationary, then its displacement in a finite time is a zero vector.

6. Unit vector : A vector divided by its magnitude is a unit vector. Unit vector for $\overrightarrow{A}$ is $\hat{A}$ (read as $A$ cap or $A$ hat)

Since, $\hat{A}=\frac{\overrightarrow{A}}{A}\Rightarrow \overrightarrow{A}=A\hat{A}$. Thus we can say that unit vector gives us the direction.

7. Position vector $(\overrightarrow{r}):$ It specifies the position of a point with reference to the given vel of coordinate axes. If we join the origin $O$ with the given point $P$, then $\overrightarrow{OP}$ is called position vector of $P$.

It is gencrally represented by $\overrightarrow{r}.\overrightarrow{r}$ means we know ’ $r$ ‘, i.e., the magnitude as well as $\theta$ i.e., the direction of the point $P$

$\therefore r=|\overrightarrow{r}|=\sqrt{x^2+y^2}$ and $\tan \theta =\frac{y}{x}$

8. Displacement vector $(\overrightarrow{s})$ : If a particle moves from position $P_1$ to position $P_2$, we call $\overrightarrow{P{P}_2}$ as displacement vector. Suppose the particle was originally at the origin $O$. It first moves to $P_1$, thereafter to $P_2$, then ${\overline{OP}}_2$ is called final displacement vector. Symbolically, we state it as $\overrightarrow{O{P}_1}+\overrightarrow{P_1{P}_2}=\overrightarrow{O{P}_2}$ or $\overrightarrow{P_1{P}_2}=\overrightarrow{O{P}_2}-\overrightarrow{O{P}_1}$ or, displacement vector $=$ Final position vector - Initial position vector. or, displacement $=$ Final position - Initial position

LAWS OF VECTOR ADDITION

Triangle law of vector addition of two vector

It states that if two vectors acting on a particle at the same time are represented in magnitude and direction by the two sides of a triangle taken in one order, their resultant vector is represented in magnitude and direction by the third side of the triangle taken in opposite order.

Let the two vector $\overrightarrow{A}$ and $\overrightarrow{B}$, inclined at an angle $\theta$ are represented in magnitude and direction by two sides $\overrightarrow{OP}$ and $\overrightarrow{PQ}$ or triangle $OPQ$, taken in order. Then resultant $R$ is represented by the third sides $\overrightarrow{OQ}$ of the triangle taken in opposite order. ( $\overrightarrow{OQ}->$ directed from tail of first vector to head of second vector.

Magnitude of $\overline{R}$ : Draw $QN$ perpendicular to $OP$ produced. In $\mathrm{△}QNP,PN=PQ\cos \theta =B\cos \theta ,QN=PQ\sin \theta =B\sin \theta$ In right angled triangle $ONQ$, we have $O{Q}^2=O{N}^2+N{Q}^2$ or $O{Q}^2=(OP+PN{)}^2+N{Q}^2$ or $R^2=(A+B\cos \theta {)}^2+(B\sin \theta {)}^2$ or $R=\sqrt{A^2+B^2+2AB\cos \theta }$

Direction of $\overrightarrow{R}$ : Let the resultant $\overrightarrow{R}$ make an angle $\beta$ with the direction of $\overrightarrow{A}$ Then from right angled triangle $QNO$,

$\tan \beta =\frac{QN}{ON}=\frac{QN}{OP+PN}=\frac{B\sin \theta }{A+B\cos \theta }$

Note: (l) Vector does not obey ordinary algebra.

(ii) In above formulation remember $\theta$ is angle between vectors you have to add angle between vector mean when tail of both vector are join (or head to head) then shorter angle formed.

In above diagram you can see $\theta$ by shifting $\overrightarrow{B}$ parallel to itself & joining its tail with $\overrightarrow{A}$.

Value of $\overrightarrow{R}(\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B})$ depends on angle between vector

$|\overrightarrow{R}|$ is maximum if $\cos \theta =1,\theta =0^{\circ }$ (parallel vector) $\stackrel{\overrightarrow{A}}{\to }\stackrel{\overrightarrow{B}}{\to }$

$R_{max}=\sqrt{A^2+B^2+2AB}=A+B$

$|\overrightarrow{R}|$ is minimum if, $\cos \theta =1,\theta ={180}^{\circ }$ (opposite vector) $\stackrel{\overrightarrow{B}}{\leftarrow }\stackrel{\overrightarrow{A}}{\to }$

$R_{min}=\sqrt{A^2+B^2-2AB}=A-B$

Thus $R\in [AB,A+B]$ i.e. $|\overrightarrow{R}|$ can be anything between including both.

Like the vector sum of the forces of $10N$ and $6N$ can be between $[4,16]$.

Note: 1. To a vector only a vector of same type can be added that represents the same physical quantity and the resultant is a vector of the same type

2. Vector addition is commutative, i.e., $\overrightarrow{A}+\overrightarrow{B}=\overrightarrow{B}+\overrightarrow{A}$
3. Vector addition is associative, i.e., $\overrightarrow{A}+(\overrightarrow{B}+\overrightarrow{C})=(\overrightarrow{A}+\overrightarrow{B})+\overrightarrow{C}$
4. If the vectors $A$ and $B$ are orthogonal, i.e., $\theta ={90}^{\circ },R=\sqrt{A^2+B^2}$

PARALLELOGRAM LAW OF VECTOR ADDITION

It states that if two vectors are represented in magnitude and direction by the two adjacent sides of a parallelogram then their resultant is represented in magnitude and direction by the diagonal of the parallelogram.

Let the two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$, inclined at angle $\theta$ are represents by sides $\overrightarrow{OP}$ and $\overrightarrow{OS}$ of parallelogram $OPQS$, then resultant vector $\overrightarrow{R}$ is represents by diagonal $\overrightarrow{OQ}$ of the parallelogram.

$R=\sqrt{A^2+B^2+2AB\cos \theta };\tan \beta =\frac{B\sin \theta }{A+B\cos \theta }$

If $\theta <{90}^{\circ }$, (Acute angle) $\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B},\overrightarrow{R}$ is called main (major) diagonal of parallelogram.

If $\theta >{90}^{\circ }$, (Obtuse angle) $\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B},\overrightarrow{R}$ is called minor diagonal.

POLYGON LAW OF VECTOR ADDITION

If a number of non zero vectors are represented by the $(n-1)$ sides of an $n$-sided polygon then the resultant is given by the closing side or the $n^{\text{th }}$ side of the polygon taken in opposite order, So,

$\begin{array}{rl}& \overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}+\overrightarrow{D}+\overrightarrow{E}\\ & \overrightarrow{OA}+\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DE}=\overrightarrow{OE}\end{array}$

Note :

1. Resultant of two unequal vectors cannot be zero.
2. Resultant of three co-planar vectors may or may not be zero.
3. Resultant of three non coplanar vectors cannot be zero, Minimum number of non coplanar vectors whose sum can be zero is four.
4. Polygon law should be used only for diagram purpose for calculation of resultant vector (For addition of more than 2 vector) we use components of vector.
5. Minimum no. of coplanar vector for zero resultant is 2 (for equal magnitude) & 3 (for unequal magnitude)

SUBTRACTION OF VECTOR

Subtraction of a vector $\overrightarrow{B}$, from a vector $\overrightarrow{A}$ is defined as the addition of vector $-\overrightarrow{B}$ (negative of vector $\overrightarrow{B}$ ) $)$, vector $\overrightarrow{A}$. Thus $\overrightarrow{A}+(-\overrightarrow{B})=\overrightarrow{A}-\overrightarrow{B}$ It $\theta$ is the angle between $\overrightarrow{A}$ an $\overrightarrow{B}$, then angle between $\overrightarrow{A}$ and $-\overrightarrow{B}$, is $({180}^{\circ }-\theta )$.

Note: 1. Addition of a viector to its own negative vector or null vector $\overrightarrow{A}+(\overrightarrow{A})=\overrightarrow{O}$ i.e. a vector with zero magnitude and an arbitrary direction.

2. Subtraction is not comnutative, i.e., $\overrightarrow{A}\cdots \overrightarrow{B}\ne \overrightarrow{B}-\overrightarrow{A}$

(a)

(b)

3. Change in a vector physical quantity means subtraction of initial vector from the final vector

COMPONENTS OF VECTOR

If $\overrightarrow{a}$ and $\overrightarrow{b}$ be any two nonzero vectors in a plane with different directions and $\overrightarrow{A}$ be another vector in the same plane. $\overrightarrow{A}$ can be expressed as a sum of two vectors-one obtained by multiplying $\overrightarrow{a}$ by a real number and the other obtained by multiplying $\overrightarrow{b}$ by another real number $\overrightarrow{A}=\lambda \dot{a}+\mu \overrightarrow{b}$ (where $\gamma$ and $\mu$ are real numbers)

We say that $\overrightarrow{A}$ has been resolved into two component vectors $\lambda \overrightarrow{a}$ and $\mu \overrightarrow{b}$ along $\overrightarrow{a}$ and $\overrightarrow{b}$ respectively. Hence one can resolve a given vector into two component vectors along a set of two vectors-all the three lie in the same plane.

The process of splitting a single vector into two or more vectors, which together produce the same effect as is produced by single vector alone, is called resolution of vectors.

The vectors into which the given single vector is splitted are called component vectors. (Resolution of a vector is just opposite to composition of vectors).

Consider a vector $\overrightarrow{A}$ that lies in xy plane, $\overrightarrow{A}={\overrightarrow{A}}_1+{\overrightarrow{A}}_2$ as shown.

The quantities $A_x$ and $A_y$ are called $x$ - and $y$-components of the vector $\overline{A}\cdot A_x$ is itself not a vector but $A_x\hat{i}$. is a vector and so is $A_y\hat{j}$.

$A_x=A\cos \theta$ and $A_y=A\sin \theta$

Its clear from above equation that a component of a vector can be positive, negative or zero depending on the value of $\theta .A$ vector $\overrightarrow{A}$ can be specified in a plane by two ways :

(a) Its magnitude $A$ and the direction $\theta$ it makes with the $x$-axis, or

(b) Its components $A_x$ and $A_y$.

$A=\sqrt{A_x^2+A_y^2},\theta ={\tan }^{-1}\frac{A_y}{A_x}$

Note: If $A=A_x\Rightarrow A_y=0$ and if $A=A_y\Rightarrow A_x=0$ i.e. components of a vector perpendicular to itself is always zero.

In three dimensions, a vector $\overrightarrow{A}$ in components along $x,y$ and $z$-axis can be written as :

$\begin{array}{c}\overrightarrow{OP}=\overrightarrow{OB}+\overrightarrow{BP}=\overrightarrow{OA}+\overrightarrow{AB}+\overrightarrow{BP}\\ \Rightarrow \overrightarrow{A}={\overrightarrow{A}}_x+{\overrightarrow{A}}_y+{\overrightarrow{A}}_z=\hat{i}{A}_x+\hat{j}{A}_y+\hat{k}{A}_z\end{array}$

See following diagram carefully

$XComp\to +A\sin \theta$

$YComp\to +A\cos \theta$

$XComp\to -A\cos \theta$

$YComp\to +A\sin \theta$

$X$ Comp $\to -A\cos \theta$

$YComp\to -\mathit{A}\sin \theta$

$X$ Comp $\to +A\cos \theta$

$Y$ Comp $\to -A\sin \theta$

MULTIPLICATION OF VECTORS

Scalar Product

The scalar product or dot product of any two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$, denoted as $\overrightarrow{A}\cdot \overrightarrow{B}$ (read $\overrightarrow{A}$ dot $\overrightarrow{B}$ ) is defined as the product of their magnitude with cosine of angle between them. Thus, $\overrightarrow{A}$. $\overrightarrow{B}=AB\cos \theta$ (here $\theta$ is the angle between the vectors)

Properties

(a) It is always a scalar.

(b) It is commutative, i.e., $\overrightarrow{A}\cdot \overrightarrow{B}=\overrightarrow{B}\cdot \overrightarrow{A}$

(c) It is distributive, i.e., $\overrightarrow{A}\cdot (\overrightarrow{B}+\overrightarrow{C})=\overrightarrow{A}\cdot \overrightarrow{B}+\overrightarrow{A}\cdot \overrightarrow{C}$

(d) As by definition $\overrightarrow{A}\cdot \overrightarrow{B}=AB\cos \theta$

(e) $\overrightarrow{A}\cdot \overrightarrow{B}=A(B\cos \theta )=B(A\cos \theta )$

The angle between the vector $\theta ={\cos }^{-1}[\frac{\overrightarrow{A}\cdot \overrightarrow{B}}{AB}]$

Geometrically, $B\cos \theta$ is the projection of $\overrightarrow{B}$ onto $\overrightarrow{A}$ and $A\cos \theta$ is the projection of $\overrightarrow{A}$ onto $\overrightarrow{B}$ as shown. So $\overrightarrow{A}.\overrightarrow{B}$ is the product of the magnitude of $\overrightarrow{A}$ and the component of $\overrightarrow{B}$ along $\overrightarrow{A}$ and vice versa.

Component of $\overrightarrow{B}$ along $\overrightarrow{A}=B\cos \theta =\frac{\overline{A}\cdot \overrightarrow{B}}{A}=\hat{A}\cdot \overrightarrow{B}$

Component of $\overrightarrow{A}$ along $\overrightarrow{B}=A\cos \theta =\frac{\overrightarrow{A}\cdot \overrightarrow{B}}{B}=\overrightarrow{A}\cdot \hat{B}$

(f) Scalar product of two vectors will be maximum when $\cos \theta =max=1$, i.e. $\theta =0^{\circ }$, i.e., vectors are parallel $\Rightarrow (\overrightarrow{A}\cdot \overrightarrow{B}{)}_{max}=AB$

(g) If the scalar product of two nonzero vectors vanishes then the vectors are orthogonal.

(h) The scalar product of a vector by itself is termed as self dot product and is given by $(\overrightarrow{A}{)}^2=\overrightarrow{A}\cdot \overrightarrow{A}=AA\cos \theta =A^2\Rightarrow A=\sqrt{\overrightarrow{A}\cdot \overrightarrow{A}}$

(i) In case of unit vector $\hat{n},\hat{n}.\hat{n}=1\times 1\times \cos 0^{\circ }=1$

$\Rightarrow \hat{n}\cdot \hat{n}=\hat{i}\cdot \hat{i}=\tilde{j}=\hat{k}\cdot \hat{k}=1$

(j) In case of orthogonal unit vectors, $\hat{i},\hat{j}$ and $\hat{k};\hat{i},\hat{j}=\hat{j}\cdot \hat{k}=\hat{k}\hat{i}=0$

(k) $\overrightarrow{A}\cdot \overrightarrow{B}=(\hat{i}{A}_x+\hat{j}{A}_y+\hat{k}{A}_z)\cdot (\hat{i}{B}_x+\hat{j}{B}_y+\hat{k}{B}_z)=[A_x{B}_x+A_y{B}_y+A_z{B}_z]$

Illustration 2 :

Prove that vectors $\overrightarrow{A}=\hat{i}+2\hat{j}+3\hat{k}$ and $\overrightarrow{B}=2\hat{i}-\hat{j}$ are perpendicular to each other.

SOLUTION : Here, $\overrightarrow{A}=\hat{i}+2\hat{j}+3\hat{k},\overrightarrow{B}=2\hat{i}-\hat{j}$

Two vectors are perpendicular to each other if, $\overrightarrow{A}\cdot \overrightarrow{B}=0$

Now $\overrightarrow{A}\cdot \overrightarrow{B}=(\hat{i}+2\hat{j}+3\hat{k})\cdot (2\hat{i}-\hat{j})\cdot (2\hat{i}-\hat{j})=0$

$=1\times 2+2\times (-1)+3\times (0)=2-2+0-0$

Since $\dot{A}\cdot \overrightarrow{B}=0$, vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ are perpendicular to each other.

CHAPTER 1

MEASUREMENT

INTRODUCTION

The process of comparing an unknown physical quantity with respect to a known quantity is known as measurement. For example, when we say that the length of our bedroom is 10 feet it implies that the bedroom is 10 times the known quantity ‘feet’. So, measurement of any physical quantity consists of two parts - (i) a numerical value and (ii) the known quantity. In above example, the numerical value is 10 and the known quantity is the ‘feet’. The known quantity is called the unit of that physical quantity. The Seet’, therefore, is the unit of length.

Physical quantities are of two types-fundamental and derived. A derived quantity can be expressed in Chers of fundamental quantities. A derived quantity expressed in terms of fundamental quantities is said to have ‘dimensions’.

Ceasurement of a quantity has in general, inaccuracy or ’errors’. In this chapter we will learn measurement, unit, dimensions and errors in measurement.

PHYSICAL QUANTITIES

Quantities which can be measured are called physical quantities. Velocity, acceleration, force, area, volume, pressure, etc. are some examples of physical quantities.

Kinds of Physical Quantities

There are two kinds of physical quantities

(i) Fundamental Physical Quantities : Fundamental physical quantities are those which do not depend on other quantities and also independent of each other. They are seven in number viz; length, mass, time, thermodynamic temperature, electric current, luminuous intensity and amount of substance.

(ii) Derived Physical Quantities : Derived physical quantities are those which are derived from fundamental physical quantities. For example, velocity is derived from the fundamental quantities length and time, hence it is a derived physical quantity.

UNIT

To measure a physical quantity it is compared with a standard quantity. This standard quantity is called the unit of that quamtity. For example, to measure the length of a desk, it is compared with the standard quantity known as ‘metre’. Thus, ‘metre’ is said to be the unit of length.

The starting point in any study of measurement is an understanding of the need for universal standardisation of the units. In early times, man used body parts as a standard of measure. With progress, these rough inaccurate measurements were discarded for accurate systems of measurement. Also, as world trade increased, there developed a need for universal standards of units of measure with opportunity for all countries to adopt such established standards.

Do you Know !!

The unit named to commemorate a scientist is not written with capital initial letter. For example, the unit of force is written as newton (and not as Newton), the unit of current is written as ampere (and not as Ampere). etc.

The symbols for units named after scientists are usually the first initial letter of their names in capital. For example, $N$ for newton, A for ampere, J for Joule etc.

Types of Units

There are two types of units :

(i) Fundamental units and

(ii) Derived units

Fundamental units

Fundamental units are those units which cannot be derived from any other unit, and they cannot be resolved into any basic or fundamental unit, for example, length, mass, time, temperature, luminous intensity and current.

Derived units

Any unit which can be obtained by the combination of one or more fundamental units are called derived unit for example, area speed, density, volume, momentum, acceleration, force, etc.

System of Units

Depending upon the units of fundamental physical quantities, there are four main systems of units, namely

1. CGS (Centimeter, Gramme or Gram, Second)
2. FPS (Foot, Pound, Second)
3. MKS (Meter, Kilogram, Second)
4. SI (Systeme International d’ Unites)

The first three of these systems recognize only three fundamental dimensions i.e. length $(L)$, mass $(M)$ and time $(I)$ while the last one recognizes seven fundamental dimensions i.e. length $(L)$, mass $(M)$, time $(T)$, electric current $(I$ or $\mathit{A})$, thermed namic temperature $(K.$ or $\theta$ ), amount of substance (mol) and luminous intensity $(I_v)$.

An international organization, the Conference Generale des Poids et Measures, or CGPM is internationally renogmeed as the authority on the definition of units. In English, this body is known as “General Conference on Wrights and Akrasun”. The Systeme International de Unites, or SI system of units, was set up in 1960 by the CGPM

Do you Know !!

The units do not have plural forms. For example, it is wong in write a forre of 10 mown (os $\overline{10N}$ ) as 10 newtons (or 10Ns) or a mass of $10kg$ as $10kgs$

No full stop is put hetween the symbols for units. l’or example, it is wrong to write new:rm metre as N.m. One should write it us Nm.

Characteristics of a Standard Unit

A standard unit must have following features to be accepted world wide.

(i) It should have a convenient size.

(ii) It should be very well defined.

(iii) It should be independent of time and place.

(iv) It should be easily available so that all laboratories can duplicate and use it as per requirement.

(v) It should be independent of physical conditions like temperature, pressure, humidity etc.

(vii) It should be easily reproducible.

Advantages of SI

(i) Coherent System : All the derived units are obtainable directly from the basic units.

(ii) Rational System : Only one unit for one physical quantity.

(iii) Metric System : This makes calculations easier.

(iv) This system is universally accepted.

Disadvantages of SI

As it is a coherent system, all the derived units are not practical e.g. 1 coulomb (unit of electric charge), 1 farad (unit of electric capacitance), 1 bel (unit of loudness of sound) are too large units to be practical.

Fundamental Units of SI

The following table shows the seven fundamental units of S.I.

S. No.	Physical quantity	Unit of measurement	Symbol for unit
1.	Length	Metre	$m$
2.	Mass	Kilogram	$kg$
3.	Time	Second	$s$
4.	Electric current	Ampere	$A$
5.	Temperature	Kelvin	$K$
6.	Luminous intensity	Candela	cd

Supplementary Units of SI

The following table shows the two supplementary units of SI

S.No.	Physical quantity	Unit of measurement	Symbol for unit
1.	Plane Angle	Redian	rad
2.	Solid Angle	Steradian	sr

Definitions of SI units

1. Metre: The metre is the length of the path travelled by light in vacuum during a time interval of $1.299,792,458$ of a second.
2. Kilogram: The kilogram is equal to the mass of the international prototype of the kilogram (a platinum-iridium alloy cylinder) kept at International Bureau of Weights and Measures, at Sevres, near Paris, France
3. Second: The second is the duration of $9,912,631,770$ periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium - 133 atom.
4. Ampere: The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 metre apart in vacuum, would produce between these conductors a force equal to $2\times {10}^{-7}$ Newton per metre of length.
5. Kelvin: The kelvin, is the $1/273.16$ fraction of the thermodynamic temperature of the triple point of water.
6. Candela: The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency $540\times {10}^{12}$ hertz and that has a radiant intensity in that direction of $1/683$ watt per steradian.
7. Mole: The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon - 12 .

FACTOR	PREFIX	SYMBOL
${10}^{24}$	yotta -	$Y$
${10}^{21}$	zeeta -	$Z$
${10}^{18}$	exa -	$E$
${10}^{15}$	peta -	$P$
${10}^{12}$	tera -	$T$
${10}^9$	giga -	$G$
${10}^6$	mega -	$M$
${10}^3$	kilo -	$k$
${10}^2$	hecto -	$h$

FACTOR	PREFIX	SYMBOL
${10}^1$	deca -	$da$
${10}^{-1}$	deci -	$d$
${10}^{-2}$	centi -	$c$
${10}^{-3}$	mili -	$m$
${10}^{-6}$	micro-	$u$
${10}^{-9}$	nano-	$n$
${10}^{-12}$	pico-	$P$
${10}^{-15}$	femto-	$f$
${10}^{-18}$	atto-	$a$
${10}^{-21}$	zepto-	$z$
${10}^{-24}$	yocto–	$y$

Some units of smaller lengths

(i) 1 centimetre $(cm)={10}^{-2}m$

(ii) 1 millimetre $(mm)={10}^{-3}m$

(iii) 1 micron $(\mu )={10}^{-6}m$

(iv) 1 nanometre $(nm)={10}^{-9}m$

(v) 1 angstrom $(\text{AA})={10}^{-10}m$

(vi) $1X$-ray unit $={10}^{-13}m$

(vii) 1 fermi (f) or femto-metre $={10}^{-15}m$

Some units of bigger lengths

1. 1 kilometre (km): This is unit for expressing terrestrial distances.

$1km={10}^{3}m$

2. Astronomical Unit (AU): It is the mean distance of the centre of sun from the centre of earth. $1AU=1.496\times {10}^{11}$ metre
3. Light Year (ly): It is distance travelled by light in vacuum in one year.

1 light year $=($ speed of light in vacuum $)\times ($ number of seconds in one year $)$

$\begin{array}{rl}& =(3\times {10}^8)\times (365\times 24\times 60\times 60)m\\ & =9.467\times {10}^{15}m\end{array}$

4. Parallactic Second (parsec) : It is the distance at which an arc of length $1AU$ subtends an angle of 1 second.

SI DERIVED UNITS WITH SPECIAL NAMES

Physical quantity			SI Unit
	Name	Symbol	Expression in terms of other units	Expression in terms of SI base units
Frequency	hertz	$Hz$		$s^{-1}$
Force	newton	$N$		$kgm{s}^{-2}$ or $kgm/s^2$
Pressure, stress	pascal	$Pa$	$N/m^2$ or $N{m}^{-2}$	$kg{m}^{-1}{s}^{-2}$ or $kg/s^{2}m$
Energy, work, quantity of heat	joule	$J$	$Nm$	$kg{m}^2{s}^{-2}$ or $kg{m}^2/s^2$
Power, radiant flux	watt	$W$	$J/s$ or $J{s}^{-1}$	$kg{m}^2{s}^{-3}$ or $kg{m}^2/s^3$
Quantity of electricity, electric charge	coulomb	$C$		As
Electric potential, potential difference, electromotive force	volt	$\bar{V}$	$W/A$ or $W{A}^{-1}$	$kg{m}^2{s}^{-3}{A}^{-1}$ or $kg{m}^2/s^{3}A$
Capacitance	farad	$\overline{F}$	$C/V$	$A^{2}s4k{g}^{-1}{m}^{-2}$
Electric resistance	ohm	$\mathrm{\Omega }$	V/A	$kg{m}^2{s}^{-3}{A}^{-2}$
Conductance	siemens	$S$	$A/V$	$m^{-2}k{g}^{-1}{s}^3{A}^2$
Magnetic flux	weber	$Wb$	$Vs$ or $J/A$	$kg{m}^2{s}^{-2}{A}^{-1}$
Magnetic field, magnetic flux density, magnetic induction	tesla	T	$Wb/m^2$	$kg{s}^{-2}{A}^{-1}$
Inductance	henry	$H$	$Wb/A$	$kg{m}^2{s}^{-2}{A}^{-2}$
Luminous flux, luminous power	lumen	$lm$		$cd/sr$
Illuminance	$lux$	$1x$	$lm/m^2$	$m^{-2}cds{r}^{-1}$

Important Conversions and Units

(A) (i) 1 Light year $=9.467\times {10}^{15}m$

(ii) 1 Astronomical unit $=1.5\times {10}^{11}m$

(iii) $1$ Parsec $=3.26$ light year $=3.08\times {10}^{16}m$

(iv) $1$ yard $=0.9144m\cong 0.91m$

(v) $1$ foot $=0.305m$

(vi) $1$ inch $=2.54cm=0.025m$

(vii) 1 mile $=1609m=1.609km$

(viii) $1ltr.=1000cc={10}^{-3}{m}^3$

(ix) $1c{m}^2={10}^{-4}{m}^2$

(x) $1mm={10}^{-3}m$

(B) (i) 1 atomic mass unit $=1amu=1.67\times {10}^{-27}kg$

(ii) $1slug=14.57kg$

(iii) 1 tonne $=10$ quintal $=1000kg$

(iv) $1kg/m^3=1000gm/c{m}^3$

(v) $1km/hr=\frac{5}{18}m/s\text{ and }1m/s=\frac{18}{5}km/hr$

(vi) $1\text{ newton }={10}^5\text{ dyne, }1kgwt=9.8N\text{ and }1gwt=981\text{ Dyne }$

(vii) $1\text{ joule }={10}^{7}erg,1eV=1.6\times {10}^{-19}J$

(viii) $1atm=76cm\text{ of }Hg=1.01\times {10}^5\frac{N}{m^2}=1.01\times {10}^6\frac{Dy}{c{m}^2}$

(ix) 1 h.p. =746 watt

(x) $1kwhr=3.6\times {10}^{6}J$

(C) (i) $1tesla=1web/m^2={10}^4\text{ gauss }$

(ii) $1\text{ curie }=3.7\times {10}^{10}\text{ disintegration }/sec$

$1\text{ rutherford }=,{10}^6\text{ disintegration }/sec$

(iii) 1 weber = ${10}^8$ maxwell

(iv) 1 degree = $\frac{\pi }{180}$ radian and 1 radian =$\frac{180}{\pi }$ degree

(v) 1 shake = ${10}^{-8}$ sec

Rules for Writing Units and Their Symbols

(i) Units named after scientists are not written with initial letter in capital, e.g. it is incorrect to write 10 Newton, the correct way is 10 newton.

(ii) But the symbols of the units named after scientists must be written in capital letter for example, $10N$.

(iii) If the symbols of unit is not from a name then small letters must be used.

(iv) For derived units index notation must be used e.g. $20m{s}^{-2}$ in place of $20m/s^2$.

(v) Units and symbol should not be written in plural form e.g. write 25 centimeter in place of 25 centimeters.

DIMENSIONS OF A PHYSICAL QUANTITY

We have seen that derived physical quantities can be obtained from fundamental units or base quantities. When we write a physical quantity in terms of base quantities it can be expressed as product of powers of the base quantity. For example :

$\begin{array}{rl}& \text{ Force }=\text{ mass }\times \text{ acceleration }\\ & \text{ Force }=\text{ mass }\times \frac{\text{ velocity }}{\text{ time }}\\ & \text{ Force }=\text{ mass }\times \frac{\text{ Length }/\text{ time }}{\text{ time }}\\ & \text{ Force }=(\text{ mass }{)}^1\times (\text{ length }{)}^1\times (\text{ time }{)}^{-2}\end{array}$

The exponent of the base quantity in the expression of a physical quantity is called dimensions of the quantity. Thus force has dimensions as 1 in mass, 1 in length and -2 in time. While finding dimensions, magnitudes of quantities are not considered. The above expression can also be written as

$[\text{ force }]=[M^1{L}^1{T}^{-2}]=M^1{L}^1{T}^{-2}$

An expression for a physical quantity in terms of fundamental quantities is known as dimensional formula.

So the dimensional formula for force is $[ML{T}^{-2}]$ and that of volume will be $[M{L}^3{T}^0]$. While writing dimensional formula we will use symbols, $M$ for mass, $L$ for length, $T$ for time, $A$ for current, $K$ for temperature, mol for amount of substance and cd for luminous intensity.

The dimensional formulae of various derived quantities can be obtained from their relations with other quantities, in the following table dimensional formulae of various physical quantities are derived.

Do you know !!

A physical quantity may be dimensionless but still may have units. For example, plane angle is dimensionless but has radian as its unit.

A physical quantity that does not have any unit must be dimensionless.

Illustration 1 :

Derive the dimensional formulae for the following quantities :

(i) Pressure (ii) Torque (iii) Angular momentum

solution: (i) Pressure $=\frac{F}{A}\{\begin{array}{l}F=ma=m\frac{v}{t}\\ A=l\times b\end{array}$

$[P]=\frac{[F]}{[A]}=\frac{[M^1{L}^1{T}^{-2}]}{[L^2]}$

$[P]=[M^1{L}^{-1}{T}^{-2}]$

(ii) Torque $=$ Force $\times$ Moment arm

$[\tau ]=[F][l]$

$[\tau ]=[M^l{L}^1{T}^{-2}][L]$

$[\tau ]=[M^1{L}^2{T}^{-2}]$

(iii) Angular Momentum $(L)=mvr$

$[L]=[M][v][r]$

$[L]=[M^l][L{T}^{-1}][L^1]$

$[L]=[M^1{L}^2{T}^{-1}]$

$✓$ CHECK POINT

• The division of energy and time is X. Among momentum, power, torque and electric field, which has the same dimensional formula of $\mathbf{X}$ ?

Check Your Answer

Power is defined as the ratio of work (or energy) to time. Therefore, $X$ has the dimensional formula as that of power.

Dimensiona of Some Physical Quantities

S.No.	Physical Quantity	Relation with other quantities	Dimensional formula	Unit
1.	Length, displacement, distance		$[M^0{L}^1{T}^0]$	$m$
2.	Area	length $\times$ breadth	$[M^0{L}^2{T}^0]$	$m^2$
3.	Volume	$V=l\times b\times h$	$[M^0{L}^3{T}^0]$	$m^3$
4.	Mass		$[M^1{L}^0{T}^0]$	$kg$
5.	Mass Density	$d=\frac{M}{V}$	$[M^1{L}^{-3}{T}^0]$	$kg{m}^{-3}$
6.	Linear Mass Density	$m=\frac{M}{l}$	$[M^1{L}^{-1}{T}^0]$	$kg{m}^{-1}$
7.	Relative Density	$\frac{\text{ density of a solid }}{\text{ density of water }}$	$[M^0{L}^0{T}^0]$	Unitless, dimensionless
8.	Specific gravity	$\frac{\text{ density of a liquid }}{\text{ density of water }}$	$[M^0{L}^0{T}^0]$	Unitless, dimensionless
9.	Speed	$\frac{\text{ distance }}{\text{ time }}$	$[M^0{L}^1{T}^{-1}]$	$m{s}^{-1}$
10.	Velocing	$\frac{\text{ displacement }}{\text{ time }}$	$[M{P}^0{L}^1{T}^{-1}]$	$m{s}^{-1}$
11.	Acceleration	$\frac{\mathrm{\Delta }v}{\mathrm{\Delta }v}$	$[M{P}^1{L}^1{T}^{-2}]$	$m{s}^{-2}$
12.	Fonce	$F=ma$	$[M^{\prime }{L}^1{T}^{-2}]$	$kgm{s}^{-2}=N$
13.	Coefficient of friction	$\mu =\frac{f}{N}$	$[M{\mathit{P}}^0{L}^0{T}^0]$	Unitless, dimensionless
14.	Work	$\mathit{W}=F.s$	$[M^d{L}^2{T}^{-2}]$	$kg{m}^2{s}^{-2}=J$
15.	Kinetic Energy	$K=\frac{1}{2}m{v}^2$	$[M^1{L}^2{T}^{-2}]$	$kg{m}^2{s}^{-2}=J$
16.	All forms of Energy	-	$[M^1{L}^2{T}^{-2}]$	$kg{m}^2{s}^{-2}=J$
17.	$\tau =F\times r$		$[M^2{L}^2{T}^{-2}]$	$kg{m}^2{s}^{-2}=J$
18 .	Linear Momentum	$P=mv$	$[M^d{L}^1{T}^{-1}]$	$kgm{s}^{-1}$
19.	Linear Impulse	$J=F\cdot \mathrm{\Delta }I$	$[M^d{L}^1{T}^{-1}]$	$kgm{s}^{-1}$
20.	Power	$P=\frac{W}{t}$	$[M^{\prime }{L}^2{T}^{-3}]$	$kg{m}^2{s}^{-3}=W$
21.	Pressure	$p=\frac{F}{A}$	$[M^{\prime }{L}^2{T}^{-3}]$	$kg{m}^2{s}^{-3}=\frac{N}{m^2}$
22.	Stress	$\sigma =\frac{F}{A}$	$[M^1{L}^{-1}{T}^{-2}]$	$kg{m}^{-1}{s}^{-2}=\frac{N}{m^2}$ or pascal
23.	Young’s Modulus	$Y=\frac{\sigma }{ϵ}=\frac{\text{ stress }}{\text{ strain }}$	$[M^{\prime }{L}^{-1}{T}^{-2}]$	$kg{m}^{-1}{s}^{-2}=\frac{N}{m^2}$ or pascal
24.	Shear Modulus	$G=\frac{F/A}{\theta }$	$[M^1{L}^{-1}{T}^{-2}]$	$kg{m}^{-1}{s}^{-2}=\frac{N}{m^2}$ or pascal
25 .	Bulk Modulus	$B=\frac{P}{\frac{-\mathrm{\Delta }v}{v}}$	$[M^{\prime }{L}^{-1}{T}^{-2}]$	$kg{m}^{-1}{s}^{-2}=\frac{N}{m^2}$ or pascal
26.	Strain	$ϵ=\frac{\mathrm{\Delta }l}{l}$	$[M^0{L}^0{T}^0]$	Unitless, dimensionless
27.	Universal constant of gravitation	$G=\frac{F{r}^2}{m_1{m}_2}$	$[M^{-1}{L}^3{T}^{-2}]$	$N{m}^{2}k{g}^{-2}$
28.	Poisson Ratio	$\frac{\text{ Lateral strain }}{\text{ Longitudinal strain }}$	$[M^0{L}^0{T}^0]$	Unitless, dimensionless
29.	Surface Tension	$S=\frac{F}{l}$	$[M^1{L}^0{T}^{-2}]$	$kg{s}^{-2}=\frac{N}{m}$
30.	Frequency	$\frac{I}{T}$	$[M^0{L}^0{T}^{-1}]$	$s^{-1}$
31.	Angular velocity	$W=\frac{\mathrm{\Delta }\theta }{\mathrm{\Delta }t}$	$[M^0{L}^0{T}^{-1}]$	$s^{-1}$
32.	Radius of gyration	length	$[M^0{L}^1{T}^0]$	$i$
33.	Moment of Inertia	$I=m{r}^2$	$[M^1{L}^2{T}^0]$	$kg{m}^2$
34.	Angular Momentum	$L=I\omega$	$[M^1{L}^2{T}^{-1}]$	$kg{m}^2{s}^{-1}$
35.	Rotational kinetic energy	$\frac{1}{2}I{\omega }^2$	$[M^1{L}^2{T}^{-2}]$	$\mathbf{J}$
36.	Wavelength	$\lambda =\frac{v}{f}$	$[M^0{L}^1{T}^0]$	$\therefore$
37.	Coefficient of visosity	$\eta =\frac{F}{6\pi rv}$	$[M^1{L}^{-1}{T}^{-1}]$	$N{m}^{-2}\mathbf{s}=$ Deca Poise
38.	Reynolds number	$R_N=\frac{\text{ Inertial force }}{\text{ Viscous force }}$	$[M^0{L}^0{T}^0]$	Unitless dimensionless
39.	Temperature		$[M^0{L}^0{T}^0{K}^1]$	$K$
40.	Heat Energy $(Q)$		$[M^1{L}^2{T}^{-2}]$	$J$
41.	Specific Heat .	$S=\frac{Q}{m\mathrm{\Delta }\theta }$	$[M^0{L}^2{T}^{-2}{K}^{-1}]$	$m^2{s}^{-2}{k}^{-1}=\frac{J}{kgK}$
42.	Heat Capacity	$C=ms=\frac{Q}{\mathrm{\Delta }\theta }$	$[M^1{L}^2{T}^{-2}{K}^{-1}]$	$kg{m}^2{s}^{-1}{K}^{-1}=J{K}^{-1}$
43.	Latent Heat of fusion or vaporisa- tion	$L=\frac{Q}{m}$	$[M^0{L}^2{T}^{-2}]$	$m^2{s}^{-2}=Jk{g}^{-1}$
44.	Gas constant	$R=\frac{PV}{nT}$	$[M^1{L}^2{T}^{-2}{K}^{-1}]$	$kg{m}^2{s}^{-2}{K}^{-1}=J{K}^{-1}$
45.	Boltzmann constant	$K=\frac{R}{N_A}$	$[M^1{L}^2{T}^{-2}{K}^{-1}]$	$kg{m}^2{s}^{-2}{K}^{-1}=J{K}^{-1}$
46.	Stefan’s constant	$\sigma =\frac{P}{A{T}^4}$	$[M^1{L}^0{T}^{-3}{K}^{-4}]$	$kg{s}^{-3}{K}^{-4}=W{m}^{-2}{K}^4$
47.	Power of a lens	$P=\frac{1}{f}$	$[M^0{L}^{-1}{T}^0]$,	$m^{-1}$
48.	Planck constant	$h=\frac{E}{v}$	$[M^1{L}^{-2}{T}^{-1}]$	$kg{m}^2{s}^{-1}=Js$
49.	Electric current		$[M^0{L}^0{T}^0{A}^1]$	A
50.	Electric charge	$q=It$	$[M^0{L}^{0}TA]$	As $=$ coulomb
51 .	Electric potential	$V=\frac{W}{q}$	$[M^1{L}^2{T}^{-3}{A}^{-1}]$	$J{C}^{-1}=$ volt
52 .	Absolute permittivity	${ϵ}_0=\frac{q_1{q}_2}{4\pi r^{2}F}$	$[M^{-1}{L}^{-3}{T}^4{A}^2]$	$k{g}^{-1}{m}^{-3}{s}^4{A}^2=C^2{N}^{-1}{m}^2$
53.	Electric field	$E=\frac{V}{l}$ or $\frac{F}{q}$	$[M^{-1}{L}^1{T}^{-3}{A}^{-1}]$	$kgm{s}^{-3}{A}^{-1}=V{m}^{-1}=N{C}^{-1}{m}^2$
54.	Electric dipole moment	$P=q\times 2l$	$[M^0{L}^1{T}^{1}A]$	$msA=Cm$
55.	Electrostatic potential Energy	$U=\frac{q_1{q}_2}{4\pi {ϵ}_{0}r}$	$[M^l{L}^2{T}^{-2}]$	$k{g}^{-1}{m}^2{s}^{-2}=J$
56.	Electric resistance	$R=\frac{V}{I}$	$[M^l{L}^2{T}^{-3}{A}^{-2}]$	$kg{m}^2{s}^{-3}{A}^2=ohm$
57.	Resistivity	$\rho =\frac{RA}{l}$	$[M^l{L}^3{T}^{-3}{A}^{-2}]$	$kg{m}^{3}s{A}^{-2}=\mathrm{\Omega }-m$
58.	Capacitance	$[M^{-1}{L}^{-2}{T}^4{A}^2]$	$k{g}^{-1}{m}^{-2}{s}^4=$ farad
59.	Magnetic field strength/ Induction	$B=\frac{F}{q}$	$[M^{-1}{L}^0{T}^{-2}{A}^{-1}]$	$kg{s}^{-2}{a}^{-1}=$ tesla
60.	Magnetic flux	$\varphi =BA$	$[M^1{L}^2{T}^{-2}{A}^{-1}]$	$kg{m}^2{s}^{-2}{a}^{-1}=t-m^2=$ weber
61.	Coefficient of self-induction	$L=\frac{\varphi }{I}$	$[M^1{L}^2{T}^{-2}{A}^{-2}]$	$kg{m}^2{s}^{-2}{a}^{-2}=$ henry (H)

Uses of Dimensional Analysis

The following are the uses of dimensional analysis.

(i) To convert a unit from one system to other

Dimensions are quite useful for finding the conversion factor for the unit of a physical quantity from one system to another. We know that

$n_1{u}_1=n_2{u}_2$

where $n_1,n_2$ are numerical values and $u_1$ and $u_2$ are their respective units, suppose dimensions of the physical quantities are $a,b,c$ in mass, length and time respectively then

$.u_1=[M_1{]}^a\mid L_1{]}^b[T_1{]}^c$

$u_2=[M_2{]}^a[L_2{]}^b[T_2{]}^c$

So, $n_2=n_1[\frac{M_1}{M_2}{]}^a[\frac{L_1}{L_2}{]}^b[\frac{T_1}{T_2}{]}^c$

Illustration 2 :

Convert 1 pascal $(N/m^2)$ into c.g.s units.

SOLUTION : Pascal is unit of pressure whose dimensional formula is $[M{L}^{-1}{T}^{-2}]$ so,

Given $a=1,b=-1,c=-2$

$n_1=1$ Pascal

$n_2=$ ?

Using the formula

$\begin{array}{rl}& n_2=n_1[\frac{M_1}{M_2}{]}^a[\frac{L_1}{L_2}{]}^b[\frac{T_1}{T_2}{]}^c\\ & n_2=[\frac{1kg}{1g}{]}^1[\frac{1m}{1cm}{]}^{-1}[\frac{1s}{1s}{]}^{-2}\end{array}$

$\begin{array}{rl}& n_2=[\frac{1000g}{1g}][\frac{100cm}{1cm}{]}^{-1}\times 1\\ & \text{ Hence, }{n}_2=10\\ & 1Pa=10cgs\text{ pressure }\end{array}$

Illustrotion 3 :

What will be the value of 100 newton in a new system which has $10gm,1000cm$ and 1 minute as fundamental units.

SOLUTION : Dimensional formula for force is $[M^1{L}^1{T}^{-2}]$

$\text{ So }$

$a=1,b=1,c=-2$

Using $n_2=n_1[\frac{M_1}{M_2}{]}^a[\frac{L_1}{L_2}{]}^b[\frac{T_1}{T_2}{]}^c$

$n_2=100[\frac{1kg}{10g}{]}^1[\frac{1m}{1000cm}{]}^1[\frac{1s}{1min}{]}^{-2}$

$\begin{array}{rl}& n_2=100[\frac{1000g}{10g}][\frac{100cm}{1000cm}{]}^1[\frac{\frac{1}{60}min}{1min}{]}^{-2}\\ & n_2=100\times 100\times \frac{1}{10}\times \frac{1}{3600}\end{array}$

$n_2=0.27$ in new units.

(ii) To check the correctness of a physical equation

According to the principle of homogeneity of dimensions, only those physical quantities can be added or subtracted which have the same dimensions. In other words an equation which is containing several terms separated from each other by the equality, plus or minus must be of the same dimensions. An equation will be dimensionally correct only and only if all the terms have the same dimensions. For example in the equation

$\begin{array}{rl}& v^2=u^2+2as\\ & [v^2]=L^2{T}^{-2}\\ & [u^2]=L^2{T}^{-2}\\ & [2as]=L{T}^2\cdot L=L^2{T}^2\end{array}$

Thus the equation $v^2=u^2+2$ as is dimensionally correct as all the terms are having the same dimensions.

Illustration 4 :

Verify dimensional accuracy of the formula $Q=\frac{\pi }{8}\frac{\mathit{P}{\mathit{r}}^4}{\eta l}$ where $Q$ is volume flow rate, $P$ is pressure, $r$ is radius, $\eta$ is coefficient of viscosity and $l$ is length.

SOLUTION : LHS $[Q]=\frac{\text{ vol }}{\text{ time }}=L^3{T}^{-1}$

RHS $[\frac{\pi }{8}\frac{P{r}^4}{\eta l}]=\frac{\text{ Pressure }\times \text{ (radius) }{)}^4}{(\text{ coefficient of viscosity) (length) }}$

$=\frac{M^1{L}^{-1}{T}^{-2}\times L^4}{M^1{L}^{-1}{T}^{-1}\times L}=L^3{T}^{-1}$

As dimensions of both sides of equation are the same, so, the formula is dimensionally correct.

(iii) To establish relation among different physical quantities

If the dependence of a physical quantity on other physical quantities is of product type then using dimensions the formula for the given physical quantity can be deduced. This method is also based on the principle of homogeneity of dimensions. For example a mass attached with a spring oscillates in the vertical direction, we assume that the time period of oscillation $(t)$ depends upon spring constant $(k)$, gravity $(g)$ and mass $(m)$, then

$t=\lambda (k{)}^a(g{)}^b(m{)}^c$

where $\lambda$ is a dimensionless constant and $a,b,c$ are to be evaluated, taking dimensions

$\begin{array}{rl}& T=[M{T}^{-2}{]}^a[L{T}^{-2}{]}^b[M{]}^c\\ & M^0{L}^0{T}^1=M^{a+c}\cdot L^b\cdot T-2^{a-2b}\\ & 0=a+c\\ & b=0\\ & 1=-2a-2b\\ & a=-1/2c=1/2\end{array}$

$\begin{array}{rl}& \text{ On comparison }0=a+c\\ & \text{ Hence, }t=\lambda \sqrt{\frac{m}{k}}\end{array}$

Thus: by dimensional analysis we have deduced that time period of a spring-mass system is independent of gravity, directly proportional to square root of mass and is inversely proportional to square root of spring constant.

Illustration 5 :

Viscous force $(F)$ on a body depends upon coefficient of viscosity $(\eta )$ of the medium, velocity $(v)$ of the body and the radius of the spherical body, deduce the formula using dimensions.

SOLUTION : $F=\lambda [\eta {]}^a[\nu {]}^b[r{]}^c,\lambda$ is dimensionless constant

Taking dimensions on both the sides

$\begin{array}{rl}& [M^1{L}^1{T}^{-2}]=[M^1{L}^{-1}{T}^{-1}{]}^a[L{T}^{-1}{]}^b[L{]}^c\\ & [M^1{L}^1{T}^{-2}]=M^a{L}^{-a+b+c}{T}^{-a-b}\\ & \text{ On comparison }a=1,-a+b+c=1,-a-b=2\end{array}$

$\begin{array}{cc}\text{ We get, }& b=1,c=1\\ \text{ So, }& F=\lambda \eta vr\end{array}$

The value of $\lambda$ can be found experimentally, it comes out to be $6\pi$.

Limitations of Dimensional Analysis

(a) For deriving a formula, we should know the quantities on which a particular quantity depends.

(b) The method works only if dependence of physical quantities is of product type only.

(c) The method does not give any information about the dimensionless constant.

(d) We cannot derive the formulae which contain exponential, logarithmic and trigonometrical functions.

(e) The method works only if there are as many equations available to us as there are unknowns. For example if in mechanics a physical quantity depends upon more than three other physical quantities then the method fails.

(f) - An equation can be dimensionally correct but otherwise incorrect. As $s=ut+a{t}^2$ is dimensionally correct but otherwise it is wrong.

Do you know !!

In composite relations, each term has the same dimensi ons. It is because, only like quantities can be added and subtracted from one another.

Types of Physical Quantities on Basis of Dimensions

According to dimensions there are four types of physical quantities.

(a) Dimensional constants : Those quantities which have fixed value but possesses dimensions. For example (i) Gravitational constants (ii) Planck’s constant (iii) Velocity of light etc.

(b) Dimensionless constant : Those quantities which have fixed value but does not have dimensions. For example (i) $\pi$ (ii) $e$ (iii) numerical constants, $1,3,6\dots$ etc.

(c) Dimensional variables : Those quantities which have dimensions but do not have fixed value. For example velocity, volume, force, etc.

(d) Dimensionless variables : Those quantities which do not have dimensions and fixed value. For example (i) strain (ii) relative density (iii) angle etc.

ERRORS IN MEASUREMENT

Measurement is an important and essential part of our life. Generally measured value of a quantity is different from the true value of the physical quantity. The difference between the true value and measured value is called error. Before we discuss about errors let us understand two important terms :

(a) Accuracy : It is the measure of how close the measured value is to the true value of the physical quantity.

(b) Precision : It tells us about the limit or resolution upto which the quantity is measured.

Suppose certain body has true value of its mass as $7.298kg$. In experiment (a) we use an instrument of resolution $0.1kg$ and measured value is $7.2kg$ and in experiment (b) its measured value is $7.48kg$ with a measuring instrument of resolution $0.01kg$. Then the experiment (a) is more accurate (as it is closer to the true value) where as experiment (b) has more precision because of higher resolution.

Types of Errors

(i) Systematic errors : Those errors which tend to be in one direction, either positive or negative, generally their cause is known. These errors can be minimised by improving experimental techniques selecting better equipments and removing personal bias. Some of the sources of systematic errors are:

(a) Instrumental errors: This type of error arises due to imperfect design or calibration of the measuring instrument, for example zero mark of varnier scale may not coincide with zero mark of main scale in a vernier callipers.

(b) Imperfection in experimental procedure : For example, measuring temperature of a human body by placing thermometer under armpit would give lower temperature than the actual body temperature, ignoring force of buoyancy during the measurement of weight a body etc.

(c) Personal error : This type of error arise due to lack of proper setting of the apparatus, individual bias, or due to carelessness while taking observation. For example if you hold your head too much towards while reading ammeter or voltmeter there will be some error due to parallax.

(d) Errors due to external factors like variation in temperature, humidity pressure wind etc. may introduce errors. For example wind may introduce error while taking the time period of a simple pendulum.

(ii) Random Errors : These arise due to unpredictable and random variations in experimental conditions like temperature, voltage supply, personal (unbiased) error by observer etc. These errors are also called ‘chance’ errors as these occurs irregular and are random with respect to sign (negative or positive) and size.

Random errors can be minimised by taking the observation several times and taking the arithmetic mean of all the observation. The error associated with the resolution of an instrument is called least count error. The smallest division on the scale of a measuring instrument is called its least countr. By using instrument of high precision and improving experimental technique we can minimise least count errors.

(iii) Gross Errors : These arise entirely due to carelessness of the observer like reading an instrument without proper setting, recording observation incorrectly etc. This type of errors can be minimised if the observer is mentally alert and sincere.

Other Types of Errors

(a) Absolute error : The magnitude of the difference between the true value and the individual measured value is called absolute error of the measurement.

Suppose individual values obtained in various observations are $a_1,a_2,a_3\dots \dots ..a_n$.

Then the true mean value is given by

$A_{\text{true }}=a_{\text{mean }}=\frac{a_1+a_2+a_3+\dots +a_n}{n}$

Then the absolute errors are :

$\begin{array}{rl}& \mathrm{\Delta }{a}_1=a_{\text{mean }}-a_1\\ & \mathrm{\Delta }{a}_2=a_{\text{mean }}-a_2\\ & \dots \dots \dots \dots \dots \\ & A_n=a_{\text{mean }}-a_n\end{array}$

Absolute errors can be negative or positive or zero also.

(b) Mean absolute error : It is the arithmetic mean of magnitudes of absolute errors in all measurements.

$\overline{\mathrm{\Delta }a}=\mathrm{\Delta }{a}_{\text{mean }}=\frac{|\mathrm{\Delta }{a}_1|+|\mathrm{\Delta }{a}_2|+\dots \dots +|\mathrm{\Delta }{a}_n|}{n}$

Final measurement in the terms of mean absolute error is expressed as $(a_{\text{mean }}\pm \mathrm{\Delta }{a}_{\text{mean }})$ or $a_{\text{mean }}+\overline{\mathrm{\Delta }a}$

Remember the mean absolute error has the same unit as that of the measured quantity.

(c) Relative or fractional error : It is equal to the ratio of mean absolute error to the mean (true) value of measured quantity.

$\begin{array}{rl}& \text{ Relative Error }=\frac{\text{ Mean Absolute Error }}{\text{ Mean or True Value }}\\ & \text{ Relative Error }=\frac{\overline{\mathrm{\Delta }a}}{a_{\text{mean }}}\end{array}$

Relative Error does not a unit.

If it is expressed in terms of percentage then it is called percentage error $(\delta a)$.

( $\delta$ a) Percentage Error $=$ Relative error $\times 100$

( $\delta a)$ Percentage Error $=\frac{\overline{\mathrm{\Delta }a}}{a_{\text{mean }}}\times 100$

Final measurement in terms of the percentage error will be expressed as ( $a_{\text{mean }}\pm \delta a$ ).

COMBINATION OF ERRORS

Suppose a physical quantity depends upon a number of other quantities which are to be measured experimentally. When we measure those quantities, there will be errors which we also affect the final measurement. In order to estimate the error in the final measurement we should know how individual errors combine in addition, subtraction, multiplication and division.

(a) Error in the sum of quantities: Let $A$ and $B$ two quantities whose measured values are $(A\pm \mathrm{\Delta }A)$ and $(B+\mathrm{\Delta }B)$ where $A$ and $B$ are their true/mean value and $\mathrm{\Delta }A$ and $\mathrm{\Delta }B$ are mean absolute errors respectively.

There is a physical quantity $P$ given as

$P=A+B$

Then, $P\pm \mathrm{\Delta }P=(A\pm \mathrm{\Delta }A)+(B\pm \mathrm{\Delta }B)$

or $P\pm \mathrm{\Delta }P=(A+B)\pm \mathrm{\Delta }(A+\mathrm{\Delta }B)$

So $\mathrm{\Delta }P=\mathrm{\Delta }A+\mathrm{\Delta }B$ (Maximum value of absolute error)

So, when two physical quantities are added the mean absolute error in the final measurement is equal to the sum of individual mean absolute errors.

(b) Error in difference of quantities : Suppose a physical quantity $P$ is equal to the difference of two quantities as :

$P=A-B$

Then $\mathrm{\Delta }P=\mathrm{\Delta }A+\mathrm{\Delta }B$

So, when two physical quantities are subtracted then the mean absolute error in the final measurement is given by the sum of individual mean absolute errors.

(c) Error in product of quantities :

Let $P=A\times B$

Then $P\pm \mathrm{\Delta }P=(A\pm \mathrm{\Delta }B)\times (B\pm \mathrm{\Delta }B)$

$P\pm \mathrm{\Delta }P=AB\pm B\mathrm{\Delta }A)\pm A\mathrm{\Delta }B\pm \mathrm{\Delta }A\mathrm{\Delta }B$

$\mathrm{\Delta }P=B\mathrm{\Delta }A+A\mathrm{\Delta }B(\mathrm{\Delta }A\mathrm{\Delta }B\text{ is quite small so it is neglected })$

$\frac{\mathrm{\Delta }P}{P}=\frac{B\mathrm{\Delta }A+A\mathrm{\Delta }B}{AB}\text{ or, }\frac{\mathrm{\Delta }P}{P}=\frac{\mathrm{\Delta }A}{A}+\frac{\mathrm{\Delta }B}{B}\text{ or }\frac{\mathrm{\Delta }P}{P}\times 100=\frac{\mathrm{\Delta }A}{P}\times 100+\frac{\mathrm{\Delta }B}{B}\times 100$

So, when two quantities are multiplied the fractional error (or percentage error) in the result is equal to the sum of individual fractional (or percentage) errors.

(d) Error in division of quantifios : Suppose a physical quantity is given as :

$$ P=\frac{A}{B} $

$P\pm \mathrm{\Delta }P=\frac{A\pm \mathrm{\Delta }A}{B\pm \mathrm{\Delta }B}=\frac{A(1\pm \frac{\mathrm{\Delta }A}{A})}{B(1\pm \frac{\mathrm{\Delta }B}{B})}$

Then

$P(1\pm \frac{\mathrm{\Delta }P}{P})=\frac{A}{B}(1\pm \frac{\mathrm{\Delta }A}{A})(1\pm \frac{\mathrm{\Delta }B}{B}{)}^{-1}$

$P(1\pm \frac{\mathrm{\Delta }P}{P})=P(1\pm \frac{\mathrm{\Delta }A}{A})(1\mp \frac{\mathrm{\Delta }B}{B})\text{ Using }(1+x{)}^n=1+nx\text{ when }x<1$

$1\pm \frac{\mathrm{\Delta }P}{P}=1\pm \frac{\mathrm{\Delta }A}{A}\mp \frac{\mathrm{\Delta }B}{B}\mp \frac{\mathrm{\Delta }A\times \mathrm{\Delta }B}{AB}$

$\frac{\mathrm{\Delta }A}{A}\cdot \frac{\mathrm{\Delta }B}{B}\text{ is small }$

hence neglected

$\frac{\mathrm{\Delta }P}{P}=\frac{\mathrm{\Delta }A}{A}+\frac{\mathrm{\Delta }B}{B}(\text{ taking maximum value) }$

Or $\frac{\mathrm{\Delta }P}{P}\times 100=\frac{\mathrm{\Delta }A}{A}\times 100+\frac{\mathrm{\Delta }B}{B}\times 100$

So, in division of quantities fractional error (or percentage error) in the final result is equal to the sum of individual fractional error (or percentage error).

(e) Errer in quantity ralsed to some power:

If $P=A^2=A\times A$

Then $\frac{\mathrm{\Delta }P}{P}=\frac{\mathrm{\Delta }A}{A}+\frac{\mathrm{\Delta }A}{A}$

$\frac{\mathrm{\Delta }P}{P}=\frac{\mathrm{\Delta }A}{A}$

In General,

If $P=\frac{A^x{B}^y}{c^z}$

then $\frac{\mathrm{\Delta }P}{P}=x\cdot \frac{\mathrm{\Delta }A}{A}+y\frac{\mathrm{\Delta }B}{B}+z\frac{\mathrm{\Delta }C}{C}$

So, the fractional error (or percentage error) in a physical quantity raised to the power is equal to power times the fractional error (or percentage error). Therefore the physical quantities which are raised to some powers should be measured with maximum accuracy (least error).

Illustration 6 :

Two resiatances $R_1=\mathbf{1}\mathbf{0}\pm 1\mathrm{\Omega }$ and and $R_2=20\pm 3\mathrm{\Omega }$ are connected in series, find equivalent resistance, express your answers in terms of mean absolute errors and percentage error.

Solution : In series combination of resistances, the resultant is given by

So,

$\begin{array}{rl}& R_S=R_1+R_2\\ & R_S=10+20& {\overline{\mathrm{\Delta }R}}_S=R_1+\mathrm{\Delta }{R}_2\end{array}$

$R_S=30\mathrm{\Omega }{\overline{\mathrm{\Delta }R}}_S=1+3=4\mathrm{\Omega }$

Resistance in terms of mean absolute error $=30\mathrm{\Omega }\pm 4\mathrm{\Omega }$

Percentage Error $=\frac{{\overline{\mathrm{\Delta }R}}_S}{R_S}=\frac{4}{30}\times 100=13.33$

Sa. resistance in terms of $$ Error $=30$

Illustration 7 :

If $L=l_1-l_2$ where $l_1\pm {\mathrm{\Delta }}_1=20\pm 0.2cm$ and $l_2+\mathrm{\Delta }{l}_2=10\pm 0.5cm$, find $L$ and express it in terms of mean absolute and percentage error

solution: As

$\begin{array}{rl}& L=l_1-l_2\\ & L=20-10=10cm\\ & \mathrm{\Delta }=\mathrm{\Delta }{l}_1+\mathrm{\Delta }{l}_2\\ & \mathrm{\Delta }I=0.2+0.5=0.7cm\end{array}$

So, the final length $10cm\pm 0.7cm$

Now, percenage error $=\frac{0.7}{10}\times 100=7$

So, final length in terms of percentage error $=10cm\pm 7$

Illustration 8 :

Leagth and breadth of a rectangle are given as $l=100cm\pm 2$ and $b=20cm\pm 4$. Find area in terms of percentage and abselute error.

solution: As in multiplication percentage errors are added so,

percentage error in area $\frac{\mathrm{\Delta }A}{A}\times 100=2+4=6$

where

$\begin{array}{rl}& A=lb\\ & A=100\times 20\\ & A=2000c{m}^2\end{array}$

Hence, area in terms of percentage error $2000c{m}^2\pm 6$

Now,

$\frac{\mathrm{\Delta }A}{2000}\times 100=6$

$\mathrm{\Delta }A=120c{m}^2$

$\Rightarrow 2000c{m}^2\pm 120c{m}^2$

Illustration 9 :

Find mean abselute error and percentiage error in $R$ if $R=\frac{V}{I}$ where $V=50\pm 2$ volt & $I=10\pm 1$ ampere.

solution:

$R=\frac{50}{10}=5\mathrm{\Omega }$

Using

$\begin{array}{rl}& \frac{\mathrm{\Delta }R}{R}=\frac{\mathrm{\Delta }V}{V}+\frac{\mathrm{\Delta }V}{I}\\ & \frac{\mathrm{\Delta }R}{R}=\frac{2}{50}+\frac{1}{10}\end{array}$

$\begin{array}{rl}& \frac{\mathrm{\Delta }R}{R}=\frac{2+5}{50}\\ & \mathrm{\Delta }R=0.7\mathrm{\Omega }\text{ (Mean Absolute Error) }\end{array}$

$\text{ Percentage Error }=\frac{\mathrm{\Delta }R}{R}\times 100=\frac{0.7}{5}\times 100=14$ %

Note : Percentage error in $R$ can also be found by adding percentage error in $V(\frac{2}{50}\times 100=4)$% and that in $I(\frac{1}{10}\times 100=10)$ %

Illustrotion 10

A physical quantity $P$ is related to four observations $a,b,c$ and $d$ as follows : $P=\frac{a^3{b}^2}{\sqrt{cd}}$

The percentage errors of measurement in $a,b,c$ and $d$ are $1$ and $2$ respectively. What is the percentage error in the quantity $P$.

solution : Given

$P=\frac{a^3{b}^2}{\sqrt{cd}}$

So,

$\begin{array}{c}\frac{\mathrm{\Delta }P}{P}\times 100=3(\frac{\mathrm{\Delta }a}{3}\times 100)+2(\frac{\mathrm{\Delta }b}{b}\times 100)\times \frac{1}{2}(\frac{\mathrm{\Delta }c}{c}\times 100)+\frac{1}{2}(\frac{\mathrm{\Delta }d}{d}\times 100)\\ \frac{\mathrm{\Delta }P}{P}\times 100=3\times 1+2\times 3+\frac{1}{2}\times 4+\frac{1}{2}\times 2\end{array}$

Hence, $\frac{\mathrm{\Delta }P}{P}\times 100=13$ %

SIGNIFICANT FIGURES

Significants digits or figures give information about the accuracy of a measurement. It tells us about the number of digits in which we have confidence. Suppose a particular measurement is reported to be $9.28cm$, then the two digits 9 and 2 are reliable and certain while the digit 8 is uncertain. The reliable and first uncertain digits are known as significant digits or figures. There are certain rules for counting significant digits which are as follows :

Rule-1. All the non-zero digits are significant-For example 2134 has four significant digits and 27184 has five significant digits.

Rule-2. All the zeros between two non-zero digits are significant, no matter where the decimal point is, if at all. For example 25089 has five significant digits, 12.0021 has six significant digits.

Rule-3. In a number which is less than one all zeros to the right of decimal point and to the left of a non-zero digit are not significant.

Rule-4. All the zeros on the right of last non-zero digits are significant in a number with a decimal point. For example in 3,500 there are four significant digits and in $\mathbf{0}\mathbf{.}\mathbf{0}\mathbf{7}\mathbf{9}\mathbf{0}\mathbf{0}\mathbf{0}$ there are five significant digits.

Rule-5. All the zeros on the right on a non-zero digit are not significant in a number without decimal point. For example 15800 has only three significant digits, 18930000 has only four significant digits.

Rule-6. All the zeros on the right on a non-zero digit are taken to be significant when these come from a measurement. For example some distance is measured to be $7890m$ then this number would have four significant digits.

Rule-7. A change of system of units does not change the number of significant digits in a measurement. Also when a number is written in scientific notation $(a\times {10}^b)$ then the powers of 10 are irrelevant to the determination of significant digits.

Illustration 11 :

Write down the number of significant figures in the following :

(i) $6729N$ (ii) 0.024 (iii) $6.0023gc{m}^{-3}$ (iv) $2.520\times {10}^{7}m$ (v) $0.08240N{m}^{-2}$ (vi) 4200 (vii) $4.57\times {10}^8$ (viii) $91.000m$

solution: (i) $6729N$ has four significant figures.

(ii) $0.024cm$ has two significant figures.

(iii) $6.0023gc{m}^{-3}$ has five significant figures.

(iv) $2.520\times {10}^{7}m$ has four significant figures.

(v) $0.08240N{m}^{-2}$ has four significant figures.

(vi) $\mathbf{4}\mathbf{2}\mathbf{0}\mathbf{0}$ has two significant figures.

(vii) $4.57\times {10}^8$ has three significant figures.

(viii) 91.000 has five significant figures.

Rules for Arithmetic Operations with significant Figures

(a) Addition and subtraction : The final result should retain as many decimal places as there are in the number with the least decimal places. For example if we add $1.269,26.57$ and 9.1 the sum will be 36.939 but according to the rule the final result must retain only one decimal place i.e. 369 .

(b) Multiplication and division : In multiplication and division the final result should retain least number of significant digits from among all the numbers. Suppose density of a material is $6.921kg/m^3$ and volume is $2.1m^3$ then it mass will be $m=6.921\times 2.11=14.60331kg$

But according to the rule of significant digits it may be $14.6kg$.

As the number 6.921 has 4 significant digits, 2.11 has 3 significant digits so the final answer must contain only ’ 3 ’ significant digits.

$✓$ CHECK POINT

Obtain the value of $(500.0m+600mm)$ with due regards to significant figures.

Check Your Answer

$500.0m+600mm=500.0+0.600m=500.600m\approx 501m$

Illustration 12 :

Each side of a cube is measured to be $7.293m$. What are the total surface area and volume of the cube to appropriate significant digits.

solution : Here, the side of cube is measured upto ’ 4 ’ significant digits, so the calculated area and volume should be rounded off to 4 singnificant digits.

Surface area $=6(\text{ side }{)}^2=6(7.293{)}^2=319.12704m^2$

Surface area $=319.1m^2$

Volume $=(\text{ side }{)}^3=(7.293{)}^3=387.8989828m^3$

Volume $=387.9m^3$

ROUNDING OFF A NUMBER

(a) Uncertainity : It is taken a half of the least count of the measuring instrument. Suppose length and breadth of a rectangular piece of paper are measured with the help of a meter scale with least count of $1mm$. If the length and breadth are $10.6cm$ and $5.1cm$ respectively then,

$\begin{array}{rl}& l=10.6\pm 0.05cm\text{ (uncertainity is half of least count which is }1mm=0.1cm\text{ ) }\\ & b=5.1\pm 0.05cm\end{array}$

If, we find the area of rectangular sheet of paper then uncertainity in the final result will equal to the square root of the sum of square of percentage uncertainities:

$l=10.6+0.05cm;b=5.1\pm 0.05cm$

$l=10.6+0.47$ % ; $b=5.1\pm 0.98$

Area $=lb=54.06\pm \sqrt{(0.47{)}^2+(0.98{)}^2}$

$lb=54.06\pm 1.1$ %

(b) Rounding off the uncertain digit :

Rule-1. Preceding digit is increased by one if the insignificant (uncertain) digit which is to be rounded off is more than 5.

For example a number $9.87\underset{―}{6}$ is to be rounded off to 3 significant digits, then rounded off number will be 9.88 as insignificant digit is more than 5 .

Rule-2. Preceding digit remains the same if the insignificant digit which is to be rounded off is less than 5. Suppose the number is 9.873 then after rounding off to 3 significant digits it will become 9.87 .

Rule-3. If the digit to be rounded off is 5 and preceding digit is even then the preceding digit is left unchanged. For example the number to be rounded off is $9.86\underset{―}{5}$ then the rounded off number will be 9.86 .

Rule-4. If the digit to be rounded off is 5 and preceding digit is odd then the preceding digit is increased by one.

For example if the number is $9.83\underset{―}{5}$ to be rounded off to three significant digits then after rounding off, the new number will be 9.84 .

Illustration 13

Round off the numbers 8121, 978.5, 12.68, 5.735, 8.925, 11.2 2

Solution :

SL. No	Number to be rounded off	Rounded off number	Rule
1.	$81\underset{―}{\underset{―}{1}}$	8100	2
2.	$978.\underset{―}{5}$	978	3
3.	$12.6\underset{―}{8}$	12.7	1
4.	$5.73\underset{―}{\underset{―}{5}}$	5.74	4
5.	$8.92\underset{―}{5}$	8.92	3
6.	$11.2\underset{―}{2}$	11.2	2

EXERCISE 1

DIRECTIONS : Complete the following statements with an appropriate word / term to be filled in the blank space(s).

1. There are fundamental and supplementary units in S.I. system.
2. The measurement of a physical quantity is the prouduct of the numerical value and the
3. In C.G.S. system, force is measured in
4. Metre per second is a of that quantity. unit.
5. Candela is a unit.
6. There are two supplementary units in S.I. system. They are and
7. Ampere is an S.I. unit for the quantity called
8. $[L{T}^{-1}]$ are the dimensions for
9. The dimensions of force are
10. The dimensional analysis can be used to check the of a given relationship.
11. If errors in the measurement of the length and the breadth of a rectangle are $1.5$ and $2$ respectively then the error in the measurement of its area is
12. The number of significant figures in 12345 is
13. The number 1040 has significant figures.
14. A physical relation which is correct is not necessarily physically correct.
15. Angstrom is a unit for
16. S.I. unit of luminous intensity is …..
17. The dimensions of ……………….. in force is -2 .
18. Instrumental error belong to ………………… errors.
19. Number of significant figures in 4.00 is
20. 1 Angstrom is equal to .. $cm$.
21. The word RADAR stands for ranging.
22. ……………….. waves are used in SONAR.
23. Mass of electron is or order of
24. S.I unit of angle is to obtain accurate value of the quantity.

DIRECTIONS : Read the following statements and write your answer as true or false.

1. Light year is a unit of time.
2. The magnitude of a quantity change with change in the unit of system.
3. A quantity has same dimensions in different system of units.
4. After rounding off the number 0.05857 to two significant figures we get 0.059 .
5. The S.I. unit of the amount of substance is mole.
6. The S.I. unit of acceleration is $m{s}^{-1}$.
7. The unit of surface tension is newton per metre $(N{m}^{-1})$.
8. The dimensions of force are $ML{T}^{-2}$.
9. The units of work and torque are same.
10. The error in the measurement in the radius of a sphere is $2$. Therefore, the error in the measurement of its volume is $8$.
11. Dimensional analysis is not applicable in case of trigonometrical and logarithmic function.
12. A constant has the dimension of length.
13. The dimensional equation for acceleration is [Acceleration] $=ML{T}^{-2}$
14. The equation $v=u+\frac{1}{2}$ at is dimensionally correct.
15. The unit of magnetic energy is same as that of the unit of mechanical energy.
16. A unitless quantity must be dimensionless.
17. If numerical value of a quantity is $x$ in cgs system and $y$ in mks system, then $x$ must be less than $y$.
18. If $x=ab$, then absolute error in $x$ is greater than absolute error in $a$ as well as in $b$.
19. If $x=\frac{a}{b}$, then absolute error in $x$ is greater than absolute error in $a$ as well as in $b$.
20. Light year is a unit of time.
21. Parsec is largest practical unit of length.
22. Two different quantities may have same dimensions.
23. Speed and velocity have same dimensions.
24. Zero error is a constant error
25. Random error can be removed completely.

**DIRECTIONS : **Each question contains statements given in two columns which have to be matched. Statements $(A,B,C,D)$ in Column I have to be matched with statements $(p,q,r,s)$ in Column II.

Column I	Column II
(Quantity)	(Unit)
(A) Area	(p) $Kg{m}^2{s}^{-2}$
(B) Speed	(q) $m{s}^{-1}$
(C) Force	(r) $m^2$
(D) Work	(s) $kg{ms}^{-2}$

Column I	Column II
(Quantity)	(Dimension)
(A) Momentum	(p) $M{L}^2{T}^{-2}$
(B) Coefficient of viscosity	(q) $ML{T}^{-2}$
(C) Force	(r) $ML{T}^{-1}$
(D) Energy	(s) $M{L}^{-1}{T}^{-1}$

Column I	Column II
(Quantity)	(C.G.S unit)
(A) Acceleration due to gravity	(p) $erg/g^{\circ }C$
(B) Specific heat capacity	(q) $gcm{s}^{-1}$
(C) Impulse	(r) $cm{s}^{-2}$
(D) Power	(s) ergs ${}^{-1}$

4. Match the quantities (Column I) with their respective dimensions (Column II).

Column I	Column II
(A) Velocity	(p) $[M^{0}L{T}^{-1}]$
(B) Force	(q) $[M^{0}L{T}^0]$
(C) Acceleration	(r) $[M^{0}L{T}^{-2}]$
(D) Displacement	(s) $[ML{T}^{-2}]$

5. Match the quantities (Column I) “with their S.I units (Column II)

Column I	Column II
(A) Mass	(p) kelvin
(B) Length	(q) kilogram
(C) Electric current	(r) metre
(D) Temperature	(s) ampere

DIRECTIONS : Give answer in one word or one sentence.

1. What is meant by unit?
2. What should we know in order to measure a physical quantity?
3. Is light year a unit of time?
4. Define light year.
5. What is the unit for measuring wavelength of light ?
6. How many times a kg is larger than a $mg$ ?
7. What are the dimensions of angular displacement?
8. What are the dimensions of angular momentum ?
9. What are the dimensional formulae for the following: (i) pressure, (ii) power, (iii) density and (iv) angle
10. Find dimensional formula of latent heat.
11. Name two physical quantities which have same dimensions as that of work.
12. Can a quantity have units but still be dimensionless ?
13. Are all constants dimensionless ?
14. What is the unit of pressure in SI?
15. Define significant figures.

DIRETIONS : Give answer in 2-3 sentences.

1. What importance do we attach to zeroes immediately after the decimal point?
2. Subtract with due regard to significant figures: $3.9\times {10}^5-2.5\times {10}^4$
3. Solve the following with due regard to significant figures: $\frac{0.9996\times 3.52}{1.758}$
4. What are the dimensions of gravitational constant ?
5. Give the dimensional formula of thermal conductivity.
6. Write four pairs of physical quantities, which have the same dimensional formula.
7. What is the advantage of expressing physical quantities in terms of dimensional equations?
8. In the equation $y=A\sin (\omega t-kx),t$ and $x$ stand for time and distance respectively. Obtain the dimensional formula of $\omega$ and $k$.
9. Derive the SI unit of joule ( $J$ ) in terms of fundamental units.
10. What is the advantage in choosing the wavelength of a light radiation as a standard of length ?

DIRECTIONS : Give answer in four to five sentences.

1. What do you mean by the unit of a physical quantity? Explain the types and systems of units.
2. What is S.I. system of units ? Write its fundamental and derived units.
3. Define the fundamental units of S.I. system.
4. What do you mean by dimension of a physical quantity? Write the uses of dimensional analysis.
5. How the dimensional method is used to convert a unit from one system to another? Explain with the help of an illustration.
6. What is significant figure? Mention the rules to write the number of significant figures.

EXERCISE

DIRECTIONS : This section contains multiple choice questions. Each question has 4 choices (a), (b). (c) and (d) out of which ONLY ONE is correct.

1. Which of the following systems of units is not based on units of mass, length and time alone?

(a) SI

(b) MKS

(b) CGS

(d) FPS

2. Unit of latent heat is

(a) $Jk{g}^{-1}$

(c) $Nk{g}^{-1}$

(b) $Jmo{l}^{-1}$

(d) $Nmo{l}^{-1}$

3. Which of the following is not a unit of time?

(a) Solar year

(b) Tropical year

(c) Leap year

(d) Light year

4. Dyne-sec is the unit of

(a) momentum

(b) force

(c) work

(d) angular momentum

5. One shake is equal to

(a) ${10}^{-8}s$

(b) ${10}^{-9}s$

(c) ${10}^{-10}s$

(d) ${10}^{9}s$

6. One torr is equal to

(a) $1cm$ of $Hg$ col.

(b) Atmosphere

(c) $1N{m}^{-2}$

(d) $1mm$ of $Hgcol$.

7. What are the units of magnetic permeability?

(a) $Wb{A}^{-1}{m}^{-1}$

(c) $WbAm$

(b) $W{b}^{-1}Am$

(d) $Wb{A}^{-1}m$

8. The ampere-second is a unit of

(a) current

(b) charge

(c) energy

(d) power

9. The SI unit of coefficient of mutual inductance of a coil is

(a) henry

(b) volt

(c) farad

(d) weber

10. SI unit of magnetic flux is

(a) gauss

(b) weber

(c) oersted

(d) ampere/metre

11. Unit of specific resistance is

(a) $ohm/m^2$

(b) ohm $m^3$

(c) ohm - m

(d) $ohm/m$

12. Light year is

(a) Light emitted by the sun in one year.

(b) Time taken by light to travel from sun to earth.

(c) The distance travelled by light in free space in one year.

(d) Time taken by earth to go once around the sun.

13. SI unit of pressure is

(a) atmosphere

(b) bar

(c) pascal

(d) $mm$ of $\mathbf{H}\mathbf{g}$

14. Electron volt is a unit of

(a) potential difference

(b) change

(c) energy

(d) capacity

15. Dimension of impulse are

(a) $[ML{T}^{-1}]$

(c) $[M{T}^{-2}]$

(b) $[ML{T}^2]$

(d) $[M{L}^{-1}{T}^3]$

16. Units of coefficient of viscosity are

(a) $Nm{s}^{-1}$

(b) $N{m}^2{s}^{-1}$

(c) $N{m}^{-2}$

(d) None of these

17. What are the dimensions of Action?

(a) $M^{2}L{T}^{-3}$

(b) $ML{T}^{-1}$

(c) $ML{T}^{-2}$

(d) $M{L}^2{T}^1$

18. Which is dimensionless?

(a) Force/acceleration

(b) Velocity/acceleration

(c) Volume/area

(d) Energy/work

19. Potential is measured in

(a) joules/coulomb

(c) newton-second

(b) watt/coulomb

(d) none of these

20. One second is defined to be equal to

(a) 1650763.73 periods of the Krypton clock

(b) $\mathbf{6}\mathbf{5}\mathbf{2}\mathbf{1}\mathbf{8}\mathbf{9}\mathbf{.}\mathbf{6}\mathbf{3}$ periods of the Krypton clock

(c) 1650763.73 periods of the Cesium clock

(d) 9192631770 periods of the Cesium clock

21. If $C$ and $L$ denote the capacitance and inductance, the units of $LC$ are

(a) $M^0{L}^0{T}^{-1}$

(c) $M^{-1}{L}^{-1}{T}^0$

(b) $M^0{L}^{-1}{T}^0$

(d) $M^0{L}^0{T}^2$

22. The dimensions of electromotive force in terms of current A are

(a) $M{T}^{-2}{A}^{-2}$

(c) $M{L}^2{T}^{-2}{A}^{-2}$

(b) $M{L}^2{T}^{-2}{A}^2$

(d) $M{L}^2{T}^{-3}{A}^{-1}$

23. The expression $[M{L}^{-1}{T}^{-2}]$ does not represent

(a) pressure

(b) power

(c) stress

(d) Young’s modulus

24. The dimensions of universal gas constant are

(a) $L^2{M}^1{T}^{-2}{K}^{-1}$

(c) $L^1{M}^1{T}^{-2}{K}^{-1}$

(b) $L^1{M}^2{T}^{-2}{K}^{-1}$

(d) $L^2{M}^2{T}^{-2}{K}^{-1}$

25. Dimensions of specific heat are

(a) $M{L}^2{T}^{-2}K$

(b) $M{L}^2{T}^{-2}{K}^{-1}$

(c) $M{L}^2{T}^2{K}^{-1}$

(d) $L^2{T}^{-2}{K}^{-1}$

26. Which physical quantities have same dimension?

(a) Moment of couple and work

(b) Force and power

(c) Latent heat and specific heat

(d) Work and power

27. Distance travelled by a particle at any instant ’ $R$ ’ can be represented as $S=A(t+B)+C{r}^2$. The dimensions of $B$ are

(a) $M^0{L}^1{T}^{-1}$

(c) $M{L}^{-1}{T}^{-2}$

(b) $M^0{L}^{0}T$

(d) $M{}^0{L}^2{T}^{-2}$

28. In the eqn. $(P+\frac{a}{V^2})(V-b)=$ constant, the units of a are

(a) dyne $\times c{m}^5$

(b) dyne $\times c{m}^4$

(c) $dyne/c{m}^3$

(d) dyne $\times c{m}^2$

29. Error in the measurement of radius of a sphere is $1$. Then error in the measurement of volume is

(a) $1$

(b) $5$

(c) $3$

(d) $8$

30. A quantity is represented by $X=M^a{L}^{b}T$. The $$ error in measurement of $M,L$ and $T$ are $\alpha$ and $\gamma$ respectively. The % error in $X$ would be

(a) $(\alpha a+\beta b+\gamma c)$

(b) $(\alpha a-\beta b+\gamma c)$

(c) $(\alpha a-\beta b-\gamma c)\times 100$

(d) None of these

31. Subtract $0.2J$ from $7.26J$, and express the result with correct number of significant figures

(a) $7.1J$

(c) $7.0J$

(b) $7.06J$

(d) $7J$

32. Multiply 107.88 by 0.610 and express the result with correct number of significant figures

(a) 65.8068

(b) 65.807

(c) 65.81

(d) 65.8

33. When 97.52 is divided by 2.54 , the correct result is

(a) 38.3937

(b) 38.394

(c) 38.39

(d) 38.4

34. The radius of a thin wire is $0.16mm$. The area of cross section of the wire in sq. $mm$ with correct number of significant figures is

(a) 0.08

(b) 0.080

(c) 0.0804

(d) 0.80384

35. S.I. unit of surface tension is

(a) degree/ $/cm$

(b) $N/m$

(c) $N/m^2$

(d) $Nm$

36. Weber $/m^2$ is equal to

(a) Tesla

(c) Watt

(b) Henry

(d) None

37. $M{L}^2{T}^{-2}$ are dimensions of

(a) Force

(b) Moment of force

(c) Momentum

(d) Power

38. Which of the following is not the name of a physical quantity?

(a) Displacement

(b) Momentum

(c) Metre

(d) Torque

39. Watt-hour meter measures

(a) current

(b) voltage

(c) power

(d) electric energy

40. If $I$ is regarded as the fourth dimension, then the dimensional formula of charge in terms of current $I$ is

(a) $[I{T}^2]$

(b) $[I^0{T}^0]$

(c) $[r^{-1}{T}^0]$

(d) $[IT]$

41. If time $T$, acceleration $\mathit{A}$ and force $F$ are regarded as base units, then the dimensional formula of work is

(a) $[FA]$

(c) $[.$ FAT ${.}^2]$

(b) $[FAT]$

(d) $[F{A}^{2}T]$

42. The number of significant figures in $0.00060m$ is

(a) 1

(b) 2

(c) 3

(d) 4

43. In a simple pendulum experiment for the determination of acceleration due to gravity, time period is measured with an accuracy of $0.2$ while length was measured with an accuracy of $0.5$. The percentage accuracy in the value of g so obtained is

(a) $0.25$

(b) $0.7$

(c) $0.9$

(d) $1.0$

44. $A$ force is given by $F=at+b{t}^2$, where $t$ is time, the dimensions of $a$ and $b$ are

(a) $[ML{T}^{-4}]$ and $[ML{T}^{-1}]$

(b) $[ML{T}^{-1}]$ and $[ML{T}^0.$

(c) $[ML{T}^{-3}]$ and $[ML{T}^{-4}]$

(d) $[ML{T}^{-3}]$ and $[ML{T}^0.$

More than One correct

DIRECTIONS : This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d) out of which ONE OR MORE may be correct.

1. Which of the following pairs have same dimensions?

(a) Light year and wavelength

(b) Torque and work

(c) Angular momentum and work

(d) Energy and Young’s modulus

2. Choose the correct statements :

(a) A dimensionally incorrect equation may be incorrect

(b) A dimensionally correct equation may be incorrect

(c) A dimensionally incorrect equation may be correct

(d) A dimensionally correct equation may be correct

3. Which of the following are not a unit of time?

(a) Light year

(b) Second

(c) Parsec

(d) Micron

4. When a wave traverses a medium the displacement of a particle located at $x$, at time $t$ is given by ; $y=a\sin (bt-cx)$

where $a,b$ and $c$ are constants of the wave.

Which of the following are dimensionless quantities?

(a) $cx$

(b) $\frac{b}{c}$

(c) $bt$

(d) $\frac{y}{a}$

5. Which of the following are the units of mass?

(a) amu

(b) Quintal

(c) $kg$-wt

(d) Metric ton

6. Which of the following are true regarding significant figures?

(a) The zeros appearing in the middle of a number are significant, while those at the end of a number without a decimal point are ambiguous.

(b) All non-zero digits are significan

(c) Greater the number of significant figures in a measurement smaller is the percentage error

(d) The power of 10 are counted while counting the number of significant figures.

where $f$ represents time in second and sepresents distance in metre. Which of the follow ing stiltement are filke?

(a) The unit of $t$ is salne is that of 4

(b) The :nit of $x$ is same as that of 4

(c) The unit of $t$ is same as that of $p$

(d) The unit of $x$ is same as that of $p$

8. Identify the pairs having ickntical dimensions

(a) Strain and ingle

(b) Planck constant and angulan momintum

(c) Linear momentun and moment of force

(d) Pressure and modulus of clasticils

Multiple Matching Questions

DIRECTIONS : Following question has lon’ velt mem. A. B. $C$ and D) given in Column I and four statcments ip, $y_1,\dots$. . Ind i) in Column II. Any given statement in Column I an have correct matching with one or more statement(s) given in Columin II. Match the entries in Column I with entries in (islumn II.

1. Match the quantities having same dimensions: Column I

Column I	Column II
(A) Stress	(p) Force
(B) Pressure	(q) Strain
(C) Tension	(r) Angle
(D) Refractive index	(s) Young’s modulus
	(t) Energy per unit volum

2. If $A$ and $B$ represent two physical quantitics and $\mathrm{\Delta }.t$ and $\mathrm{\Delta }B$ represent their respective mean absolute errors, then match the following.

Column I	Column II
(A) $x=A+B$	(p) $\pm \mathrm{\Delta }x=\pm (\mathrm{\Delta }A+\mathrm{\Delta }B)$
(B) $x=A-B$	(q) $\pm \frac{\mathrm{\Delta }x}{x}=\pm (\frac{\mathrm{\Delta }P}{P}+\frac{\mathrm{\Delta }B}{B})$
(C) $x=AB$	(r) $\pm \mathrm{\Delta }x=:(.+\wedge B+B\wedge A)$
(D) $x=\frac{A}{B}$	(s) $t\frac{\mathrm{\Delta }x}{x}=\pm (\frac{\mathrm{\Delta }P+\mathrm{\Delta }B}{A+B})$
	(t) $\pm \frac{\mathrm{\Delta }x}{x}=\pm (\frac{14+\mathrm{\Delta }B}{A-B})$

Fill in the Passage

DIRECTIONS : Complete the following passage(s) with an appropriate word/term to be filled in the blank spaces.

minimised, instrumental errors, environmental errors, Systematic errors. random errors, personal errors.

I. No measurement can be perfect, some errors are always associated with the measurement. The errors cause of which are known ____ (1) ____ . This type of errors can again of three types. The errors caused due to faulty instrument are called ____ (2) ____ the crrors caused due to carelessness of the observer are called ____ (3) ____ and the errors caused due to environmental factors are called ____ (4) ____ The other category of errons fier which cause of errors is not known are called ____ (5) ____ This type of errors cannot be eluminated completely but can be ____ (6) ____ by taking more and more observations .

kg. S.I system. seven, independent, second, ampere,Fundamental, derived

II. Physical quantities are mainly of two types ____ (1) ____ quantities are the quantities which are basic in nature and cannot be derived from other quantities. There are a total of ____ (2) ____ such quantities. On the other hand ____ (3) ____ quantities are the physical quantities which can be derived from other quantities. The chose of basic quantities is not, i.e. any quantity can be taken a basic but all such quantities must be ____ (4) ____ of each other. There are different systems of units used for measurement of physical quantities but the system which is internationally accepted is called ____ (5) ____ In this system, unit of time is ____ (6) ____ unit of electric current is ____ (7) ____ and unit of mass is ____ (8) ____

Passage Based Questions

DIRECTIONS: Stuly the given paragraph(s) and answer the following yuctions.

Passage - I

Dimensions are the powers to which fundamental quantities must be raised to represent a given physical quantity. Again the choice of fundamental quantities is not unique. A set of physical quantities which are independent of each other may be taken as fundamental quantities. Based on the above information and knowledge of dimensions, answer the following.

1. Which of the following set of quantities cannot be taken as fundamental?

(a) force, length, time

(b) momentum, length, time

(c) acceleration, length, time

(d) mass, velocity, acceleration.

2. If momentum $(P)$, velocity $(V)$ and Time $(T)$ is taken as fundamental quantities, the dimensional formula for mass will be

(a) $[P{V}^{-1}T]$

(c) $[P^{0}V{T}^{-1}]$

(b) $[P{V}^{-1}{T}^0]$

(d) $[PV{T}^0]$

3. In the previous case, the dimensional formula for force will be

(a) $[PVT]$

(c) $[P{V}^{-1}{T}^0]$

(b) $[P{V}^0{T}^{-1}]$

(d) $[P{V}^{-1}{T}^2]$

Passage II

the mas of a cube measured with a balance of least cont $11.2g$ is lound to be $5.0g$ and its length measured w ith the le.lp of a vernier calliper of least count $0.01cm$ is found to he $10cm$. Then.

4. The pereentage error in the measurement of mass of the cule is

(a) $20$

(b) 4”:"

(c) $7$

(d) $10$

5. Nevolute ctror in the measurement of volume is

(i) $0.03c{m}^3$

(b) $0.3c{m}^3$

(c) $0.1c{m}^3$

(d) $0.15c{m}^3$

6. Wholute error in the measurement af density is

(i) $11.0.5g/c{m}^3$

(i) $0.25g/c{m}^3$

(b) $0.15\cdot g/c{m}^3$

(d) $0.35g^{\prime }c{m}^3$

Assertion & Reason

DIRECTIONS : Each of these questions contains an Assertion followed in Reason. Redl them carefully and answer the question an the basis of following options. You have to select the one that best describes the two statements.

(a) II both Assertion and Reason are correct and Reason is the correct explanation of Assertion

(b) If hoth Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

(c) If Assertion is correct but Reason is incorrect.

(d) If Assertion is incorrect but Reason is correct.

1. Assertion : Light year and wavelength both measure distance.

Reason : Both have dimensions of time.

2. Assertion : Force cannot be added to pressure.

Reason : Their dimensions are different.

3. Assertion : Density is a derived physical quantity.

Reason : Density cannot be derived from the fundamental physical quantitics.

4. Assertion : The graph between $P$ and $Q$ is straight line When $P/Q$ is constant.

Reason: The straight line graph means that $P$ is proportional to $Q$ or $P$ is equal to constant multipled by $Q$.

5. Assertion : Number of significant figures in 0.005 is one and that in 0.500 is three.

Reason : This is because zeros are not significant.

6. Assertion : Radian is the unit of distance.

Reason : One radian is the angle subtended at the centre of a circle by an arc equal in length to the radius of the circle.

Hots subjective Questions

DIRECTIONS: Answer the following questions.

1. Each side of a cube is measured to be $7.203m$. What are the total surface area and the volume of the cube to appropriate significant figures?
2. 5.74 g of a substance occupiev $12c{m}^{6}1$ express its density by keeping the significant figures in view.
3. The mass of a box measured by a grocer’s balance is $2.300kg$. Two hold peaces of masses $20.15g$ and $20.17g$ are added we the box. What in a the lotal mass of the box. (b) the difference in the masses of the pieces to correct signiticant figures?
4. Find the dimensions of $a$ and $b$ in the Van der waal’s equation $(P+\frac{a}{v^2})(V-b)=RT$
5. The formula $/$ " is pressure and $V$ is solution of a problem Use dimensions to find whether this is a reasonable solution is a velocity, m are masses/ is a length and g is gravitational accelaration).

D diffusion coefficients $n_1$ and $n_2$, numper of molecules in unit volume along $v_1$, and v. Which represents distances where V is number of molecules passing through per unit area per unit time calculate dimensional question of D

7. A gas bubble from an explosion under water, oscillates with a period T proportionaly to P d F where p isthe state pressure, d isthe density of water and b is the total energy of explosion. Find the values of a, b and c
8. If velocity. force and time are taken to be furdiamental quantities, find dimensional formula tor (a) Masses and (b) Energy.
9. Let us consider an equation $\frac{-1}{2}m{v}^{\ast }$ mgh where $m$ is the mass of the body, its velocity g is the accelaration due to gravity and $h$ is the height check whether this equation is dimensionally correct
10. A calorie is a unit of heat or energy and it equals about 4.2 J where 1J is $1kg{m}^2{s}^2$ . Suppose we employ a system of units in which the unit of mass equals $\alpha$ kg, the unit of the length equals $\beta$m. the unit oftime is $\gamma$s.Show that a caloriw has a magnitude $4.2{\alpha }^1{\beta }^2{\gamma }^2$m terms ofthe new units.
11. What is the dimensions of physical quantity 1 m the equation Force $=\frac{X}{\text{ Density }}$ ?
12. If force, acceleration and time are taken as fundamental quantities, then find the dimensions of length.
13. Check the dimensional consistency of the equation :

$Fs=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2$

where $s$ is distance moved, $u$ and $v$ are the initial and final velocities of a body of mass $m$ and $F$ is the force acting on it.

14. Check the correctness of the relation by dimensional analysis :

$\lambda =\frac{h}{mv}$

where the letters have their usual meanings.

EXERCISE - I

FILL IN THE BLANKS

1. seven, two
2. unit
3. dyne
4. derived
5. fundamenta
6. radian, steradian
7. electric current
8. velocity
9. $ML{T}^{-2}$
10. correctness
11. 3.5 %

Hint : Area $=$ length $\times$ breadth

$\therefore$ Error in measurement of area $=$ Error in measurement of length + Error in measurement of breadth.

12. five
13. three
14. dimensionally
15. length

Hint : Angstrom $(\text{AA})$ is the unit of length. It is related to metre $(m)$ as $1m={10}^{10}\text{AA}$

16. Candela
17. time
18. systematic
19. three
20. ${10}^{>}$
21. radio
22. ultrasonic
23. ${10}^{-30}kg$
24. radian
25. subtracted

TRUE/FALSE

1. False
2. True
3. True
4. True
5. True
6. False
7. True
8. True
9. True
10. False

Hint: Volume $\frac{4}{3}\pi R^3$

$\Rightarrow$ Error in the measurement of volume $3\times$ Error in measurement of R = $(3\times 2)$ % = 6%

11. True
12. False

Hint: A constant has no dimension

13. True
14. True
15. True

Hint : All kind of energy has same unit.

16. True 17. False 18. True
17. False 20. False 21. True
18. True 23. True 24. True
19. False

MATCH THE COLUMNS

1. $A\to (r);B\to (q);C\to (s);D\to (p)$
2. $A\to (r);B\to (s);C\to (q);D\to (p)$
3. $A\to (r);B\to (p);C\to (q);D\to (s)$
4. $A\to (p);B\to (s);C\to (r);D\to (q)$
5. $A\to (q);B\to (r);C\to (s);D\to (p)$

VERY SHORT ANSWER QUESTIONS

1. The unit of a physical quantity is a standard of the same kind chosen in order to measure that quantity.

2. Its unit and the number of times the unit is contained in the physical quantity.

3. No, it is a unit of distance.

4. One light year is defined as the distance travelled by light in one year.

5. Angstrom $(\text{AA})$. One angstrom is equal to ${10}^{-10}m$.

6. $\frac{1kg}{1mg}=\frac{{10}^{3}g}{{10}^{-3}g}={10}^6$

7. Since angular displacement $=\frac{\text{ length of arc }}{\text{ radius }}$ it is dimensionless.

8. $[M{L}^2{T}^{-1}]$

9. (i) $[M{L}^{-1}{T}^{-2}]$, (ii) $[M{L}^2{T}^{-3}]$, (iii) $[M{L}^{-3}{T}^0]$ and (iv) dimensionless.

10. [Latent heat] $=[\frac{\text{ quantity of heat }}{\text{ mass }}]=[\frac{\text{ energy }}{\text{ mass }}]$ $=\frac{[M{L}^2{T}^{-2}]}{[M]}=[M^0{L}^2{T}^2]$

11. Torque and energy

12. Yes. For example, unit of angle is radian, but it is a dimensionless quantity.

13. No. For example, $G$ (gravitational constant); $h$ (Planck’s constant): $k$ (Boltzmann constant); etc are constants, but possess dimensions.

14. $N{m}^{-2}$

15. The digits, whose values are accurately known in a particular measurement, are called its significant figures.

SHORT ANSWER QUESTIONS

1. All zeros to the right of a decimal point and to the left of a non-zero digit are never significant. 0.00967 contains three significant figures. The single zero conventionally placed to the left of the decimal point in such an expression is also never significant. All zeroes to the right of a decimal point are significant, if they are not followed by a non-zero digit. 30.00 contains four significant figures. All zeros to the right of the last non-zero digit after the decomal point are significant. 0.07800 contains four significant figures.

2. Now, $3.9\times {10}^5-2.5\times {10}^4=390000-25000=365000$ $=3.65\times {10}^5$

Since each of $3.9\times {10}^5$ and $2.5\times {10}^4$ has got two significant figures, the result should also have two significant figures. Rounding off two significant figures, we have

$3.9\times {10}^5-2.5\times {10}^4=3.7\times {10}^5$

3. $\frac{0.9996\times 3.52}{1.758}=2.00147$ (by actual calculations)

3.52 has least number of significant figures i.e., three. Therefore, rounding off the result to three significant figures,

we have $\frac{0.9996\times 3.52}{1.758}=2.00$

4. We know, $F=G\frac{M_1{M}_2}{r^2}$

$\therefore [G]=[\frac{F\times r^2}{M_1\times M_2}]=\frac{[ML{T}^{-2}]\times [L{]}^2}{[M]\times [M]}=[M^{-1}{L}^3{T}^{-2}]$

5. The coefficient of thermal conductivity is given by

$k=\frac{Qd}{A({\theta }_1-{\theta }_2)t^{\prime }}$

Where, $Q,d,A({\theta }_1-{\theta }_2)$ and $t$ stand for energy, distance, area, temperatue difference and time.

$\therefore [k]=\frac{[M{L}^2{T}^{-2}]\times [L{]}^2}{[L{]}^2\times [K]\times [T]}=[ML{T}^{-1}{K}^{-1}]$

Note: Here, the coefficient of thermal conductivity has been denoted by $k$ instead of $K$ in order to distinguish it from $K$ (kelvin), the unit of temperature.

6. 1. Work and energy

2. Pressure and stress

3. Velocity gradient and frequency

4. Angular momentum and Planck’s constant.

7. The dimensional formula of a physical quantity indicates the fundamental units on which the physical quantity depends. It further tells the powers of the fundamental units, on which the given physical quantity depends.

8. Since, angle is dimensionless, both $\omega ,t$ and $k,x$ have no dimensions.

Therefore, $\omega =\frac{\text{ dimensionless quantity }}{t}=\frac{1}{[T]}=[M^0{L}^0{T}^{-1}]$

and $k:=\frac{\text{ dimensionless quantity }}{x}=\frac{1}{[L]}=[M^0{L}^{-1}{T}^0]$

9. Joule is unit of work. We know that

work $=$ force $\times$ distance $=$ mass $\times$ acceleration $\times$ distance

$=$ mass $\times \frac{\text{ distance }}{{\text{ time }}^2}\times$ distance $=\frac{\text{ mass }\times {\text{ distance }}^2}{{\text{ time }}^2}$

$\therefore J=\frac{kg\times m^2}{s^2}=kg{m}^2{s}^{-2}$

10. The advantages in choosing the wavelength of a light radiation as a standard of length are as follows :
11. The wavelength of light is not affected by time and environment.
12. This standard of length does not undergo any change with place.

EXERCISE - 2

MULTIPLE CHOICE QUESTIONS

1. (a) $SI$ is based on seven fundamental units.

2. (a) $L=\frac{Q}{m}=\frac{J}{kg}=Jk{g}^{-1}$

3. (b) Tropical year is the year in which there is total solar eclipse. Light year represents distance.

4. (a) As force $=$ change in momentum/time.

$\therefore \text{ force }\times \text{ time }=\text{ change in momentum }$

5. (a)
6. (d)
7. (a) From Biot Savart’s law

$B=\frac{{\mu }_0}{4\pi }\frac{idl\sin \theta }{r^2}$

${\mu }_0=\frac{4\pi B{r}^2}{idl\sin \theta }=\frac{Wb{m}^{-2}{m}^2}{Am}=Wb{A}^{-1}{m}^{-1}$

8. (b) charge $=$ current $\times$ time
9. (a)
10. (b) According to Faraday’ Law $\in =\frac{d\varphi }{dt}$

so dimensionally $[\varphi ]=[E.M.F][T]=[W/q][T]$

$=[M{L}^2{T}^{-2}/AT][T]=[M{L}^2{T}^{-2}{A}^{-1}]$

and the S.I unit of magnetic flux is volt $\times$ sec., which is known as weber ( $Wb)$.

11. (c) $\rho =\frac{RA}{\ell }=\frac{\text{ ohm }{m}^2}{m}=ohm-m$
12. (c) 1 light year speed of light in vacuum no. of seconds in one year $(3\times {10}^8)\cdot (365\cdot 24\cdot 60\cdot 60)$ $=9.467\cdot {10}^{15}m$
13. (C) 1 pascal “1 N: $m^2$
14. (c) Electron volt is a unitof energy &

$IeV=1.6\times {10}^{-19}$ joule

15. (a) Impulse - force $\times$ time $=ML{T}^2\times 1|l/1|$

16. (d) coefficient of viscosity ( $\eta$ )

Force required per unit area to maintain unit velocity gradient

$N/[m^2\times (m/s)/m]$

$N/m^2{s}^{1}N{m}^{-2}|M{I}^1{T}^1|$

17. (C) Action means, force & the dimension of force is $ML{T}^{-2}(\because F$ mu. where a is acceleration)
18. (d) Both energy and work have same unit energy/work is a pure number
19. (a) Potential is work done per unit charge
20. (d)
21. (d) 1 to $mv=\frac{1}{2\pi \sqrt{LC}}$

$I.C=\frac{1}{(2\pi )},\text{ II }1,2=I^2|M^0{I}^{\prime \prime }{I}^2|$

22. (d) I:lectromotse lisec pulential dillewence

$I=\frac{W}{q}=\frac{M{I}^2{\eta }^2}{A}=|M{I}^2|$

23. (h) $P$ ‘wer $=\frac{\text{ energy }}{\text{ timc }}=\frac{M{I}^{-}-?}{j}=|M{I}^{\prime }|$ ‘?
24. (a) $k=\frac{PV}{\mu T}=\frac{W}{\mu T}=\frac{M{L}^2{T}^2}{molK}$

Where $\mu$ is number of mole of the gas $[M^1{L}^2{T}^{-2}{K}^{-1}mo{l}^{-1}]$

25. (d) $\frac{Q}{m\theta }=\frac{M{L}^2{T}^{-2}}{MK}=[L^2{T}^2{K}^{-1}]$

26. (a) Moment of couple = force $\times$ distance $|M^1{L}^2{T}^2$ work = force $\times$ distance $=|M^1{L}^2{T}^2|$

27. (b) InS $=A(t\cdot B)+C{t}^2:B$ is added to time t therefore dimensions of $B$ are those of time

28. (b) $s\frac{a}{T^2}\ne P$,

$I=PVc{m}^2(c{m}^3{)}^2=\text{ dyne }-cm{}^4$

29. (C) $1\frac{4}{3}\pi r^3$

$.11<100<\frac{\mathrm{\Delta }}{r})\cdot 100=3\times 1$ 31). (i) 1 I $M^{\prime }{I}^n{I}^{\prime }$

$(a\alpha \cdot h\beta )\cdot (\gamma {)}^{\prime \prime }=$

31. (a) Subtraction is correct upto one place of decimal, correspending to the least number of decinal places. $7.2611.27.1$.
32. (d) Number of significant figures in multiplication is three, corresponding to the minimum number $107.88\cdot 0.610-65.8068-65.8$
33. (d) $\frac{97.52}{2.54}=38.393=38.4$ (with least number of signticant figures, 3 ).
34. (b) Radius,r 0.16 mm

Area of cross section $\pi r^2$

$=3.14\times (0.16{)}^2$

$=0.08038-0.080$

35. (b) Surface tension $T=\frac{Force}{Length}=\frac{N}{m}$
36. (a) Weber $/m^2=$ Tesla

38. (c) The meter is a unit and not a physical quantity?
39. (d) W hr is a unit of energy
40. (d) $t=\frac{q}{t}$
41. (C) $|A|[L{T}^2]$or $[L]=[A{T}^2$ or [Work] = [Force $\times$ Distance] = [E] [FAT^{2}]$
42. (b) According to rules of significant figures.
43. (c) $T=2\pi \sqrt{\frac{1}{g}},g,\frac{t}{T^2}$

$\therefore \frac{\mathrm{\Delta }g}{g}\times 100=0.5$ % $+2\times 0.2$ %=0.9 %

44. (c) |at| - [F] amd $[b{t}^2]$ [F]$

$\Rightarrow |a|=ML{T}^3\text{ and }[b]ML{T}^4$

MORE THAN ONE CORRECT

1. (a. b)
2. (a. b. d)
3. (a. c. d)
4. (a. c. d)

$[h]|t^{\prime }|$ & $|C|$ & $x^{\prime }|$ & $|y||x|=|a|$

凡. $$ \frac{y}{a}$ are dumensionless

5. (a, b, c, d)
6. (a, b, c)
7. (a, b, d)

[P] [t] & |q|, $|x^1|$

8. (a, b, d)

MULTIPLE MATCHING QUESTIONS

1. (A) $\to (s,t);(B)\to (s,t);(C)\to (p);(D)\to (q,r)$
2. (A) $\to (p,s):(B)\to (p,1);(C)\to (q,r);(I)\to (q)$

FILL IN THE PASSAGE

I. (1) Systematic errors (2) instrumental errors (3) personal errors (4) environmental errors (5) random crrors (6) minimised.

II. (1) Fundamental (2) seven (3) derived (4) independent (5) S.I system (6) second (7) ampere (8) kg.

PASSAGE BASED QUESTIONS

1. (c) Acceleration can be derived from length and time, i.e. $[a]=[L{T}^{-2}]$. Hence acceleration, length and time cannot be taken as set of fundamental quantities.
2. (b) We have, momentum $P=mV$

or $m=\frac{P}{V}$ or $[m]-[P^{-1}{T}^0]$

3. (b) We have

$Force=\frac{\text{ Change in momentum }}{Time}$

or $[F]=\frac{[P]}{t}=[P{V}^{\prime },t^1]$

4. (c) Here, mass of cube, $m=(5.0\pm 0.2)g$

$l=(1.0\pm 0.01)cm$

$\therefore V=l^3=1.0c{m}^3$

$\frac{\mathrm{\Delta }V}{V}=3\frac{\mathrm{\Delta }l}{l}=0.03c{m}^3$

$\therefore V=(1.0\pm 0.03)c{m}^3$

$\therefore P=\frac{m}{V}=\frac{5.0}{1.0}=5.0c{m}^{-3}$

$\frac{\mathrm{\Delta }P}{P}=\frac{\mathrm{\Delta }m}{m}+\frac{\mathrm{\Delta }V}{V}=\frac{0.2}{5.0}+\frac{0.03}{1.0}$

or $\frac{\mathrm{\Delta }P}{P}=0.07=7$

5. (a) $\mathrm{\Delta }V=0.03c{m}^3$
6. (d) $\mathrm{\Delta }P=0.07\times 5.0=0.35g/c{m}^3$

ASSERTION & REASON

1. (c)
2. (a) Quantities having different dimensions cannot be added or subtracted.
3. (c)
4. (d) Graph between $P$ and $Q$ is straight line if.
5. (c)

$\frac{\mathrm{\Delta }P}{\mathrm{\Delta }Q}=$ constant

(d) Radian is unit of distance

HOTS SUBJECTIVE QUESTIONS

1. The number of sigmticant figures in the measured length is 4. The calculated area and the volume should therefore be rounded off to 4 , significant figures.

Surface area of the cube $6(7.203{)}^2{m}^2$

$311.299254m^2=311.3m^2$

volume of the cube $=(7.203{)}^3{m}^3=373.714754m^3$

$373.7m^3$

2. There are 3 significant figures in the measured mass whereas there are only 2 significant figures in the measured volume. Hence the density should be expressed to only 2 significant figures.

Density $\frac{5.74}{1.2}gc{m}^{-3}=4.8gc{m}^{-3}$

3. (a) Total mass $=2.3403kg=2.3kg$,

(b) Difference $=20.17g-20.15g$

4. Dimensionally $P=\frac{a}{V^2}$ [Principle of homogenity]

$.\Rightarrow M{L}^1{T}^2=\frac{a}{V^2}\Rightarrow a-[M{L}^5{T}^2]$

Also dimensionally V - b [ By principle of homogenity]

$\therefore b\cdot |L^3|$

5. Dimension on left $[L{T}^1{]}^2[L^2{T}^2]$

Dimension on right $=\frac{[M][L{T}^2][L]}{[M]}=[L^2{T}^{-2}]$,

$\therefore$ formula is reasonable

6. By Homogencity theory of I Dimension

Dimension of $(N)$

$=$ Dimension of $D\times \frac{\text{ Dimension of }(n_2-n_1)}{\text{ Dimension of }(x_2-x_1)}$

$\frac{1}{L^{2}T}=\text{ Dimension of }D\times \frac{L^{-3}}{L}$

$\Rightarrow$ Dimension of ’ $D$ ’ $=\frac{L}{L^{-3}\times L^{2}T}=\frac{L^2}{T}=L^2{T}^{-1}$

7. Given that, $T\propto p^a{d}^b{E}^r$

Equating both sides dimensionally,

$[T]=[M{L}^1{T}^2{]}^a[M{L}^3{]}^b[M{L}^2{T}^2{]}^c$

$[M^0{L}^{0}T]=[M^{a+b+c}{L}^{a-3b+2c}{T}^{-2,2-c}]$

Equating the exponents of similar quantities,

$a+b+c=0,a3b+2c=0$ and $2a2c=1$

Solving these equations, we get

$a=-\frac{5}{6}:b=\frac{1}{2}\text{ and }c=\frac{1}{3}$

8. Let the quantity be $Q$ then, $Q=f(v,F,T)$

Assuming that the function is the product of power functions of $v,F$ and $T$,

$Q=K{v}^x{\mu }^y{\eta }^z$

where $K$ is a dimensionless constant of proportionality.

The above equation dimensionally becomes.

$[Q]=[L{T}^{-1}{]}^x[ML{T}^{-2}{]}^y[T{]}^z$

i.e., $[Q]=[M^N][L^{x+y}{T}^{-x-2y+z}]$

Now, (a) $Q=$ mass i.e., $[Q]=[M]$

So, Eq. (2) becomes $[M]=[M^y{L}^{x+y}{Y}^{-x-2y+z}]$ it dimensional correctness requires $y=1,x+y=0$ and $-x-2y+z=0$.

which on solving yields $x=-1;y=1$ and $z=1$

Substituting it in Eqn. (1), we get $Q=K{v}^{-1}FT$

(b) $Q=$ energy i.e., $[Q]=[M{L}^2{T}^{-2}]$

So Eqn. (2) becomes, $[M{L}^2{T}^{-2}]=[M^y{L}^{x+y}{T}^{-x-2y+z}]$

which in the light of principle of homogeneity yields

$y=1,x+y=2\text{ and }-x-2y+z=-2$

Which on solving yields, $x=y=z=1$

So Eqn. (1) becomes, $Q=KvFT$.

9. The dimensions of LHS are

$[M][L{T}^{-1}{]}^2=[M][L^2{T}^{-2}]=[M{L}^2{T}^{-2}]$

The dimensions of RHS are

$[M][L{T}^{-2}][L]=[M][L^2{T}^{-2}]=[M{L}^2{T}^{-2}]$

The dimensions of LHS and RHS are the same and hence the equation is dimensionally correct.

10. $1cal=4.2kg{m}^2{s}^{-2}$.

$\begin{array}{cc}\text{ SI }& \text{ New system }\\ n_1=4.2& n_2=?\\ M_1=1kg& M_2=\alpha kg\\ L_1=1m& L_2=\beta \text{ metre }\\ T_1=1s& T_2=\gamma \text{ second }\end{array}$

Dimensional formula of energy is $[M{L}^2{T}^{-2}.$ ]

Comparing with $[M^a{L}^b{T}^c]$, we find that $a=1,b=2$, $c=-2$

Now, $n_2=n_1[\frac{M_1}{M_2}{]}^a[\frac{L_1}{L_2}{]}^b[\frac{T_1}{T_2}{]}^c$

$=4.2[\frac{1kg}{\alpha kg}{]}^1[\frac{1m}{\beta m}{]}^2[\frac{1s}{\gamma s}{]}^{-2}$

$=4.2{\alpha }^{-1}{\beta }^{-2}{\gamma }^2$

11. $X$ has dimensions $ML{T}^{-2}\frac{M}{L^3}=M^2{L}^{-2}{T}^{-2}$

12. $L=F^x{A}^y{T}^z$

$M^0{L}^1{T}^0\equiv [ML{T}^{-2}{]}^x[L{T}^{-2}{]}^y{T}^z=M^x{L}^{x+y}{T}^{-2x-2y+z}$

$x=0,x+y=1,-2x-2y+z=0$

$x=0,y=1,z=2$

Hence, $L=A{T}^2$

13. We have, $Fs=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2$,

Dimensional formula of $Fs=[ML{T}^{-2}]\times [L]=[M{L}^2{T}^{-2}]$

Dimensional formula of $\frac{1}{2}m{v}^2=[M][L{T}^{-1}{]}^2$

$=[M{L}^2{T}^{-2}]$

Dimensional formula of $\frac{1}{2}m{u}^2=[M][L{T}^{-1}{]}^2=[M{L}^2{T}^{-2}]$ As the dimensional formula of all the terms in the given equation is same, the equation is a correct one.

14. Here $\lambda =\frac{h}{mv}$,

Dimensional formula of wavelength $\lambda =[L]$

Therefore, dimensional formula of L.H.S.

$=[L]=[M^{0}L{T}^0]$

Dimensional formula of Plank’s constant $h$.

$[\frac{\text{ energy }}{\text{ frequency }}]=\frac{[M{L}^2{T}^{-2}]}{[T^{-1}]}=[M{L}^2{T}^{-1}]$

Dimensional formula of mass $\mathit{m}=[M]$

Dimensional formula of velocity $v=[L{T}^{-1}]$

Therefore, dimensional formula of R.H.S.

$=\frac{[M{L}^2{T}^{-1}]}{[M]\times [L{T}^{-1}]}=[M^{0}L{T}^0]$

As dimensional formula of L.H.S. is the same as that of R.H.S, the given relation is correct.

CHAPTER 2

The car is moving in a straight line

Motion in a Straight Line

INTRODUCTION

Motion is everywhere. It is fundamental to our human existence. We need motion for growing, for learning, for thinking and for enjoying life. We use motion for walking through a forest, for listening to its noises and for talking about all this.

Like all animals, we rely on motion to get food and to survive dangers. Like all living beings, we need motion to reproduce, to breathe and to digest; like all objects, motion keeps us warm.

Motion in an object can take place in different ways and style. When an object moves in one direction it is called motion in a straight line or motion in one dimension. When an object moves along two directions at the same time, it is called motion in a plane or motion in two dimensions and when an object moves on a circular path, the motion is called circular motion. This chapter deals with the different aspects of the motion in one dimension.

REST AND MOTION

Rest : An object is said to be at rest if it does not change its position with respect to its surroundings with the passage of time Motion : A body is said th he in motion if its position changes continuously with respect to the surroundings (or with respect to an observer) with the passage of time.

We know that earth is rotating about its axis and revolving around the sun. The stationary objects like your class-room, a tree and the lamp posts etc. do not change their position with respect to each other i.e. they are at rest. Although carth is in motion. To an observer situated outside the earth say in a space ship, your classroom, trees etc. would appear to be in motion. Therefore, all motions are relative. There is nothing like absolute motion. If you move with book in your hand, book is not moving with respect to you

Rest and Motion are Relative Terms

Rest and motion are relative terms. A particle at rest with respect to an observer can be in motıon with respect to another observer. To the passengers in a moving bus or train, trees, buildings and people on the roadsides observe that the bus or the train and its passengers are moving in the forward direction. At the same time, each passenger in a moving bus or train finds that fellow passengers are not moving, as the distance between them is not changing.

If you will observe the man moving on moving flat car from ground your observation will be different from what man himself will observe. Similarly, if you will observe pendulum in moving car from ground your observation will be different from what person inside car will observe.

Frame of Reference

To locate the position of object we need a frame of reference. A convenient way to set up a frame of reference is to choose three mutually perpendicular axes and name them $x-y-z$ axes. The co-ordinates $(x,y,z)$ of the particle then specify the position of object w.r.t. that frame. If any one or more co-ordinates change with time, then we say that the object is moving w.r.t. this frame.

DISTANCE AND DISPLACEMENT

Motion is related to change of position. The length traveled in changing position may be expressed in terms of distance, the actual path length between two points. Distance is a scalar quantity, which has only a magnitude with no direction.

The direct straight line pointing from the initial point to the final point is callectadisplacement (change in position). Displacement only measures the change in position, not the details involved in the change in position Displacement is a vector quantity. which has both magnitude and direction. In the figure shown, an object goes from point $A$ to point (’ by following paths $AB$ and $BC$. The distance traced is $3.0m+4.0m=7.0m$, and the displacement is $5.0m$ in the direction of the arrow.

If one states ’the car has travelled $200m$ ‘, it means that the distance travelled hy the car is $200m$. But if one states ’the car has travelled $200m$ due east’ it means that the displacement of the car is $200m$ towards cast

The displacement can be zero even if the distance is not %ero. For example when a body is thrown vertically upwards from a point on the ground, after sometime it returns back to the same point, then the displacement of the body is zero but the distance travelled by the body is not zero, it is $2h$ if $h$ is the maximum height attained hy the body.

Similarly, if a body is moving in a circular or closed path and reaches its original position after one rotation, then the displacement in one rotation is zero, but the distance travelled is equal to the circumference of the circular path $=2\pi r$ if $r$ is the radius of the circular path.

Do you know !!

The actual distance travelled by an object in a given time interval can he equal to or greater than the magnitude of displacement. It can never be less than the magnitude of displacement.

The displacement of an object in a given time interval can be positive, zero or negative. However. distance covered by the object in a given time interval is always positive.

AVERAGE SPEED

Average speed is defined as the distance traveled divided by the time interval to travel that distance. Average speed $=\frac{d}{t},d$ is distance traveled, and $t$ is time interval (change in tme ).

It is a scalar quantity.

The Cheetah averages $70m/s$ for 30 seconds

Instantaneous Speed

Instantaneous speed is the speed at a particular time instant ( $t$ is infinitesimal small or close to zero). It is also a scalar quantity.

Uniform and Non-uniform Speed

A body is said to be moving with uniform speed if it covers equal distances in equal time intervals and with non-uniform or variable speed if covers unequal distances in the same time intervals.

Do you know !!

The velocity of an object may be positive, zero or negative, but the speed of an object can never be negative.

AVERAGE VELOCITY

Average velocity is defined as the ratio of change in position or displacement to the time taken.

$\overline{v}=v_{a\nu }=\frac{x_2-x_1}{t_2-t_1}=\frac{\mathrm{\Delta }x}{\mathrm{\Delta }t}$

Here $x_1$ and $x_2$ are the positions of the particle at time $t_1$ and $t_2$ respectively.

Also, $\mathrm{\Delta }x=x_2-x_1=$ change in position and $\mathrm{\Delta }t=t_2-t_1=$ change in time

It is a vector quantity, its unit is $m{s}^{-1},cm{s}^{-1}$ or $kmh{r}^{-1}$.

$✓$ CHECK POIIT

• Under what condition is the average velocity equal to instantaneous velocity?

Check Your Answer

• When the displacement-time curve is a straight line, i.e., when the body is moving with uniform velocity.

Illustration 1 :

A particle moved from point $A$ to point $B$, travelling some distance with speed $60kmh{r}^{-1}$. Then it moves back from $B$ to $A$ with speed $40km/h$. Find displacement, distance covered, average velocity and average speed for the entire journey.

SOLUTION : Let us take, $AB=s$

(i) As particle comes back to the same point, displacement $=0$

(ii) Total distance covered = length of path $=2d=2AB$

(iii) Average velocity $=\frac{0}{\mathrm{\Delta }t}=0$

(As displacement is zero)

For average speed, total time $=\frac{d}{60}+\frac{d}{40}$ and total distance $=d+d=2d$

So, average speed $=\frac{2d}{\frac{d}{40}+\frac{d}{40}}=48kmh{r}^{-1}$

(i) If a particle covers two consecutive equal distances with speeds $v_1$ and $v_2$ then,

$\text{ Average speed }=\frac{2v_1{v}_2}{v_1+v_2}$

(ii) If a particle covers three consecutive equal distances with speeds $v_1,v_2$ and $v_3$ then,

$\text{ Average speed }=\frac{3v_1{v}_2{v}_3}{v_1{v}_2+v_2{v}_3+v_3{v}_1}$

(iii) If a particle has speed $v_1$ for time $t_1$ and speed $v_2$ for time $t_2$ then,

$\text{ Average speed }=\frac{v_1{t}_1+v_2{t}_2}{t_1+t_2}$

Illustration 2 :

A particle moves in a circular path of radius $1m$ with uniform speed and takes 4 seconds to complete anc circalar path. Pind distance, displacement, average speed and average velocity for : (i) $A$ to $B$ (ii) $A$ to $C$ (iif) $A$ to $D$ and (bv) $A$ to $A$.

SOLUTION :

**For $A$ to $B:$Distance covered $=\frac{2\pi r}{4}=\frac{\pi r}{2}=\frac{\pi }{2}m=1.57m$

Displacement $=$ shortest distance between $A$ and $B$

$=\sqrt{2}r=\sqrt{2}m=1.41m$

Average velocity $=\frac{\text{ displacement }}{\text{ time }}=\frac{\sqrt{2}m}{1s}=\sqrt{2}\frac{m}{s}=1.41m{s}^{-1}$

Average speed $=\frac{\text{ total distance }}{\text{ total time }}$

$=\frac{\pi /2}{1}=1.57m{s}^{-1}$

(ii) For $\mathit{A}$ to $\mathit{C}$ : Distance covered $=\pi r=3.14\times 1=3.14m$

Displacement $=2r=2\times 1=2m$

Average velocity $=\frac{2}{2}=1m{s}^{-1}$

Average speed $=\frac{3.14}{2}=1.57m{s}^{-1}$

(iii) For $\mathit{A}$ to $\mathit{D}$ : Distance covered $=\frac{3\pi r}{2}=4.71m$

Displacement $=\sqrt{2}r=1.41m$

Average velocity $=\frac{\sqrt{2}}{3}=0.47m{s}^{-1}$

Average speed $=\frac{4.71}{3}=1.57m{s}^{-1}$

(iv) For $\mathit{A}$ to $\mathit{A}$ : Distance covered $=2\pi r=6.28m$

Displacement $=0$

Average velocity $=0$

Average speed $=\frac{6.28}{4}=1.57m{s}^{-1}$

Instantaneous Velocity

Velocity of a body at a particular instant or moment of time is called instantaneous velocity.

Then the instantaneous velocity will be given as

$\begin{array}{rl}{v}_{\text{inst }}& =\underset{\mathrm{\Delta }t\to 0}{lim}\frac{\mathrm{\Delta }x}{\mathrm{\Delta }t}\\ \text{ or }{v}_{\text{inst }}& =\frac{dx}{dt}\end{array}$

Instantaneous velocity is a vector quantity, as it has direction as well as magnitude, its unit is $m/s,cm/s,Am$ to Instantaneous velocity at that moment is a scalar quantity. Remember that the magnitude of instantannus intwity an motant would always be equal to the instantaneous speed that instant.

Illustration 3

Displacement $x$ of a particle is given by the equation $x=3t^2+4t+1$, where $x$ is in metre and $t$ is in second. Find the instantanerous velocity of the particle at (i) $t=1s$; (ii) $t=5s$. Also find average velocity of the particle between $t=1s$ and $t=5s$.

Solution : We know that, $v_{\text{inst. }}=\frac{dx}{dt}$

So,

$v_{\text{inst. }}=(6t+4)$

(i) at $t=1s$,

$v_{\text{inst. }}=6\times 1+4$

(ii) at $5s,v_{\text{inst. }}=10m/s$

ss, $$ inst. $6\times 5+4$

$v_{\text{inst. }}=34m/s$

(iii) For average velocity we use, $v_{\text{av. }}=\frac{x_2-x_1}{t_2-t_1}$

where

$\begin{array}{rl}{r}_2=& \text{ displacement at }{t}_2=5,=3(5{)}^2+4\times 5+1=75+20+1=96m\\ r_1=& \text{ displacement at }{t}_1=1.=3(1{)}^2+4(1)+1=8m\end{array}$

$v_{\text{av. }}=\frac{96-8}{5-1}=\frac{88}{4}=22m/s$

AVERAGE ACCELERATION

Average acceleration is defined as the change in velocity divided by the time interval to make the change $a=\frac{\mathrm{\Delta }v}{\mathrm{\Delta }t}=:\frac{v-v_0}{t-t_0}$, where $a$ is as crage acceleration, $\mathrm{\Delta }v$ is change in velocity, and $\mathrm{\Delta }t$ is time interval.

Instantaneous Acceleration

Instantaneous acceleration of the particle is the acceleration at particular instant, mathematically, it will be

$a_{\text{inst. }}=\underset{\mathrm{\Delta }\to \to 0}{lim}\frac{\mathrm{\Delta }v}{\mathrm{\Delta }t}=\frac{dv}{dt}$

Instantaneous acceleration is also referred to as ‘acceleration’,

Types of Acceleration :

(i) Positive acceleration : If the velocity of an object increases in the same direction, the object has a positive acceleration

(ii) Negative acceleration (Retardation) : If the velocity of a body decreases in the same direction, the body has a negative acceleration or it is said to be retarding e.g : A train slows down.

IDEA BOX

Misconception

A common misconception about velocity and acceleration has to do with their directions. Since vekecity has benth masyintoute and direction, a change in either magnilude (specl) and/or direction will result in a change in velecily, therefure an acceleratum. We can accelerate objects either by speeding them up on down (change magninde) and/on by changing lineer directune of travel.

For motion in one-dimension, when the velocity and acceleration of an whject are in the arene directum they hase the same directional signs), the velocity incieases and the object speeds up (acceleratim). When the velusity and seceration are in opposite direction, the velocity decreases and the object slows down deceleration.

When velocity of a particle increases with time, it is said to he accelerated motion i.e both aceceleration and selkeity will be positive and speed (magnitude of velocity) would increase.

When both acceleration and velocity are negative, that would mean that the direction of inotion is in the opprsite direction but in this case also speed of particle would increase with time.

If acceleration and velocity are of opposite signs, in that case speed of the particle would decrease. Iheceration is equivalent to negative of acceleration.

Factors Characterising the Motion of a Particle

Time ( $t$ ): How much time t has elapsed since some initial time. The initial time is often referred to as “the start of observations” and even more often assigned the value 0 . We will refer to the amount of time $t$ that has elapsed since time zero as the stopwatch reading. A time interval $\mathrm{\Delta }t$ (to be read “delta $t$ “) can be referred to as the difference between two stopwatch readings.

Position $(x)$ : Where the object is along the straight line. To specify the position of an object on a line, one has to define a reference position (the start line) and a forward direction. Having defined a forward direction, the backward direction is understood to be the opposite direction. It is conventional to use the symbol $x$ to represent the position of a particle.

Speed is constant. Direction is changing Hence velocity is not constant

The values that $x$ have units of length. The SI unit of length is the meter. The symbol for the meter is $m$. The physical quantity $x$ can be positive or negative where it is understood that a particle which is said to be minus five meters forward of the start line (more concisely stated as $x=-5m$ ) is actually five meters behind the start line.

Velocity $(v)$ : How fast and which way the particle is going. We use the symbol $v$ for this and call it the velocity of the object. Because we are considering an object that is moving only along a line, the “which way” part is either forward or backwand. Since there are only two choices, we can use an algebraic sign ("+” or “-”) to characterize the direction of the velocity. By convention. a positive value of velocity is used for an object that is moving forward, and, a negative value is used for an object that is moving backward. Velocity has both magnitude and direction. The magnitude of a physical quantity that has direction is how big that quantity is. regardless of its direction. So the magnitude of the velocity of an object is how fast that object is going. regandless of which way it is going. Consider an object that has a velocity of $5m/s$. The magnitude of the velocity of that object is $5m/s$. Now consider other object that has a velocity of $-5m/s$. (It is going backward at $5m/s$.) The magnitude of its velocity is also $5m$ s. Another the name for the magnitude of the velocity is the speed. In both of the cases just considered, the speed of the object is $s$ ms despite the fact that in one case the velocity was $-5m/s$. To understand the “how fast” part, just imagine that the object whose motion is under study has a built-in speedometer. The magnitude of the velocity, i.e. the speed of the object, is simply the specthoneter reading. Acceleration (a): Next we have the question of how fast and which way the velocity of the object is changing. We call this the acceleration of the object. Instrumentally, the acceleration of a car is indicated by how fast and which way the tip of the spectometer needle is moving. In a car, it is determined by how far down the acceleration pedal is pressed, or, in the case of car that is slowing down, how hard the driver is pressing on the brake pedal. In the case of an object that is moving along a straight line, if the object has some acceleration, then the speed of the object is changing.

$✓$ CHECK POINT

• Can the velocity of an object ever be in the direction other that the direction of the acceleration of the object?

Check Your Answer

Yes, instantaneous acceleration is independent of instantaneous velocity. So the direction of velocity has no relation to the direction of its acceleration.

UNIFORM AND NON-UNIFORM MOTION

Uniform Rectilinear Motion

It is a motion in which a material point moves in a straight line and covers equal distances in equal intervals of time.

The path length of a body in a uniform rectilinear motion is equal to the magnitude of the displacement. Consequently, the path length in the motion is equal to the magnitude of the velocity multiplied by the time : $s=vt$.

$x=x_0+s=x_0+vt$

Do you know !!

No force is required to keep an object in uniform motion. When an object has uniform motion along a straight line in a given direction, the magnitude of displacement is equal to actual distance covered.

Non-uniform motion

It is a motion in which the velocity varies with time.

The change in the velocity of a material point in nonuniform motion is characterized by acceleration.

Uniformly variable motion is a motion with a constant acceleration.

Uniformly variable motion can be curvilinear like circular motion.

If a uniformly variable motion is rectilinear, i.e., the velocity $v$ changes only in magnitude, it is convenient to take the straight line in which a material point moves as one of the coordinate axes (say, the $x$-axis).

Non-uniform acceleration

If during motion of a body its velocity increases by unequal amounts in equal intervals of time.

IDEA BOX

General method of approaching numerical problems.

1. Draw a ‘sketch’ diagram wherever possible.
2. Copy down the numerical information given in the question.
3. Write down the relevant formula.
4. Substitute the given values into the formula.
5. Calculate the answer, remembering to show all steps in the working out and giving the correct units for our fioal answer.

Illustration 4 :

An object moving to the right has a decrease in velocity from $5.0m/s$ to $1.0m/s$ in $2.0s$. What is the average acceleration ? What does your result mean?

SOLUTION : Given $v_0=+5.0m/s,v=+1.0m/s,t=2.0s$

Find $\overrightarrow{a}$

According to the definition of average acceleration,

$\overrightarrow{a}=\frac{\mathrm{\Delta }v}{\mathrm{\Delta }t}=\frac{v-v_0}{t}=\frac{+1.0m/s-(+5.0m/s)}{2.0s}=\frac{-4.0m/s}{2.0s}=-2.0m/s^2$

The negative sign means the acceleration is opposite to velocity (deceleration). The result means that the object decreases its velocity by $2.0m/s$ every $s$ or $2.0m/s^2$.

Illustration 5

A car covers the $1^{\text{st }}$ half of the distance between two places at a speed of $40km/hr$ and the $2nd$ half at $60km/hr$. What is the average speed of the car ?

SOLUTION : Suppose the total distance covered is 25 . Then time taken to cover first’s distance with speed $40km/hr.t_1=\frac{s}{40}hrs$.

Time taken to cover second $s$ distance with speed $60km/hrs$., $t_2=\frac{s}{60}hr$.

$\begin{array}{rl}{v}_{av}& =\frac{\text{ Total distance }}{\text{ Total time }}=\frac{s+s}{(\frac{s}{40}+\frac{s}{60})}\\ \text{ Or }{v}_{av}& =\frac{2s}{(\frac{3s+2s}{120})}=\frac{2s}{5s}\times 120\\ v_{av}& =48km/hr.\end{array}$

Illustration 6 :

The table below shows the distance in $cm$, travelled by the objects $A,B$ and $C$ during each second.

Time	Distance (in cm) covered in each second by $\mathit{A},\mathit{B}$ and $\mathit{C}$
	Object $\mathit{A}$	Object $\mathit{B}$	Object $\mathit{C}$
1st second	20	20	20
2nd second	20	36	60
3rd second	20	24	100
4th second	20	40	140
5th second	20	48	180

(I) Which object is moving with constant speed ? Give a reason for your answer.

(ii) Which object is moving with a constant acceleration? Give a reason.

(iii) Which object is moving with irregular acceleration?

SOLUTION : The object $A$ is moving with constant speed. The reason is that it covers equal distance $=20cm$ in each second.

(i) The object $A$ is moving with constant speed. The reason is that it covers equal distance $=20cm$ in each second.

(ii) The object $C$ is moving with a constant acceleration. The reason is that for the object $C$, the distance covered increases by the same amount in each second. It can further be verified by drawing graph between $s$ (total distance covered) and $t^2$ (square of time taken). The graph will be a straight line.

$t^2$	1	4	9	16	25
$s$	20	80	180	320	500

(iii) The object $b$ is moving with irregular acceleration.

EQUATIONS OF MOTION (KINEMATIC EQUATIONS)

Kinematic equations can be used to describe the motion with constant acceleration.

The symbols used in the kinematic are : $v_0$ or $u$ initial velocity; $v$, final velocity; $a$, acceleration; $x$, displacement; $t$, time interval. Be aware that the terms initial and final are relative. The end of one event is always the beginning of another. There are three general equations and two algebraic combinations of these equations that provide calculation convenience.

$x-\overrightarrow{v}t,$ displacement $=$ average velocity $\times$ times interval

$\overline{v}=\frac{\nu +u}{2}$, average velocity $=($ final velocity + initial velocity $)/2$

First equation : $v:\cdot u+at$, final velocity - initial velocity + acceleration $\times$ time interval,

By definition, Acceleration $=\frac{\text{ change in velocity }}{\text{ time taken }}=\frac{\text{ final velocity }-\text{ initial velocity }}{\text{ time taken }}$

$\begin{array}{rl}& \text{ or }a=\frac{v-u}{t}\\ & \text{ or }at=v-u\\ & \text{ or }v=u+at\end{array}$

Second equation : $s=ut+\frac{1}{2}a{t}^2$, displacement $=$ initıal velocity $\times$ time interval $+\frac{1}{2}\times$ acceleration $\times$ time interval squared

Distance travelled - tverage velocity $\times$ time $=(\frac{\text{ Initial velocity }+\text{ final velocity }}{2})\times$ time

But from eq. (1), $v=u+a$

$s=\frac{u+v}{2}\times t$

$\begin{array}{cc}\therefore & s=\frac{u+(u+at)}{2}\times :\\ \text{ or }& s=\frac{2u+at}{2}\times t\text{ or }s=ut+\frac{1}{2}a{t}^2\end{array}$

Third equation : $v^2=u_0^2+2as$, final velocity squared $=$ initial velocity squared $+2\times$ acceleration $\times$ displacement

Distance travelled - Average velocity $\times$ time

$s=\frac{u+v}{2}\times t$

But from eq. (1),v:: $u+at$

$\begin{array}{c}\text{ or }t:=\frac{v-u}{a}\therefore s=\frac{u+v}{2}+\frac{v-u}{a}\\ \text{ or },=\frac{v^2}{2a}\text{ or }{v}^2-u^2=2as\text{ or }{v}^2=u^2+2as\end{array}$

Among the three cyuations listed can be used to solve the majority of kinematic problems.

Which equation should you select for a particular problem? The equation you select must have the unknown quantity in it and everything clse musi be gıven, because we can only solve for one unknown in one equation.

Distance Covered by Body in $n^{\text{th }}$ Second :

$\begin{array}{rl}& s=ut,\frac{1}{2}a{t}^2\text{, is the distance covered by a body in }t\text{ sec. or }\\ & S_n=un+\frac{1}{2}a{n}^2\dots \text{ (i) distance covered by a body along straight line in }n\text{ sec. }\\ & S_{n-1}=u(n1)+\frac{1}{2}a(n-1{)}^2\dots .\text{ (ii) distance covered by a body along straight line in }(n-1)\text{ sec. }\\ & \therefore \text{ The distance covered by the body in }{n}^{\text{th }}\text{ second will be }{S}_{nth}-S_{n-1}\\ & \therefore S_{nth}=un+\frac{1}{2}a{n}^2\cdot u(n-1)+\frac{1}{2}a(n-1{)}^2\\ & S_{nth}=un+\frac{1}{2}a{n}^2\dots un-u+\frac{1}{2}a(n+1-2n)\\ & \Rightarrow un+\frac{1}{2}a{n}^2,nu-u+\frac{a{n}^2}{2}+\frac{a}{2}-an\\ & \Rightarrow \frac{1}{2}a{n}^2-nu+u-\frac{a{n}^2}{2}-\frac{a}{2}+an=u+a(n-\frac{1}{2})=n+a(\frac{2n-1}{2})\\ & \Rightarrow S_{nth}=u+\frac{a}{2}(2n-1)\end{array}$

Tips to solve problem on kinematic equations :

1. Make a drawing to represent the situation being studied.
2. Decide which directions are to be called positive $(+)$ and negative $(-)$ relative to a conveniently chosen coordinate origin. Do not change your decision during the course of a calculation.
3. In an organized way, write down the values (with appropriate plus and minus signs) that are given for any of the five kinematic variables $(x,a,v,v_0.$ and $.t)$.
4. Before attempting to solve a problem, verify that the given information contains values for at least three of the five kinematics variables.
5. When the motion of an object is divided into segments, remember that the final velocity of one segment is the initial velocity for the next segment.

Illustration 7 :

An automobile accelerates uniformby from reat to $25m/s$ while traveling $100m$. What he the acecleration of the automoblle?

SOLUTION

Given : $v_0=0$ (rest), $v=25m/s,s=100m$. Find $a$.

Since $v^2=v_0^2+2as,a=\frac{v^2-v_0^2}{2s}=\frac{(25m/s{)}^2-(0{)}^2}{2(100m)}=3.1m/s^2$

Since $a$ is positive, it is in the direction of the velocity or motion.

Illustration 8 :

A car is moving at a speed $50km/h$. Two seconds there after it is moving at $60km/h$. Calculate the acceleration of the car.

solution : Here, $\nu 0=50km/h=50\times \frac{5}{18}m/s=\frac{250}{18}m/s$

and $v=60km/h=60\times \frac{5}{18}=\frac{300}{18}m/s$

Since $a=\frac{v-v_0}{t}=\frac{\frac{300}{18}-\frac{250}{18}}{2}=\frac{\frac{50}{18}}{2}=\frac{50}{36}=1.39m/s^2$

Illustration 9 :

A bus moving with a veloeity of $60km/h$ is brought to rest in 20 seconds by applying brakes. Find lts acceleration.

solution : $v_0=60km/h=60\times \frac{5}{18}=\frac{50}{3}m/s$

$v=0\text{ (as bus comes to rest) }$

as $a=\frac{v-v_0}{t}\therefore a=\frac{0-\frac{50}{3}}{20}=\frac{-50}{3\times 20}=\frac{-5}{6}=-0.83m/s^2$

Illustration 10 :

A bullet moving with $10m/s$ hits a wooden plank. The bullet is stopped when it penetrates the plank $20cm$ deep. Calculate the retardation of the bullet.

solution :

$v_0=10m/s,v=0$ & $s=20cm=\frac{2}{100}=0.02m$

$\text{ Using, }{v}^2-v_0^2=2as$

$0-(10{)}^2=2a(0.2)\Rightarrow \frac{-100}{2\times 0.02}=a\text{ or }a=-2500m/s^2$

$\text{ retardation }=2500m/s^2$

Illustration 11 :

A body covers a distance of 20m in the 7th second and 24m inthe 9th second. How much distance shall it cover in 15th sec.

SOLUTION : $s_7$ th $=u+\frac{a}{2}(2\times 7-1)$ hut $s_7$ th $=20m$

$\therefore 20=u+\frac{a}{2}\times 13\Rightarrow 20=u+\frac{13a}{2}$

also $s_9$ th $=24m$

$\therefore 24=u+\frac{17a}{2}$

From (1) equation $u=20-\frac{13a}{2}$

Substituting this value in (2),

$24=20-\frac{13a}{2}+\frac{17c}{2}$

$\begin{array}{rl}& 24-20=\frac{17a}{2}-\frac{13a}{2}\\ & 4=\frac{4a}{2}\Rightarrow 4=2a\Rightarrow a=\frac{4}{2}=2m/s^2\end{array}$

Use this value in (3).

$u=20\cdots \frac{13a}{2}$

$\therefore u=20-\frac{13\times 2}{2}\Rightarrow u=20-13\therefore u=7m/s$

Now, $s_{{15}^{\text{th }}}\cdot u+\frac{a}{2}(2\times 15-1)=7+\frac{2}{2}(29)=7+29=36m$

Illustration 12

A body with an initial velocity of $18km/hr$ accelerates uniformly at the rate of $9cm{s}^{-2}$ over a distance of $200m$. Calcalate: (i) the acceleration in ${\mathbf{m}\mathbf{s}}^{-2}$ (ii) its final velocity in ${\mathbf{m}\mathbf{s}}^{-1}$

SOLUTION :

(i) Acceleration $=9cm{s}^{-2}=\frac{9}{100}m{s}^{-2}=0.09m{s}^{-2}$

(ii) Initial velocity $u=18km{h}^{-1}=\frac{18000m}{60\times 60s}=5m{s}^{-1}$

Acceleration, $a:0.09m{s}^{-2}$

and distance $s=200m$

From equation of motion

$v^2=u^2+2as$

$v^2=(5{)}^2+2\times 0.09\times 200$

$v^2=25+36=61\therefore v=\sqrt{61}=7.81m{s}^{-1}$. Thus, final velocity $=7.81m{s}^{-1}$.

GRAPHICAL INTERPRETATION OF MOTION IN A STRAIGHT LINE

In physics we often use graphs ds important tools for picturing certain concepts. Below concepts of displacement, velocity and acceleration.

Displacement-Time Graphs

Belon is a graph showing the displationent of the cyclist from $A$ to $C$

This graphs shows us how. in III seconds lime, the cyclist has moved from $t$ to $C$. We know the gradient (slope) of a graph is defined as the change in , diskd hy the change in $1.1.9\frac{.1}{\mathrm{\Delta }x}$. In this graph the gradient of the graph is just $\frac{\mathrm{\Delta }\overrightarrow{s}}{\mathrm{\Delta }s}$ and this is just the expression for velocity The slope of a displacemint-timic graph gives the velocity.

Do you know !!

The $x-1$ graph of an object having uniform motion is a straight line inclined to the time-cxis. The slope of straight line $x-t$ graph gives velocity of the uniform motion of the object.

The slope is the same all the way from $A$ to $C$, so the cyclist’s velocity is constant over the entire displacement he travels. Observe the following displacement-time graphs.

(a) Shows the graph for an object stationary over a period of time. The gradient is zero, so the object has zero velocity.

(b) Shows the graph for an object moving at a constant velocity. You can see that the displacement is increasing as time goes on. The gradient, however, stays constant (remember is the slope of straight line) so the velocity is constant. Here the gradient is positive, so the object is moving in the direction we have defined as positive.

(c) Shows the graph for an object moving at a constant acceleration. You can see that both the displacement and the velocity (gradient of the graph) increase with time. The gradient is increasing with time, thus the velocity is increasing with time and the object is accelerating.

Velocity-Time Graphs

Look at the velocity-time graph below:

This is the velocity-time graph of a cyclist traveling from $A$ to $B$ at a constant acceleration, i.e. with steadily increasing velocity. The gradient of this graph is just $\frac{\mathrm{\Delta }\overrightarrow{v}}{\mathrm{\Delta }t}$ and this is just the expression for acceleration. Because the slope is the same at all points on this graph, the acceleration of the cyclist is constant.

Do you know !!

The $v$-t graph of an object having uniform motion is a straight line parallel to time-axis. The area between $v$-t graph of an object and time-axis is numerically equal to distance covered by it.

The slope of a velocity-time graph gives the acceleration.

We can also calculate displacement traveled from velocity-time graph.

This graph shows an object moving at a constant velocity of $10m/s$ for a duration of $5s$. The area between the graph and the time axis of the above plot will give us the displacement of the object during this time. In this case we just need to calculate the area of a rectangle with width $5s$ and height $10m/s$. area of rectangle $=$ height $\times$ width

$=\overrightarrow{v}\times t$

$\begin{array}{rl}& =10m/s\times 5s\\ & =50m=\overrightarrow{s}=\text{ displacement }\end{array}$

(a) Shows the graph for an object moving at a constant velocity over a period of time. The gradient is zero, so the object is not accelerating.

(b) Shows the graph for an object which is decelerating. You can see that the velocity is decreasing with time. The gradient, however, stays constant (remember: its the slope of a straight line), so the acceleration is constant. Here the gradient is negative, so the object is accelerating in the opposite direction to its motion, hence it is decelerating.

(a)

(b)

Acceleration-Time Graphs

In this chapter on rectilinear motion we will only deal with objects moving at a constant acceleration, thus all acceleration-time graphs will look like these two:

(a)

(b)

Deseription of the graphs

(a) Shows the graph for an object which is either stationary or traveling at a constant velocity. Either way, the acceleration is zero over time.

(b) Shows the graph for an object moving at a constant acceleration. In this case the acceleration is positive - remember that it can also be negative.

We can obtain the velocity of a particle at some given time from an acceleration time graph-it is just given by the area between the graph and the time-axis. In the graph below, showing an object at a constant positive acceleration, the increase in velocity of the object after 2 seconds corresponds to the portion.

$\text{ Area of rectangle }=\overrightarrow{a}\times t=5\frac{m}{s^2}\times 2s=10\frac{m}{s}=\overrightarrow{v}$

Its useful to remember the set of graphs below when working on problems. Figure shows how displacement, velocity and time relate to each other. Given a displacement-time graph like the one on the left, we can plot the corresponding velocity-time graph by remembering that the slope of a displacement-time graph gives the velocity. Similarly, we can plot an acceleration-time graph from the gradient of the velocity-time graph.

Fig. A relationship between displacement, velocity and acceleration

$✓$ CHECK POINT

• A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is $1m$ long and requires $1s$. Plot the $x-t$ graph of his motion. Determine graphically, how long the drunkard takes to fall in a pit $13m$ away from the start.

Check Your Answer

Here, length of each step $=1m$

time required to take one step $=1s$

It follows that when the drunkard takes 5 steps forward, he will cover a distance of $5m$ in fonvard direction in a time interval of $5s$. On the other hand, when he takes 3 steps backwards, he will cover a distance of $sm$ in hackward direction in a time interval of $3s$. Therefore, $x-t$ graph for the drunkard, till he falls in the pit $13m$ away will be as shown in fig. As is evident from the graph, the drunkard will take $37s$ to fall in the pit.

GRAPHICAL DERIVATION OF EQUATION OF MOTION

(i) First Equation

$v=u+at$

It can be derived from $u-t$ graph, as shown in graph.

From line $PQ$. The slope of the line $=$ acceleration $a$

$\begin{array}{rl}& a=\frac{QR}{RP}=\frac{SP}{RP}\\ & \text{ or }SP=aRP=at\\ & \text{ As }OS=OP+PS\\ & \text{ Putting values, }v=u+at\end{array}$

(ii) Second Equation

$s=ut+\frac{1}{2}a{t}^2$

It can also be derived from $u-t$ graph as shown in figure.

From relation,

Distance covered $=$ Area under $u-t$ line

$s=$ Area of trapezium $OPQS$

$=$ Area of rectangle $OPRS+$ Area of triangle $PQR$

$=OP\times PR+\frac{RQ\times PR}{2}$

Putting values,

(iii) Third equation

$=u\times t+\frac{1}{2}at\times t(\because v-u=at)$

or $s=ut+\frac{1}{2}a{t}^2$

$v^2=u^2+2as$

From above graph $OP=u,SQ=v,OP+SQ=u+v$

$a=\frac{QR}{PR}\text{ or }PR=\frac{QR}{a}=\frac{v-u}{a}$

$s=$ Area of trapezium $OPQS=\frac{OP+SQ}{2}\times PR$

On putting the values, $s=\frac{v+u}{2}\times \frac{v-u}{a}=\frac{v^2-u^2}{2a}$ or $v^2=u^2+2as$

Illustration 13 :

Given the displacement-time graph below, draw the corresponding velocity-time and acceleration-time graphs, and then describe the motion of the object.

SOLUTION

Step 1 : Decide what information is supplied

The question explicitly gives a displacement-time graph.

Step 2 : Decide what is asked?

3 things are required:

1. Draw a velocity-time graph
2. Draw an acceleration-time graph
3. Describe the behavior of the object

Step 3 : Velocity-time graph - 0-2 seconds

For the first 2 seconds we can see that the displacement remains constant - so the object is not moving, thus it has zero velocity during this time. We can reach this conclusion by another path too: remember that the gradient of a displacement velocity. For the first 2 seconds we can see that the displacement-time remer that the gradient of a displacement-time graph is the the velocity during this time is zero and the object is stationary.

Step 4 : Velocity-time graph - 2-4 seconds

For the next 2 seconds, displacement is increasing with time so the object is moving. Looking at the gradient of the displacement graph we can see that it is not constant. In fact, the slope is getting steeper (the gradient is increasing) as time goes on. Thus, remembering that the gradient of a displacement-time graph is the velocity, the velocity must be increasing with time during this phase.

Step 5 : Velocity-time graph - 4-6 seconds.

For the final 2 seconds we see that displacement is still increasing with time, but this time the gradient is constant, so we know that the object is now travelling at a constant velocity, thus the velocity-time graph will be a horizontal line during this stage.

So our velocity-time graph looks like this one below. Because we haven’t been given any values on the vertical axis of the displacement-time graph, we cannot figure out what the exact gradients are and hence what the values of the velocity are. In this type of question it is just important to show whether velocities are positive or negative, increasing, decreasing or constant.

Once we have the velocity-time graph its much easier to get the acceleration-time graph as we know that the gradient of a velocitytime graph is the just the acceleration.

Step 6 : Acceleration-time graph - 0-2 seconds

For the first 2 seconds the velocity-time graph is horizontal at zero, thus it has a gradient of zero and there is no acceleration during this time. (This makes sense because we know from the displacement time graph that the object is stationary during this time, so it can’t be accelerating).

Step 7 : Acceleration-time graph - 2-4 seconds

For the next 2 seconds the velocity-time graph has a positive gradient. This gradient is not changing (i.e. its constant) throughout these 2 seconds so there must be a constant positive acceleration.

Step 8 : Acceleration-time graph - 4-6 seconds

For the final 2 seconds the object is traveling with a constant velocity. During this time the gradient of the velocity-time graph is once again zero, and thus the object is not accelerating.

The acceleration-time graph looks like this:

Step 9 : A description of the object’s motion

A brief description of the motion of the object could read something like this:

At $t=0s$ and object is stationary at some position and remains stationary until $t=2s$ when it begins accelerating. It accelerates in a positive direction for 2 seconds until $t=4s$ and then travels at a constant velocity for a further 2 seconds.

Illustration 14 :

The graph represents the velocity of a particle as a function of time.

(a) What is the acceleration at $1.0s$ ?

(b) What is the acceleration at $3.0s$ ?

(c) What is the average acceleration between 0 and $5.0s$ ?

(d) What is the average acceleration for the $8.0s$ interval ?

(e) What is the displacement for the $8.0s$ interval ?

solution :

(a) Acceleration is the slope of the line

$a=\frac{\mathrm{\Delta }v}{\mathrm{\Delta }t}=\frac{20m/s-0m/s}{2.0s-0s}=10m/s^2$

(b) The slope of the line is zero and $a=0$.

(c) $\overrightarrow{a}=\frac{\mathrm{\Delta }v}{\mathrm{\Delta }t}=\frac{10m/s-0m/s}{5.0s-0s}=2.0m/s^2$

(d) $a=\frac{\mathrm{\Delta }v}{\mathrm{\Delta }t}=\frac{-20m/s-0m/s}{8.0s-0s}=-2.5m/s^2$

(e) The net area equals the displacement.

The area of a rectangle is length $\times$ width and area of a triangle is $\frac{1}{2}\times$ base $\times$ height

$\begin{array}{rl}& \mathrm{\Delta }{x}_{0-2}=\frac{1}{2}(2.0s-0s)(20m/s)=20m;\mathrm{\Delta }{x}_{2-4}=(4.0s-2.0s)(20m/s)=40m\\ & \mathrm{\Delta }{x}_{4-6}=\frac{1}{2}(6.0s-4.0s)(20m/s)=20m;\mathrm{\Delta }{x}_{6-8}=\frac{1}{2}(8.0s-6.0s)(-20m/s)=-20m\\ & \text{ So, }\mathrm{\Delta }x=20m+40m+20m+(-20m)=60m\end{array}$

Illustration 15

A train starts from rest and accelerates uniformly at $100m$ minute ${}^{-2}$ for 10 minutes. It then maintains a constant velocity for 20 minutes. The brakes are then applied and the train is uniformly retarded. It comes to rest in 5 minutes. Draw a velocity-time graph use it to find : (i) the maximum velocity reached (ii) the retardation in the last 5 minutes (iii) total distance travelled, and (iv) the average velocity of the train

solution : The velocity-time graph is shown in figure.

Acceleration $=\frac{\text{ Final velocity }-\text{ Initial velocity }}{\text{ Time interval }}=\frac{\text{ Final velocity }-0}{\text{ Time interval }}$

or

Final velocity $=$ acceleration $\times$ time interval

$=\frac{100m}{{\text{ minute }}^2}\times 10\text{ minute }=1000m{\text{ minute }}^{-1}$

(I) The maximum velocity reached $=1000m$ minute ${}^{-1}$.

(ii) The retardation in the last 5 minutes $=$ slope of the line $BC$.

$=\frac{BE}{EC}=\frac{(0-1000)mminut{e}^{-1}}{(35-30)\text{ minute }}=\frac{-1000mminut{e}^{-1}}{5\text{ minute }}=-200m{\text{ minute }}^{-2}$

(iii) Total distance travelled $=$ Area of trapezium $OABC$

$=\frac{1}{2}(OC+AB)\times AD=\frac{1}{2}(35+20)\times 1000=55\times 500=27500m(\text{ or }27.5km)$

(iv) Average velocity $=\frac{\text{ Total distance travelled }}{\text{ Total time of travel }}=\frac{27500m}{35\text{ minute }}=785.7mminut{e}^{-1}$

MOTION UNDER GRAVITY

It is a common experience that when a body is dropped form a certain height it experiences acceleration due to gravity and its motion is in a straight path. Sthilarly, when a body is thrown vertically up, it goes to a certain height and then starts falling again, experiencing acceleration due to gravity throughout the motion. The value of acceleration due to gravity $(g)$ is taken as $9.8m/s^2$, $980cm/s^2$ or $32ft/s^2$. Let us consider the three cases discussed below.

Case-I : Body thrown downward :

In this case, initial motion of the body is downward so according to the sign convention, downward direction will be taken as positive and upward direction as negative. So, the kinematic equations will be :

$\begin{array}{rl}& v=u+gt\\ & h=ut+\frac{1}{2}g{t}^2\\ & v^2=u^2+2gh\\ & h^{\text{nth }}=h+\frac{1}{2}g(2n-1)\end{array}$

In a special case when the body is dropped/let fall/allowed to fall we will take the initial velocity $(u)$ as zero, then equation becomes

$v=gt;h=\frac{1}{2}g{t}^2;v^2=2gh;h^{\text{nth }}=\frac{1}{2}g(2n-1)$

Case-II : Body thrown upward : If a body is thrown vertically up with an initial velocity $(u)$. Hence $a=-g$. (i) $v=u-gt$ (ii) $h=ut-\frac{1}{2}g{t}^2$ (iii) $v^2-u^2=-2gh$ (iv) $h_n=u-g(n-\frac{1}{2})$

Maximum height reached by the body

$H=\frac{u^2}{2g}$

$\therefore$ Therefore, the maximum height reached by the body is directly proportional to the square of the initial velocity.

Time of Ascent : The time taken by body thrownup to reach maximum height ’ $h$ ’ is called its time of ascent.

$t_a=\frac{u}{g}$

Hence time of ascent $t_a$ is directly proportional to the initial velocity $u$.

Time of Descent : The time taken by a freely falling body to reach the ground is called the time of descent.

$\begin{array}{rl}& t_d=\sqrt{\frac{2h}{g}}\\ & h=\frac{v^2}{2g},td=\sqrt{\frac{2\times v^2}{2g\times g}}=\frac{v}{g}\end{array}$

and

But, we know that $u=v$ i.e., projected velocity of a body is equal to the velocity of the body on reaching the ground.

$t_d=\frac{u}{g}=\text{ time of ascent }(t_a)\therefore \text{ Time of ascent }=\text{ time of descent }$

Time of flight : Time of flight is the time for which a body remains in the air and is given by sum of time of ascent and time of descent.

Therefore,

$\begin{array}{rl}& t=t_a+t_d\\ & t_d=\frac{u}{g}+\frac{u}{g}\end{array}$

$\therefore$ Time of flight, $t=\frac{2u}{g}$

Velocity on reaching ground : When a body is dropped from a height $h$, its initial velocity is zero. Late fimel vecthy the ground be $v$. For a freely falling body.

but

$\begin{array}{rl}& v^2-u^2=2gh\\ & u=0\\ & v^2-0=2gh;v=\sqrt{2gh}\end{array}$

Case-III : Body projected vertically up from the top of a tower:

If a body is projected vertically up from the top of a tower of height ’ $h$ ’ with velocity ’ $u$ ‘. Then

displacement after time $t$ is $s=ut-\frac{1}{2}g{t}^2$

velocity after time $t$ is $v=u-gt$.

Its velocity on reaching the ground is $\sqrt{u^2+2gh}$

Its maximum height above the ground is $h+(u^2/2g)$

Illustration 16 :

A body is allowed to fall from a height of $98m$. Find the time taken by the body to hit the ground, its veluciny befince bungs the ground and the distance travelled by it in the last second of motion. $(g=9.8m/s^2)$

solution :

Given

$\begin{array}{rl}& g=9.8m/s^2\\ & h=98m\\ & u=0\text{ (allowed to fall) }\end{array}$

Since, the body is falling downward we will use the following equations.

For velocity,

$\begin{array}{rl}& h=ut+\frac{1}{2}g{t}^2\\ & 98=0+\frac{1}{2}\times 9.8\times t^2\\ & t^2=20\Rightarrow t=4.47s\end{array}$

$v=u+gt$

$\begin{array}{rl}& v=0+9.8\times 4.47\\ & v=43.83m/s\end{array}$

For distance travelled in the last second of motion

$\begin{array}{rl}& h_{\text{nth }}=u+\frac{1}{2}g(2n-1)\\ & h_{\text{nth }}=0+\frac{1}{2}\times 9.8(2\times 4.47-1)\\ & h_{\text{nth }}=38.91m\end{array}$

Illustration 17 :

A body is thrownward from a tower of height $192m$ with initial velocity of $2m/s$. Find the time takten hy in is the the gramed velocity just before it hits the ground and its velocity and height for the ground after 2 sec after the fall $(0m=10m/s^3.$ )

solution : Given $u=2m/s$

$h=192m$

For time taken to hit the ground, we use

$\begin{array}{rl}& h=ut+\frac{1}{2}g{t}^2\\ & 192=2t+\frac{1}{2}\times 10\times t^2\\ & 5t^2+2t-192=0\end{array}$

$t=\frac{-2\pm \sqrt{(2{)}^2-(4)(5)(-192)}}{2\times 5}$

$t=\frac{-2\pm 62}{10}$

$t=6,\frac{-64}{10}$

So, $t=6$ seconds

Velocity just before hitting the ground

$v=u+gt$

$v=2+10\times 6\Rightarrow v=62m/s$

Now, velocity after two seconds

$v=2+10\times 2$

$v=22m/s$

(Using $h=ut+g{t}^2$)

Then $x=2\times 2+\frac{1}{2}\times 10\times 4$

$x=24m$

Illustration 18 :

A person sitting on the top of a tower drops ball at regular intervals of one second. Find the position of 1st, 3rd and 5th ball when the sixth ball is being dropped. (Use $g=10m/{\mathbf{s}}^2$ )

solution : You can see from the adjoining figure that when the $6^{\text{th }}$ ball is about to be dropped then the $1^{\text{st }}$ ball has already fallen for 5 seconds, so distance of $1^{\text{st }}$ ball, $h_1=0+\frac{1}{2}\times 90(5{)}^2$

$h_1=125m\text{ (from the top) }$

The third ball has fallen for three seconds

$\begin{array}{rl}& h_3=0+\frac{1}{2}\times 10(3{)}^2\\ & h_3=45m\text{ (from the top) }\end{array}$

The fifth ball has fallen for one second only

$\begin{array}{rl}& h_5=\frac{1}{2}\times 10\times (1{)}^2\\ & h_5=5m\text{ (from the top) }\end{array}$

Illustration 19

A body is projected vertically upwards. If $t_1$ and $t_2$ be the times at which its height is $h$ above the point of projection, while ascending and descending respectively. Find $h$, the velocity of projection, maximum height reached by the body and total time of filight in terms of $g,{\mathit{t}}_1$ and ${\mathit{t}}_{\mathbf{2}}$.

solution : As the initial direction of motion is upward it will be taken as positive and downward direction as negative. So using

$h=ut-\frac{1}{2}g{t}^2$

This equation is a quadratic equation which has two roots $t_1$ and $t_2$ for the same displacement $h$

$\begin{array}{rl}& g{t}^2-2ut+2h=0\\ & t_1+t_2=-(\frac{-2u}{g})\begin{array}{c}\text{ For }a{x}^2+bx+c=0\\ \text{ Some of roots, }{x}_1+x_2=-\frac{b}{a}\\ \text{ Product of roots }{x}_1{x}_2=\frac{c}{a}\end{array}\end{array}$

So, the velocity of projection, $u=\frac{g}{2}(t_1+t_2)$

Now the product of rook $t_1{h}_2=\frac{2h}{g}\Rightarrow h=\frac{g{h}_2}{2}$

Fer maxion heride $H$

$\begin{array}{rl}& 0=r^2-2gH(\text{ Using }{r}^2=r^2-2g{r}^2)\\ & 2gH={\frac{g}{2}(t_1+t_2)}^2\\ & H=\frac{g(t_1+t_2{)}^2}{2}\end{array}$

So the neal time of firis $=t_1+t_2$

Illustration 20

is:

spositive. Sa

$h=r_1+\frac{1}{2}{g}_1^2$

downond so $k$ will be then megaive, grvity will be also megive heace

Let $T$ be the tine ulven, when the partick is dropped, so

$h=0+\frac{1}{2}g{r}^2$

Non, aniphing (i) by $r^2$ and (ii) by $t_1$ ad then suberacting

$m_1=m_1{s}_2+\frac{1}{2}{s}_1^2{t}_2$
$m_2=m_1{t}_2-\frac{1}{2}s{t}_2^2{t}_1$
$+-+$
$m=\frac{1}{2}{s}_1{z}_2$

Substing tis $\mathit{h}$ in (iii)

$\begin{array}{rl}& g{r}^2=gn2\\ & T=\sqrt{P_2}\text{ (Requmer tis resin) }\end{array}$

Illustration 21

Fer the then :

$h=sen-\frac{1}{2}se{n}^2$

For the stone thrown downward :

downward direction +ve as it is the initial direction of motion upward direction -ve, so

$400-h=20t+\frac{1}{2}g{t}^2$ …(ii)

Adding (i) and (ii)

$400=100t\Rightarrow t=4sec$

Substituting in (i)

$\begin{array}{rl}h& =80\times 4-\frac{1}{2}\times 10\times 16\\ h& =320-80\\ h& =240m\end{array}$

Illustration 22

A ball is thrown vertically upwards with a velocity of $20m/s$ from the top of a tall building of height $25m$. (a) how long will it go before the ball hits the ground? and (b) How high will the ball rise? ( $g=10m/s^2$ )

Solution We solve this problem by two methods, so that your concepts become crystal at the same time you will be able to use sign conventions properly so that a problem can be solved in the shortest time possible.

Method-1 : Upward motion $B$ to $C$

time to reach $C$,

$\begin{array}{rl}& t=\frac{u}{g}\\ & t=\frac{20}{10}=2s\end{array}$

downward motion $C$ to $B$

$t=2s\text{ (As time of ascent and descent are equal) }$

Now, the motion from $B$ to $A:$ At $B$ the ball would have the same speed as it was thrown with i.e. $20m/s$, the only change in the direction, now the ball in moving downward so,

$25=20t+1/2\times 10t^2$

$t^2+4t-5=0$

or

So, the total time $=2+2+1=5$ seconds

Method-2 : Now let us take the point of projection to be the $(0,0)$, as the initial direction of motion is upward, the upward direction will be taken as +ve and the downward direction as -ve. So displacement BA will be -25 (as it is from $B$ to $A$ ) and gravity (g) will be negative.

$\begin{array}{rl}& -25=20t-\frac{1}{2}\times 10\times t^2\\ & t^2-4t-5=0\\ & t=5sec\\ & t=-1\text{ (discard) }\end{array}$

( $h=ut+1/2g{t}^2,+$ ve sign is due to the initial direction of motion being downward in the motion $BA$ )

EXERCISE 1

Fill in the blanks

DIRECTIONS : Complete the following statements with an appropriate word / term to be filled in the blank space(s).

1. Distance traveled divided by elapsed time gives ____
2. If a car starts at rest and accelerates uniformly, the distance it travels is proportional to the ____ of the time it travels.
3. All objects in free fall at a given place have the same All ____
4. If a car is going.northward and the driver jams on its brakes, the direction of its acceleration is ____
5. When an object is going in a circular path at constant speed, the direction of its acceleration is ____
6. The length of second’s hand in a watch is $1cm$. The change in velocity of its tip in 15 second is ____
7. A truck travelling due north at $20m/s$ turns left and travels at the same speed. Then the change in velocity is ____
8. A particle is moving eastward with a velocity of $5m/s$. In 10 second the velocity changes to $5m/s$ northward. The average acceleration in this time is ____
9. If a particle moves in a circle describing equal angles in equal interval of times, its velocity vector ____
10. A particle moves with a velocity $v$ in a circle of radius $r$, then its angular velocity is equal to ____ and acts along the ____
11. The ratio of angular speeds of minute hand and hour hand of watch is ____
12. A ball thrown vertically upwards return to its starting point in $4s$. If $g=10m/s^2$, its initial speed was ____
13. A body falling freely from the rest has a velocity $v$ after it has fallen through a distance $h$. The distance it has to fall down further for its velocity to become $2v$ is ____ times $h$.
14. A body, dropped from a tower with zero velocity, reaches the ground in $4sec$. The height of the tower is about ____ $m$
15. The magnitude of average velocity ____ equal to the average speed.

True and false

DIRECTIONS : Read the following statements and write your answer as true or false.

1. Area under velocity-time graph shows displacement.
2. Magnitude of displacement can be equal to or lesser than distance.
3. If particle speed is constant, acceleration of the particle must be zero
4. A particle moving with a uniform velocity must be along a straight line.
5. A particle is known to be at rest at time $t=0$. If its acceleration at $t=0$ must be zero.
6. The equation $s=ut+\frac{1}{2}a{t}^2$ with the usual notation is vectorial in nature.
7. A ball is thrown vertically upwards in vacuum. Then the time of ascent is equal to the time of descent.
8. In a journey, numerical value of displacement $\le$ distance.
9. An object covers distances in direct proportional to the square of the time elapsed. Its acceleration is increasing.
10. A particle in one dimensional motion with positive value of acceleration must be speeding up.
11. In circular motion the centripetal and centrifugal forces acting in opposite directions balance each other and the net force on the revolving particle is zero.
12. Magnitude of acceleration is constant in the rotating motion along a circular path.
13. Forces responsible for uniform circular motion are called centrifugal force.
14. Centrifugal force is the reaction of the centripetal force.
15. If a body is moving on a curved path with constant speed, then its acceleration is perpendicular to the direction of motion.

MATCH THE COLUMN

DIRECTIONS : Each question contains statements given in two columns which have to be matched. Statements $(A,B,C,D$ ) in Column I have to be matched with statements $(p,q,r,s)$ in Column II.

1.

Column I	Column II
(A) Uniform speed	(p) Unequal distance in equal time
(B) Constant speed	(q) Zero acceleration
(C) Uniform acceleration	(r) Unequal velocity in equal time change
(D) Non-uniform acceleration	(s) Equal distance in equal time
(E) Non-uniform speed	(t) Equal velocity change in equal time

2.

Column I	Column II
(A) Average velocity	(p) $\frac{v-u}{T}$
(B) Acceleration	(q) D/T
(C) Final velocity	(r) $rt+\frac{1}{2}a{t}^2$
(D) Distance	(s) $\frac{v+u}{t}$
(E) Speed	(t) u+at

3.

Column I	Column II
(A) Slope of displacement-time graph	(p) acceleration
(B) Slope of velocity-time graph	(q) velocity
(C) Area under velocity-time graph intercepted with time-axis.	(r) change in velocity
(D) Area under acceleration-time graph intercepted with time-axis.	(s) Displacement

Very short answer question

DIRECTIONS : Give answer in one word or one sentence.

1. What does the speedometer record - the average speed or instantaneous speed?
2. Can a body moving with a uniform velocity be in equilibrium?
3. Two particles $A$ and $B$ are moving along the same straight line with $B$ ahead of $A$. Velocity remaining unchanged, what would be the effect on the magnitude of relative velocity, if $A$ is ahead of $B$ ?
4. Under what condition the average velocity of a body is equal to its instantaneous velocity?
5. When the magnitude of average velocity is same as that of average speed?
6. Identify the type of motions? (a) a carom coin is striking against the side of a board and not rebounding smoothly but hoping up as it rebounds, (b) a car going along a zigzag path on a road.
7. Can a particle has varying speed but a constant velocity?
8. What is the acceleration of a particle moving with uniform velocity?
9. Under what condition will the distance and displacement of a moving object have the same magnitude?
10. What does the slope of position and time graph represent for uniform motion?
11. If the velocity of a particle is non-zero, can its acceleration ever be zero? Explain.
12. If the velocity of a particle is zero, can its acceleration ever by non-zero? Explain.
13. If a car is traveling eastward, can its acceleration be westward? Explain.
14. What does slope of $v-t$ graph represent ?
15. A ball is thrown straight up. What is its velocity and acceleration at the top?

Short answer question

DIRECTIONS : Give anss ’er in 2-3 sentences.

1. A car, $A$, travelling with a speed of $60km/hr$ on a straight road is ahead of another car $B$ travelling with a speed of $40km/hr$. What would be relative velocity of $A$ with respect to $B$ ? Would it be changed if $B$ is ahead of $A$ ?
2. A body moving along $X$ direction has at any instant its $x$ coordinate given by $x=a+bt+c{t}^2$. What will be acceleration of the particle?
3. A stone released with zero velocity from the top of the tower reaches the ground in 4 second. What is the approximate height of the tower?
4. A particle moves along $x$-axis in such a way that its coordinate $x$ varies with time according to the equations $=2-5t+6t^2$. What is the initial velocity of the particle?
5. In the previous example, what is the acceleration of the particle?
6. The displacement of a body is given to be proportional to the cube of time elapsed. What will be the acceleration of the body?
7. What is the distance travelled by a body falling freely from rest in the first, second and third seconds.
8. A stone is thrown upwards with a velocity $v$ from the top of a tower. It reaches the ground with a velocity $3v$. What is the height of the tower?
9. A body is projected vertically upwards with a velocity of $96f/s$. What will the total time for which the body remains in air? Assume, acceleration due to gravity, $g=32ft{s}^{-2}$.
10. The velocity of a body moving with a uniform acceleration of $2m{s}^{-2}$ is $10m{s}^{-1}$. What is its velocity after an interval of 4 second?.
11. A motor car moving with a uniform velocity of $20m{s}^{-1}$ comes to a stop, on the application of brakes, after travelling a distance of $10m$. What is its acceleration?
12. A wooden block of mass $10g$ is dropped from the top of a cliff $100m$ high. Simultaneously, a bullet of mass $10g$ is fired from the foot of the cliff upwards with a velocity $100m{s}^{-1}$. After what time, the bullet and the block meet?
13. Draw the position time graph for particle moving with positive and negative velocities.
14. A car travels half the distance with constant velocity $30km{h}^{-1}$ and another half with a constant velocity of $40km{h}^{-1}$. What is the average velocity of the car?
15. For a particle in one dimensional motion, the instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?

Long Answer Questions

DIRECTIONS : Give answer in four to five sentences.

1. Establish the three equations of uniformly accelerated motion graphically.
2. Draw velocily-lime graph of a uniformly accelerated mintion
3. Draw position-time graph of an uniformly accelerated motion.
4. Derive the distance travelled by body performing a motion with constant acceleration at the $n^{\text{th }}$ second.
5. Define (a) instantaneous velocity (b) instantaneous acceleration (c) average velocity and (d) average acceleration.
6. What do you mean by the distance and displacement covered by a particle? Explain with examples.

EXAMPLE 2

Multiple choice Questions

DIRECTIONS : This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct

1. If the displacement of a particle varies with time as $\sqrt{x}-1+7$, the

(a) velocity of the particle is inversely proportional to $t$

(b) velocity of the particle is proportional to $t$

(c) velocity of the particle is proportional to $\sqrt{t}$

(d) the particle moves with a constant acceleration

2. The initial velocity of a particle is $u($ at $t=0)$ and the acceleration a is given by $ft$.

Which of the following relation is valid

(a) $1,u\cdot t{t}^2$

(b) $v=u+f{t}^2/2$

(c) , $11\cdot 11$

(d) $v=u$

3. The poution, “l’ a particle varies with time $(t)$ as 1: $A{t}^{2}B{I}^{\ast }$ The acceleration at time $t$ of the particle will be equal to $/$ ero

(i) $2A$

(h) $A$

(c) $\frac{A}{3B}$

(d) terr

4. If a car at rest accelerates uniformly to a speed of $144km/h$ in 20 sec.. il covers a distance of

(a) $20cm$

(b) $400m$

(c) $1440cm$

(d) $2980cm$

5. The relation between time $f$ and distance $x$ is

$t=\alpha x^2+\beta x$

where $\alpha$ and $\beta$ are constants The retardation is

(a) $2a{v}^3$

(b) $2b{v}^3$

(c) $2abr$

(d) $2b^2{v}^3$

6. The coordinates of a moving particle at time $t$ are given by $x=c{t}^2$ and $y=b{t}^2$. The speed of the particle is given by

(a) $2t(c+b)$

(c) $t\sqrt{(c^2+b^2)}$

(b) $2t\sqrt{(c^2-b^2)}$

(d) $2t\sqrt{(c^2+b^2)}$

7. The displacement of a particle is given by

$y=a+bt+c{t}^2-d{t}^A$

The initial velocity and acceleration are respectively

(a) $b,-4d$

(b) $-b,2c$

(c) $b,2c$

(d) $2c,-4d$

8. The displacement $x$ of a particle moving along a straight line at time $t$ is given by

$x=a_0+a_{1}t+a_2{t}^2$

What is the acceleration of the particle?

(a) $a_1$

(c) $2a_2$

(b) $a_2$

(d) $3a_2$

9. The displacement-time graphs of two particles $A$ and $B$ are straight lines making angles of respectively ${30}^{\circ }$ and ${60}^{\circ }$ with the time axis. If the velocity of $A$ is $v_A$ and that of $B$ is $v_B$, the value of $v_A/v_B$ i

(a) $1/2$

(b) $1/\sqrt{3}$

(c) $\sqrt{3}$

(d) $1/3$

10. A car accelerates from rest at a constant rate $\alpha$ for some time after which it decelerates at a constant rate $\beta$ to come to rest. If the total time elapsed is $t$, the maximum velocity acquired by the car is given by

(a) $(\frac{{\alpha }^2+{\beta }^2}{\alpha \beta })\hat{i}$

(b) $(\frac{{\alpha }^2-{\beta }^2}{\alpha \beta })$ t

(c) $(\frac{\alpha +\beta }{\alpha \beta })t$

(d) $(\frac{\alpha \beta }{\alpha +\beta })t$

11. A person travels along a straight road for the first half time with a velocity $v_1$ and the second half time with a velocity $v_2$. Then the mean velocity $\overline{v}$ is given by

(a) $\overline{v}=\frac{v_1+v_2}{2}$

(b) $\frac{2}{v}=\frac{1}{v_1}+\frac{1}{v_2}$

(c) $\overline{v}=\sqrt{\frac{v_2}{v_1}}$

(d) $\overline{v}=\sqrt{\frac{v_2}{v_1}}$

12. A passenger travels along the straight road for half the distance with velocity $v_1$ and the remaining half distance with velocity $v_2$. Then average velocity is given by

(a) $v_1{v}_2$

(b) $v_2^2/v_1^2$

(c) $(v_1+v_2)/2$

(d) $2v_1{v}_2/(v_1+v_2)$

13. A point moves with uniform acceleration and $v_1,v_2$ and $v_3$ denote the average velocities in $t_1,t_2$ and $t_{3}sec$. Which of the following relation is correct?

(a) $(v_1-v_2):(v_2-v_3)=(t_1-t_2):(t_2+t_3)$

(b) $(v_1-v_2):(v_2-v_3)=(t_1+t_2):(t_2+t_3)$

(c) $(v_1-v_2):(v_2-v_3)=(t_1-t_2):(t_1-t_3)$

(d) $(v_1-v_2):(v_2-v_3)=(t_1-t_2):(t_2-t_3)$

14. A bus starts moving with acceleration $2m/s^2$. A cyclist 96 $m$ behind the bus starts simultaneously towards the bus at $20m/s$. After what time will he be able to overtake the bus?

(a) $4sec$

(b) $8sec$

(c) $12sec$

(d) $16sec$

15. When the speed of a car is $v$, the minimum distance over which it can be stopped is $s$. If the speed becomes $nv$, what will be the minimum distance over which it can be stopped during same retardation

(a) $s/n$

(b) $ns$

(c) $s/n^2$

(d) $n^{2}s$

16. The velocity of a particle at an instant is $10m/s$. After 5 $sec$, the velocity of the particle is $20m/s$. The velocity at 3 seconds before from the instant when velocity of a particle is $10m/s$.

(a) $8m/s$

(b) $4m/s$

(c) $6m/s$

(d) $7m/s$

17. A rubber ball is dropped from a height of 5 metre on a plane where the acceleration due to gravity is not known. On bouncing, it rises to a height of $1.8m$. On bouncing, the ball loses its velocity by a factor of

(a) $\frac{3}{5}$

(b) $\frac{9}{25}$

(c) $\frac{2}{5}$

(d) $\frac{16}{25}$

18. A stơne thrown vertically upwards with a speed of $5m/sec$ attains a height $H_1$. Another stone thrown upwards from the same point with a speed of $10m/sec$. attains a height $H_2$. The correct relation between $H_1$ and $H_2$ is

(a) $H_2=4H_1$

(c) $H_1=2H_2$

(b) $H_2=3H_1$

(d) $H_1=H_2$

19. A particle covers half of the circle of radius $r$. Then the displacement and distance of the particle are respectively

(a) $2\pi r,0$

(b) $2r,\pi r$

(c) $\frac{\pi r}{2},2r$

(d) $\pi r,r$

20. A bod ${}^2$ covers $26,28,30,32$ meters in ${10}^{\text{th }},{11}^{\text{th }},{12}^{\text{th }}$ and ${13}^{\text{th }}$ seconds respectively. The body starts

(a) from rest and moves with uniform velocity

(b) from rest and moves with uniform acceleration

(c) with an initial velocity and moves with uniform acceleration

(d) with an initial velocity and moves with uniform velocity

21. A ball released from a height falls $5m$ in one second. In 4 seconds it falls through

(a) $20m$

(b) $1.25m$

(c) $40m$

(d) $80m$

22. A food packet is released from a helicopter rising steadily at the speed of $2m/sec$. After 2 seconds the velocity of the packet is $(g=10m/se{c}^2)$

(a) $22m/sec$

(b) $20m/sec$

(c) $18m/sec$

(d) nonc of the above

23. The speed of the car is $v$, the minimum distance over which it can be stopped is $x$. If the speed becomes $nv$, then the minimum distance in which it can be stopped in same time is

(a) $x/n$

(b) $nx$

(c) $x/n^2$

(d) $n^{2}x$

24. A stone thrown upward with a speed $u$ from the top of the tower reaches the ground with a velocity $3u$. The height of the tower is

(a) $3u^2/g$

(c) $6u^2/g$

(b) $4u^2/g$

(d) $9u^2/g$

25. A rifle bullet loses $1/20$ th of its velocity in passing through a plank. The least number of such planks required just to stop the bullet is

(a) 5

(b) 10

(c) 11

(d) 20

26. A stone is dropped into a well in which the level of water is $h$ below the top of the well. If $v$ is velocity of sound, the time $T$ after which the splash is heard is given by

(a) $T=\frac{2h}{v}$

(b) $T=\sqrt{(\frac{2h}{g})}+\frac{h}{v}$

(c) $T=\sqrt{(\frac{2h}{v})}+\frac{h}{g}$

(d) $T=\sqrt{(\frac{h}{2g})}+\frac{2h}{v}$

27. A body dropped from a height ’ $h$ ’ with an initial speed zero, strikes the ground with a velocity $3km/$ hour. Another body of same mass dropped from the same height ’ $h$ ’ with an initial speed $u$ ’ $=4km/$ hour. Find the final velocity of second mass, with which it strikes the ground

(a) $3km/hr$

(b) $4km/hr$

(c) $5km/hr$

(d) $6km/hr$

28. The acceleration of a particle is increasing linearly with time $t$ as $bt$. The particle starts from the origin with an initial velocity $v_0$. The distance travelled by the particle in time $t$ will be

(a) $v_{0}t+\frac{1}{3}{b}^2$

(b) $v_{0}t+\frac{1}{3}{b}^3$

(c) $v_{0}t+\frac{1}{6}b{t}^3$

(d) $v_{0}t+\frac{1}{6}{b}^3$

29. A ball is dropped downwards, after $1sec$ another ball is dropped downwards from the same point. What is the distance between them after $3sec$ ?

(a) $25m$

(b) $20m$

(c) $50in$

(d) $9.8m$

30. Two trains are each $50m$ long meviny parallel towards each other at speeds $10m/s$ and $15m/s$ respectively. After what time will they pass wach other?

(a) $5\sqrt{\frac{2}{3}}sec$

(b) $4sec$

(c) $2sec$

(d) $6sec$

31. If a ball is thrown vertically upwards with a velocity of $40m/s$, then velocity of the ball after two seconds is : $(g=$ $10m/se{c}^2$ )

(a) $15m/s$

(b) $20m/s$

(c) $25m/s$

(d) $28m/s$

32. A car moving with a vpecd of $40km/$ hour can be stopped by applying brakes after at least $2m$. If the same car is moving with a speed of $80km/hour$, what is the minimum stopping distance

(a) $8m$

(b) $6m$

(c) $4m$

(d) $2m$

33. A stone is thrown vertically upwards. When the particle is at a height half of its maximum height, its speed is $10m/sec$; then maximum height attained by particle is ( $g=10m/se{c}^2$ )

(a) $8m$

(b) $10m$

(c) $15m$

(d) $20m$

34. The acceleration due to gravity on planet $A$ is nine times the acceleration due to gravity on planet $B$. $A$ man jumps to a height $2m$ on the surface of $A$. What is height of jump by same person on planet $B$ ?

(a) $2/3m$

(b) $2/9m$

(c) $18m$

(d) $6m$

35. A ball is released from the top of a tower of height $h$ meters. It takes $T$ seconds to reach the ground. What is the position of the ball at $\frac{T}{3}$ second

(a) $\frac{8h}{9}$ metres from the ground

(b) $\frac{7h}{9}$ metres from the ground

(c) $\frac{h}{9}$ metres from the ground

(d) $\frac{17h}{18}$ metres from the ground

36. An automobilc travelling with a speed of $60km/h$, can brake to stop within a distance of $20m$. If the car is going twice as fast $1.c.120km/h$, the stopping distance will be

(a) $60m$

(b) $40m$

(c) $20m$

(d) $80m$

37. The motion of a particle is described by the equation $u=$ $at$. The distance travelled by particle in first $4sec$ is

(a) $4a$

(b) $12a$

(c) 6 m

(d) $x_1$

38. For the velucity tumc graph ,hown in the figure below the distance coveral hy the turly in the last two seconds of its motion is what fraction of the total distance travelled by it in all the seven seconds?

(a) $\frac{1}{2}$

(b) $\frac{1}{4}$

(c) $\frac{2}{3}$

(d) $\frac{1}{3}$

39. A stone is just released from the window of a train moving along a horizontal straight track. The stone will hit the ground following a

(a) straight line path

(b) circular path

(c) parabolic path

(d) hyperbolic path

40. A particle is moving eastwards with a velocity of $5m{s}^{-1}$. In 10 seconds the velocity changes to $5m{s}^{-1}$ northwards. The average acceleration in this time is

(a) $\frac{1}{2}m{s}^{-1}$ towards north

(b) $\frac{1}{\sqrt{2}}m{s}^{-2}$ towards north-east

(c) $\frac{1}{\sqrt{2}}m{s}^{-2}$ towards north-west

(d) zero

41. A parachutist after bailing out falls $50m$ without friction. When parachute opens, it decelerates at $2m/s^2$. He reaches the ground with a speed of $3m/s$. At what height, did he bail out?

(a) $182m$

(b) $91m$

(c) $111m$

(d) $293m$

More than one correct

DIRECTIONS : This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d) out of which ONE OR MORE may be correct.

1. A vector quantity is a physical quantity which needs

(a) magnitude

(b) direction

(c) distance

(d) time

2. Which of the following are examples of uniform velocity?

(a) Motion of moon around earth

(b) Motion of planet around sun

(c) Motion of car on crowded road

(d) Motion of a moving fan

3. Average velocity can be calculated by

(a) $\frac{\text{ Distance travelled along given direction }}{\text{ Time taken }}$

(b) $\frac{\text{ Initial velocity }}{2}$

(c) $\frac{\text{ Initial velocity + Final velocity }}{2}$

(d) $\frac{\text{ Final velocity }}{2}$

4. Which of the following are vector quantities ?

(a) Speed

(b) Distance

(c) Velocity

(d) Acceleration

5. If a body starts from rest, its

(a) $u=0$

(b) $a=0$

(c) velocity increases

(d) velocity decreases

6. If a body moves with uniform velocity, its

(a) $u=v$

(b) $v=0$

(c) $u=0$

(d) $a=0$

7. Velocity time graph of a body moving with variable acceleration is

8. Which of the following is correct about the given below graph?

(a) Velocity is zero

(b) Velocity is constant

(c) Acceleration is zero

(d) Acceleration is constant

9. Which of the following statements are true for displacements?

(a) It can be zero

(b) It cannot be zero

(c) Its magnitude is greater than distance travelled

(d) Its magnitude is lesser than distance travelled

10. The speed of an object is

(a) distance per unit tims

(b) a scalar quantity

(c) displacement per unit times

(d) a vector quantity

Multiple matching questions

DIRECTIONS : Following question has four statements (A, $B,C$ and $D)$ given in Column I and four statements $(p,q,r,s,\dots )$ in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II. Match the entries in Column I with entries in Column II.

Column I	Column II
(A) $v$	(p) $ut+1/2a{t}^2$
(B) $u$	(q) $v-u/t$
(C) $a$	(r) Final Velocity
(D) $s$	(s) $u+at$
	(t) Initial Velocity
	(u) $\sqrt{u^2+2as}$

Column I	Column II
(A) Acceleration	(p) $m/s^2$
(B) Speed	(q) centripetal force
(C) Circular motion	(r) $m$
(D) Displacement	(s) $m/s$
	(t) scalar quantity
	(u) vector quantity

3. A body is in motion for some time, then at a centain insent of time.

Column I	Column II
(A) Distance	(p) may be positive
(B) Displacement	(q) may be negative
(C) Speed	(r) may be zero
(D) Velocity	(s) may be increasing
	(t) may be decreasing

4. Match the following for the $v-t$ graph shown in the figure.

Column I	Column II
(A) At point A	(p) Speed is increasing
(B) At point B	(q) Acceleration is positive
(C) At point C	(r) Acceleration is negative
(D) At point $D$	(s) Acceleration is zero
	(t) Speed is zero

Fill in the Passage

DIRECTIONS : Complete the following passage(s) with an appropriate word/term to be filled in the blank spaces.

I. Straight, distance, equal, directly, distance

When a body moves with uniform speed, it travels equal ____ (1) ____ in ____ (2) ____ intervals of time. This means the ____ (3) ____ travelled is ____ (4) ____ proportional to time. Therefore the graph obtained is a ____ (5) ____ line passing through the origin.

II. speed, direction, centrifugal, radial, centripetal

When a body moves in a circular path, its ____ (1) ____ changes at every moment. ____ (2) ____ of the body may remain constant or change. When the force exerted is such that it keeps the body in a circular path, it is called ____ (3) ____ or ____ (4) ____ force whereas if the force exerted throws the body out of circular motion, it is called ____ (5) ____ force.

III. displacement, vector, scaler, less than, zero, distance, velocity, speed,

When a body moves from one point to another, the total length of path covered by the body is called and the shortest distance between the initial and the final position is called ____ (1) ____. Distance is a ____ (2) ____ quantity while displacement is a ____ (3) ____ quantity. The magnitude of distance can never be ____ (4) ____ that of displacement. Displacement of the body in a round trip is always ____ (5) ____ The time rate of change of distance is called ____ (6) ____ while that of displacement is called ____ (7) ____

IV. zero, straight line, parabola, increasing, decreasing, acceleration.

Equations of motion are valid only if the ___ (1) ____ of the body is constant. For such a motion, velocity-time graph is ____ (2) ____ and it passes through origin if initial velocity of the body is ____ (3) ____. Displacement time graph for such a motion is ____(4) ____. If the acceleration of the body is positive w.r.t its velocity, slope of displacement time graph is ____ (5) ____ and if its acceleration is negative w.r.t velocity, slope of displacement-time graph is ____ (6) ____

Passage based questions

DIRECTIONS : Study the given paragraph(s) and answer the following questions.

PASSAGE - 1

Consider the motion of a batsman in a cricket game. The length of the pitch is $18m$. Suppose the batsman complete one run and he is now $18m$ away from his batting crease or his starting point. Then he turns back and gets run out, when he is exactly mid way through his second run.

1. The total distance travelled by batsman is

(a) $18m$

(b) $36m$

(c) $9m$

(d) $27m$

2. How far is the batsman from his starting point ?

(a) $9m$

(b) $0m$

(c) $18m$

(d) $27m$

3. What is the net displacement of the batsman ?

(a) $9m$

(b) $0m$

(c) $18m$

(d) $27m$

PASSAGE - II

The velocity of any moving body, in ordinary circumstances, does not always remain constant. For example, a bus gains velocity while leaving a station and loses velocity while approaching a station. Similarly, when a store is dropped from a certain height, its velocity goes on increasing as it comes to the ground. But if a stone is thrown upwards its velocity goes on decreasing, till it becomes zero and then its velocity starts increasing as it approaches the ground.

4. The rate of change of velocity with time is known as

(a) Acceleration

(b) Speed

(c) Intial velocity

(d) Final velocity

5. What can be concluded about acceleration if velocity increases continously?

(a) Acceleration is positive

(b) Acceleration is constant

(c) Acceleration is negative

(d) Acceleration is zero

6. Which of the following case represents a negative acceleration?

(a) Car starting from rest

(b) A stone falling from height

(c) Train coming to halt

(d) Bus moving with uniform velocity

PASSAGE-III

A ball is projected upwards from ground. It is observed from a $73.5m$ high window, twice at the interval of $2s$. Answer the following for the motion of the ball. (Take $g=9.8m{s}^{-2}$ )

7. What is the velocity of projection of the ball?

(a) $19.6m{s}^{-1}$

(b) $39.2m{s}^{-1}$

(c) $58.8m{s}^{-1}$

(d) $78.4m{s}^{-1}$

8. The maximum height attained by the ball above the ground

(a) $78.4m$

(c) $83.3m$

(b) $176.4m$

(d) $88.2m$

9. Total time of flight of the ball is

(a) $4s$

(b) $6s$

(c) $8s$

(d) $12s$

PASSAGE-IV

A thief is running on a motorcycle at a constant speed of 25 $m{s}^{-1}$. A police jeep starts chasing from a point $1.25km$ behind him with a uniform acceleration of $2m{s}^{-2}$.

10. After how much time will the police catch the thief?

(a) 1 minute

(b) 1.5 minute

(c) 2 miniutes

(d) 50 seconds

11. How much distance will the jeep cover to reach the thief ?

(a) $2.5km$

(b) $3.75km$

(c) $5km$

(d) $1.25km$

12. What should be the minimum acceleration of the thief to escape from police.

(a) $1m{s}^{-2}$

(b) $1.25m{s}^{-2}$

(c) $1.5m{s}^{-2}$

(d) $1.75m{s}^{-2}$

Assertion and Reason

DIRECTIONS : Each of these questions contains an Assertion followed by Reason. Read them carefully and answer the question on the basis of following options. You have to select the one that best describes the two statements.

(a) If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

(b) If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

(c) If Assertion is correct but Reason is incorrect.

(d) If Assertion is incorrect but Reason is correct.

1. Assertion : The distance-time graph of uniform motion is a straight line.

Reason : Independent variable is taken along $x$-axis and dependent variable along $y$-axis.

2. Assertion : The velocity of a body is a vector quantity.

Reason : A vector quantity has only magnitude and no direction.

3. Assertion : Motion of moon around earth is a non-uniform motion.

Reason : The size of moon is smaller than that of earth.

4. Assertion: If a body moves with uniform velocity, its acceleration is zero.

Reason : Rate of change of velocity is zero in case of body moving with uniform velocity.

5. Assertion : Instantaneous speed is the speed of a body over a long period of time.

Reason : The graph representing non-uniform speed will be a curve with increasing or decreasing slope.

6. Assertion: Displacement of a body may be zero when distance travelled by it is not zero.

Reason: The displacement is the longest distance between initial and final positive ions.

7. Assertion: The displacement time graph of a body moving with uniform acceleration is a straight line.

Reason: The displacement is proportional to square of time for uniformly accelerated motion.

Hots Subjectve Questions

DIRECTIONS : Answer the following questions.

1. Sailing fun, especially on a windy day. consider the top views of the two boats betow, one sailing with the wind and the other across the wind. Which can sail faster than wind speeds?
2. What will be the acceleration of a rock thrown straight upward at the moment it reaches the tippity-top of its trajectory?
3. Two balls of different masses are thrown vertically upwards with the same speed. They pass through the point of projection in their downward motion with the same speed (Neglect air resistance). Is it possible?
4. A particle is moving eastwards with a velocity of $5m/s$. In $10s$ the velocity changes to $5m/s$ northwards. Find the average acceleration in this time.
5. Two bodies, $A$ (of mass $1kg$ ) and $B$ (of mass $3kg$ ), are dropped from heights of $16m$ and $25m$. respectively. What is the ratio of the time taken by them to reach the ground?
6. A body starts from rest and moves with a uniform acceleration. Find the ratio of the distance covered in the $n$th sec to the distance covered in $n$ sec.
7. A body is moved along a straight line by a machine delivering constant power. How is the distance moved by the body proportional to time $t$ ?
8. A stone is dropped into a well in which the level of water is $h$ below the top of the well. If $v$ is velocity of sound, find the time $T$ after which the splash is heard?
9. A rocket is fired upward from the earth’s surface such that it creates an acceleration of $19.6m{s}^{-2}$. If atter $5s$, its engine is switched off, what would be the maximum height of the rocket from earth’s surface?
10. When a particle is thrown vertically upward with some initial velocity, draw velocity-time and acceleration-time graph.
11. Acceleration-time graph of a particle moving in a straight line is given as shown in fig. Plot its velocity-time graph. What is the maximum velocity of the particle. Initially particle was at rest.

EXERCISE - I

Fill in the Blanks

1. Average speed
2. Square
3. Accelaration
4. South
5. Centripetal
6. $(\frac{2\pi }{30\sqrt{2}})cm/sec$
7. $20\sqrt{2}$ m/sec, south-west
8. $\frac{1}{\sqrt{2}}m/s^2$ north-west
9. changes in direction
10. $\frac{v}{r}$, axis of notation
11. $12:1$
12. $20m/s$
13. 3
14. 80
15. may or may not be

True/Flase

1. True	2. True	3. Flase
4. True	5. False	6. True
7. True	8. True	9. False
10. False	11. False	12. True
13. False	14. True	15. True

Match The Columns

1. (A) $\to$ (s) : (B) $\to$ (q) : (C) $\to$ (t) :(D) $\to$ (r):(E) $\to$ (p)
2. (A) $\to$ (s): (B) $\to$ (p); (C) $\to$ (t): (D) $\to$ (r); (E) $\to$ (q)
3. (A) $\to$ (q); (B) $\to$ (p);(C) $\to$ (s); (D) $\to$ (r)

Very Short Answer Questions

1. Instantaneous speed.
2. Yes, since the net force acting on it is zero.
3. No effect.
4. When the body is moving with constant velocity, the average velocity of a body is equal to its instantaneous velocity.
5. When the body is moving with a constant velocity, the magnitude of average velocity is same as that of average speed.
6. (a) three dimensional motion (b) two dimensional motion
7. No since both speed and velocity remain constant in uniform motion.
8. Zero.
9. If the object is moving along a straight line.
10. Velocity.
11. Acceleration 15 Velocity at the top $=0$ and accelerntion at the top $=9.8m$ $s^{-2}$ (downwands).

Short Answer Questions

1. Relative velocity of $A$ with respect to $B$ is $20km/hr$ and it will ins change
2. Ditferentiating the given equation, $a/d=b+2c$

Difterentiating again, frodor $=2c$

Therefive, anceleration $=2c$.

3. $x=0,a=10m{s}^3,t=4s$

Therefive, $h=\frac{1}{2}\times 10\times (4{)}^2=80m$

4. $s=2-st+a{r}^2$

Therefore, after differentiating, we get the velocity, v=ds/dt = -5 + 12t

For initial velocity, $t=0,\therefore v=-5m{s}^{-1}$

5. $a=dv/dt=d/dt(-5+12t)=12m{s}^{-2}$

6. $s=k{t}^3$, velocity $=3k{t}^2$, acceleration $=6kt$, which shows the acceleration is increasing with time.

7. Distance travelled in first second of motion

$=\frac{1}{2}\times 10\times 1\times 1=5m$

Distance travelled in first two seconds of motion

$=\frac{1}{2}\times 10\times 2\times 2=20m$

Distance travelled in first three seconds of motion

$=\frac{1}{2}\times 10\times 3\times 3=45m$

Distance travelled in 2nd second $=(20-5)m=15m$

Distance travelled in 3rd second $=(45-20)m=25m$

The required ratio will be $5:15:25=1:3:5$.

8. We know, $v^2-u^2=2as$.

From this relation, we get, $9v^2-v^2=2gh$.

Therefore, $h=8v^2/2g=4v^2/g$

9. $0=-96+32t$ or, $t=3s$. Total time $=(3+3)s=6s$.
10. $v=10+2\times 4=18m{s}^{-1}$.
11. $u=20m{s}^{-1},s=10m$

We know, $v^2-u^2=2$ as.

Therefore, $a=-400/(2\times 10)m{s}^{-2}=-20m{s}^{-2}$

12. Let the block and the bullet meet at a height $x$ from the ground. Considering direction as +ve and the upward direction as negative, we get

for block, $u=0,s=100-x,a=+g$

For bullet, $u=-100m{s}^{-1},a=+g,s=-x$

We get. $-x=-100t+\frac{1}{2}g{r}^2$

$100-x=\frac{1}{2}g{r}^2$

Therefore, $2-x=-100x+100-x$

Therefore, $t=1$ s.

13. The $x-t$ graph for an object moving with a positive velocity.

The $x-t$ graph for an object moving with a negative velocity.

14. Let $s$ be the total distance covered by the car; and $t_1$ and $t_2$ be the time taken by the car to cover first half and second half respectively. Then, total time taken, $t=t_1+t_2$

$=\frac{S/2}{30}+\frac{S/2}{40}=\frac{S}{60}+\frac{S}{80}=\frac{7S}{240}h$

Therefore, average velocity,

$v_{av}=\frac{S}{t}=\frac{S}{7S/240}=\frac{240}{7}=34.3km{h}^{-1}$

15. In an accelerated motion, the velocity of an object always keeps on changing. Therefore, one has to measure the instantaneous velocity. However, when accelerated motion takes place along a straight line, the velocity of the body changes only due to change in magnitude of velocity. Therefore, the instantaneous speed is always equalto the magnitude of instantaneous velocity of the particle in one dimensional motion.

EXERCISE - 2

Multiple Choice Questions

1. (b) $\sqrt{x}=(t+7)$ or $x=(t+7{)}^2$

$\frac{dx}{dt}=2(t+7),\therefore$ velocity $\propto$ time

2. (b) $a=ft,a=\frac{dv}{dt}=ft$ at $t=0$, velocity $=u$ ${\int }_u^{v}dv={\int }_0^{t}ftdt,v-u=f\frac{t^2}{2}\Rightarrow v=u+f\frac{t^2}{2}$

Note : Do not use $v=u+$ at directly because the acceleration is not constant.

3. (c) Given that $x=A{t}^2-B{t}^3$

$\therefore$ velocity $=\frac{dx}{dt}=2At-3B{t}^2$

and acceleration $=\frac{d}{dt}(\frac{dx}{dt})=2A-6Bt$

For acceleration to be zero $2A-6Bt=0$.

$\therefore t=\frac{2A}{6B}=\frac{A}{3B}$

4. (b) $v=[144\times 1000/(60\times 60)]m/sec$.

$v=u+at$ or $(144\times 1000)(60\times 60)=0+a\times 20$

$\therefore a=\frac{144\times 1000}{60\times 60\times 20}=2m/se{c}^2$

Now $s=ut+\frac{1}{2}a{t}^2=0+\frac{1}{2}\times 2\times (20{)}^2=400m$

5. (a) Since $\alpha x^2+\beta x-t=0$ & $x=\frac{-\beta \pm \sqrt{{\beta }^2+4\alpha t}}{2\alpha }$

so $\frac{dx}{dt}=\pm \frac{1}{\sqrt{{\beta }^2+4\alpha t}}=v$

and $\frac{d^{2}x}{d{t}^2}=\pm [\frac{-1}{2}\frac{1\times 4\alpha }{({\beta }^2+4\alpha t{)}^{3/2}}]$

$=\pm \frac{-2\alpha }{({\beta }^2+4\alpha t{)}^{3/2}}=-2\alpha v^3$

6. (d) $\frac{dx}{dt}=2ct$ and $\frac{dy}{dt}=2bt$

$\therefore v=\sqrt{(\frac{d{x}^2}{dt})+(\frac{dy}{dt}{)}^2}$

$=\sqrt{(2ct{)}^2+(2bt{)}^2}=2t\sqrt{(c^2+b^2)}$

7. (c) $v=\frac{dy}{dt}=b+2ct-4d{t}^3$

$v_0=b+2c(0)-4d(0{)}^3=b$

( $\because$ for initial velocity, $t=0$ )

Now $a=\frac{dv}{dt}=2c-12d{t}^2$

$\therefore a_0=2c-12d(0{)}^2=2c,($ at $t=0)$

8. (c) $v=\frac{dx}{dt}=a_1+2a_{2}t\therefore a=\frac{dv}{dt}=2a_2$
9. (d) $v_A=\tan {30}^{\circ }$ and $v_B=\tan {60}^{\circ }$

$\therefore \frac{v_A}{v_B}=\frac{\tan {30}^{\circ }}{\tan {60}^{\circ }}=\frac{1/\sqrt{3}}{\sqrt{3}}=\frac{1}{3}$

10. (d) Let the car accelerates for a time $t_1$ and travels a distance $s_1$. Suppose the maximum velocity attained by the car be $v$. Then

$s_1=\frac{1}{2}\alpha t_1^2$ and $v=\alpha t_1,t_1=v/\alpha$

$\therefore s_1=\frac{1}{2}\times \alpha \times (v^2/{\alpha }^2)=\frac{v^2}{2\alpha }$

Let the car decelerates for a time $t_2$ and travels a distances $s_2$. Then

$s_2=v{t}_2-\frac{1}{2}\beta t_2{}^2$ and $0=\nu -\beta t_2$ or $t_2=\frac{\nu }{\beta }$ $\therefore s_2=\nu \times (\frac{v}{\beta })-\frac{1}{2}\beta (\frac{v^2}{{\beta }^2})$

or $s_2=\frac{v^2}{\beta }-\frac{v^2}{2\beta }=\frac{v^2}{2\beta }$

11. (a) Let for the first half time $t$, the person travels a distance $s_1$. Hence $v_1=\frac{s_1}{t}$ or $s_1=v_{1}t$

For second half time, $v_2=\frac{s_2}{t}$ or $s_2=v_{2}t$

Now, $\overline{v}=\frac{\text{ Total displacement }}{\text{ Total time }}=\frac{s_1+s_2}{2t}$

$=\frac{v_{1}t+v_{2}t}{2t}=\frac{v_1+v_2}{2}$

12. (d) $\frac{\frac{x}{2}+\frac{x}{2}}{\frac{x}{2v_1}+\frac{x}{2v_2}}=\frac{1}{(\frac{v_2+v_1}{2v_1{v}_2})}=\frac{2v_1{v}_2}{v_1+v_2}$
13. (b) Let $u$ be the initial velocity

$\therefore v_1^{\prime }=u+a{t}_1,v_2^{\prime }=u+a(t_1+t_2)$

and $v_3^{\prime }=u+a(t_1+t_2+t_3)$

Now $v_1=\frac{u+v_1^{\prime }}{2}=\frac{u+(u+a{t}_1)}{2}=u+\frac{1}{2}a{t}_1$

$v_2=\frac{v_1^{\prime }+v_2^{\prime }}{2}=u+a{t}_1+\frac{1}{2}a{t}_2$

$v_3=\frac{v_2^{\prime }+v_3^{\prime }}{2}=u+a{t}_1+a{t}_2+\frac{1}{2}a{t}_3$

So, $v_1-v_2=-\frac{1}{2}a(t_1+t_2)$

and $v_2-v_3=-\frac{1}{2}a(t_2+t_3)$

$\therefore (v_1-v_2):(v_2-v_3)=(t_1+t_2):(t_2+t_3)$

14. (b) Let after a time $t$, the cyclist overtake the bus. Then

$96+\frac{1}{2}\times 2\times t^2=20\times t$ or $t^2-20t+96=0$

$\therefore t=\frac{20\pm \sqrt{400-4\times 96}}{2\times 1}$

$=\frac{20\pm 4}{2}=8sec$. and $12sec$

15. (d) $v^2=u^2+2as$ or $v^2-u^2=2as$

Maximum retardation, $a=v^2/2s$

When the initial velocity is $nv$, then the distance over which it can be stopped is given by

$s_n=\frac{u_0^2}{2a}=\frac{(nv{)}^2}{2(v^2/2s)}=n^{2}s$ 16. (b) $u=10m/s,t=5sec,v=20m/s,a=$ ?

$a=\frac{20-10}{5}=2m{s}^{-2}$

From the formula $v_1=u_1+at$, we have

$10=u_1+2\times 3$ or $u_1=4m/sec$.

17. (c) Downward motion

$v^2-0^2=2\times 9.8\times 5\Rightarrow v=\sqrt{98}=9.9$

Also for upward motion

$0^2-u^2=2\times (-9.8)\times 1.8\Rightarrow u=\sqrt{3528}=5.94$

Fractional loss $=\frac{9.9-5.94}{9.9}=0.4$

18. (a) From third equation of motion $v^2=u^2+2ah$

In first case initial velocity $u_1=5m/sec$

final velocity $v_1=0,a=-g$

and max. height obtained is $H_1$, then, $H_1=\frac{25}{2g}$

In second case $u_2=10m/sec,v_2=0,a=-g$

and max. height is $H_2$ then, $H_2=\frac{100}{2g}$.

It implies that $H_2=4H_1$

19. (b) When a particle cover half of circle of radius $r$, then displacement is $AB=2r$

distance $=$ half of circumference of circle $=\pi r$

20. (c) The distance covered in $n^{\text{th }}$ second is

$S_n=u+\frac{1}{2}(2n-1)a$

where $u$ is initial velocity & $a$ is acceleration then

$26=u+\frac{19a}{2}$

$28=u+\frac{21a}{2}$

$30=u+\frac{23a}{2}$

$32=u+\frac{25a}{2}$

From equations (1) & (2) we get $u=7m/sec$, $a=2m/se{c}^2$

$\therefore$ The body starts with initial velocity

$u=7m/sec$

and moves with uniform accoleration $a=2m/se{c}^2$

21. (d) Since, $s=ut+\frac{1}{2}g{t}^2$

where $u$ is initial velocity a is acceleration.

In this case $u=0$ & $a=g$

so, distance travelled in $4sec$ is

$s=\frac{1}{2}\times 10\times 16=80m$

22. (c) The food packet has an initial velocity of $2m/sec$ in upward direction, therefore

$v=-u+gt$ or $v=-2+10\times 2=18m/sec$

23. (b) In first case retardations,

$a=\frac{0-v}{t}=-\frac{v}{t}\therefore x=vt+\frac{1}{2}a{t}^2=vt-\frac{vt}{2}=\frac{vt}{2}$

In second case retardation is,

$a^{\prime }=\frac{0-nv}{t}=-\frac{nv}{t}\therefore s=nvt-\frac{1}{2}\frac{nv}{t}{t}^2=\frac{nvt}{2}$

but $\frac{vt}{2}=x\therefore s=nx$

24. (b) The stone rises up till its vertical velocity is zero and again reached the top of the tower with a speed $u$ (downward). The speed of the stone at

the base is $3u$. Hence $(3u{)}^2=(-u{)}^2+2gh$ or $h=\frac{4u^2}{g}{\downarrow }_{v,g,h}^{+}$

25. (c) Let $x$ be the thickness of each plank and $v$ be the initial velocity. Let $F$ be the resistance of the plank

$\therefore \frac{1}{2}m{v}^2-\frac{1}{2}m(\frac{19v}{20}{)}^2=Fx$.

suppose the velocity becomes zero after passing through $\mathit{n}$ such planks. Then

$\frac{1}{2}m(\frac{19v}{20}{)}^2-0=nFx$.

or $\frac{1}{2}m(\frac{19v}{20}{)}^2=n[\frac{1}{2}m{v}^2-\frac{1}{2}\frac{m{v}^2\times 361}{400}]$

$\therefore n=\frac{\frac{1}{2}(361m{v}^2/400)\times 400}{\frac{1}{2}m{v}^2[400-361]}=\frac{361}{39}=9.2$

$\therefore$ A total number of planks that the bullet will have to pass through before coming to rest $=(1+9.2)=(10.2)\approx 11$.

26. (b) Time taken by the stone to reach the water level

$t_1=\sqrt{\frac{2h}{g}}$

Time taken by sound to come to the mouth of the well,

$t_2=\frac{h}{v}$

$\therefore$ Total time $t_1+t_2=\sqrt{\frac{2h}{g}}+\frac{h}{v}$

27. (c) From third equation of motion, $v^2=u^2+2as$

where $\underset{―}{v}$ & $u$ are final & initial velocity, a is acceleration, $s$ is distance.

For first case $v_1=3km/$ hour

$u_1=0,a_1=g$ & $s_1=$ ?

$s_1=\frac{9\times 100}{36\times 36\times 20}$ metre

For second case $v_2=?,u_2=4km/hour$,

$a_2=g=10m/sec$

& $s_1=s_2=\frac{9\times 100}{36\times 36\times 20}$

so $v_2^2=\frac{16\times 1000\times 1000}{3600\times 3600}+\frac{2\times 10\times 9\times 100}{20\times 36\times 36}$

or $v_2=5km/hour$

28. (c) $a=bt$ or $\frac{dv}{dt}=bt$. Integrating, we get

$v=\frac{b{t}^2}{2}+c$, where $c$ is a constant of integration.

At $t=0,v=v_0$. Thus, $v_0=c$.

Now, $v=\frac{ds}{dt}=\frac{b{t}^2}{2}+v_o\therefore ds=(\frac{b{t}^2}{2}+v_o)dt$

Integrating we get, $s=\frac{b{t}^3}{6}+v_{o}t$

29. (a) $s=ut+\frac{1}{2}a{t}^2$ here $a=g$

For first body $u_1=0\Rightarrow s_1=\frac{1}{2}g\times 9$

For second body $u_2=0\Rightarrow s_2=\frac{1}{2}g\times 4$

So difference between them after $3sec$.

$=s_1-s_2=\frac{1}{2}g\times 5$

If $g=10m/se{c}^2$ then $s_1-s_2=25m$.

30. (b) Relative speed of each train with respect to each other be, $v=10+15=25m/s$

Here distance covered by each train $=$ sum of their lengths $=\mathbf{5}\mathbf{0}+\mathbf{5}\mathbf{0}=\mathbf{1}\mathbf{0}\mathbf{0}m$

$\therefore$ Required time $=\frac{100}{25}=4sec$.

31. (b) From first equation of motion $v=u+at$

here $u=40,a=g=-10,t=2$

so, $v=40-10\times 2=20m/sec$

32. (a) From third equation of motion : $v^2=u^2+2as$ for first case $u=\frac{40\times 10}{36}m/sec$, $v=0,a=?,s=2m$

So, $a=(\frac{40\times 10}{36}{)}^2\frac{1}{4}m/se{c}^2$

for second case $u=\frac{80\times 10}{36}m/sec,v=0$,

So $s_2=(\frac{80\times 10}{36}{)}^2/2\times \frac{1}{4}\times (\frac{40\times 10}{36}{)}^2=8$ meter

33. (b) From third equation of motion

$v^2=u^2-2gh(\because a=-g)$

Given, $v=10m/sec$ at $\frac{h}{2}$, But $v=0$, when particle attained maximum height $h$.

Therefore $(10{)}^2=u-\frac{2gh}{2}$

or $100=2gh-2gh/2(\because 0=u^2-2gh)$

$\Rightarrow h=10m$

34. (c) Since the initial velocity of jump is same on both planets

So, $0=u^2-2g_A{h}_A$

$0=u^2-2g_B{h}_B$

or $\frac{g_A\times h_A}{g_B}=h_B\Rightarrow h_B=\frac{9}{1}\times 2=18m$

35. (a) $h=\frac{1}{2}g{T}^2$

now for $t=\frac{T}{3}$ second vertical distance moved is

given by, $h^{\prime }=\frac{1}{2}g(\frac{T}{3}{)}^2\Rightarrow h^{\prime }=\frac{1}{2}\times \frac{g{T}^2}{9}=\frac{h}{9}$

$\therefore$ position of ball from ground $=h-\frac{h}{9}=\frac{8h}{9}$

36. (d) Speed $v_1=60\times \frac{5}{18}m/s=\frac{50}{3}m/s$

$d_1=20m,v_1^{\prime }=120\times \frac{5}{18}=\frac{100}{3}m/s$

Let deceleration be a

$\therefore 0=v_1^2-2a{d}_1$

or $v_1^2=2a{d}_1$

$(2v_1{)}^2=2a{d}_2$

(2)divided by (1) gives,

$4=\frac{d_2}{d_1}\Rightarrow d_2=4\times 20=80m$

37. (d) Equation of motion is $u=at$

we know that $u=\frac{ds}{dt}\Rightarrow \frac{ds}{dt}=at$ or $ds=atdt$

integrating it we get, ${\int }_0^{s}ds={\int }_0^{4}tdt$

$s=\frac{a}{2}[t^2{]}_0^4=8a$

38. (b) Distance in last two second

$=\frac{1}{2}\times 10\times 2=10m$

Total distance $=\frac{1}{2}\times 10\times (6+2)=40m$.

39. (c) The horizontal velocity of the stone will be the same as that of the train. In this way, the horizontal motion will be uniform motion. The vertical motion will be controlled by the force of gravity, i.e., vertical motion is accelerated motion. Thus the resultant motion will be along a parabolic trajectory.
40. (c) Average acceleration $=\frac{\text{ change in velocity }}{\text{ time interval }}=\frac{\mathrm{\Delta }\overline{v}}{t}$

$\overrightarrow{v_1}=5\hat{i},\overline{v_2}=5\hat{j}$

$\mathrm{\Delta }\overline{y}=({\overrightarrow{v}}_2-{\overrightarrow{v}}_1)$

$=\sqrt{v_1^2+v_2^2+2v_1{v}_2\cos 90}$

$=\sqrt{5^2+5^2+0}[.$ As $.|v_1|=|v_2|=5m/s]$

$=5\sqrt{2}m/s$

Average acceleration $=\frac{\mathrm{\Delta }\overline{v}}{t}=\frac{5\sqrt{2}}{10}=\frac{1}{\sqrt{2}}m/s^2$

$\tan \theta =\frac{5}{-5}=-1$

which means $\theta$ is in the second quadrant.

41. (d) Initial velocity of parachute after bailing out,

$u=\sqrt{2gh}$

$u=\sqrt{2\times 9.8\times 50}=14\sqrt{5}$

The velocity at ground,

$v=3m/s$

$S=\frac{v^2-u^2}{2\times 2}=\frac{3^2-980}{4}\sim 243m$

Initially he has fallen $50m$.

$\therefore$ Total height from where he bailed out $=243+50=293m$

More Than One Correct

1. (a,b)	2. (a,b)	3. (a,b)	4. (c,d)
5. (a,c)	6. (a,d)	7. (a,d)	8. (b,c)
9. (a,d)	10. (a,b)

Multiple Matching Questions

1. (A) $\to$ (r, s, u) ; (B) $\to$ (t); (C) $\to$ (q); (D) $\to$ (p)
2. (A) $\to$ (p, u); (B) $\to$ (s, t); (C) $\to$ (q); (D) $\to$ (r, u)
3. (A) $\to$ (p, s) ;(B) $\to$ (p, q, r, s, t); (C) $\to$ (p, r, s, t);(D) $\to$ (p, q, r, s, t)
4. (A) $\to$ (p, r) ;(B) $\to$ (s) ;(C) $\to$ (q, t); (D) $\to$ (p, q)

Fill In The Passage

I. (1) distance

(2) equal

(3) distance

(4) directly

(5) straight

II. (1) direction

(2) Speed

(3) centripetal

(4) radial

(5) centrifugal

III. (1) distance

(2) displacement

(3) scalar

(4) vector

(5) less than

(6) zero

(7) speed

(8) velocity

IV. (1) acceleration

(2) straight line

(3) zero

(4) parabola

(5) increasing,

(6) decreasing

Passage based questions

1. (d) The total distance travelled is $(18+9)m=27m$
2. (a) $9m$
3. (a) Net displacement $=9m$
4. (a)
5. (a)
6. (c)
7. (b) Let $v$ be the velocity of the ball at the instant when it crosses the window, then using $s=ut+\frac{1}{2}a{t}^2$, We get

$0=v(2)-\frac{1}{2}(2{)}^2$

$v=9.8m{s}^{-1}$

Now using

$v^2-u^2=2$ as

We get

$(9.8{)}^2-u^2=2(-g)(73.5)$

or $u=39.2m{s}^{-1}$

8. (a) $h=\frac{u^2}{2g}=\frac{(39.2{)}^2}{2\times 9.8}=78.4m$
9. (c) $T=\frac{2u}{g}=\frac{2\times 39.2}{9.8}=8s$
10. (d) Initial velocity of police jeep w.r.t thief’s motorcycle, $u_{pt}=u_p-u_t=-25m{s}^{-1}$

Acceleration of police jeep w.r.t thief’s motorcycle, $a_{pt}=a_p-a_t=2m{s}^{-2}$

Using, $s=ut+\frac{1}{2}a{t}^2$, we get

$1250=-25t+t^2$

or $t^2-1250=0$

or $t=50s,-25s$

but time cannot be -ve, $\therefore t=50s$

11. (a) Distance covered by jeep in $50s,s=ut+\frac{1}{2}a{t}^2=2500$ $m=2.5km$.
12. (b) Let $a$ be the acceleration of thief, then

$a_{pt}=a_p-a_t=2-a$

We have $u_{pt}{}^2-u_{pt}{}^2=2a_{pt}{s}_{pt}$

$u_{pt}{}^2=u_{pt}{}^2+2a_{p{r}_{pt}}$

Thief will escape from police if, $v_{pt}{}^2\le 0$

$u_{pt}{}^2+2a_{pr}{s}_{pt}\le 0$

or $625+2(2-a)\times 1250\le 0$

or $a\ge \frac{5}{4}$ or $a\ge 1.25m{s}^{-2}$

Assertion and Reason

1. (b)
2. (c)
3. (d)
4. (a)
5. (d)
6. (c) Displacement is shortest possible distance between initial and final position.
7. (d) $s\times t^2$, hence $s-t$ graph for uniformly accelerated motion is parabola.
8. (a)
9. (d) Motion of body is uniform only if direction of motion is same or opposite to the initial velocity.
10. (b)
11. (a) ${\overrightarrow{V}}_{AB}={\overrightarrow{V}}_A-{\overrightarrow{V}}_B$

If ${\overrightarrow{V}}_A$ and ${\overrightarrow{V}}_B$ are in opposite directions,

$|{\overrightarrow{V}}_{AB}|=|{\overrightarrow{V}}_A|+|{\overrightarrow{V}}_B|$

12. (d) Velocity of the bus will change as direction of motion is changing.

Hots Subjective Questions

1. The boat that sails directly with the wind can sail no faster than wind speed. Why? Even sailing as fast as the wind, there would be no wind impact against the sail. It would sag. But when sailing crosswind, there would still be wind impact against the sail, and speeds greater than wind speed can be achieved.
2. Although its speed and velocity at the top will both instantaneously be zero, its acceleration will be $\mathbf{g}$ or $9.8m/s^2$. Remember, acceleration is not speed or velocity ____ it is the rate at which velocity changes. A moment before or after the rock reaches the top, it is moving, which is evidence that its velocity is changes at every instant. The rock undergoes a change as its passes through the zero value of velocity just as it undergoes the same rate of change passing through any other value of velocity. Or look at it via Newton’s $2^{\text{nd }}$ law. At the top or anywhere in its path, the rock has both weight and mass, and

$a=\frac{F}{m}=\frac{mg}{m}=g$

3. Yes.

When the two balls are thrown vertically upwards with the same speed $u$ then their final speed $v$ at the point of projection is $v^2-u^2=2\times g\times s$

Here, $s=0\therefore v=u$ for both the cases

Thus, we find that final velocity is independent of mass.

4. Average acceleration

$\overrightarrow{a}=\frac{\overrightarrow{v_f}-\overrightarrow{v_i}}{t}=\frac{\overrightarrow{v_f}+(-\overrightarrow{v_i})}{t}=\frac{\overrightarrow{v}}{t}$

To find the resultant of $\overrightarrow{v_f}$ and $-\overrightarrow{v_i}$, we draw a diagram

$|\overrightarrow{v}|=\sqrt{v_f^2+v_i^2}=\sqrt{5^2+5^2}=5\sqrt{2}m/s$

Since, $|\overline{v_f}|=|\overline{-v_i}|$

$\therefore \overline{v}$ is directed in between $\overline{v_f}$ and $-\overline{v_i}$

Therefore, $\overrightarrow{v}$ is directed towards $N-W$.

$\overrightarrow{a}=\frac{5\sqrt{2}}{10}=\frac{1}{\sqrt{2}}$

5. Let be time taken to reach the ground then from the formula,

$h=\frac{1}{2}g{t}^2\text{, }$

For first body, $16=\frac{1}{2}g{t}_1{}^2$

For second body, $25=\frac{1}{2}g{t}_2{}^2$

$\therefore \frac{16}{25}=\frac{t_1^2}{t_2^2}\Rightarrow \frac{t_1}{t_2}=\frac{4}{5}$.

6. $s_n=u+\frac{a}{2}(2n-1)$ or $s_n=0+\frac{a}{2}(2n-1)$

Further distance covered in $\mathit{n}$ seconds is

$s=ut+\frac{1}{2}a{t}^2=0+\frac{1}{2}a{n}^2$

$\therefore \frac{s_n}{s}=\frac{\frac{a}{2}(2n-1)}{(a{n}^2/2)}=\frac{2}{n}-\frac{1}{n^2}$

7. $P=F\times v=$ mav. Also $s=\frac{1}{2}a{t}^2$

$\therefore a=\frac{2s}{t^2}$ and $v=$ at $=\frac{2s.t}{t^2}=\frac{2s}{t}$;

$\therefore P=m(\frac{2s}{t^2})(\frac{2s}{t})$

or $s^2=\frac{P}{4m}{t}^3$ or $s\propto t^{3/2}$

8. Time taken by the stone to reach the water level

$t_1=\sqrt{\frac{2h}{g}}$

Time taken by sound to come to the mouth of the well, $t_2=\frac{h}{v}$

$\therefore$ Total time $t_1+t_2=\sqrt{\frac{2h}{g}}+\frac{h}{v}$

9. Velocity when the engine is switched off $\nu =19.6\times 5=98m{s}^{-1}$

$h_{max}=h_1+h_2$ where

$h_{max}=\frac{1}{2}\times 19.6\times 5\times 5+\frac{98\times 98}{2\times 9.8}$

11. Between 0 to 5 sec. $a=2t,\frac{dv}{dt}=2t$,

$\int dv=\int 2tdt\Rightarrow v=t^2$

Velocity at $t=5,v=25m/s$

Between 5 to 10 second

$u=25m/s,a=10$,

so, $v=25+10\times 5=75m/s$

CHAPTER 3

Motion in a Plane

INTRODUCTION

In last chapter we studied the motion in a straight line. This chapter is all about a different mode of motion in which the object does/does not move in a straight line but it can be assumed to be moving along two perpendicularly placed straight lines at the same time. Such type of motion is named as the motion in a plane or motion in two dimensions. The motion of a swimmer in a river is an expmple of this type of motion. Oher common examples of such a motion are the circular motion and the projectile motion.

Actually, motion in a plane is a combination of two motion in a straight line. Therefore, to study the motion in a plane we don’t need extra knowledge or equations but we just apply the knowledge or - motions which we studied in the last chapter on motion in a straight line. This chapter will be rstudied under the parts - (i) projectile motion, (ii) circular motion and (iii) relation motion.

MOTION IN A PLANE AS A COMBINATION OF TWO MOTIONS IN A STRAIGHT LINE

When position of a moving particle can be completely described by two dimensions it is said to be a two dimensional motion. If a particle is at $A$ at $t=0$ and reaches at $B$ after time ’ $t$ ’ experiencing accelerations $a_x$ and $a_y$ along $x$ and $y$ axes respectively then we can use the following equations.

For motion along $x$-axis

(i) $v_x=u_x+a_{x}t$

(ii) $x=u_{x}t+\frac{1}{2}{a}_x{t}^2$

(iii) $v_x^2=u_x^2+2a_x\cdot x$

Here,

$u_x=$ initial velocity along $x$-axis

$x=$ displacement along $x$-axis

$v_x=$ final velocity along $x$-axis

$a_x=$ acceleration along $x$-axis

For motion along $y$-axis

(i) $v_y=u_y+a_{y}t$

(ii) $y=u_{y}t+\frac{1}{2}{a}_y{t}^2$

(iii) $v_y^2=u_y^2+2a_y\cdot y$

Here,

$u_y=$ initial velocity along $y$-axis

$y=$ displacement along $y$-axis

$v_y=$ final velocity along $y$-axis

$a_y=$ acceleration along $y$-axis

In vector form :

$\begin{array}{cc}\overrightarrow{r}=x\hat{i}+y\hat{j}& |\overrightarrow{r}|=\sqrt{x^2+y^2}\\ \overrightarrow{v}=v_x\hat{i}+v_y\hat{j}& |\overrightarrow{v}|=\sqrt{v_x^2+v_y^2}\end{array}$

Examples of motion in two dimension are (i) projectile motion (ii) circular motion and (iii) relative motion. Now we will discuss projectile motion in detail.

Projectile motion

Projectile is the name given to a body thrown with some initial velocity in any arbitrary direction an the influence of a constant acceleration. The motion of a projectile is called projectile motion.

Example : A football kicked by the player, a stone thrown from the top of building motion. The path followed by the projectile is called a trajectory.

Terms Related to Projectile Motion

1. Trajectory : The path followed by a projectile is called its trajectory. Generally, the trajectory of a projectile is parabolic

2. Maximum Helght $(\mathbf{H})$ : When a projectile moves it covers a maximum distance in vertical direction. This maximum distance is called the maximum height attained by the projectile.

3. Horizontal Range (R) : The horizontal distance between the point of projection and the point of landing of a projectile is called its horizontal range.

4. Time of Flight $(\mathbf{T})$ : The time taken by the projectile to reach the point of landing from the point of projection is called the time of flight. It is generally represented by $T$.

Study of Motion of a Projectile

Let us consider a projectile projected from a point $O$ on a level ground with a velocity $v$ at an angle $\alpha$ to the horizontal, reaching ground level again at a point $A$.

The different kinematic parameters along horizontal and vertical directions are given below.

Horizontal direction

Initial velocity, $u_x=v\cos \alpha$

Acceleration, $a_x=0$

Vertical direction

Initial velocity, $u_y=u\sin \alpha$

Acceleration, $a_y=-g$

Do you know !!

A projectile returns to ground at the same angle and with the same velocity with which it is projected.

The time of flight (T):

This is the time taken for the particle to travel along its path from $O$ to $A$.

At any time $t,y=(u\sin \alpha )t-\frac{1}{2}g{t}^2$

When the particle is at $A,y=0$

Therefore, $(u\sin \alpha )t-\frac{1}{2}g{t}^2=0\Rightarrow t=0$ or $t=\frac{2u\sin \alpha }{g}$

i.e., the time of flights is $T=\frac{2u\sin \alpha }{g}$

The maximum height $(\mathbf{H})$ : This is given as $H$ in the diagram and is the height at the midpoint of the path.

At any time $t,y=u\sin \alpha -gt$

When the particle is at $B$, it is moving horizontally, i.e. $y=0$

So $u\sin \alpha -gt=0\Rightarrow t=\frac{u\sin \alpha }{g}$ (Note that this is half the total time of flight)

Then, $y=(u\sin \alpha )t-\frac{1}{2}g{t}^2$ gives $h=\frac{u^2{\sin }^2\alpha }{g}-\frac{u^2{\sin }^2\alpha }{2g}$

i.e., the greatest height is given by $H=\frac{u^2{\sin }^2\alpha }{2g}$

The horizontal range (R) : This is the distance from the initial position to the limal position on a horrzntal planc through the point of projection, i.e. $OA$.

At any time $t,x=ut\cos \alpha$ but at $A,t=\frac{2u\sin \alpha }{g}$

So, for $OA,x=\frac{2u^2\sin \alpha \cos \alpha }{g}=\frac{u^2\sin 2\alpha }{g}$ i.e., the range is $R=\frac{u^2\sin 2\alpha }{g}$

Do you know !!

When a projectile is at the highest point of its trajactory.

(i) It possess velocity only along horizontal.

(ii) the velocity and acceleration of the projectile are perpendicular to each other:

Equation of trajectory : L.et us say particle is at $P$ at any time $/$.

$x=|\cos \alpha t$

$y=u\sin \alpha t-\frac{1}{2}g{t}^2=u\sin \alpha \frac{x}{u\cos \alpha }-\frac{1}{2}g\frac{x^2}{u^2{\cos }^2\alpha }$

$y=x\tan \alpha -\frac{g{x}^2}{2u{\cos }^2\alpha }$

Equation is of type $y=ax\cdot h{x}^2$ where $a=\tan \alpha :b=\frac{g}{2u{\cos }^2\alpha }$

This represents a parabola

Another form of the equation of trajectory: Multiply by $\sin \alpha$ in numerator & denominator in second term

$y=x\tan \alpha -\frac{g{x}^2\sin \alpha }{2u^2\cos \alpha \sin \alpha \cos \alpha }x\tan \alpha -\frac{x^2\tan \alpha }{(\frac{2u^2\cos \alpha \sin \alpha }{g})}$

$y=x\tan \alpha -\frac{x^2\tan \alpha }{R}=x\tan \alpha (1-\frac{x}{R})$

Various Common Results

1. Velocity at any time $t$ : Let us say particle projected with a velocity $u$ is at $P$ at any time $t$. To calculate velocity at $P$, calculace its $x$ and $y$ component.

In projectile motion $x$-component of velocity remains constant

$v_x=u\cos \theta ,v_y=u\sin \theta -g\prime$

$\overrightarrow{v}=v_x\hat{i}+v_y\cdot \hat{i}=u\cos \theta \hat{i}+(u\sin \theta -gt)\hat{j}$

Magnitude : $|\overrightarrow{v}|=\sqrt{(u\cos \theta {)}^2+(u\sin \theta \cdot gt{)}^2}$

$=\sqrt{u^2{\cos }^2\theta +u^2{\sin }^2\theta +g^2{t}^2-2ugt\sin \theta }=\sqrt{u^2+g^2{t}^2-2ugt\sin \theta }$

Direction : $\tan \alpha =\frac{v_y}{v_x}=\frac{u\sin \theta -gt}{u\cos \theta }[\alpha \ne \theta ]$

2. Velocity at any position P(x, y): $v_x^2=u^2{\cos }^2\theta ,v_y{}^2=u^2{\sin }^2\theta -2gv[\because v^2=u^2+2gh]$ $v^2=v_x^2+v_y{}^2=u^2{\cos }^2\theta +u^2{\sin }^2\theta -2g{v}^{\prime }=u^2-2gv$

$\Rightarrow v=\sqrt{u^2-2gy}$ independent of $x$ co-ordinate.

IDEA BOX

When a projectile is thrown at ${45}^{\circ }$ with horizontal,

(i) its range is maximum and is equal to $u^2/g$.

(ii) the maximum height attained by the proiectile is equal to one fourth of its maximum range i.e. equal to $U^2/4g$.

3. Maximum Range : For given speed, range varies with angle of projection. It is maximum at $\theta ={45}^{\circ }.R=\frac{u^2\sin 2\theta }{g}$ $n\to$ constant, $R\to$ maximum, $\sin 2\theta \to$ maximum $(\sin 2\theta {)}_{max}=1$

$2\theta ={90}^{\circ }\Rightarrow \theta ={45}^{\prime \prime }$

In many practical applications (javelin throw, shot put etc.), the initial and final elevations may not be equal, and other consideration are important. For example, in the shot put. the hall ends its flight when it hits the ground, but it is projected from an initial height of about $2m$ above the ground. This causes the range to be maximum at an angle somewhat loue than ${45}^{\circ \prime }$ (app. ${42}^{\circ }$ in practice). When calculating the range of artillery shells, air resistance must be taken into account to predict the range accurately. As expected, air resistance reduces the range for a given angle of projection. It also decreases the optimum angle of projection slightly. If the initial and final elevations were the same the $4.5“$ trajectory would have the greater range.

4. Two angles of projection for same range :

If speed is constant then for angle of projection ${\theta }_1,R_1=\frac{u^2\sin 2{\theta }_1}{g}$; and for angle of projection ${\theta }_2,R_2=\frac{u^2\sin 2{\theta }_2}{g}$

$R_1=R_2$

$\sin 2{\theta }_1=\sin 2{\theta }_2$

$\sin 2{\theta }_1=\sin ({180}^{\circ }-2{\theta }_2)$

$2{\theta }_1={180}^{\circ }-2{\theta }_2\Rightarrow {\theta }_1={90}^{\circ }-{\theta }_2={\theta }_1+{\theta }_2={90}^{\circ }$

Thus for given $u$ there are two angle of projection whose sum is ${90}^{\circ }$ for same range,

like $\alpha ,90-\alpha \Rightarrow {30}^{\circ },{60}^{\circ };{45}^{\circ }-\alpha ,{45}^{\circ }+\alpha$

For these two angles of projection range is same but maximum height and time of flight will be different.

$✓$ CHECK POINT

1. As a projectile moves in its parabolic path, is there any point along its path where the velocity and acceleration vectors are (a) perpendicular to each other? (b) parallel to each other?
2. In the absence of air resistance, why does the horizontal component of a projectile’s motion not change, while the vertical component does?

Check Your Answer

1. (a) At the top of its flight, $v$ is horizontal and $a$ is vertical. This is the only point at which the velocity and acceleration vectors are perpendicular.

(b) If the object is thrown straight up or down, then $v$ and $a$ will be parallel throughout the downward motion. Otherwise, the velocity and acceleration vectors are never parallel.

2. In the absence of air resistance, the force acting on the projectile is only the force of gravity acting vertically downward. No component of the force due to gravity acts in the horizontal direction therefore the horizontal component of the projectile motion remains unchanged.

Whereas the vertical component of motion undergoes changes. The vertical component of motion decreases while rising against gravity and increases during downward motion.

Illustration 1 :

A ball is thrown at an angle of ${30}^{\circ }$ with horizontal with a speed of $30m{s}^{-1}$. Calculate maximum height, time of flight and horizontal range. (Use $g=10m/s^2$ ).

Solution : Given $u=30m{s}^{-1}$

${\theta }_0={30}^{\circ }$

Using $h=\frac{u^2{\sin }^2{\theta }_0}{2g};h=\frac{(30{)}^2\times {\sin }^2{30}^{\circ }}{2\times 10}=11.25m$

Time of flight, $T=\frac{2u\sin {\theta }_0}{g};T=\frac{2\times 30\times \sin {30}^{\circ }}{10}=3sec$

Horizontal Range, $R=\frac{u^2\sin 2{\theta }_0}{g};R=\frac{(30{)}^2\times \sin {60}^{\circ }}{10}=77.85m$

Illustration 2 :

A body is projected at an angle of ${45}^{\circ }$ if its horizontal range is $\mathbf{4}\mathbf{0}\mathbf{0}m$, find corresponding maximum height.

Solution : At ${45}^{\circ }$ a projectile has maximum range ( $R$ ) and the corresponding height is, $h=\frac{R}{4}:h=\frac{400}{4}=100m$

Illustration 3 :

A particle projected with some velocity at an angle ${30}^{\circ }$ with horizontal and another particle with the same speed but an angle of ${60}^{\circ }$. Find the ratio of the range and maximum height.

Solution : As ${30}^{\circ }$ and ${60}^{\circ }$ are complimentary angles their horizontal range will be the same so, $\frac{R_1}{R_2}=1:1$

For vertical heights, we can use $\frac{H_1}{H_2}={\tan }^2\theta$, where $\theta ={30}^{\circ }$

So,

$\frac{H_1}{H_2}={\tan }^2{30}^{\circ }=\frac{1}{3}=1:3$

Illustration 4 :

A body, when projected at angles of ${30}^{\circ }$ and ${60}^{\circ }$ the corresponding time of flights are $10s$ and $20s$, find horizontal range of projectile. (Use $g=10m/s^2$ )

Solution: Remember, $R=\frac{g{T}_1{T}_2}{2}$ where $T_1$ and $T_2$ are time of flights

$R-\frac{10\times 10\times 20}{2}=1000m$

Illustration 5 :

A staircase contains 3 steps cach $10cm$ high and $20cm$ wide. What should be the minimum horizontal velocity of a ball rolling off the uppermosi plane so as to hit the lowest plane.

Solution : In order to hit the lower most plane and ’ $u$ ’ be the minimum, let us assume that the ball almost touches the 2 nd staircase and then hits the 3rd step.

$\begin{array}{rl}& \text{ So, }h=2\times 10=20cm=0.2m\\ & \text{ and }x=2\times 20=40cm=0.4m\\ & \text{ Now using }h=\frac{1}{2}\frac{g{x}^2}{u^2}\\ & 0.2=\frac{1}{2}\frac{10\times 0.4\times 0.4}{u^2}\Rightarrow u=2m/s\end{array}$

Note : If $y$ is height of each step and $x$ is width then the minimum speed for a ball to hit the $n^{\text{th }}$ step is given by $u=\sqrt{\frac{g(n-1)x^2}{2y}}$.

HORIZONTAL PROJECTILE

Suppose a projectile is projected horizontally with initial speed ’ $u$ ’ from height $H$.

Here $u_{x}u$

$a_{x}0$ (No acceleration along $x$-axis)

$u_{y}0$ (Initial vertical velocity is zero as projectile is projected horizontally)

$a_{y}g$

Suppose it reaches the ground in time $T$ with verticul velocity $v_y$, then

$x=u\cdot r$

and $H{=}_2^{1}y{T}^{\prime }$

$H\frac{1}{2}g\frac{x^2}{u^2}$

$\overline{v}=v,i,v_v\hat{j}\because \dot{v}\cdot u\hat{i}+gı\hat{j}\text{ or }|\overrightarrow{v}|=\sqrt{u^2+g^2{t}^2}$

Angle with hori/untal $u={\tan }^{-1}(\begin{array}{c}gt\\ u\end{array})$

$✓$ CHECK POINT

• A bullet is fired horizontally towards a target directly in front of the gun. The target starts falling freely the moment bullet is fired, will the bullet hit the target.

Check Your Answer

Suppose the bullet takes ’ $t$ ’ time to reach to the point $B$ after being fired from the gun at $A$. Both the bullet and the target have their initial vertical velocity as zero. The vertical distance covered by both the bullet and the falling target will be $\frac{1}{2}g{t}^2$. Hence the bullet will surely hit the target.

Illustration 6 :

An aeroplane flying horizontally at an altitude of $2km$ with a speed of $288km/hr$. When it is just above a target it drops a bomb find by what distance, the bomb will miss the target and the time taken by it to hit the ground. (Take $g=10m/s$ )

Solution :

$\begin{array}{rl}u& =288km/h\\ u& =288\times \frac{5}{18}=80m/s\\ h& =2km=2000m\end{array}$

Using, $h=\frac{1}{2}g\frac{x^2}{u^2}$

$\begin{array}{rl}{x}^2& =\frac{2h{u}^2}{g}=\frac{2\times 2000\times 80\times 80}{10}\\ x& =1600m\end{array}$

For time $t=\frac{x}{u};t=\frac{1600}{80}=20sec$.

CIRCULAR MOTION

Motion of a particle (small body) along a circle (circular path), is called a circular motion. If the body covers equal distances along the circumference of the circle, in equal intervals of time, the motion is said to be a uniform circular motion, $\wedge$ uniform circular motion is a motion in which speed remains constant but direction of velocity changes.

Explanation :

Consider a boy running along a regular hexagonal track (path) as shown in l’ig. As the hoy runs along the sidc of the hexagon at a uniform speed, he has to take a turn at each corner changing direction but keeping the speed same. In onc round he has to take six turns at regular intervals. If the same boy runs along the side of a regular octagonal track with same uniform speed, he will have to take eight turns in one round at regular intervals but the interval, will become smaller.

By increasing the number of sides of the regular polygon, we find the number of turns per round becomes more and the interval between two turns become still shorter. A circle is a limiting case of a polygon with an infinite number of sides On the circular track, the turning becomes a continuous process without any gap in between the boy running along the sides of such a track will be performing a circular motion. Hence, circular motion is the motion of a body along the sides of a polygon of infinite number of sides with uniform speed, the direction changing continuously.

Examples of uniform circular motion are

(i) Motion of moon around the earth.

(ii) Motion of satellite around its planet.

Angular Position and Angular Displacement

Angular position : Angle made by radius to the particle with a reference radius is known as angular position. In figure $\theta$ is known as angular position. $OA$ is known as reference radius, with respect to which the angle is measured.

Angular displacement : Angular displacement is defined as the angle traced out by the radius vector at the axis of the circle in a given time. ${\theta }_1\to$ angular position at time $t_1,{\theta }_2\to$ angular position at time $t_2$. Time taken $\mathrm{\Delta }t=t_2-t_1$.

Here $\mathrm{\Delta }\theta ={\theta }_2-{\theta }_1$ is known as angular displacement. It is a vector quantity, provided $\mathrm{\Delta }\theta$ is small because commutative law of vector addition is not valid for large $\mathrm{\Delta }\theta$.

Its units are degree, radian revolution, etc

Angular Velocity

It is defined as the rate of change of angular displacement. If $\mathrm{\Delta }\theta$ be the change in angular displacement during time $\mathrm{\Delta }t$ then angualr velocity $(\omega )$ is given by $\omega =\frac{\mathrm{\Delta }\theta }{\mathrm{\Delta }t}$

Unit of $\omega$ : Its units are radian/second (rad/s), rotations per minute (rpm), rotations per second (rps), etc.

Some Terms Related to Circular Motion

Time period : It is defined as the time taken by the particle to complete one revolution on its circular path. It is denoted by $T$.

Frequency : It is defined as number of revolutions completed by particle in its circular path in unit time It is denoted by $f$. Its unit is $s^{-1}$ or $Hz$.

Relation between time period and frequency : $T=\frac{1}{f}$

Relation between $\omega ,f$ and $T:\omega =\frac{2\pi }{T}=2\pi f$

Acceleration and Force in Uniform Circular Motion

Any change of velocity-speeding up, slowing down, or turning a corner is an acceleration. We distinguish these different kinds of accelerated motion by assigning a direction to the acceleration. In other words, acceleration is a vector.

If you are traveling in a straight line and speed up, your change in velocity its $(\mathrm{\Delta }\overrightarrow{v})$ is in the direction you are going. Then your acceleration is also in that direction. This is the first rule for the direction of an acceleration : when the acceleration is in the same direction as the velocity, the result is an increase in speed.

Now suppose you do not change speed but are making a turn. This is a change in only the direction of velocity, since velocity is a vector, this is surely a change in velocity-an acceleration. Figure shows how to find the direction of this acceleration. The dotted line shows what the path of the car would be if its motion were not accelerated that is, if it traveled at constant speed in a straight line.

The arrow is a vector representing the change in the velocity of the car, that is, the vector that must be added to the old velocity to get the new velocity. The acceleration is in the same direction as the change in velocity. When the car turns to the right, its acceleration is to the right, perpendicular to its velocity, this acceleration is called centripetal acceleration and its value is $\frac{v^2}{r}$ (you will learn derivation of the formula in higher class), where $v$ is the magnitude of the velocity and $r$ the radius of the path.

Do you know !!

1. If circular motion of the object is uniform, the object will possess only centripetal acceleration.
2. If circular motion of the object is non-uniform, the object will possess both centripetal and transverse accelerations.

IDEA BOX

Nothing will follow a circular path unless it is forced to.

If $m$ be the mass of the object then it experiences a force which directs towards the centre of the path and has a magnitude given by

$F=\frac{m{v}^2}{r}$

This force is known as the centripetal force.

Illustration 7 :

The planet Neptune travels in a nearly circular orbit of radius, $r=4.5\times {10}^{9}km$, about the sun. It takes Neptune $165y$ to make a complete trip around the sun. How fast (in $km/h$ ) does not Neptune travel in its orbit?

Solution : $v=\frac{2\pi r}{T}=\frac{2(3.14)(4.5\times {10}^{9}km)}{(165d)(365d/1y)(24h/1d)}=2.0\times {10}^{4}km/h$

Illustration 8 :

A grinding wheel (radius $7.6cm$ ) is rotating at $1750rpm$.

(a) What is the speed of a point on the outer edge of the wheel?

(b) What is the centripetal acceleration of the point?

Solution : $v=\frac{2\pi }{i}$

The period of motion is $T=(1/1750)(60s)=3.43\times {10}^{-2}s$

Su, $1\frac{2\pi (7.6\times {10}^{-2}m)}{3.43\times {10}^{-2}s}=14m/s$

$a_1\cdot \frac{v^2}{r}=\frac{14m/s^2}{7.6\times {10}^{-2}m}=2.6\times {10}^{3}m/s^2$

Illustration 9 :

Rishabh, a $20kg$ child on his bicycle moving with a speed of $10m{s}^{-1}$ takes a turn on a circular turning of radius $20m$. Calculate (i) the centripetal acceleration and (ii) the centripetal force acting on Rishabh.

Solution : Here, $m=20kg$

1. $10ms{}^{-1}$

$r=20m$

(i) Centripetal acceleration, $a=\frac{{\nu }^2}{r}=\frac{(10{)}^2}{20}=\frac{100}{20}=5m{s}^{-2}$

(ii) (entripetal force, $F=\frac{m{v}^2}{r}=ma=(20\times 5)N=100N$

Relative Motion

The motion of an object $A$ as observed by another object $B$ is said to be the relative motion of $A$ with respect to $B$. Let us consider a car moving with a uniform velocity $v$ on a straight road. There are two

observers $A$ and $B.A$ is at rest on the road and $B$ is inside the moving car. Observer $A$ will see the car moving with a uniform velocity but whserver $B$ will see the car to be at rest. So, relative motion depends on the observer.

Relative Velocity

The velocity of point $A$ relative to point $B$ is the velocity with which $A$ appears to be moving w.r.t an observer who is moving with the velocity of $B$. The velocity of a body A relative to a body $B$ is represented by ${\text{dotv}}_{AB}$ and is given by

${\overrightarrow{v}}_{AB}={\overrightarrow{v}}_A-{\overrightarrow{v}}_B$

Let two objects $A$ and $B$ be moving with velocities $A$ and $B$ at an angle $\theta$ as shown in the figure. The velocity of $A$ relative to $B$ is given by ${\text{ddotv}}_{AB}={\overrightarrow{v}}_A-{\overrightarrow{v}}_B$

To find the magnitude and direction of the velocity ${\overrightarrow{v}}_{AB}$, we apply the law of vector addition. First we draw the vectors ${\overline{v}}_A$ and $(-{\overrightarrow{v}}_B)$ as shown in the figure.

Now, we complete the parallelogram $OPQR$ and apply the rule of finding the resultant as per the parallelogram law of vecter addition.

The resultant of ${\overrightarrow{v}}_A$ (along $OP$ ) and $-{\overrightarrow{v}}_B$ (along $OR$ ) will be represented by the diagonal $OQ$. Therefore, the relative velocity

${\overline{v}}_{AB}$ is represented by $OQ$ in magnitude and direction,

$\begin{array}{c}{v}_{AB}=\sqrt{v_A^2+v_B^2+2v_A\cdot v_B\cdot \cos ({180}^{\circ }-\theta )}=\sqrt{v_A^2+v_B^2-2v_A\cdot v_B\cdot \cos \theta }\text{ and }\\ \tan \alpha =\frac{v_A\sin ({180}^{\circ }-\theta )}{v_B+v_A\cos ({180}^{\circ }-\theta )}=\frac{v_A\sin \theta }{v_B-v_A\cos \theta }\end{array}$

$✓$ CHECK POINT

• When will the relative velocity of two moving objects be zero ?

Check Your Answer

When the two objects are moving with equal velocities (i.e. same speed and in the same direction.)

Do you know !!

The relative velocity of an object A w.r.t. another object B is equal to resultant of the velocity of A and negative of the velocity of B. mathematically ${\overrightarrow{v}}_{AB}={\overrightarrow{v}}_A+(-{\overrightarrow{v}}_B)$

Illustration 10 :

A man standing on a road has to hold his umbrella at ${30}^{\circ }$ with the vertical to keep the rain away. He throws the umbrella and starts running at $10km/hr$. He finds that rain drop are hitting his head vertically. Find the speed of raindrops with respect to (a) road (b) the moving man.

Solution : Given that the velocity of rain drops with respect to road is making an angle ${30}^{\circ }$ with the vertical, and the velocity of the man is $10kph$, also the velocity of rain drops with respect to main is vertical. We have

$v_{RM}=v_R-v_M$

hence $v_R=v_{RM}-v_M$

The situation is shown in velocity triangle in figure.

It shows clearly that, $v_R=V_{M}cosec\theta =10\times 2=20kph$

and

$V_{RM}=V_M\cos \theta =10\times \sqrt{3}=10\sqrt{3}kph$

Illustration 11

A plane is to fly due north. The speed of the plane relative to the air is $200km/h$, and the wind is blowing from west to east at $90km/h$. (a) In which direction should the plane head ? (b) How fast does the plane travel relative to the ground ?

Solution : Since the wind is blowing toward the east, the plane must head west of north as shown in figure. The velocity of the plane relative to the ground ${\overline{v}}_{pg}$ will be the sum of the velocity of the plane relative to the air ${\overrightarrow{v}}_{pa}$ and the velocity of the air relative to the ground ${\overrightarrow{v}}_{ag}$

(a) 1. The velocity of the plane relative to the ground is given by equation: ${\overrightarrow{v}}_{pg}={\overrightarrow{v}}_{pa}+{\overrightarrow{v}}_{ag}$

2. The sine of the angle $\theta$ between the velocity of the plane and north equals the ratio of $v_{Qg}$ and $v_{pa}$.

$\sin \theta =\frac{v_{ag}}{v_{pa}}=\frac{90km/m}{200km/h}=0.45$

(b) Since $v_{ag}$ and $v_m$, are perpendicular, we can use the Pythagorean theorem to find the magnitude of $v_p$

$v_{pg}^2={12}_{\rho g}+v_{pg}^2$

$v_{pg}=\sqrt{v_{pa}^2-v_{ag}^2}=\sqrt{(200km/h{)}^2-(900km/h{)}^2}=179km/h$

Illustration 12 :

A man swims at an angle $\theta ={120}^{\circ }$ to the direction of water flow with a speed $v_{\text{mw }}=5km/hr$ relative to water. If the speed of water $v_w=3km/hr$, find the speed of the man.

Solution :

$\begin{array}{rl}& {\overrightarrow{v}}_{mw}={\overrightarrow{v}}_m-{\stackrel{harpoonup}{v}}_w\\ & {\overrightarrow{v}}_m={\overline{v}}_{mw}+{\overline{v}}_w\\ \Rightarrow & v_m=|{\overrightarrow{v}}_{mw}+{\overline{v}}_w|=\sqrt{v_{mw}^2+v_w^2+2v_{mw}\cdot v_w\cdot \cos \theta }\\ \Rightarrow & v_m=\sqrt{5^2+3^2+2(5)(3)\cos {120}^{\circ }}\\ \Rightarrow & v_m=\sqrt{25+9}-15=\sqrt{19}m/sec\end{array}$

Illustration 13

On a two-lane road, car $A$ is travelling with a speed of $36km{h}^{-1}$. Two cars $B$ and $C$ approach car $A$ in opposite directions with a speed of $54km{h}^{-1}$ each. At a certain instant, when the distance $AB$ is equal to $AC$, both being $1km.B$ decindes te overtake $\mathit{A}$ before $\mathit{C}$ does. What minimum acceleration of car $B$ is required to avoid an accident :

Solution :

$v_A=36km{h}^{-1}=36\times \frac{5}{18}m{s}^{-1}=10m{s}^{-1}$

$v_B=v_C=54km{h}^{-1}=54\times \frac{5}{18}m{s}^{-1}=15m{s}^{-1}$

Relative velocity of $B$ w.r.t. $A,v_{BA}=5m{s}^{-1}$

Relative velocity of $C$ w.r.t. $A,v_{CA}=5m{s}^{-1}$

Time taken by $C$ to cover distance $AC=\frac{1000m}{25m{s}^{-1}}=40s$

Now, for $B,1000=5\times 40+\frac{1}{2}a\times 40\times 40$

On simplification, $a=1m{s}^{-2}$.

Illustrotion 14 :

A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial spend the cam. rymal $5m{s}^{-1}$ and the boy again throws the ball up with the maximum speed he can starts moving up with a uniform sund of hands?

Solution :

$\begin{array}{rl}& v(0)=49m{s}^{-1},a=9.8m{s}^{-2},t=?,v(t)=0,\\ & v(t)=v(0)+at\\ & 0=49-9.8t\text{ or }9.8t=49\text{ or }t=\frac{49}{9.8}s=5s\end{array}$

This is time taken by the ball to reach the maximum height. The time of descent is also $5s$. So. the thral time atter whinh the hall comes back is $5s+5s$ i.e., 10 s.

The uniform velocity of the lift does not change the relative motion of ball and lift. So, the hall woukd take the sanke notal time i.c. it would come back after 10 second.

EXERCISE 1

Fill in the blanks

DIRECTIONS : Complete the following statements with an appropriate word / term to be filled in the blank space(s).

1. The maximum height attained by a projectile is maximum when the angle of projection is ____
2. The range of a projectile is maximum when the angle of projection is ____
3. The maximum horizontal range for a projectile thrown with velocity $u$ at angle $\theta$ is ____
4. Horizontal range is same for a projectile at the angle of projection $\theta$ and ____
5. The maximum height of a projectile thrown with velocity $u$ at angle ${45}^{\circ }$ is ____
6. The angle between the velocity and acceleration of a projectile at the highest point is ____
7. Time taken by an object to complete one round of the circular path on which it is moving is called its
8. The S.I. unit of angular speed is ____
9. Centripetal acceleration always acts towards the ____ of the circular path.
10. Centrifugal force acts ____ the circular path.
11. The horizontal range of a projectile is maximum when the angle of projection is ____
12. A projectile thrown with a velocity of $10m{s}^{-1}$ at an angle of ${30}^{\circ }$ with the horizontal has time of light ____ seconds.
13. Two buses $A$ and $B$ are approaching each other with velocities $60km/h$ and $40km/h$ respectively. The velocity of each bus with respect to other at the time of approach is ____
14. The maximum height attained by a projectile at a fixed angle of projection varies directly with the ____ of velocity of projection.
15. A projectile projected at some angle with the horizontal has an acceleration ____ along horizontal.

True/False

DIRECTIONS : Read the following statements and write your answer as true or false.

1. In uniform circular motion, the speed is constant but the velocity changes.
2. The acceleration in non-uniform circular motion directs towards the centre of the path.
3. The relative velocity of a car $A$ moving with a velocity in east with respect to $carB$, also moving in east with the same velocity is zero.
4. The relative velocity of a $carA$ moving with a velocity $u$ in east with respect to a car $B$, moving in west with same velocity is $2u$.
5. A bus $M$ is moving with a velocity $20m/s$ in the north Another bus $B$ is moving with a velocity $40m/s$ in the west. When the two buses approach each other, passengers in each bus feels to move with a velocity of $60m/s$.
6. The velocity of a projectile at the highest point is along the vertical.
7. The velocity of a projectile at the highest point is along the horizontal.
8. The horizontal distance covered by a projectile during its time of flight is known as its trajectory.
9. Projectile is a body which moves under the action of a variable acceleration.
10. The relative velocity of car moving with a velocity $50km/h$ w.r.t. a car moving with a velocity $30km/h$ in the same direction is $80km/h$.
11. Projectiles projected at angles of projection ${30}^{\circ }$ and ${60}^{\circ }$ have same horizontal range.

Match the Columns

DIRECTIONS : Each question contains statements given in two columns which have to be matched. Statements $(A,B,C,D)$ in Column I have to be matched with statements $(p,q,r,s)$ in Column II.

1. A body of mass $m$ is projected in air with a velocity $u$ at an angle $\theta$ with the horizontal . It covers a maximum height $H$, horizontal range $R$ and its time of flight is $T$. Match the columns given below.

Column I	Column II
(A) $H$	(p) $\frac{2u\sin \theta }{g}$
(B) $T$	(q) $\frac{u^2\sin 2\theta }{g}$
(C) $R$	(r) $\frac{u^2{\sin }^2\theta }{2g}$
(D) $T\cos \theta$	(s) $\frac{u\sin 2\theta }{g}$

2. A particles completes one and a half-round of a circular track of radius $14m$ with a uniform speed of $33m{s}^{-1}$ Match the following for the motion of the particle.

Column I	Column II
(A) Distance covered	(p) 4 S.I units
(B) Displacement covered	(q) 7 S.I units
(C) Time taken	(r) 28 S.I units
(D) Average velocity	(s) 132 S.I units

3. A particle is projected from ground with a speed $20m{s}^{-1}$ making an angle ${30}^{\circ }$ with the horizontal. For the motion of the particle, match the following :

Column I	Column II
(A) Time of flight	(p) $2S.Iunits$
(B) Horizontal range	(q) $5S.Iunits$
(C) Maximum height	(r) $10\sqrt{3}S.Iunits$
(D) Speed of the particle at the highest point of its trajectory	(s) $20\sqrt{3}S.Iunits$

Very Short Answer Questions

DIRECTIONS : Give answer in one word or one sentence.

1. Does a uniform acceleration in two-dimensional motion necessarily imply motion along a straight-line path?
2. What will be the effect on horizontal range of a projectile when its initial velocity is doubled, keeping the angle of projection identical?
3. What will be the effect on maximum height of a projectile if its angle of projection is changed from ${30}^{\circ }$ to ${60}^{\circ }$ keeping the same initial velocity of projection?
4. Is the maximum height attained by projectile is largest when its horizontal range is maximum?
5. If the velocity at the maximum height of a projectile is half its initial velocity of projection $u$, find out its range on the horizontal plane.
6. Write down the expression for the equation of trajectory of a projectile.
7. What is the nature of the trajectory of a particle having a uniformly accelerated motion in a plane?
8. If the two bodies are projected at an angle $\theta$ and $({90}^{\circ }-\mathit{\theta })$ to the horizontal with the same speed, then find out the ratio of their time of flight.
9. If the two bodies are projected at the same initial velocity at angles $\theta$ and $({90}^{\circ }-\theta )$ to the horizontal. What is the ratio of their (i) maximum height attained by them and (ii) of horizontal ranges?
10. In a projectile motion the range $R$ is ’ $n$ ’ times that of its maximum height $\mathit{H}$, find its angle of projection.
11. A body of mass $m$ is thrown with velocity $v$ at an angle ${30}^{\circ }$ to the horizontal and another body $B$ of the same mass is thrown with velocity $v$ at an angle of ${60}^{\circ }$ to the horizontal, find the ratio of the horizontal range and maximum height of $A$ and $B$.
12. If the greatest height to which a man can throw a body is ’ $h$ ‘, what will be the greatest distance up to which he can throw the body?
13. Define angular speed. What is its SI unit?
14. A particle moves along a circle of radius $r$ where $v$ and $\omega$ are the linear and angular speed of the particle. What is the relation between $\nu$ and $\omega$ ?
15. Two cars are going around two concentric circular paths at the same angular speed. Does the inner or the outer car have the larger linear speed?
16. Among the angular velocities of hour hand and that of earth, which will be greater?
17. The range of a particle when launched at angle of ${15}^{\circ }$ with the horizontal is $1.5km$. What is the range of the particle when launched at angle of ${45}^{\circ }$ with the horizontal.
18. Give two examples of circular motion with constant speed.
19. What is the direction of velocity vector of a particle in circular motion?
20. What furnishes centripetal acceleration for earth to go round the sun?
21. What provides the centripetal force to satellite revolving round the earth?

Short Answer Questions

DIRECTIONS : Give answer in 2-3 sentences.

1. Justify the statement that a uniform circular motion is an accelerated motion.
2. For uniform circular motion, does the direction of the centripetal force depend upon the sense of rotation?
3. A stone tied to the end of a string is whirled in a circle. If the string breaks, the stone flies away tangentially. Why?
4. Passengers are thrown outwards, when the bus takes a circular turn. Why?
5. The outer rail of a curved railway track is generally raised over the the inner. Why?
6. A bucket containing water is rotated in vertical circle Explain, why the water does not fall?
7. A body is dropped freely from the window of a train. Will the time of the free fall be equal, if the train is stationary, the train moves with a constant velocity, the train moves with an acceleration?
8. A person sitting in a moving train throws a ball vertically upwards. How does the ball appear to move to an observer (i) inside the train (ii) outside the train?
9. A ship is sailing westward at $8m{s}^{-1}$. While trying to fix a bolt at the top of the mast, the sailor drops the bolt. If the mast of the ship is $19.6m$ high, where will the bolt hit the deck?
10. While firing, one has to aim a little above the target and not exactly on the target. Why ?
11. The magnitude of the velocity of projection being fixed, how does the range of projectile depend upon its angle of projection? Is there any optimum value for the angle of projection so that range may be maximum ?

Long Answer Questions

DIRECTIONS : Give answer in four to five sentences.

1. Find the time of flight, maximum height, horizontal range of projectile projected with speed $u$ and making an angle $\theta$ with the horizontal direction from ground.
2. What do you mean by a projectile? A projectile is fired with velocity $u$ making an angle $\theta$ with the horizontal. Show that its path is parabolic. Also, find the expressions for (i) maximum height attained and (ii) time of flight.
3. A projectile is fired making an angle $\theta$ with the vertical.

(a) Find mathematically the nature of its path

(b) Derive relations for its horizontal range and maximum horizontal range.

4. Show that there are two angles of projection for a projectile to have the same horizontal range. What will be the maximum heights attained in the two cases? Compare the two heights for $\theta ={30}^{\circ }$ and ${60}^{\circ }$.
5. Show that the velocity with which a projectile hits the ground is same with which it was initially projected and the angle made by the trajectory with the horizontal is same as the angle of projection.
6. Define angular displacement, angular speed, angular acceleration, time period and frequency as regards circular motion. How are linear velocity and linear acceleration related to angular speed and angular acceleration respectively?
7. What are centripetal acceleration and centripetal force ? Derive expressions for them.

EXAMPLE 2

Multiple choice Questions

DIRECTIONS: This section contains multiple choice questions Each. question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct

1. A projectile is projected with a kinetic energy $E$. Its range is $R$. It will have the minimum kinetic energy, after covering a horizontal distance equal to

(a) $0.25R$

(c) $0.75R$

(b) $0.5R$

(d) $R$

2. The range of a particle when launched at an angle of ${15}^{\circ }$ with the horizontal is $1.5km$. What is the range of the projectile when launched at an angle of ${45}^{\circ }$ to the horizontal?

(a) $1.5km$

(c) $6.3km$

(b) $3.0km$

(d) $0.75km$

3. A gun fires two bullets at ${60}^{\circ }$ and ${30}^{\circ }$ with horizontal. The bullets strike at some horizontal distance. The ratio of maximum height for the two bullets is in the ratio

(a) $2:1$

(c) $4:1$

(b) $3:1$

(d) $1:1$

4. Two balls are projected at an angle $\theta$ and $({90}^{\circ }-\theta )$ to the horizontal with the same speed. The ratio of their maximum vertical heights is

(a) $1:1$

(b) $\tan \theta :1$

(c) $1:\tan \theta$

(d) ${\tan }^2\theta :1$

5. A body is projected at an angle of ${30}^{\circ }$ to the horizontal with speed $30m/s$. What is the angle with the horizontal after 1.5 seconds? Take $g=10m/s^2$.

(a) $0^{\circ }$

(b) ${30}^{\circ }$

(c) ${60}^{\circ }$

(d) ${90}^{\circ }$

6. If range is double the maximum height of a projectile, then $\theta$ is

(a) ${\tan }^{-1}4$

(b) ${\tan }^{-1}1/4$

(c) ${\tan }^{-1}1$

(d) ${\tan }^{-1}2$

7. A body is projected such that its $KE$ at the top is $3/4$ th of its initial KE. What is the angle of projectile with the horizontal?

(a) ${30}^{\circ }$

(b) ${60}^{\circ }$

(c) ${45}^{\circ }$

(d) ${120}^{\circ }$

8. A projectile can have the same range $R$ for two angles of projection. If $t_1$ and $t_2$ be the times of flight in two cases, then what is the product of two times of flight?

(a) $t_1{t}_2\propto R$

(c) $t_1{t}_2\propto 1/R$

(b) $t_1{t}_2\propto R^2$

(d) $t_1{t}_2\propto 1/R^2$

9. A particle having a mass $0.5kg$ is projected under gravity with a speed of $98m/sec$ at an angle of ${60}^{\circ }$. The magnitude of the change in momentum (in $N$-sec) of the particle after 10 seconds is

(a) 0.5

(b) 49

(c) 98

(d) 490

10. The velocity of projection of a body is increased by $2$. Other factors remaining unchanged, what will be the percentage change in the maximum height attained?

(a) $1$

(b) $2$

(c) $4$

(d) $8$

11. A body is thrown with a velocity of $9.8m{s}^{-1}$ making an angle of ${30}^{\circ }$ with the horizontal. It will hit the ground after a time

(a) $3.0s$

(b) $2.0s$

(c) $1.5s$

(d) $1s$

12. If the horizontal range of a projectile is equal to the maximum height reached, then the corresponding angle of projection is

(a) ${\tan }^{-1}(1)$

(b) ${\tan }^{-1}(\sqrt{3})$

(c) ${\tan }^{-1}(4)$

(d) ${\tan }^{-1}(12)$

13. The co-ordinates of a moving particle at any time ’ $t$ ’ are given by $x=\alpha t^3$ and $y=\beta t^3$. The speed of the particle at time ’ $t$ ’ is given by

(a) $3t\sqrt{{\alpha }^2+{\beta }^2}$

(b) $3t^2\sqrt{{\alpha }^2+{\beta }^2}$

(c) $t^2\sqrt{{\alpha }^2+{\beta }^2}$

(d) $\sqrt{{\alpha }^2+{\beta }^2}$

14. Which of the following statements is FALSE for a particle moving in a circle with a constant angular speed?

(a) The acceleration vector points to the centre of the circle

(b) The acceleration vector is tangent to the circle

(c) The velocity vector is tangent to the circle

(d) The velocity and acceleration vectors are perpendicular to each other.

15. A body moves $6m$ north $8m$ east and $10m$ vertically upwards, the resultant displacement from its initial position is

(a) $10\sqrt{2}m$

(c) $\frac{10}{\sqrt{2}}m$

(b) $10m$

(d) $20m$

16. In uniform circular motion, the velocity vector and acceleration vector are

(a) perpendicular to each other

(b) same direction

(c) opposite direction

(d) not related to each other

17. The circular motion of a particle with constant speed is

(a) periodic but not simple harmonic

(b) simple harmonic but not periodic

(c) periodic and simple harmonic

(d) neither periodic nor simple harmonic

18. Two projectiles are fired from the same point with the same speed at angles of projection ${60}^{\circ }$ and ${30}^{\circ }$ respectively. Which one of the following is true?

(a) Their maximum height will be same

(b) Their range will be same

(c) Their landing velocity will be same

(d) Their time of flight will be same

19. From a $10m$ high building a stone $A$ is dropped, and simultaneously another identical stone $B$ is thrown horizontally with an initial speed of $5m{s}^{-1}$. Which one of the following statements is true?

(a) It is not possible to calculate which one of the two stones will reach the ground first

(b) Both the stones ( $A$ and $B$ ) will reach the ground simultaneously

(c) A stone reaches the ground earlier than $B$

(d) B stone reaches the ground earlier than $A$

20. An aeroplane flying at a constant speed releases a bomb.

As the bomb moves away from the aeroplane, it will

(a) always be vertically below the aeroplane only if the aeroplane was flying horizontally

(b) always be vertically below the aeroplane only if the aeroplane was flying at an angle of ${45}^{\circ }$ to the horizontal

(c) always be vertically below the aeroplane

(d) gradually fall behind the aeroplane if the aeroplane was flying horizontally.

21. The time of flight of a projectile on an upward inclined plane depends upon

(a) angle of inclination of the plane

(b) angle of projection

(c) the value of acceleration due to gravity

(d) all of these

22. At the highest point on the trajectory of a projectile, its

(a) potential energy is minimum

(b) kinetic energy is maximum

(c) total energy is maximum

(d) kinetic energy is minimum.

23. Two bullets are fired horizontally, simultaneously and with different velocities from the same place. Which bullet will hit the ground earlier?

(a) It would depend upon the weights of the bullets.

(b) The slower one.

(c) The faster one.

(d) Both will reach simultaneously.

24. In the case of a projectile fired at an angle equally inclined to the horizontal and vertical with velocity $u$, the horizontal range is

(a) $\frac{u^2}{g}$

(b) $\frac{u^2}{2g}$

(c) $\frac{u^2}{3g}$

(d) $\frac{u^2}{4g}$

25. A bomb is released by a horizontal flying acroplane. The trajectory of bomb is

(a) a parabola

(b) a straight line

(c) a circle

(d) a hyperbola

26. The greatest height to which a man can throw a ball is $h$. What is the greatest distance to which he can throw the ball?

(a) $h/4$

(b) $h/2$

(c) $h$

(d) $2h$

27. The $x$ and $y$ coordinates of a particle and any time $t$ are given by $x=2t+4t^2$ and $y=5t$, where $x$ and $y$ are in metre and $t$ in second. The acceleration of the particle at $t=5s$ is

(a) $40m{s}^{-2}$

(b) $20m{s}^2$

(c) $8m{s}^{-2}$

(d) zero.

28. A cricket ball is hit with a velocity $25m{s}^{-1},{60}^{\circ }$ above the horizontal. How far above the ground, ball passes over a fielder $50m$ from the bat (consider the ball is struck very close to the ground). Take $\sqrt{3}=1.7$ and $g=10m{s}^{-2}$

(a) $6.8m$

(b) $7m$

(c) $5m$

(d) $10m$

29. A plane flying horizontally at a height of $1500m$ with a velocity of $200m{s}^{-1}$ passes directly overhead on antiaircraft gun. Then the angle with the horizontal at which the gun should be fired for the shell with a muzzle velocity of $400m{s}^{-1}$ to hit the plane, is

(a) ${90}^{\circ }$

(b) ${60}^{\circ }$

(c) ${30}^{\circ }$

(d) ${45}^{\circ }$

30. A projectile is thrown horizontally with a speed of $20m{s}^{-1}$. If $g$ is $10m{s}^{-2}$, then the speed of the projectile after 5 second will be nearly

(a) $0.5m{s}^{-1}$

(b) $5m{s}^{-1}$

(c) $54m{s}^{-1}$

(d) $500m{s}^{-1}$

31. A ball is projected at such an angle that the horizontal range is three times the maximum height. The angle of projection of the ball is

(a) ${\sin }^{-1}(\frac{3}{4})$

(b) ${\sin }^{-1}(\frac{4}{3})$

(c) ${\cos }^{-1}(\frac{4}{3})$

(d) ${\tan }^{-1}(\frac{4}{3})$

32. If a particle is projected at ${45}^{\circ }$, then

(a) $R=4H$

(c) $2H=R$

(b) $4R=H$

(d) none of these

33. The equation of trajectory of projectile is given by

$y=\frac{x}{\sqrt{3}}-\frac{g{x}^2}{20}\text{, where }x\text{ and }y\text{ are in metre. }$

The maximum range of the projectile is

(a) $\frac{8}{3}m$

(b) $\frac{4}{3}m$

(c) $\frac{3}{4}m$

(d) $\frac{3}{8}m$

34. A projectile can have the same range for two angks of projection. If $h_1$ and $h_2$ are maximum heights when the range in the two cases is $R$, then the relatien tewween $R,h_1$ and $h_2$ is

(a) $R=4\sqrt{h_1{h}_2}$

(b) $R=2\sqrt{h{h}_2}$

(c) $R=\sqrt{h_1{h}_2}$

(d) none of these

35. A bullet is fired with a speed of $1500m/s$ in order to hit a target $100m$ away. If $g=10m/s^2$. The gun should be aimed

(a) $15cm$ above the target

(b) $10cm$ above the target

(c) $2.2cm$ above the target

(d) directly towards the target

36. A wheel rotates with constant acceleration of $2.0rod/s^2$, if the wheel starts from rest the number of revolutions it makes in the first ten seconds will be approximately.

(a) 32

(b) 24

(c) 16

(d) 8

37. A projectile of mass $m$ is thrown with a velocity $v$ making an angle ${60}^{\circ }$ with the horizontal. Neglecting air resistance, the change in momentum from the departure $A$ to its arrival at $B$, along the vertical direction is

(a) $2mv$

(b) $\sqrt{3}mv$

(c) $mv$

(d) $\frac{mv}{\sqrt{3}}$

38. A projectile is thrown in the upward direction making an angle of ${60}^{\circ }$ with the horizontal direction with a velocity of $147m{s}^{-1}$. Then the time after which its inclination with the horizontal is ${45}^{\circ }$, is

(a) $15s$

(b) $10.98s$

(c) $5.49s$

(d) $2.745s$

39. A bomb is dropped from an aeroplane moving horizontally at constant speed. If air resistance is taken into consideration, then the bomb

(a) falls on earth exactly below the aeroplane

(b) falls on the earth exactly behind the aeroplane

(c) falls on the earths ahead of the aeroplane

(d) flies with the aeroplane

40. A cyclist moving at a speed of $20m/s$ takes a turn, if he doubles his speed then chance of overturn

(a) is doubled

(b) is halved

(c) becomes four times

(d) becomes $1/4$ times

41. A projectile can have the same range ’ $R$ ’ for two angles of projection. If ’ $t$ ’ and ’ $t$ ’ be the times of flights in the two cases, then the product of the two time of flights is proportional to

(a) $\frac{1}{R^2}$.

(b) $R^2$

(c) $R$

(d) $\frac{1}{R}$

42. A person swims in a river aiming to reach exactly on the opposite point on the bank of a river. His speed of swimming is $0.5m/s$ at an angle of ${120}^{\circ }$ with the direction of flow of water. The speed of water is

(a) $1.0m/s$

(b) $0.5m/s$

(c) $0.25m/s$

(d) $0.43m/s$

43. The relative velocity $v_{AB}$ or $v_{BA}$ of two bodies may be

(1) greater than velocity of body $A$

(2) greater than velocity of body $B$

(3) less than the velocity of body $A$

(4) less than the velocity of body $B$

(a) (1) and (2) only

(b) (3) and (4) only

(c) (1), (2) and (3) only

(d) (1), (2), (3) and (4).

More Than One Correct

DIRECTIONS : This section contains multiple choice questions. Each question has 4 choices (a). (b), (c) and (d) own of which ONE OR MORE may be correct.

1. The following quantities may remain constant during uniform circular motion :

(a) magnitude of acceleration

(b) acceleration

(c) speed

(d) velocity

2. A cart moves with a constant speed along a horizontal circular path. From the cart, a particle is thrown up vertically with respect to the cart. Then the particle will :

(a) land outside the circular path

(b) land somewhere on circular path

(c) follow a parabolic path

(d) follow an elliptical path

3. A man on a moving cart, facing in the direction of motion, throws a ball straight upwards with respect to himself. Then the ball will

(a) fall behind him if the cart moves with some acceleration

(b) return to him if the cart moves with a constant velocity

(c) never return to him

(d) always return to him

4. Which of the following represent the projectiles ?

(a) A stone thrown horizontally from the top of the tower

(b) A bullet fired from the gun

(c) A rocket fired into space

(d) A ball thrown upwards

5. A particle is acted upon by a force of constant magnitudk which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that:

(a) it moves in a circular path

(b) velocity is constant

(c) acceleration is constant

(d) kinetic energy is constant

6. A bullet fired with a certain muzzle speed at an angle of ${30}^{\circ }$ with the horizontal hits the ground $2\sqrt{3}km$ away. Which of the following statements are correct? For the same muzzle speep, by act justing the angle of projection, the bullet cannot be made to hit a target?

(a) $5km$ away

(b) $3km$ away

(c) $6km$ away

(d) $4km$ away

7. A ball is projected upwards at a certain angle with the horizontal. Which of the following statements are correct, at the highest point?

(a) The acceleration of the projectile is vertically downwards.

(b) The velocity of the projectile is along the horizontal direction

(c) The velocity of the projectile is zero

(d) The acceleration of the projectile is zero

8. A body is moving in a circle of radius $r$ with a uniform speed $v$, angular frequency $\omega$, time period $T$ and frequency $v$. The centripetal acceleration is given by :

(a) $\frac{4{\pi }^{2}r}{T}$

(b) $\omega v$

(c) $\frac{v^2}{r}$

(d) $4{\pi }^{2}r{v}^2$

9. Two vectors of the same physical quantity are unequal if :

(a) they have the same magnitude but different directions

(b) they have different magnitudes and different directions

(c) they have different magnitudes but the same direction

(d) they have the same magnitude and the same direction

10. Two particles are projected from the same point with the same speed, at different angles ${\theta }_1$ and ${\theta }_2$ to the horizontal. They have the same horizontal range. Their time of flight are $t_1$ and $t_2$ respectively. Then

(a) ${\theta }_1+{\theta }_2={90}^{\circ }$

(b) $\frac{t_1}{\sin {\theta }_1}=\frac{t_2}{\sin {\theta }_2}$

(c) $\frac{t_1}{t_2}=\tan {\theta }_2$

(d) $\frac{t_1}{t_2}=\tan {\theta }_1$

11. Let $a_r$ and $a_{\text{, represent radial and tangential acceleration. }}$ The motion of a particle may be circular if

(a) $a_r\ne 0$ and $a_1\ne 0$

(b) $a_r\ne 0$ and $a_t=0$

(c) $a_r=a_1=0$

(d) $a_r=0$ and $a_t\ne 0$

12. The maximum height attained by the projectile depends upon the following factors

(a) mass of the projectile

(b) acceleration of the projectile

(c) angle of projectile

(d) magnitude of initial velocity

13. A particle is moving along a circular path. The angular velocity, linear velocity, and centripetal acceleration of the particle at any instant respectively are $\overrightarrow{\omega },\overrightarrow{v},\overrightarrow{a}$ and ${\overrightarrow{a}}_c$. Which of the following relations are correct ?

(a) $\overrightarrow{\omega }\perp \overline{v}$

(b) $\overline{v}\perp {\overrightarrow{a}}_c$

(c) $\overrightarrow{\omega }\perp {\overrightarrow{a}}_c$

(d) None

14. Two balls $A$ and $B$ are projected horizontally from the top of a tower with velocities $V_A$ and $V_B$ respectively such as $V_A>V_B$. If $T_A$ and $T_B$ be their respective time of flights and $R_P$ and $R_B$ be their horizontal ranges, then

(a) $T_A=T_B$

(b) $T_A<T_B$

(c) $R_A=R_B$

(d) $R_A>R_B$

15. For a projectile motion, which of the following depend on angle of projection?

(a) Time of flight

(b) Horizontal range

(c) Horizontal component of velocity

(d) Acceleration

Multiple Matching Questions

DIRECTIONS : Following question has four statements (A, $B,C$ and D) given in Column I and four statements ( $p,q,r,s...$. in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II. Match the entries in Column I with entries in Column II.

1. Match the following for a uniform circular motion.

Column I	Column II
(A) Speed	(p) Constant
(B) Direction of velocity	(q) Variable
(C) Magnitude of acceleration	(r) Towards the centre acceleration of the circular path
(D) Direction of acceleration	(s) Along the tangent to the circular path
	(t) Circular Path
	(u) $\frac{v^2}{r}$

2. A particle is projected from ground at a certain angle. Match the following for the motion of the particle.

Column I	Column II
(A) Speed of the particle	(p) Constant
(B) Acceleration of the particle	(q) Variable
(C) Horizontal component of velocity	(r) First increases then decreases
(D) Vertical displacement	(s)First decreases then increases
	(t) Always donwards

Fill in the Blanks

DIRECTIONS : Complete the following passage(s) with an appropriate word/term to he filled in the blank spaces.

(I) non-uniform, centre, centripetal acceleration, magnitude, direction, perpendicular, uniform.

When a body exccutes circular motion, its speed is (1) ____ but its velocity is ____ (2) ____ The acceleration of the particle is always directed towards ____ (3) ____ of the circular path, hence is always ____ (4) ____ to its velocity. The acceleration responsible for change in direction of velocity during circular motion is called ____ (5) ____, it is constant in ____ (6) ____ but varies in ____ (7) ____.

(II) verticily &uwnwards. constant, horizontal, minimum, horizonui range, parabola, equal to.

When a particle is projected from ground making some angle with the horizontal. its acceleration is ____ (1) ____ and is directed ____ (2) ____. At the highest point of its trajectory, the velocity of the particle is along ____ (3) ____ and its magnitude is ____ (4) ____ at that point. The straight line distance between the point of projection and the point of landing is called ____ (5) ____. The speed with which the particle strikes the ground is ____ (6) ____ the speed of proiection. The path of projectile is always a ____ (7) ____.

Passage Based Questions

DIRECTIONS : Study the given paragraph(s) and answer the following yuestions

PASSAGE - I

A boy moves along a straight line with a uniform speed of $5m{s}^{-1}$ and takes a 90 ${}^{\circ }$ left turn after every $10s$. Answer the following for the motion of the hoy.

1. The displacement covered by the boy from the beginning to the instant when it takes sixth turn is :

(a) $50m$

(b) $50\sqrt{2}m$

(c) $100m$

(d) $150\sqrt{2}m$

2. The average relocity of the boy between the instant of begining of its motion to the instant when it takes fifth turn is :

(a) $1ms1$

(b) $5m{s}^{-1}$

(c) $2.5ms1$

(d) zero

3. The average acceleration of the body from the beginning to the instant just after it takes second turn is :

(a) $0.5m{s}^{-2}$

(b) $1m{s}^{-2}$

(c) $2m{s}^{-1}$

(d) zero

PASSAGE - II

A particle is projected from ground with an initial speed $10m{s}^{-1}$ making an angle ${45}^{\circ }$ with the horizontal.

Then (take $g=10m{s}^{-2}$ )

4. The time of flight of the particle is :

(a) $1s$

(b) $\sqrt{2}s$

(c) $2s$

(d) $2\sqrt{2}s$

5. The horizontal range of the particle is :

(a) $5m$

(b) $10m$

(c) $20m$

(d) $25m$

6. Velocity of the particle at the highest point of its trajectory is:

(a) $5m{s}^{-1}$

(b) $5\sqrt{2}m{s}^{-1}$

(c) $10m{s}^{-1}$

(d) zero

PASSAGE - III

A particle is projected horizontally with a speed of $29.4m{s}^{-1}$ from the top of a tower of height $78.4m$. If $g=9.8m{s}^{-2}$, then answer the following questions.

7. How much time will the particle take to reach the ground?

(a) $2s$

(b) $2\sqrt{2}s$

(c) $4s$

(d) $8s$

8. At what distance from the bottom of the tower will the particle hit the ground?

(a) $58.8m$

(c) $117.6m$

(b) $78.4m$

(d) $147m$

9. The speed with which the particle hit the ground is :

(a) $29.4m{s}^{-1}$

(b) $39.2m{s}^{-1}$

(c) $49m{s}^{-1}$

(d) $78.4m{s}^{-1}$

PASSAGE - IV

A particle is moving along a circular track of radius $3.5m$ with a constant speed $4m{s}^{-1}$. Answer the following for the motion of the particle.

10. How much time will it take to complete one complete round?

(a) $5.5s$

(c) $22s$

(b) $11s$

(d) $3.5s$

11. What is the average velocity of the particle in one complete round?

(a) $4m{s}^{-1}$

(c) $2m{s}^{-1}$

(b) $2.6m{s}^{-1}$

(d) Zero

12. The acceleration of the particle is

(a) $\frac{42}{7}m{s}^{-2}$

(b) $\frac{32}{7}m{s}^{-2}$

(c) $\frac{18}{7}m{s}^{-2}$

(d) 0

Assertion and Reason

DIRECTIONS : Each of these questions contains an Assertion followed by reason. Read them carefully and answer the question on the basis of following options. You have to select the one that best describes the two statements.

(a) If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

(b) If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

(c) If Assertion is correct but Reason is incorrect

(d) If Assertion is incorrect but Reason is correct

1. Assertion : When a body is dropped or thrown horizontally from the same height, it would reach the ground at the same time.

Reason : Horizontal velocity has no effect on the vertical motion.

2. Assertion : Horizontal range of a projectile is same for angles of projection $\theta$ and $({90}^{\circ }-\theta )$ for same initial speed.

Reason : Horizontal range does not depend on angle of projection.

3. Assertion : If the rain is falling vertically, to a man walking on a horizontal road, the apparent speed of rain to the man will always be greater than its actual speed.

Reason : As velocity of the rain and the man are mutually perpendicular.

$v_{rm}=\sqrt{v_r^2+v_m^2}$

4. Assertion : In a uniform circular motion, acceleration of the body is constant.

Reason : In uniform circular motion, the body has only centripetal acceleration.

5. Assertion : Horizontal component of velocity of projectile is constant.

Reason : Acceleration of the projectile is along the vertical.

6. Assertion : In projectile motion, acceleration is always perpendicular to the velocity.

Reason - Path of projectile is a parabola.

Hots Subjective Questions

DIRECTIONS : Answer the following questions.

1. A ball is thrown horizontally from the top of a tower with a speed of $50m/s$. Find the velocity and position at the end of 3 second. $[g=9.8m/s^2]$
2. A ball is projected from ground with a velocity of $19.6m/s$ at an angle of ${30}^{\circ }$ with the horizontal. Calculate its time of flight and horizontal distance it travels.
3. A particle is projected “ith a velocity " " that its horizontal range is twice the greatest height attained. Find the horizontal range of it
4. The angular velocity of a particle moving in a circle of radius $50m$ is increased in a mınutes from 100 revolutions per minute to $\mathbf{4}\mathbf{0}\mathbf{0}$ revolution per minute. Find (a) angular acceleration (b) linear acceleration
5. A body is moving in a circle of radius $100cm$ with a time period of $2s$. Find the acceleration
6. A stone tied to the end of string $80cm$ long $$ whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 seconds. what is the magnitude and direction of the acceleration of the stone "
7. A wheel is $0.60m$ in radius s moving with a speed of $10m{s}^{-1}$. Find the angular speed
8. An aircraft executes a horizontal loop of radius $1km$ with a steady speed of $900km{h}^1$. Compare its centripetal acceleration with the acceleration due to gravity.
9. A cyclist is riding with a speed of $27km{h}^{-1}$. As he approaches a circular turn on the road of radius $80m$. he applies brakes and reduces his speed at the constant ratc of $0.5m{s}^{-1}$. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?
10. What is the centripetal acceleration of the carth is it moves in its orbit around the sun?
11. The height $y$ and horizontal distance covered by a particle is given by $x=12t$ and $y=10t-5t^2$, find maximum height and initial speed of projection.
12. How does the vertical distance a projectile falls helow an otherwise straight-line path compare with the vertical distance it would fall from rest in the same time?
13. In the absence of air resistance, why does the horizontal component of a projectile’s motion not change, while the vertical component does?
14. At what point in its trajectory does a batted baseball have its minimum speed? If air resistance can be neglected, how does this speed compare with the horizontal component of its velocity at other points?
15. Two golfers each hit a ball at the same speed. one at ${60}^{\circ }$ to the horizontal and the other at ${30}^{\circ }$. Which ball goes farther? Which hits the ground first? (Ignore air resistance)
16. When you jump upward, your hang time is the time you feet are off the ground. Does hang time depend on the vertical component of your velocity when you jump, on the horizontal component of your velocity, or on both? Defend your answer.
17. A projectile can have the same range for two angles of projection. If $h_1$ and $h_2$ are maximum heights when the range in the two cases is $R$, then what is the relation between $R,h_1$ and $h_2$ ?

SOLUTIONS

EXERCISE - 1

FILL IN THE BLANKS

1. ${90}^{\circ }$
2. ${45}^{\circ }$
3. ${90}^{\circ }-\theta$
4. angular speed
5. $\frac{u^2}{4g}$
6. $\frac{u^2}{g}$
7. ${90}^{\circ }$
8. $rad{s}^{-1}$
9. away
10. ${45}^{\circ }$
11. centre
12. 1.25

Hint :

$\begin{array}{rl}T& =\frac{u^2{\sin }^2\hat{ȷ}}{2g}=\frac{(10{)}^2\times {\sin }^2{30}^{\circ }}{2\times 10}=\frac{(10{)}^2\times \frac{1}{4}}{2\times 10}\\ & =\frac{10}{8}=1.25sec\end{array}$

13. $100km/h$

Hint :

Velocity of $A$ w.r.t. $B=$ Velocity of $B$ w.r.t. $A$

14. Square

$=60+40=100km/h$

Hint :

$H=\frac{u^2}{}\frac{\sin 2\theta }{\mu }$

Here. $\theta$ and $g$ are constants; therefore $H\propto u^2$.

15. zero.

TRUE/FALSE

1. True
2. True
3. False
4. True
5. True

Hint : Relative velocity of passengers in each bus $=(40+2n)m/s=6()m/s$

6. False
7. True
8. False
9. False

Hint :

Projectile moves under the action of a constant acceleration

10. False

Hint :

Required relatwe selocity (50 30$)km/h=20km/h$

11. True

Hint :

The horizontal range of the projectiles is same at two angles of projection $\theta$ and $({90}^c-\theta )$.

Match The Column

1. (A) $\to$ (r) ;(B) $\to$ (p) ;(C) $\to$ (q); (D) $\to$ (s)

Hint :

$T\cos \theta =\frac{2u\sin \theta }{g}\times \cos \theta$

$=\frac{4(2\sin \theta \cdot \cos \theta )}{g}=\frac{u\sin 2\theta }{g}$

2. (A) $\to$ (s) ;(B) $\to$ (r) ;(C) $\to$ (p) ;(D) $\to$ (q)

Distance $=\frac{3}{2}\times 2\pi r=3\times \frac{22}{7}\times 14=132m$

Displacement $=2r=28m$

Time taken $=\frac{\text{ Distance }}{\text{ Speed }}=\frac{132}{33}=4s$

Average velocity $=\frac{\text{ Displacement }}{\text{ Time taken }}=\frac{28}{4}=7m{s}^{-1}$

3. $(A)\to (p);(B)\to (s);(C)\to (q);(D)\to (r)$

Time of flight, $T=\frac{2u\sin \theta }{g}=\frac{2\times 20\times \frac{1}{2}}{10}=2s$

Horizontal range,

$R=\frac{u^2\sin 2\theta }{g}=\frac{(20{)}^2\times \sqrt{3}/2}{10}=20\sqrt{3}m$

Maximum height, $H=\frac{u^2\sin \theta }{2g}=\frac{(20{)}^2(1/2{)}^2}{2\times 10}=5m$

At the highest point of trajectory,

$v-u\cos \theta =20\times \frac{\sqrt{3}}{2}=10\sqrt{3}m{s}^{-1}$

Very Short Answer Questions

1. No

2. It will be four times the initial horizontal range.

3. It will be three times the initial vertical height.

4. Maximum height will be attained at $\theta ={90}^{\circ }$.

5. $u\cos \theta =u/2$ i.e. $\cos \theta =\frac{1}{2}$ or, $\theta ={60}^{\circ }$.

$\therefore R=u^2/g\sin 2({60}^{\circ })=u^2/g(\sqrt{}3/2)=\sqrt{3}{u}^2/2g$

6. The equation of trajectory of a projectile having initial speed $v$ at angle $\theta$ with the horizontal is,

$y=x\tan \theta -g(x/v\cos \theta {)}^{2/2}$

7. Parabolic.

8. $T_1=2u\sin \theta /g$ and $T_2=2u\sin ({90}^{\circ }-\theta )/g$

$\therefore T_1/T_2=\sin \theta /\sin ({90}^{\circ }-\theta )=\sin \theta /\cos \theta =\tan \theta$

9. Horizontal range, $R=(u^2/g)\sin 2\theta$ and maximum height, $H=u^2{\sin }^2\theta /2g$

(i) when $\theta =\theta ,R_1=(u^2/g)\sin 2\theta$ and $H_1=u^2{\sin }^2\theta /2g$

(ii) when $\theta =({90}^{\circ }-\theta ),R_2=(u^2/g)\sin 2({90}^{\circ }-\theta )$ $=u^2/g\sin 2\theta$ and $H_2=u^2{\sin }^2({90}^{\circ }-\theta )/2g$ $=u^2{\cos }^2\theta /2g$

$\therefore R_1/R_2=1$ and $H_1/H_2=u^2{\sin }^2\theta /2g/u^2{\cos }^2\theta /2g$ $={\sin }^2\theta /{\cos }^2\theta ={\tan }^2\theta$

10. Here, $R=nH$ so, $(u^2/g)\sin 2\theta =n{u}^2{\sin }^2\theta /2g$, $\therefore \tan \theta =4/n,\theta ={\tan }^{-1}(4/n)$

11. Horizontal range, $R=(u^2/g)\sin 2\theta$ and maximum height, $H=u^2{\sin }^2\theta /2g$

(i) when $\theta ={30}^{\circ },R_A=(u^2/g)\sin 2({30}^{\circ })=u^2/2(\sqrt{}3/2)$

and $H_A=u^2/2g{\sin }^2({30}^{\circ })=u^2/2g(1/4)$

(ii) when $\theta ={60}^{\circ },R_B=(u^2/g)\sin 2({60}^{\circ })=u^2/2(\sqrt{3}/2)$ and $H_B=u^2/2g{\sin }^2({60}^{\circ })=u^2/2g(3/4)$

$\therefore R_A/R_B=1:1$ and $H_A/H_B=u^2/2g(1/4)/u^2/2g(3/4)$

$=1/3=1:3$

12. We know, maximum height, $H=u^2{\sin }^2\theta /2g$, if $\theta ={90}^{\circ }$, $\sin \theta =1$ (maximum), then, $H_{max}=u^2/2g=h$

Similarly, maximum horizontal range, $R_{max}=u^2/g=2h$

13. The angular displacement per unit time is known as angular speed. Mathematically, angular speed = angular displacement / time taken

14. $v=r\omega$

15. The linear speed is given by $v=r\omega$. Since the angular speed $\omega$ is same for both the cars, the linear speed $v$ is more for the outer car for having higher radius $r$.

16. For hour hand of watch, time period, $T_h=12h$ for earth, $T_e=24h$,

Angular velocity, $\omega =2\pi /T$

$\therefore {\omega }_h/{\omega }_e=T_e/T_h=24/12=2$

$\therefore {\omega }_h=2{\omega }_e$

17. Here, $(u^2/g)\sin 2({15}^{\circ })=1.5$, or, $u^2/g(1/2)=1.5$

or, $u^2/g=3km$

Horizontal range, $R=(u^2/g)\sin 2({45}^{\circ })=(u^2/g)\sin ({90}^{\circ })$

$=(u^2/g)=3km$

18. 1. The motion of the hour or the minute hand of a watch. 2. The motion of the blades of the fan, when the electric fan has attained constant speed.

19. At every point along tangent to the circular path.

20. The gravitational pull of the sun on the earth.

21. Garavitational force of attraction on the satellite due to the earth.

Short Answer Questions

1. In uniform circular motion, the speed of the body remains the same but the direction of motion of the body changes continuously. Since velocity is a vector quantity, the continuous change in the direction of motion of.the body implies continuous change in the velocity of the body. As the rate of change of velocity is acceleration. a uniform circular motion is an accelerated motion.

2. The direction of the centripetal force does not depend, whether the body is moving in clockwise or anticlockwise direction. It is always directed along the radius and towards the centre of the circle.

3. The instantaneous velocity of the stone going around the circular path is tangential to the circle. When the string breaks, the centripetal force ceases to act. Due to inertia. the stone continuous its motion along the tangent to the circular path. That is why, it flies away tangentially.

4. When the bus takes a turn, the centripetal acceleration of the bus acts towards the centre of the circular turn and along its radius. Due to inertia, the passengers are thrown outwards. The force experienced by the passengers is called the centrifugal force.

. When the outer rail of a curved railway track is raised, the weight of the train gives a component of its weight along the radius of the curved track. This component of the weight provides the train the necessary centripetal force so as to enable it to move along the curved path.

6. Fig shows a bucket containing water and whirled in a vertical circle.

Let $v$ be the velocity of the bucket at the top of the circular path. For rotating the bucket of water along a certical circle, it requires a centripetal force $M,\therefore$, where,$M$ is the mass of water and $r$, the radius of the cincular path The weight of the water provides the centripetal fine and hence the water does not fall. In other wonds, the weight due to which the water can fall is used up in providing the necessary centripetal force.

7. The motion of the train only alfects the magnitude of the horizontal component of velocity and the acceleration of the body along the horizontal, if any. It does not atficy the nature of the motion along the vertical. Therefior, in all the three cases, the time of thee fall of the body will be cqual.

8. (i) To the observer inside the train, the ball will appear to move straight vertically upwards and then downwards

(ii) The observer outside the train, the ball will appear to move along the parabolic path.

9. The bolt will drop just at the foot of the mast. It is because, when the bolt drops, it will possess the following two motions :

(i) Motion towards west with uniform velocity of $8m{s}^{-1}$ i.e. equal to that of the ship.

(ii) Vertically downwards motion with initial velocity zero and acceleration equal to $g$. Thus,

4 0: a ${}^{2}9.8ms{}^2,x=19.6m$

Now $,ut,\frac{1}{2}a{t}^2$

$19.6(1\cdot t^1-9.8\cdot t^2-4.9t^2.$ or $t=2s$

In 2s, the bolt will thavel a distance of $8\times 2=16m$ westward liom the mast and the mast of the ship will also travel a distance of $16m$ westward and will therefore meet the bolt at its foot

10. During the time interval, the bullet moves up to the target, it falls through some vertical distance under the effect of gravity. As a result, it will not hit the target at the point, it wals aimed at It will hit a little below. Therefore, so as to hit the target exactly at that point, one has to aim a little above that point on the target.
11. The horizontal range of a projectile is given by

$R=\frac{u^2\sin 2\theta }{g}$

As the magnitude of the velocity is fixed and $g$ is constant at a given place, therefore

$R\propto \sin 2\theta$

i.e., range of the projectile is directly proportional to sin of the twice of the angle of projection.

For $\sin 2\theta =1$ i.e., $\theta {45}^{\circ }$, the range of the projectile is maximum.

EXERCISE - 2

Multiple Choice Questions

1. (b) K.E. is minimum at the highest point. So, the horizontal distance is half of the range $R$ i.e., $0.5R$.

2. (b) $R_1=\frac{u^2\sin 2\theta }{g}=\frac{u^2\sin {30}^{\circ }}{g}$

or $1.5=\frac{u^2}{2g}$ or $\frac{u^2}{g}=3$

$R_2=\frac{u^2\sin {90}^{\circ }}{g}=\frac{u^2}{g}=3km$

3. (b) The bullets are fired at the same initial speed

$\frac{H}{H^{\prime }}=\frac{u^2{\sin }^2{60}^{\circ }}{2g}\times \frac{2g}{u^2{\sin }^2{30}^{\circ }}=\frac{{\sin }^2{60}^{\circ }}{{\sin }^2{30}^{\circ }}$

$=\frac{(\sqrt{3}/2{)}^2}{(1/2{)}^2}=\frac{3}{1}$

4. (d) $\frac{H_1}{H_2}:=\frac{u^2{\sin }^2\theta /2g}{u^2{\sin }^2({90}^{\circ }-\theta )/2g}={\tan }^2\theta$

5. (a) $v_x=30\cos {30}^{\circ }=30\sqrt{3}/2$,

$v_y=30\sin \theta \cdots gt$

$v_y=30\sin {30}^{\circ }-10\times 1.5=0$

As vertical velocity $=0$,

Angle with horizontal $\alpha =0^{\circ }$

It is a state, when a particle reach to a highest point of its path.

6. (d)

$\frac{u^{2}2\sin \theta \cos \theta }{g}=2\times \frac{u^2{\sin }^2\theta }{2g}$ or $\tan \theta =2$

7.

(a) $\frac{1}{2}m(u\cos \theta {)}^2=\frac{3}{4}\times \frac{1}{2}m{u}^2$

or ${\cos }^2\theta =\frac{3}{4}$ or $\cos \theta =\frac{\sqrt{3}}{2}=\cos {30}^{\circ }$.

8. (a) $t_1=\frac{2u\sin \theta }{g}$ and

$t_2=\frac{2u\sin (90–\theta )}{g}=\frac{2u\cos \theta }{g}$

$\therefore t_1{t}_2=\frac{4u^2\cos \theta \sin \theta }{g^2}=\frac{2}{g}[\frac{u^2\sin 2\theta }{g}]=\frac{2}{g}R$,

where $R$ is the range. $t_1{t}_2\propto R$

9. (b) There is no change in horizontal velocity, hence no change in momentum in horizontal direction. The vertical velocity at $t=10sec$ is

$v=98\times \sin {60}^{\circ }-(9.8)\times 10=-13.13m/sec$

so change in momentum in vertical direction is

$=(0.5\times 98\times \sqrt{3}/2)-[-(0.5\times 13.13)]$

$=42.434+6.56=48.997\approx 49$

10. (c) We know that, $y_m=H=\frac{(u\sin \theta {)}^2}{2g}=\frac{u^2{\sin }^2\theta }{2g}$

$\therefore \frac{\mathrm{\Delta }H}{H}=\frac{2\mathrm{\Delta }u}{u}$. Given $\frac{\mathrm{\Delta }u}{u}=2$

$\therefore \frac{\mathrm{\Delta }H}{H}=2\times 2=4$

11. (d) Time of flight $=\frac{2u\sin \theta }{g}$

$=\frac{2\times 9.8\times \sin {30}^{\circ }}{9.8}=2\times \frac{1}{2}=1sec.$

12. (c) $\frac{u^{2}2\sin \theta \cos \theta }{g}=\frac{u^2{\sin }^2\theta }{2g}$ or $\tan \theta =4$.

13. (b) $v_x=\frac{dx}{dt}=3\alpha t^2,v_y=\frac{dy}{dt}=3\beta t^2$ $v=\sqrt{v_x^2+v_y^2}=3t^2\sqrt{{\alpha }^2+{\beta }^2}$

14. (b) Acceleration vector is always radial (i.e . towards the centre) for uniform circular motion.

15. (a)

Now, taking initial position at origin. Final position in vector form is given by $\overrightarrow{r}=6\hat{j}+8\hat{i}+10\hat{k}$ Now magnitude of distance is $|\overline{r}|=\sqrt{6^2+8^2+{10}^2}=\sqrt{36+64+100}=10\sqrt{2}m$

16. (a) In uniform circular motion speed is constant. So, ne tangential acceleration.

It has only radial acceleration $a_R=\frac{v^2}{R}$ [directed towards center]

and its velocity is always in tangential direction. So these two are perpendicular to each other.

17. (a) In circular motion of a particle with constant speed, particle repeats its motion after a regular interval of time but does not oscillate about a fixed point. So, motion of particle is periodic but not simple harmonic.

18. (b) Given, $u_1=u_2=u,{\theta }_1={60}^{\circ },{\theta }_2={30}^{\circ }$

In lst case, we know that range

$\begin{array}{rl}& R_I=\frac{u^2\sin 2({60}^{\circ })}{g}=\frac{u^2\sin {120}^{\circ }}{g}=\frac{u^2\sin ({90}^{\circ }+{30}^{\circ })}{g}\\ & =\frac{u^2(\cos {30}^{\circ })}{g}=\frac{\sqrt{3}{u}^2}{2g}\end{array}$

In IInd case when (1) in then

$R_2=\frac{u^2\sin {60}^{\circ }}{g}\frac{u^2\sqrt{3}}{2g}:k_1=R_2$

(we get same value of innues).

19. (b) $s=\frac{1}{2}g{t}^2,x$ and $g$ are same for both the balls, so time of fall ’ $t$ ’ will alse be the same for both of them ( $s$ is vertical height).

20. (c) Since horizontal component of the velocity of the bomb will be the same as the velocity of the aeroplane. therefore horizontal displacements remain the same at any instant of time.

21. (d) $T=\frac{2u\sin (\theta -\alpha )}{g\cos \alpha }$

22. (d) Velocity and hence kinetic energy is minimum at the highest point.

$K.E=\frac{1}{2}m{v}^2{\cos }^2\theta$

23. (d) The initial velocity in the vertically downward direction is zero and same height has to be covered.

24. (a) $0={45}^{\circ },k=\frac{u^2}{g}$

25. (a) A parabola

26. (d) $u=\sqrt{2gh},R_{max}=\frac{u^2}{g}=\frac{2gh}{g}=2h$

27. (c) $v_x=\frac{d}{dt}\cdot (2t+4t^2)=2+8t$

$a_x=\frac{d}{dt}(2+8t)=x$

$v_y=\frac{d}{dt}(5t)=5,a_v=\frac{d}{dt}(v_y)=0$

$\Rightarrow$ acceleration is along $x$-axis and its value is $8m{s}^{-2}$.

28. (c) $y=x\tan \theta -\frac{1}{2}\frac{g{x}^2}{u^2{\cos }^2\theta }$

$y=50\tan {60}^{\circ }-\frac{10\times 50\times 50}{2\times 25\times 25\times {\cos }^2{60}^{\circ }}=5m$

29. (b) Horizontal distance covered should be same for the lime ol collision

$400\cos \theta =200$ or $\cos \theta =\frac{1}{2}$ or $\theta ={60}^{\circ }$

30. (c) Even after 5 second, the horizontal velocity $v_x$ will be $20m{s}^{-1}$. The vertical velocity $v_y$ is given by

$v_y=0+10\times 5=50m{s}^{-1}$

Now, $v=\sqrt{v_x^2+v_y^2}=\sqrt{{20}^2+{50}^2}\approx 54m{s}^{-1}$

31. (d) Given $\frac{u^2\sin 2\theta }{g}=3\frac{u^2{\sin }^2\theta }{2g}$

$\Rightarrow \theta ={\tan }^{-1}(\frac{4}{3})$

32. (a) We have,

$R=\frac{u^2\sin 2\theta }{g}=\frac{u^2}{g}(\because \theta ={45}^{\circ })$

$H=\frac{u^2{\sin }^2\theta }{2g}=\frac{u^2}{4g}(.$ for $.\theta ={45}^{\circ })$

$\Rightarrow R=411$

33. (b) Comparing the given equation with the equation of trajectory of a projectile,

$y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }$

we get, $\tan \theta =\frac{1}{\sqrt{3}}\Rightarrow \theta ={30}^{\circ }$

and $2u^2{\cos }^2\theta =20\Rightarrow u^2=\frac{20}{2{\cos }^2\theta }$

$=\frac{10}{{\cos }^2{30}^{\circ }}=\frac{10}{(\frac{\sqrt{3}}{2}{)}^2}=\frac{40}{3}$

Now, $R_{max}=\frac{u^2}{g}=\frac{40}{3\times 10}=\frac{4}{3}m$

34. (a) $h_1=\frac{u^2{\sin }^2\theta }{2g}$

$h_2=\frac{u^2{\sin }^2(90-\theta )}{2g},R=\frac{u^2\sin 2\theta }{g}$

Range $R$ is same for angle $\theta$ and $({90}^{\circ }-\theta )$

$\therefore h_1{h}_2=\frac{u^2{\sin }^2\theta }{2g}\times \frac{u^2{\sin }^2(90-\theta )}{2g}$

$=\frac{u^4({\sin }^2\theta )\times {\sin }^2(90-\theta )}{4g^2}[\because \sin (90-\theta )=\cos \theta ]$

$=\frac{u^4({\sin }^2\theta )\times {\cos }^2\theta }{4g^2}[\because \sin 2\theta =2\sin \theta \cos \theta ]$

$=\frac{u^4(\sin \theta \cos \theta {)}^2}{4g^2}=\frac{u^4(\sin 2\theta {)}^2}{16g^2}$

$=\frac{(u^2\sin 2\theta {)}^2}{16g^2}=\frac{R^2}{16}$

or, $R^2=16h_1{h}_2$ or $R=4\sqrt{h_1{h}_2}$

35. (c) The bullet performs a horizontal journey of $10cm$ with constant velocity of $150(0)m/s$. The bullet alse performs a vertical journey of $h$ with zero initial velocity and downward acceleration

$\therefore$ For horizontal journey, time (f) $\approx \frac{\text{ Distance }}{\text{ Velocity }}$

$\therefore t=\frac{100}{1500}=\frac{1}{15}sec$

The bullet performs vertical journey for this time.

For vertical journey, $h=ut+\frac{1}{2}g{t}^2$

$h=0+\frac{1}{2}\times 10\times (\frac{1}{15}{)}^2$

or, $h=\frac{10}{2\times 15\times 15}m=\frac{10\times 100}{2\times 15\times 15}cm$

or, $h=\frac{20}{9}cm=2.2cm$

The gun should be aimed $(\frac{20}{9})cm$ above the urgee.

36. (c) For circular angular motion, the formula for angular displacement $\mathit{\theta }$ and angular acceleration $\alpha$ is

$\theta =\omega t+\frac{1}{2}\alpha t^2$

where $\omega =$ initial velocing

or $\theta =0+\frac{1}{2}\alpha t^2$ or $\theta =\frac{1}{2}\times (2)(10{)}^2$

or $\theta =100$ radian

$2\pi$ radian are covered in 1 revolution

$\therefore 1$ radian is covered in $\frac{1}{2\pi }$ revolution

or 100 radian are covered in $\frac{100}{2\pi }$ revolution

$\therefore$ Number of revolution $=\frac{50}{3.14}=16$

37. (b)

As the figure drawn above shows that at pouns in and $B$ the vertical component of velocity is $v\sin 0^2$ their directions are opposite.

Hence, change in momentum is given by

$\mathrm{\Delta }p=mv\sin {60}^{\circ }-(-mv\sin {60}^{\circ })=2mv\sin {60}^2$

$=2mv\frac{\sqrt{3}}{2}=\sqrt{3}mv$

38. (c)

Velocity of projectile $u=147m{s}^{-1}$

angle of projection $\alpha ={60}^{\circ }$

Let, the time taken by the projectile from $O$ to $A$ be $t$ where direction $\beta ={45}^{\circ }$. As horizontal component of velocity remains constant during the projectile motion.

$\Rightarrow v\cos {45}^{\circ }=u\cos {60}^{\circ }$

$\Rightarrow v\times \frac{1}{\sqrt{2}}=147\times \frac{1}{2}\Rightarrow v=\frac{147}{\sqrt{2}}m{s}^{-1}$

For vertical motion, $v_y=u_y-gt$

$\Rightarrow v\sin {45}^{\circ }=45\sin {60}^{\circ }-9.8t$

$\Rightarrow \frac{147}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=147\times \frac{\sqrt{3}}{2}-9.8t$

$\Rightarrow 9.8t=\frac{147}{2}(\sqrt{3}-1)\Rightarrow t=5.49s$

39. (b) If there is no resistance, bomb will drop at a place exactly below the flying aeroplane. But when we take into account air resistance, bomb will face deceleration in its velocity. So, it will fall on the earth exactly behind the aeroplane.

40. (d) When a cyclist moves on a circular path, it experiences a centrifugal force which is equal to $m{v}^2/r$. It tries to overturn the cyclist in outward direction. If speed increases twice, the value of centrifugal force too increases to 4 times its earlier value. Therefore the chance of overturning is $1/4$ times.

41. (c) The time of flight of a projectile is given as,

$t=\frac{2u\sin \theta }{g}$

A projectile can have same range if angle of projection are complementary i.e., $\theta$ and $(90-\theta )$

$\therefore t_1=\frac{2u\sin \theta }{g}$

and $t_2=\frac{2u\sin (90-\theta )}{g}=\frac{2u\cos \theta }{g}$

$\therefore t_1{t}_2=\frac{4u^2\sin \theta \cos \theta }{g^2}$

$=\frac{4u^2}{g^2}\times \frac{\sin 2\theta }{2}=\frac{2u^2}{g^2}\sin 2\theta =\frac{2R}{g}$

where range $(R)=\frac{u^2\sin 2\theta }{g}$

42. (c) Here $v=0.5m/sec,u=$ ?

so $\sin \theta =\frac{u}{v}\Rightarrow \frac{u}{5}=\frac{1}{2}$ or $u=0.25m{s}^{-1}$

43. (d) All options are correct :

(i) When two bodies move in opposite directions then relative velocity between either $v_{AB}$ or $v_{BA}$ both are greater than .

(ii) When two bodies move in parallel direction then $v_{AB}=v_A-v_B\Rightarrow v_{AB}<v_A$ $v_{BA}=v_B-v_A\Rightarrow v_{BA}<v_B$

More Than One Correct

1. $(a,c)$

In uniform circular motion, magnitude of acceleration and velocity is constant but not their directions.

2. $(a,c)$

The particle has initially horizontal as well as vertical velocity hence its motion is projectile.

3. $(a,b)$
4. $(a,b)$
5. (a,d)

Motion of the particle will be uniform circular.

6. (b, d)

For a projectile motion,

$R=\frac{u^2\sin 2\theta }{g}$

or $2\sqrt{3}\times {10}^3=\frac{u^2\times \frac{\sqrt{3}}{2}}{10}$ or $u=200m{s}^{-1}$

$\therefore R_{max}=\frac{u^2}{g}=\frac{(200{)}^2}{10}=4000m=4km$

7. $(a,b)$
8. $(b,c,d)$

Centripetal acceleration,

$a_c=\frac{v^2}{r}={\omega }^{2}r=4{\pi }^2{v}^{2}r=\omega \nu =\frac{4{\pi }^{2}r}{T^2}$

9. $(a,b,c)$
10. $(a,b,d)$

For same horizontal range

${\theta }_1+{\theta }_2={90}^{\circ }$

$\therefore \frac{f_1}{f_2}=\frac{\sin {\theta }_1}{\sin {\theta }_2}=\frac{\sin {\theta }_1}{\cos {\theta }_1}=\tan {\theta }_1$ or $\frac{f_1}{\sin {\theta }_1}=\frac{f_2}{\sin {\theta }_2}$

11. $(a,b)$

For circular motion $a_r\ne 0$

12. (b, c, d)

$H=\frac{u^2{\sin }^2\theta }{2g}$

13. $(a,b,c)$

In uniform circular motion, velocity, angular velocity and centripetal acceleration are mutually perpendicular.

14. (a, d)

For horizontal projection,

$T=\sqrt{\frac{2h}{g}}$ is independent of velocity of projection

$R=u\sqrt{\frac{2h}{g}}$

As $v_A>v_B,R_A>R_B$

15. $(a,b,c)$

Multiple Matching Questions

1. (A) $\to$ (p); (B) $\to$ (q, s) ;(C) $\to$ (p, t); (D) $\to$ (q, r)
2. (A) $\to$ (q, s) ;(B) $\to$ (p, t) ;(C) $\to$ (p) ;(D) $\to$ (q, r)

Fill In The Passage

(I)

(1) uniform

(2) non-uniform

(3) centre

(4) perpendicular

(5) centripetal acceleration

(6) magnitude

(7) direction.

(II)

(1) constant

(2) vertically downwards

(3) horizontal

(4) minimum

(5) horizontal range

(6) equal to

(7) parabola

Passage Based Questions

1. (b) Let the boy starts from point $A$, then at the instant of sixth turn, he will be at point $C$ then

Displacernent $=AC=\sqrt{A{B}^2+B{C}^2}=50\sqrt{2}m$

$[\because AB=5\times 10=50m]$

2. (a) At the instant of fith turn, boy will be at point $B$,

$\therefore$ Displacement $=AB=50m$. & time $=10\times 5=50s$.

$\therefore$ Average velocity $=\frac{\text{ Displacement }}{\text{ Time }}=\frac{50}{50}=1m{s}^{-1}$

3. (a) Change in velocity, $\mathrm{\Delta }v=5-(-5)=10m{s}^{-1}$ and time, $f=10\times 2=2as$

$\therefore$ Average acceleration, $\overline{a}=\frac{\mathrm{\Delta }v}{f}=\frac{10}{20}=0.5m{s}^{-2}$

4. (b) Time of flight

$T=\frac{2x\sin \theta }{g}=\frac{2\times 10\times \frac{1}{\sqrt{2}}}{10}=\sqrt{2}s$

5. (b) Horizontal range,

$R=\frac{u^2\sin 2\theta }{g}=\frac{(10{)}^2\times 1}{10}=10m$

6. (b) At the highest point of trajectory.

$v=u\cos \theta =10\times \frac{1}{\sqrt{2}}=5\sqrt{2}m{s}^{-1}$

7. (c) Time of flight,

$T=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 78.4}{9.8}}=4s$

8. (c) Horizontal range,

$R=u\sqrt{\frac{2h}{g}}=29.4\sqrt{\frac{2\times 78.4}{9.8}}=117.6m$

9. (c) Speed with which the particle hits the ground

$v=\sqrt{u^2+2gh}=\sqrt{(29.4{)}^2+2\times 9.8\times 78.4}=49m{s}^{-1}$

10. (a) $t=\frac{2\pi r}{v}=\frac{2\times \frac{22}{7}\times 3.5}{4}=5.5s$

11. (d) Displacement of the particle in one complete round is zero, hence average velocity of the particle in one complete round is zero.

12. (b) Acceleration

$a=\frac{v^2}{r}=\frac{4^2}{3.5}=\frac{32}{7}m{s}^{-2}$

Assertion And Reasons

1. (a) Time of flight depends only on vertical motion and not on the horizontal motion.
2. (c) Horizontal range depends on angle of projection and is equal for two angles of projection ${\theta }_1$ and ${\theta }_2$ if ${\theta }_1+{\theta }_2={90}^{\circ }$.
3. (a) Both assertion and reason are true and reason is correct explanation of assertion.
4. (c) Centripetal acceleration is not constant as it varies in direction.
5. (a) Acceleration due to gravity acts vertically downwards hence horizontal velocity does not change.
6. (d) In projectile motion, acceleration is perpendicular to velocity only at the heighest point of its trajectory.

Hots Subjective Questions

1. Ans : $58m/s,{30}^{\circ }{27}^{\prime }$ with horizontal, $44.1m$ below and $150m$ horizontally away from the starting point

Hint : $u=50m/s,t=3s,x=ut=150m$;

$\begin{array}{rl}& y=\frac{1}{2}g{t}^2=44.1m\\ & v_x=u_x+a_{x}t=50m/s;v_y=u_y+a_{y}t=29.4m/s;\\ & v=\sqrt{v_x^2+v_y^2}=\sqrt{{50}^2+(29.4{)}^2}=58m/s\end{array}$

$\tan \theta =\frac{v_y}{v_x}=\frac{29.4}{50}=0.588=\tan {30}^{\circ }{27}^{\prime }$

2. Ans : $2s,33.95m$

Hint : $T=\frac{2u\sin \theta }{g}=\frac{2\times 19.6\times \sin {30}^{\circ }}{9.8}=2s$;

$R=\frac{u^2\sin 2\theta }{g}=\frac{(19.6{)}^2\times \sin 2\times {30}^{\circ }}{9.8}=33.95m$

3. Ans : $\frac{4u^2}{5g}$

Hint : $R=2H\Rightarrow \frac{u^2\sin 2\theta }{g}=2\times \frac{u^2{\sin }^2\theta }{2g}$

$\Rightarrow 2\sin \theta \cos \theta ={\sin }^2\theta \Rightarrow \tan \theta =2$

$\therefore \sin \theta =\frac{2}{\sqrt{5}}$ and $\cos \theta =\frac{1}{\sqrt{5}}$

$\therefore R^2=\frac{2u^2}{g}\cdot \frac{2}{\sqrt{5}}\times \frac{1}{\sqrt{5}}=\frac{4u^2}{5g}$

4. Ans. $\frac{\pi }{30}rad/s^2,\frac{5\pi }{3}cm/s^2$

Hint : (a) $\alpha =\frac{\omega -{\omega }_0}{t}=\frac{(400-100)\times 2\pi }{60\times 5\times 60}=\frac{\pi }{30}rad/s^2$

5. Ans. $987.7cm/s^2$

Hint : $T=2s,r=100cm;\omega =\frac{2\pi }{T}=\frac{2\pi }{2}=\pi rad/s$;

$\nu ={\omega }^{2}r={\pi }^2\times 100=(3.14{)}^2\times 100=987.7cm/s^2$

6. Here, $r=80cm;T=\frac{25}{14}s$

Now, $\omega =\frac{2\pi }{T}=\frac{2\pi }{25/14}=\frac{28\pi }{25}rad{s}^{-1}$

The acceleration of uniform circular motion is given by

$a={\omega }^{2}r=(\frac{28\pi }{25}{)}^2\times 80=990.4cm{s}^{-2}$

The acceleration is directed along the radius of the circular path and towards the centre of the circle.

7. Here, $r=0.60m;v=10m{s}^{-1}$

We know, $v=\omega r$

$\therefore \omega =\frac{v}{r}=\frac{10}{0.60}=16.67rad{s}^{-1}$

8. Here, $r=1km=1000m;v=900km{h}^{-1}=250m{s}^{-1}$

Now, the centripetal acceleration is given by

$a=\frac{v^2}{r}=\frac{(250{)}^2}{1000}=62.5m{s}^{-2}$

Now, $\frac{a}{g}=\frac{62.5}{9.8}=6.38$

or $a=6.38g$

9. Here, $v=27km{h}^{-1}=7.5m{s}^{-1};r=80m$

Centripetal acceleration, $a_c=\frac{v^2}{r}=\frac{(7.5{)}^2}{80}=0.7m{s}^{-2}$

Suppose that the cyclist applies brakes at the puint $A$ of the circular turn. Then, retardation produced due to the brakes, say $a_T$ will act opposite to the velocity $v$ [Fig]

Thus, $a_T=-0.5m{s}^{-2}$

Therefore, total acceleration is given by

$a=\sqrt{a_c^2+a_T^2}=\sqrt{(0.7{)}^2+(-0.5{)}^2}$

$=\sqrt{0.49+0.25}=\sqrt{0.74}=0.86m{s}^{-2}$

If $\theta$ is the angle between the total acceleration and the velocity of the cyclist, then,

$\tan \theta =\frac{a_c}{a_T}=\frac{0.7}{0.5}=1.4$

or, $\theta ={54}^{\circ }-{28}^{\prime }$

10. $a_c=\frac{v^2}{r}=\frac{(\frac{2\pi r}{T}{)}^2}{r}=\frac{4{\pi }^{2}r}{T^2}$

$\begin{array}{rl}& =\frac{4{\pi }^2(1.5\times {10}^{11}m)}{(1yr{)}^2}(\frac{1yr}{3.15\times {10}^{7}s}{)}^2\\ & =6.0\times {10}^{-3}m/s^2\end{array}$

11. $\frac{dx}{dt}=u_x=12,\frac{dy}{dt}=u_y=10-10t$

at $t=0,u_x=12,u_y=10$

so $\overline{v_0}=12i+10j|{\psi }_0|$

$\begin{array}{rl}& =\sqrt{144+100}=15.6\frac{m}{s}\\ & \text{ At maximum height }{u}_y=0.0=10-10t\Rightarrow t=1s\\ & y_{max}=10\times 1-5\times 1^2=5m\end{array}$

12. The vertical distance a projectile falls below an otherwise straight line path is the same vertical distance it would fall from rest in the same time because the distance covered in vertical direction in both the cases is given by $d=\frac{1}{2}g{t}^2$, where $t$ is the elapsed time, and $g$ is the acceleration due to gravity.

13. In the absence of air resistance, the force acting on the projectile is only the force of gravity acting vertically downward. No component of the force due to gravity acts in the horizontal direction therefore the horizontal component of the projectile motion remains unchanged.

Whereas the vertical component of motion undergoes changes. The vertical component of motion decreases while rising against gravity and increases during downward motion.

14. A batted baseball has the minimum speed at the top of its trajectory. The vertical component of velocity is zero at the top. The baseball has only the horizontal component of velocity which remains the same at any point on the trajectory.

15. Both of the two golf balls will hit the ground at the same distance from the point of projection because the same range is obtained from two different launching angles when the angles add up to ${90}^{\circ }$ (launching speed being the same in the two cases).

The ball thrown at an angle of ${30}^{\circ }$ to horizontal will hit the ground first. This is because the object remains in the air for a shorter time for the smaller angle of projection.

16. When anybody jumps upwards the hang time in air depends on the vertical component of velocity not on the horizontal component of velocity.

This is because the vertical component and horizontal component of velocity are perpendicular to each other and hence are independent of each other.

The vertical component of velocity matters only for the motion in vertical direction and there is no contribution of the horizontal component of velocity.

17. $h_1=\frac{u^2{\sin }^2\theta }{2g}$

$h_2=\frac{u^2{\sin }^2(90-\theta )}{2g},R=\frac{u^2\sin 2\theta }{g}$

Range $R$ is same for angle $\theta$ and $({90}^{\circ }-\theta )$

$\therefore h_1{h}_2=\frac{u^2{\sin }^2\theta }{2g}\times \frac{u^2{\sin }^2({90}^{\circ }-\theta )}{2g}$

$=\frac{u^4({\sin }^2\theta )\times {\sin }^2({90}^{\circ }-\theta )}{4g^2}[\because \sin ({90}^{\circ }-\theta )=\cos \theta ]$

$=\frac{u^4({\sin }^2\theta )\times {\cos }^2\theta }{4g^2}[\because \sin 2\theta =2\sin \theta \cos \theta ]$

$=\frac{u^4(\sin \theta \cos \theta {)}^2}{4g^2}=\frac{u^4(\sin 2\theta {)}^2}{16g^2}$

$=\frac{(u^2\sin 2\theta {)}^2}{16g^2}=\frac{R^2}{16}$

or, $R^2=16h_1{h}_2$ or $R=4\sqrt{h_1{h}_2}$

CHAPTER 4

FORCE AND LAWS OF MOTION

INTRODUCTION

In last couple of chapters we dealt with different kinds of motion. We didn’t bother about their cause. This chapter is concerned about the cause of motion (i.e. force) and its effect (i.e. acceleration) and their Melationship. We will read different kinds of force.

When force is applied on a body, some acceleration is produced in it. These two quantities-force and acceleration are closely related. The existence of one causes the existence of other.

The great physicist sir Isaac Newton formulated three laws which govern the motion and its cause. This chapter deals with all the three laws of motion given by Newton and their applications in day to day life.

The chapter also gives us detailed knowledge of linear momentum, its conservation, impulse and their practical applications.

Force

In our daily life we find that a body at rest can be moved and one that is moving can be stopped. The doors and windows of a house are pushed to open and pulled to shut them. In games like cricket, hockey and football, players on the ground sometimes stop a moving ball or manage to deflect it to some other direction. The boatsman pulls a boat with a rope in order to bring it to the shore or pushes it to get it into the water. This push or pull is called the force in every day language.

Force is a vector quantity (it has both magnitude and direction). There must be a net force (unbalanced force) acting on an object for the object to change its velocity (either magnitude and/or direction), or to accelerate.

If we want the car to stop, we have to do something to it. That is what our brakes are for : to exert a force that decreases the car’s velocity.

In a basketball, a player launches a shot by pushing on the ball.

Thus it can be said that the force is a physical influence which can change the state of motion or state of rest of a body. A force can change the direction of motion of a body

Therefore, a force can change the speed and direction of motion of a body. A force is that physical quantity which tries to change or changes the state of rest or of uniform motion of a body.

To obtain a complete information about the force acting on an object one should know

(i) the point of application of force

(ii) the magnitude of force

(iii) the direction in which the force acts.

$✓$ CHECK POINT

• If an object experiences no acceleration, is no force acting on it?

Check Your Answer

If an object has no acceleration, you cannot conclude that no forces act on it. In this case, you can only say that the net forceon the object is zero.

Some Common Forces

(i) Contact forces

When a body $M$ is in contact with body $N,M$ can exert force on $N$ and $N$ can exert a force on $M$. These forces are called contact forces. Push or pull by a person, force by wind, force by a weight on the head of a porter, frictional force, normal reaction force, tension in strings, forces exerted during collision are the examples of contact forces.

Practical examples

(a) The pulling of a trolley by a coolie.

(b) The pulling of a cart by a horse.

(c) The pushing of a door to close it.

(d) The stretching of a spring by suspending a load.

(e) The squeezing of a gum tube to extract the gum.

(ii) Elastic recoil

The special property of solids, in contrast to liquids and gases, is that they resist any change in their shape. When we push on a solid, it pushes back. The plank exerts an upward force against the hand because the hand is changing the shape of the board. The floor pushes up when our weight pushes down on it, because our weight changes its shape

(iii) Tension

This is a special type of elastic recoil, resulting from stretching something. In a tug-of-war, the rope is under great tension, so it pulls back on the teams at the opposite ends. The spring scale is actually a tension-measuring device. The harder you pull on both ends of the spring, the more it stretches, and the harder it pulls back.

(iv) Compression

This is the opposite of tension. Pushing on a window pole shortens the pole-a little-and the pole pushes back. Actually, the bending board exerts its recoil because its upper surface is compressed and its lower surface is tensed.

(v) Buoyancy

Anything submerged in a liquid or a gas experiences an upward force. A rock under water feels lighter than the same rock in the air. If buoyancy is greater than weight, as for a cork under water or a helium-filled balloon in the air, the object rises. If we are under water, we can rise to the top until we emerge into the air. Then, buoyancy and gravity are equal, and we float.

(vi) Viscous drag

Anything moving through a liquid or gas feels a retarding, friction-like force. That is why a boat needs an engine, why we cannot swim fast enough to get into the Olympics, why automobiles are streamlined, and why a parachute is advisable if we plan to fall out of an airplane.

(vii) Normal force

If contact force between the bodies is perpendicular to the surface in contact, the force is known as normal force. Let us consider a block on a table. The table pushes the block upwards and block pushes the table down wards. Then forces are perpendicular to the surfaces of block and table. Thus the table applies a normal force on block in the upward direction and block applies a normal force on table in downward direction.

(viii) Friction

Two bodies placed in contact can also exert forces parallel to the surfaces in contact, such a force is called force of friction or simply friction. Suppose a body is placed on the table. Following three forces acts on it.

(1) Force by earth in downward direction

(2) Normal force due to table in upward direction.

(3) Applied force towards right.

Body is not moving, so all the forces must be balanced, normal force due to table and force by earth balanced each other. To balance the applied force there must be an equal and opposite force. This force is known as force of friction. If we increase the applied force the body is still at rest. It means force of friction is also increased till it is balanced by the applied force. The force of friction is self adjusting force. On increasing the applied force, the force of friction will increase up to a limit. It is known as limiting friction. After it on increasing the applied force, the body will start to move.

(ix) Spring Force

A spring is made of a coiled metallic wire. A spring has a definite length when it has been neither pushed nor pulled. The length is called natural length. At natural length the spring does not exert any force on the objects attached to its ends. If the spring is pulled at the ends its length becomes larger than natural length. It is known as stretched or extended spring. Extended spring pulls objects attached to its ends.

If the spring is pushed at the ends its length becomes less than natural length. It is known as compressed spring. A compressed spring pushes the block attached to its ends.

Weight

The earth attracts all the bodies towards its centre. The force exerted by the earth on the body is known as the weight of the body. It acts in vertically downward direction. These forces are not contact forces. If mass of the body is $m$ and gravitational acceleration is $g$, then the weight of body is $mg$, here $g=9.8m/se{c}^2$.

Balanced and Unbalanced Forces

One effect of a force is to alter the dimension or shape of a body on which the force acts; like by loading a spring, there occurs an increase in its length. By hammering a small piece of silver sheet, thin foil is made. The steam pushing out from a pressure cooker occupies a large volume. On pressing a piece of rubber, its shape changes. In a cycle pump, when the piston is lowered, the air is compressed to occupy less volume. Another is to alter the state of motion of the body. A player applies force with a hockey stick to change the speed and direction of motion of a ball. When more force is applied on the pedal by a cyclist, speed of the cycle increases. In many situations we may find that a body remains at rest or moves uniformly even if it is acted on by a force ! For example try to push a heavy stone or a heavy iron safe, probably we may not be able to move it. Does it mean that something is wrong with the definition of force? What happens actually is that there exists another force which is acting on the body in a direction opposite to your push and exactly compensating it. In effect there is no net force acting on the body. This force which we have not taken into account is the force of friction.

When several forces act on a body simultaneously, their effects can compensate one another with the result there is no change in the state of rest or motion. When this is the case, the body is said to be in equilibrium. Equivalently, one can say that the net force acting on the body is zero for balanced forces. Balanced forces do not change the speed. This means that the body as a whole either remains at rest or moves in a straight line with constant rate.

Fig. : Balanced forces

An object in equilibrium may or may not be at rest. A parachutist, descending at constant speed, is in equilibrium. His weight is just balanced by the viscous drag on the parachute-which is why he put it on in the first place. $\wedge$ heavier parachutist falls a little faster, his speed increases until the viscous drag just balances his weight.

Balancing the vertical forces is not enough to produce equilibrium. An airplane traveling at constant speed, as in figure, is in equilibrium under the influence of four forces, two vertical and two horizontal.

Vertical : gravity (down) is just balanced by the lift produced by the flow of air across the wing. I/orizontal : viscous drag is just balanced by the thrust of the engines. Both the vertical and the horizontal velocities are constant.

When the resultant force or net force acting on a body is not zero we say that it is acted on by an unbalanced force. A net or unbalanced force when acts on a body it changes its state of rest or uniform motion.

To imagine unbalanced force, imagine see saw. If the people on either side push down with the same force, they cancel each other. The see-saw is balanced and does not move [(fig. (a)]. In fig (b), there is an unbalanced force of $100N$. This moves the see-saw down at one end.

(b)

Component of a Force

The crate of figure is being dragged along the floor by means of a rope, which is not horizontal. The rope makes an angle $\theta$ to the floor.

The tension in the rope, acting on the crate, does two things to it. First, it drags the crate across the floor. Second, it tends to lift the crate off the floor. The smaller the angle $\theta$, the larger the effective force that is dragging the crate, and smaller the effective force that is lifting it. When $\theta =0^{\circ }$, the entire force is dragging and there is no lifting at all. Conversely, when $\theta ={90}^{\circ }$, the entire force is lifting the crate.

How can you find out how much force is being used to drag the crate? Force is a vector, and it obeys the same mathematical rules as velocity vectors and displacement vectors. The dragging force is the component of the tension in the rope acting parallel to the floor, that is, the horizontal component.

$F_{\text{horizontal }}=F\cos \theta$

Similarly, the component of the force tending to lift the crate is the vertical component and is given by

$F_{\text{vertical }}=F\sin \theta$

Equilibrium of a body with several forces

For the purpose of making a complete analysis of the forces acting on an object, a vector diagram is a useful device.

Figure is a vector diagram showing the forces on a brick being dragged along a tabletop. Four forces act: gravity (weight), friction, elastic recoil of the tabletop, and tension in the cord. Each force is represented by a vector, drawn at the correct angle and with its length proportional to the force. For the purpose of analysis, all vectors are represented by components along a pair of axes perpendicular to each other. In this case, we elect to use horizontal and vertical axes, since three of the forces are already on these axes. To do the analysis, we have to resolve the tension vector into its components on the vertical and horizontal axes. Then we can write two equations: one says that there is no net vertical force, and the other says that there is no net horizontal force.

$✓$ CHECK POINT

At the moment an object that has been tossed upward into the air reaches its highest point, is it in equilibrium? Defend your answer.

Check Your Answer

When an object reaches at the highest point of its path, when it is tossed upwards in air, its velocity is zero. The force due to gravity acts in the downward direction which balances upward force with which it is thrown. So it is in equilibrium..

Units of Force

(i) The S.I. unit of force is newton.

One newton is the force which when acts on a body of mass $1kg$, produces an acceleration of $1m{s}^{-2}$, i.e.,

$1\text{ newton }(N)=1kg\times 1m{s}^{-2}$

(ii) In C.G.S. system, the unit of force is dyne.

One dyne is the force which when acts on a body of mass 1 gramme, produces an acceleration of $1cm{s}^{-2}$, i.e.,

$1\text{ dyne }=1g\times 1cm{s}^{-2}$

(iii) Units of force in terms of force due to gravity

In MKS system, the unit of force is the kilogramme force (kgf).

One kilogramme force is the force due to gravity on 1 kilogramme mass.

Thus, $1kgf=$ force due to gravity on $1kg$ mass

$=1kg$ mass $\times$ acceleration due to gravity $gm{s}^{-2}$

$=g$ newton

Since the average value of $g$ is $9.8m{s}^{-2}$

$\therefore 1kgf=9.8$ newton (or $9.8N$ )

Illustration 1

The 2.5 kilogram brick of figure in above section is being pulled by a cord that makes an angle of ${20}^o$ with thehorizontal and has 7.0 N of tension in it. Find (a)the force of friction (b) the elastic recoil of the tabletop.

Solution : (a) On the horizontal axis, the friction must be equal to the horizontal component of the tension, so $F=(7.0N)(\cos {20}^{\circ })$

$F=6.6N$

(b) On the vertical axis, the downward force (weight) must equal the sum of upward forces, so $mg=E+T\sin \theta$

$E=mg-T\sin \theta$

$\begin{array}{rl}& E=(2.5kg)(9.8\frac{m}{s^2})-(7.0N)(\sin {20}^{\circ })\text{ [value are seen from trigonometric table] }\\ & E=24.5N-2.4N=22N\end{array}$

Illustration 2

A block of weight $5N$ is placed on a horizontal table. A person push the block from top by exerting a downward force of $3N$ on it. Find the force exerted by the table on the block.

Solution : There are three forces on the body:

(i) $5N$, downward by earth

(ii) $3N$, downward by the person

(iii) $F$, upward by the table

As the block is at rest, the resultant force on it must be zero. The total downward force is $5N+3N=8N$. Hence the upward force $F$ should be $8N$. So the force exerted by the table on the block is $8N$ in the upward direction.

NEWTON’S LAWS OF MOTION

Newton, Sir Isaac (1642-1727), mathematician and physicist, one of the foremost scientific intellects of all time. Born at Woolsthorpe, near Grantham in Lincolnshire, where he attended school, he entered Cambridge University in 1661; he was elected a Fellow of Trinity College in 1667, and Lucasian Professor of Mathematics in 1669 . He remained at the university, lecturing in most years, until 1696. Of these Cambridge years, in which Newton was at the height of his creative power, he singled out 1665-1666 (spent largely in Lincolnshire because of plague in Cambridge) as “the prime of my age for invention”. During two to three years of intense mental effort he prepared Philosophiae Naturalis Principia Mathematica (Mathematical Principles of Natural Philosophy commonly known as the Principia, although this was not published until 1687 .

Newton’s First Law of Motion

It gives qualitative definition of force. It states : An object continues in a state of rest or in a state of motion at a constant speed along a straight line, unless compelled to change that state by a net force. In other words, if a body is in a state of rest, it will remain in the state of rest and if it is in the state of motion, it will remain moving in the same direction with the same velocity unless an external force is applied on it.

Do you know !!

A common misconception about Newton’s first law is that a force is required to keep an object in motion. This is not so. Experiments done on air tracks (where there is a negligible friction) show that no force is required to keep an object moving with constant velocity. We get this misconception because friction is always present in our everyday lives

Do you know !!

To maintain constant velocity of a moving car; you have to push the accelerator. Does this contrudict Newton’s first law? No! It is seen clearly in the diagram on the right that the forward force on the car is equal in magnitude and opposite to the backward friction force (air friction plus ground friction) and so they cancel each other. resulting in a zero net force. Therefore, there is still no net force and the car moves away with constant velocity.

Inertia and mass

A greater net force is required to change the velocity of some objects than of others. For instance, a net force that is just enough to cause a bicycle to pick up speed will cause an imperceptible change in the motion of a freight train. In comparison to the bicycle, the train has a much greater tendency to remain at rest. Accordingly, we say that the train has more inertia than the bicycle. Quantitatively, the inertia of an object is measured by its mass. Inertia is the natural tendency of an object to remain at rest or in motion at a constant speed along a straight line. The mass of an object is a quantitative measure of inertia. The greater for the mass, the greater is the inertia of body. The definition of inertia and mass indicates why Newton’s first law is sometimes called the law of inertia.

Although the law of inertia was first clearly enunciated by Galileo, the English natural philosopher Issac Newton (1642-1727) incorporaled the law into the solid logical basis on which he founded the science of mechanics in his great 1687 work, commonly called Principia.

Do you know !!

According to Aristotle, an external force is required to keep a body in motion. His observation was based on common day experience. He was led to this wrong conclusion due to the reason that friction was not known in those days!

Examples :

(i) Coin drops into the glass when sudden force is applied on the cardboard.

(ii) Slow continue force breaks the upper string while if a sudden jerk is applied lower string breakout.

(iii) During the downward motion hammer’s sudden stop tighten the hammerhead.

(iv) Too little force, too little time to overcome “inertia” of tableware.

Galileo Galilei, bom in Pisa, Italy in 1564 was a key figure in the Scientific Revolution in Europe about four centuries ago. Galileo invented the concept of acceleration. From experiments on motion of bodies on inclined planes or falling freely, he contradicted the Aristotelian notion that a force was required to keep a body in motion, and that heavier hodies fall down faster under gravity. The law of inertia he thus arrived at was the starting point of the subsequent epochal work of Isaac Newton.

With Galileo came a turning point in the very method of scientific inquiry, Science was no longer merely observations of nature and logicul inferences from them. Science incant devising and doing experiments to verify or refute theories. Science meant measurement of quantities and a search for mathematical relations between them. Not undeservedly, many regard Galileo as the father of modern science.

The Law of Inertia

Galileo studied the motion of objects on an inclined plane. Objects moving down an inclined plane accelerate, while those going up the plane suffer retardation. Motion on a horizontal plane is an intermediate situation. Galileo concluded that at ubject moving on a horizontal plane must neither have acceleration nor retardation, i.e. it should move with constant velocity.

Another experiment of Galileo leading to the same conclusion involves a double inclined plane. A ball released from rest on one of the planes rolls down and climbs up the other. If the two planes are not very rough. the final height of the ball is nearly the same (a little less but never greater) as the initial height. In the ideal situation, when friction is completely eliminated, the final height of the ball should equal its initial height?

If now the slope of the second plane is decreased and the experiment repeated, the ball will still reach the same height, but in doing so it will travel a longer distance. In the limiting case, when the slope of the second plane is zero (i.e. it is a borizontal plane) the ball travels an infinite distance. In other words, its motion will never cease. This is, of course, an idealised situation. In practice, the ball does come to a stop after in motion continues to move with uniform velocity. This property of every object in nature is called inertia. Inertia means ‘resistance to change]. A body does not change its state of rest or of uniform motion, unless an external force compels it to change that state.

Types of Inertia

(I) Inertia of rest

The tendency of the body to continue in state of rest even when some external unbalanced force is applied on it is called inertia of rest.

Example 1 :

When a carpet is suddenly jerked the dust fly off, because due to the sudden moment the carpet moves but the dust on account of inertia of rest is left behind.

Example 2 :

The passenger standing in a bus tends to fall backwards when the bus suddenly starts, this is because his feet are in direct contact with the floor of the bus and the friction at the contact is high this friction does not allow the feet to slip on the floor, the feet therefore move forward with the floor the upper part of the body is still at rest for a while thus the passenger gets a jerk.

Example 3 :

On shaking a tree, the fruits fall down. The reason is that when the stem or branches are shaken, they come in motion while the fruits remain in the state of rest due to the inertia of rest. Thus the fruits get detached from the branches and fall down due to the pull of gravity.

(ii) Inertia of motion

The tendency of the body to continue in its state of motion even when some unbalanced force is applied on it is called the inertia of motion.

Example 1 :

It is dangerous to jump out of a moving vehicle (bus/train), because the jumping man, who is moving with the high speed of the vehicle would tend to move with the high speed of the vehicle. On reaching the ground his feet come to rest but upper part of the body continues to move with the speed of vehicle and the person falls forward on the ground. It is dangerous to jump out of a moving train and it is better to come out when it halts. However if in case of some emergency if some person wants to jump safely from a moving vehicle he should run for quite a while in the direction of motion of the vehicle after the jump so that his entire body remains in motion for sometime.

Example 2 :

When a running car stops suddenly, the passenger is jerked forward. The reason is that in a running a car, the whole body of passenger is in the state of motion. As the car stops suddenly, the lower part of his body being in contact with the car, comes to rest but his upper part remains in the state of motion due to the inertia of motion. Thus he gets jerked forward.

Example 3 :

In an event of long jump, an athlete runs fast before making the jump. Due to inertia of motion he is able to jump to a longer distance.

Example 4 :

A person riding a bicycle along a level road does not come to rest immediately after he stops pedaling. The bicycle continues to move forward due to the inertia of motion. But eventually it comes to rest as a result of the retarding action of friction etc.

Example 5 :

When we shake a wet piece of cloth, cloth as well as water in it comes to motion but when cloth comes to rest, the water is still moving due to the inertia of motion and falls forward on the front body of the person.

Example 6 :

When a person jumps out of a moving train, he falls down. This is so because inside the train, his complete body was in a state of motion with the train. On jumping out of the moving train as soon as his feet touch the ground, the lower part of his body being in contact with the car, comes to rest but his upper part remains in the state of motion due to the inertia of motion. Thus he gets jerked forward.

$✓$ CHECK POINT

Is it possible to have motion in the absence of a force? Is it possible to have force in the absence of motion?

Check Your Answer

Motion requires no force. Newton’s first law says an object in motion continues to move by itself in the absence of external forces. It is possible for forces to act on an abject with no resulting motion if the forces are balanced.

MOMENTUM

Most of us know from our experience of playing cricket that a cricket ball moving with a high speed could be fatal. Is it due to the velocity of the ball only? Had it been so, a dust particle moving with the same velocity would have been equally fatal. On the other hand, it could not be only due to the mass of the ball as we can safely hold a stationary (or a slowly moving) cricket ball in our hand without any fear. From this and many such examples we conclude that a quantity consisting the product of mass and velocity (mv) is of much significance. This quantity is so important in physics that it is given its own name and symbol. We call this quantity as momentum and represent it by symbol p.

Momentum is a measure of the quantity of motion in a body.

Example: Consider, a truck and a rickshaw moving with same velocities, heading towards each other and eventually ending up in a head on collision. Needless to say, that the rickshaw might get deformed to such an extent, that it would be difficult for us to make it, were it was a rickshaw before ! However, the truck might get some minor damages. Why is it so ? The first thing that comes to mind, is because the truck has more mass. Exactly! But in this context, it is more wise to say that, it all happens because the truck has more quantity of motion (i.e., momentum).

The momentum of a moving body is defined as the product of its mass and velocity. If we represent the mass and velocity of a body by $m$ and $\overline{v}$. respectively, then momentum is a vector quantity given by

$\overrightarrow{p}=m\overrightarrow{v}$

The direction of momentum of a body is same as that of its velocity. If only the magnitude is considered, then

$p=mv$

The SI unit of momentum is kilogram meter per second $(kgm/s)$.

Illustration 3

A ball of mass $500g$ is thrown with a velocity of $72km/hr$. What is its initial momentum ?

Solution : Given mass of the ball $=500g=\frac{1}{2}kg$ and velocity (initial) of the ball $=72km/hr=20m/s$

$\therefore$ Momentum of the ball $p=mv=\frac{1}{2}\times 20=10kgm/s$

Newton’s Second Law of motion

It states that the rate of change of momentum of a body is directly proportional to the applied unbalanced force.

Rate of change in momentum $\propto$ Force applied

Do you know !!

When a body is travelling at constant speed, the net force on it is zero.

If a body is moving with initial velocity $u$ and after applying a force $F$ on it. Its velocity becomes $v$ in time $t$.

Initial momentum of the body $p_1=mu$

Final momentum of body $p_2=mv$

Change in momentum in time $t=mv-mu$

So, rate of change of momentum $=\frac{m\nu -mu}{t}$

But according to Newton’s second law

$\frac{mv-mu}{t}\propto F$ or $F\propto \frac{m(v-u)}{t}$

Here $\frac{v-u}{t}=a$ (acceleration)

So $F\propto ma$ or $F=kma$, where $k$ is proportionality constant.

If $1N$ force is applied on a body of mass $1kg$ and the acceleration produced in the body is $1m/se{c}^2$ then $1=k\times 1\times 1$ or $k=1$

Hence, $F=ma$

So, the magnitude of the resultant force acting on a body is equal to the product of mass of the body and the acceleration produced. Direction of the force is same as that of the acceleration.

Newton’s First and Second Laws : The logical connection between them

Newton’s first law asserts that a body tends to move with constant velocity. Newton’s second law provides a quantitative measure of the force that will produce a given acceleration of the mass.

At first glance, it may seem that the Newton’s first law is not an independent law of nature at all, but merely the special case of Newton’s second law, $F=ma$, in which the net external force is zero. If $F=0$, we have $a=F/m=0$ and therefore $v=$ constant, as the first law states. Is this all there is to Newton’s first law ? No, because this interpretation of Newton’s first law is based on an incomplete statement of the law.

Do you Know !!

$F=$ ma equation was first written down by the Swiss mathematician Euler in 1747 after 20 years the death of Newton, to whom it is usually and falsely described, it was Euler not Newton, who first understood that this definition of force is useful in every case of motion.

Illustration 4

A force acts for $0.1s$ on a body of mass $1.2kg$ initially at rest. The force then ceases to act and the body moves through $2m$ in the next one second. Find the magnitude of force.

Solution : When force ceases to act the body shall move with the constant velocity since it moves a distance $2m$ in $1s$ therefore its uniform velocity $=2m{s}^{-1}$. Thus under the influence of force the body acquires a velocity $2m{s}^{-1}$ in $0.1s$.

i.e. $u=0,v=2m{s}^{-1}$ and $t=0.1s$

Acceleration $a=\frac{\text{ Change in velocity }}{\text{ Time }}=\frac{\nu -u}{t}=\frac{2-0}{0.1}=20m{s}^{-2}$

From the relation $F=ma$

Force $1.2\times 20=24N$.

Illustration 5

Calculate the force required to impart to a car a velocity of $30m/s$ in 10 seconds starting from rest. The mass of the car is $1500kg$.

Solution : Here, Mass, $m=1500kg$

Let us calculate the value of acceleration by using the first equation of motion.

Now, Initial velocity, $u=0$ (car starts from rest)

Final velocity, $v=30m/s$

And, Time taken, $t=10s$

Now, putting these values in the equation :

$v=u+at$

We get, $30=0+a\times 10$

$10a=30$ or $a=\frac{30}{10}m/s^2$

or acceleration, $a=3m/s^2$

Now, putting $m=1500kg$ and $a=3m/s^2$ in equation :

$F=m\times a$

We get, $F=1500\times 3N=4500N$

Thus, the force required in this case is of 4500 newtons.

Illustration 6

A motorcar is moving with a velocity of $108km/hr$ and it takes $4s$ to stop after the brakes are applied. Calculate the force exerted by the brakes on the motorcar if its mass along with the passengers is $1000kg$.

Solution : The initial velocity of the motorcar

$u=108km/hr=108\times 1000m/(60\times 60s)=30m{s}^{-1}$

and the final velocity of the motorcar $v=0m{s}^{-1}$

The total mass of the motorcar along with its passengers $=1000kg$ and the time taken to stop the motorcar, $t=4s$.

We have the magnitude of the force $(F)$ applied by the brakes as $\frac{m(v-u)}{t}$

On substituting the values, we get

$F=1000kg\times (0-30)m{s}^{-1}/4s$

$F=-7500kgm{s}^{-2}\text{ or }-7500N$

The negative sign tells us that the force exerted by the brakes is opposite to the direction of motion of the motorcar.

Illustration 7

A force of $20N$ acts on a body of mass $5kg$ for $5sec$. Find

(i) the acceleration of the body’

(ii) velocity at the end of $5sec$, and

(iii) displacement at the end of $5sec$.

Solution : Given : $F=20N,m=5kg$ and $t=5sec$

From $\overrightarrow{F}=m\overrightarrow{a}\Rightarrow 20=5\times a$

(i) Thus, acceleration $a=4m/s^2$

(ii) $u=0,a=4,t=5,v=$ ?

Using $v=u+at$

$v=0+5\times 4=20m/s$

(iii) $u=0,a=4,t=5,s=$ ?

From $s=ut+\frac{1}{2}a{t}^2=0+\frac{1}{2}\times 4\times 25=50m$

Illustration 8

A ball of mass $10g$ is initially moving with a velocity of $50m{s}^{-1}$. On applying a constant force on ball for $2.0s$, it acquires a velocity of $70m{s}^{-1}$. Calculate :

(i) the initial momentum of ball

(iii) the rate of change of momentum

Solution : Given : $m=10g=\frac{10}{1000}kg=0.01kg$

$u=50m{s}^{-1},t=2.0s,v=70m{s}^{-1}$.

(i) Initial momentum of ball $=$ mass $\times$ initial velocity $=mu=0.01kg\times 50m{s}^{-1}=0.5kgm{s}^{-1}$.

(ii) Final momentum of the ball $=$ mass $\times$ final velocity $=mv=0.01kg\times 70m{s}^{-1}=0.7kgm{s}^{-1}$.

(iii) Rate of change of momentum $=\frac{\text{ Final momentum }-\text{ Initial momentum }}{\text{ Time interval }}=\frac{(0.7-0.5)kgm{s}^{-1}}{2.0s}=0.1kgm{s}^{-2}$ (or $0.1N$ )

(iv) Acceleration $a=\frac{v-u}{t}=\frac{70-50}{2}=10m{s}^{-2}$

(v) Force $=$ mass $\times$ acceleration $=ma=0.01kg\times 10m{s}^{-2}=0.1N$

Newton’s Third Law Of Motion

Newton’s first two laws of motion describe what happens to a single object that has forces acting on it while Newton’s third law deals with the relationship between the forces objects exert on each other.

If we push on a wall it pushes back. This doesn’t hurt if you push gently, but if you punch a wall hard it hurts very much. Newton hypothesized that any time two objects interact in such a way that a force is exerted on one of them, there is always a force that is equal in magnitude exerted in the opposite direction on the other object. This hypothesis is called Newton’s third law.

Newton’s third law of motion states that ‘if a body $A$ exerts a force $+F$ on a body $B$, then body $B$ exerts a force $-F$ on $A$, that is a force of the same size and along the same line of interaction but in the opposite direction’. This law says that forces always occur in pairs as the result of the interaction between two objects. Note that two objects are involved and the two forces each act on a different object. The two opposing forces are sometimes called the action and reaction forces.

Motor cars are able to move along a road because the reaction of the road pushes the car along in response to the action of the wheels pushing on the road.

(i) A swimmer pushes (or applies force) the water with his hands and feet to move in the forward direction in water. It is the reaction to this force that pushes the swimmer forward.

(ii) The propellers of an aeroplane pushes the air backwards and the forward reaction of the air makes the aeroplane move forward.

(iii) When a bullet is fired from a gun, the force sending the bullet forward is equal to the force sending the gun backward. But due to the high mass of the gun, it moves only a little distance backward and gives a kick to the shoulder of the gunman. The gun is said to have recoiled.

(iv) Consider, a horse cart being driven by a horse. The horse pulls forward, the cart with a force $F_{CH}$. The cart also pulls the horse with an equal force $F_{HC}$, but in opposite direction. But then how does the complete system (horse + cart) moves forward ‘? The reason is that there are yet some other forces active on the system, which lend to the movement of the horse-cart system.

The forward thrust, $F_G$ offered by the frictional force between ground and horse and the frictional force acting backwards, between ground and cart $F_B$.

Thus, the net force on the system, acting forwards is given by

Now, if $F_G>F_B$, then $F>0$

$\begin{array}{rl}& F=F_G-F_{HC}+F_{CH}-F_B\\ & F=F_G-F_B[\therefore F_{HC}=F_{CH}]\end{array}$

and hence the system would start moving with an acceleration

$a=\frac{F}{m}=\frac{F_G-F_B}{m}$

where $m$ is the mass of the system.

But if $F_G<F_B$, no motion is possible

(v) Action and reaction are equal in magnitude but both act on the body in opposite direction.

For an orbiting satellite action of earth on the satellite is the force exerted on the satellite by the gravitational pull of the earth. $({\overline{F}}_{SE})$

Reaction of the satellite on the earth. It is the gravitational pull of the satellite on the earth. $({\overrightarrow{F}}_{ES})$

(vi) Hammer pushes on stake then stake pushes on hammer and the hammer acts, the stake re-acts.

Do you know !!

The paired forces (called action and reaction) always act on different bodies. There is no way one of them can balance the other one!

(vii) When car moves tyre pushes the road and road pushes on the tyre as a reaction force.

Misconcept

Consider a book on the table. The table pushes the book upward i.e. normal force. Student feels weight and normal force are action and reaction but they aren’t action and reaction. Reason : The book does not fly up because there is another force on the book pulling it downward this is the force exerted by the earth i.e. the weight of the book as book does not accelerate so, we conclude that the forces acting on the book are balanced but they are not action and reaction pair because the action and reaction acts on different bodies.

There is another misconception concerning the third law. The third law states that the two forces Weight example, an egg and a stone collide with each other, the egg breaks and the stone is intact.

Since the egg breaks, we often conclude that the force by the stone on the egg is greater than the force by the egg on the same. This is not so. The forces are always equal. The egg breaks because it is simply easier to break. It takes a smaller force to break the egg then to break the stone.

Newton’s Third Law from Newton’s Law

Let us consider two bodies 1 and 2. Let $F_{21}$ be the force exerted by secoed body on fins and $F_{12}$ the farce ecerted by the finity one second.

Extermal force $=0$

$\therefore {\overline{F}}_{2}1+{\overline{F}}_{1}2=0\Rightarrow {\overline{F}}_{2}1=-{\overline{F}}_{1}2$

Experiment : The ring of a spring balance $B$ is attached to the hook fixed in a wall and then the hook of acthr spring bakmoce $A$ is whached to the hook of the spring balance $B$. Now the ring of balance $A$ is pulled. We find that both the bahaces represent fine sune reating. The reason is that the balance $A$ pulls the balance $B$ due to which we gat some reading in $B$. Buat se sen reding in $A$ shows that the balance $B$ also pulls the balance $A$ by the same force of reaction. This concludes the 70 every activen, thene is $=$ eqpal and opposite reaction” (i.e., $F_{AB}=F_{BA}$ but in opposite direction).

Fig. : Demonstration of action and reaction

Guidelines to solve problem based on NLM

(i) Draw a free-body diagram for each object involved in the analysis.

(ii) Select a rectangular coordinate system. The solutions will be much easier if you seloct the $+x-axis$ in the dircation of mexeliersint and the $+y$-axis perpendicular to the $x$-axis.

(iii) Resolve all forces not pointing in the $x$ or $y$ directions to their $x$ and $y$ components, respectively.

(iv) Add, algebraically, all the $x$ components and $y$ components of the forces, respectively.

(v) Set $\mathrm{\Sigma }{F}_x=$ ma and $\mathrm{\Sigma }{F}_y=0$ and solve for the unknown quantities.

Since you have choosen the $+x$-axis in the direction of acceleration, the objoct will not socelarate in the y-diroctice So is acceleration in the $y$ direction is zero and $\mathrm{\Sigma }{F}_y=m{a}_y=0$.

$✓$ CHECK POINT

• Mayank strongman and some small pull on opposite ends of rope in a tag of war. Who exerts the greater force on the rope.

Check your answer

Both of them would exert the same force on the rope. Mayank can pull no harder on the rope than Sonu. Rope tension is the same all along the rope, including the ends. Just as a wheel on ice can exert no more force on the ice than the ice exerts on the wheel and just as one cannot punch an empty paper bag with any more force than the bag can exert on the puncher. Mayank can exert no more force on his end of the rope than Sonu exerts on his end.

Do you know !!

Atwood’s machine gives us a direct way of demonstrating Newton’s laws of motion in the laboratory.

Illustration 9

A water skier is towed by a motorboat at a constant velocity of magnitude $15km/h$. The boat speeds up, and after a short interval the skier is towed at a new constant velocity of magnitude $20km/h$. What is the net force on the skier when she is moving at $15km/h$ ? at $20km/h$ ?

Solution : Two major forces act on the water skier, one of these forces is that exerted on her hands by the towrope. The other is the resistance of the water (and to a lesser extent the air). When the skier is moving in a straight line at a constant $15km/h$, her velocity is constant. According to Newton’s first law, the net force on the skier (and the skis) must be zero. Indeed, the law itself is the basis for asserting that the force exerted by the water and the air on the skier is exactly equal in magnitude, and opposite in direction, to that exerted on her by the towrope.

When the skier is moving at $20km/h$, the force exerted on her by the towrope is greater in magnitude than that at $15km/h$.

But so is the resistive force. Again, the net force on the skier must be zero because her velocity is constant.

Illustration 10

What thrust is needed to fire a 350-kilogram rocket straight up with an acceleration of 8.0 meters per second squared?

Solution : The net force needed to produce this acceleration is

$F=ma=(350kg)(8.0m/s^2)=2800N$

However, there is a downward force acting on it as well, equal to

$W=mg=(350kg)(9.8m/s^2)=3430N$

The net force is the thrust minus the weight, or

thrust $-3430=2800N$

so the rocket engine must product a thrust of $6230N$.

Illustration 11

A pile driver of mass $150kg$. falls from a height of $5m$ above the pile and is brought to rest in $0.5s$. Ignoring the motion of the pile, calculate the average force exerted on the pile.

Solution : $m=150kg,h=5m,v_1=?,t=0.5s,v_1=0m/s,u=0m/s$

$v^2=u^2+2gh$

$v^2=2\times 10\times 5=100$

$v=10m/s$

$Ft=mv1-mv$

$F\times 0.5=0-150\times 10$

$F=-\frac{1500}{0.5}=-3000N$

$F$ is the force exerted by the pile on the driver. Thus the force the driver exerts on the pile is $+3000N$.

Illustrotion 12

A force of $20N$ acting on a mass $m_1$, produces an acceleration of $4m/se{c}^2$. The same force is applied on mass $m_2$ then the acceleration produced is $0.5m/se{c}^2$. What acceleration would the same force produce, when both masses are tied together.

Solution : For mass $m_1:F=20N,a=4m/se{c}^2$

then

$m_1=\frac{F}{a}=\frac{20}{4}=5kg$

For mass $m_2:F=20N,a=0.5m/se{c}^2$

then

$m_2=\frac{F}{a}=\frac{20}{0.5}=40kg$

When $m_1$ and $m_2$ tied together :

$\text{ mass }=m_1+m_2=45kg,F=20N$

then

$a=\frac{F}{(m_1+m_2)}=\frac{20}{45}=0.44m/se{c}^2$

Free Body Diagram

To study the motion of an individual body of a system, a diagram is drawn which represents different types of interactions with surrounding in terms of forces. This diagram is said to be the Free Body Diagram (FBD) of that body. To draw the free body diagram, the object is first made isolated from its surroundings and then all forces acting on it are represented by \tos $(arrow)$. To make it clear, let us see the following examples.

Illustration 13

Two blocks of mass ${\mathit{m}}_A$ and ${\mathit{m}}_B$ are arranged in the diagram as shown in Fig. Draw the free body diagram of

1. Block ${\mathbf{m}}_A$
2. Block ${\mathbf{m}}_B$
3. Pulley 1
4. Pulley 2

Solution :

Illustration 14

Three blocks $A,B$ and $C$ of masses $m_A,m_B$ and $m_C$ respectively are arranged in the diagram as shown in Fig. Draw the free body diagram of

1. Block A
2. Block B
3. Block C
4. Pulley

Solution :

Illustration 15

Find acceleration and contact force between the two bodies

Solution:

Illustration 16

Find acceleration of each mass and tension in the string ( $g=10m/s_2$ )

Solution :

(Both masses will have the same acceleration as string is inelastic)

$T=10a$ … (i)

$5g-T=5a$ … (ii)

From (i) and (ii) $5g=10a+5a$

$a=\frac{5g}{15}=\frac{g}{3}=\frac{10}{3}m/s^2$

$T=10\times \frac{10}{3}=\frac{100}{3}N$

NOTE : For this system, $T=(\frac{m_1{m}_{2}g}{m_1+m_2})a=(\frac{m_{2}g}{m_1+m_2})$

Illustration 17

Draw free-body diagram of each body in the system

Solution :

Illustration 18

Find acceleration of masses and tension in the strings, pulley is massless and frictionless and strings are massless and inelastic.

Solution :

$T_3=2T_2$ … (i)

$T_2-10g=10a$ … (ii)

$T_1+15g-T_2=15a$ … (iii)

$5g-T_1=5a$ … (iv)

Adding (ii), (iii) and (v)

$30a=10g{T}_1=5g-5\times \frac{g}{3}$

$a=\frac{g}{3},T_1=\frac{10g}{3}$

$T_2=10g+10\times \frac{g}{3}=\frac{40g}{3}$ and $T_3=\frac{80g}{3}$

Illustration 19

Draw F.B.D. of the system shown

Solution :

Impulse

From Newton’s second law, $\overrightarrow{F}=\frac{\mathrm{\Delta }\overline{p}}{\mathrm{\Delta }t}$ we can derive the impulse-momentum theorem. This theorem states that impulse is equal to the change in momentum, or $\overrightarrow{F}\mathrm{\Delta }t=\mathrm{\Delta }\overrightarrow{p}=\overrightarrow{p}-{\overrightarrow{p}}_0$ where $\overrightarrow{F}\mathrm{\Delta }t$ is called impulse, $\overrightarrow{F}$ is the average force and $\mathrm{\Delta }t$ is the time interval the force is in action).

Impulse-momentum is very useful in explaining some everyday phenomena like a cricket player while taking a catch always moves his hands backward, because the total change in momentum $(mv-mu)$ remains constant for a moving ball. The change in momentum is numerically equal to $F\times t$. Here $F$ is force applied by the hands for time $t$. Now the player moves his hands backward so that the value of $t$ will increase. With the result the value of applied force $F$ will decrease.

Do you know !!

The change in momentum of a body depends on the magnitude and direction of the applied force and the period of time over which it is applied; i.e. it depends on its impulse.

Thus the hands of the player are not hurt while taking a catch. Similarly the idea of the impulse of a force is important when considering a hammer driving a nail into a block of wood, a boy kicking a football and a girl striking a hockey ball.

Example :

(i) A girl standing at rest with her right foot on a skateboard thrusts backwards on the ground with her left foot as Figure. The push of the ground on her left foot will increase the forward momentum of the girl and skateboard. From $Ft=mv-0$, the velocity $v$ acquired by the girl will depend on the force $F$ which can be applied and the time $t$ for which it acts.

(ii) A body throwing a javelin is shown in figure. For a javelin of a given mass, the distance it will travel depends upon the force exerted by the boy’s arm and the time for which it is exerted.

(iii) During the collision of a truck with the wall, wall exerts great impulse on truck

Illustration 20

A 650-kilogram rocket is to be speeded up from 440 meters per second to 520 meters per second in outer space. If the thruet of the engine is 1200 newtons, for how long must the engine be fired ?

Solution : The change in the momentum of the rocket is $(650kg)(520m/s)-(650kg)(440m/s)=52000kg\cdot m/s$. This must be equal to the impulse, so $(1200N)(\mathrm{\Delta }t)=52000kg\cdot m/s$

$\mathrm{\Delta }t=43s$

IIlustration 21

A 0.10 kilogram ball is dropped onto a table top. The speeds of the ball right before hitting the table top and right after hitthy the table top are $5.0m/s$ and $4.0m/s$, respectively. If the collision between the ball and the tabletop lasts $0.15s$, what in the average force exerted on the ball by the table top ?

Solution : Given : $m=0.10kg,v_0=-5.0m/s$ (downward), $v=+4.0m/s$ (upward), $\mathrm{\Delta }t=0.15s$. Find $\overrightarrow{F}$.

The velocity of the ball before the collision is downward and the velocity of the ball after the collision is upward. Since these two velocities are opposite, we have to assign signs (+ or - ) to them to find the change in velocity, $\mathrm{\Delta }v$, properly. Conventionally, we choose the upward direction as positive so the downward direction is negative.

From impulse-momentum theorem, we have $\overrightarrow{F}\mathrm{\Delta }t=\mathrm{\Delta }\overrightarrow{p}=\overrightarrow{p}-{\overrightarrow{p}}_0$

So, $\overrightarrow{F}=\frac{\overrightarrow{p}-{\overrightarrow{p}}_0}{\mathrm{\Delta }t}=\frac{mv-m{v}_0}{\mathrm{\Delta }t}=\frac{(0.10kg)(4.0m/s)-(0.10kg)(-5.0m/s)}{0.15s}=+6.0N$

The positive sign indicates that the force on the ball by the tabletop is upward, which makes sense.

Illustration 22

A boy of mass $58kg$ jumps with a horizontal velocity of $3m{s}^{-1}$ onto stationary skateboard of mass $2kg$. What is his velocity as he moves off on the skateboard?

Solution : Assume there is zero unbalanced horizontal force in the horizontal direction and that left to right is the positive (+) direction.

Momentum before interaction $=58\times 3+2\times 0$

Momentum after interaction ${}^{\circ }=58\times v+2\times v=60v$

Equating momenta $60v=58\times 3$

$v=\frac{58\times 3}{60}=+2.9m{s}^{-1}\text{. }$

Conservation of Momentum

The greatest value of the concept of momentum is in the calculation of what happens when two or more objects interact. It turns out that we can often find out the results of a collision, for example, without knowing anything at all about the for interact. It turns out long they persist.

The explanation is this : Think of two objects, $A$ and $B$, colliding. Object $A$ exerts a force $F$ on object $B$, and $B$ exerts a force $-F$ on $A$. The negative sing indicates that the two forces are in opposite directions. They have equal magnitudes. Since the duration of the impact, $\mathrm{\Delta }t$, is the same for both, the impulse that $B$ exerts on $A$ is the negative of the impulse $A$ exerts on $B$.

Now, impulse is equal to change in momentum. It follows that, in the collision, the change in the momentum of $A$ is the negative of the change in the momentum of $B$. The sum of the two changes is thus zero. In other words, during the collision, the total momentum does not change! The increase in the momentum of the other. This gives us the very fundamental and important law of nature called the law of conservation of momentum: In an isolated system, the total momentum does not change.

Think of using this rule whenever dealing with an interaction between two objects. For example, a little girl is standing on a wagon, and jumps off to the rear of it. To do so, she has to kick the wagon so that it moves forward. Before she jumped, the total momentum of the system was zero, so it must continue to forward.

In other words, the principle of conservation of momentum states that “if there is a direction in which there is zero unbalanced force acting on a system then the total momentum of that system in that direction is constant even if the bodies act on each other’.

Note that other forces, e.g. the pull of the Earth, do act on the bodies, but the result can still be used if there is a direction in which the external forces are balanced.

The law of conservation of Linear momentum by third law of motion :

Suppose two objects $A$ and $B$ of mass $m_1$ and $m_2$ are moving in the same direction with velocity $u_1$ and $u_2$ respectively $(u_1>u_2)$. Object $A$ collides with object $B$ and after time $t$ both moves in the original direction with velocity $v_1$ and $v_2$ respectively.

The change in momentum of object $A$ is $m_1{v}_1-m_1{u}_1$

The force on $B$ by $A$ is

$F_1=\frac{\text{ Change in momentum }}{\text{ time }}$

$F_1=\frac{m_1{v}_1-m_1{u}_1}{t}$ … (i)

IDEA BOX

The change in momentum of object $B$ is $m_2{v}_2-m_2{u}_2$

The force on $A$ by $B$ is

$F_2=\frac{\text{ Change in momentum }}{\text{ time }}$

$F_2=\frac{m_2{v}_2-m_2{u}_2}{t}$ … (2)

By Newton’s third law

$\begin{array}{rl}& F_1=-F_2\\ & \frac{m_1{v}_1-m_1{u}_1}{t}=-(\frac{m_2{v}_2-m_2{u}_2}{t})\\ \Rightarrow & m_1{v}_1-m_1{u}_1=m_2{u}_2-m_2{v}_2\\ \text{ or }& m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2\end{array}$

$\text{ or Initial momentum }=\text{ Final momentum }$

Example :

(1) A bomb shell initially at rest is shown in figure. Now the bomb shell suddenly explodes in two fragments having masses in the ratio $2:1$. As there is no external force involved in the process of explosion the momentum of the system should be conserved. In this case as the bomb is initially at rest the initial momentum of system is zero. The conservation of momentum demands that the final momentum should also be zero. The two fragments should carry equal and opposite momentum to make total momentum zero. Thus the two fragments will move along the same straight line but opposite in direction.

Let the masses of the two fragments be $2m$ and $m$ and their velocities be $v_1$ and $v_2$. According to the law of conservation of momentum. $2(m)v_1+m(v_2)=0$

or $v_2=-2v_1$ [in magnitude $v_2=2v_1$ ]

Thus the smaller fragment of mass $m$ will move with double the speed of larger fragment i.e. with a velocity of magnitude $2v_1$ in a direction opposite to that of heavier fragment of mass $2m$.

(2) Recoil of the Gun : Initially, before a bullet is fired from the gun, both, the bullet and the gun are at rest. So, before a bullet is fired, the initial momentum of the system (gun + bullet) is zero.

Now, when a bullet is fired it leaves the barrel with some velocity, that is it has some momentum. According to the law of conservation of momentum, the final momentum should be equal to the initial momentum which is in this case is zero. The final momentum could be zero only if the gun has a momentum equal and opposite to that of the bullet. The gun is said to be recoiled. Although their momenta are equal but as the mass of the gun is much higher than that of the bullet the recoil velocity of the gun is much smaller compared to the velocity of the bullet.

(3) Jet Engines and Rockets : The working of jet engines and rockets can also be described on the basis of the law of conservation of momentum. A rocket standing at the launching pad has produced.

Theses gases pass out through the tail nozzle of the rocket in downward direction in the form of jet with tremendous velocity. Therefore the rocket moves up with such a velocity so as to make the momentum of the system (rocket + emitted gases) zero again.

Think it Overt

Tie the mouth of a balloon tightly to a small piece of discarded ballpoint refill or a plastic tube having a narrow bore and inflate it. Close the opening of the tube with your finger to prevent the air to escape (fig. (a)). Now let the air escape from the opening of the tube (fig. (b)). In which direction does the balloon move? Why does it move in a direction opposite to the direction in which the air escapes?

In doing this activity you may also fix the inflated balloon on the top of (fig. c). You will notice that the toy car moves in the direction opposite to the direction a trolley before you let the air escape principle involved in the working of jet engines and rookets. In jet engines, a large volume which the air escapes. This is the basic of fuel is allowed to escape through a jet and as a result, it moves the forward.

Illustration 23

A bullet of mass $\mathit{m}$ is fired from a gun of mass $M$ with a horizontal velocity of $v$. What is the recoil velocity of the gun ?

Solution : Choose a suitable direction (the horizontal direction) in which the effect of the external forces is zero. Choose a sign convention for the direction of the velocities : here left to right is positive ( + ). Let $v_R$ be the Momentum before gun is fired $=m\times 0+M\times 0=0$ (gun at rest)

Momentum after gun is fired $=m\times (+v)+M\times (-v_R)$

Momentum before firing $=$ momentum after firing

$\begin{array}{rl}0& =mv-M{v}_R\\ M{v}_R& =mv\\ v_R& =\frac{mv}{M}\end{array}$

In hand-held guns the recoil can be absorbed by the person holding the gun. In very large guns which fire massive shells at high velocity the recoil could rip the gun from its mounting unless special provision is made to absorb the force.

Illustration 24

A 1.2-kilogram basketball traveling at 7.5 meters per second hits the back of a 12-kilogram wagon and bounces off at 3.8 meters per second, sending the wagon off in the original direction of travel of the ball. How fast is the wagon going?

Solution :

The momentum of the ball before it hit has to equal the sum of the momenta of the two objects after the collision. Let’s call the original direction of travel of the ball positive. Then the equation becomes

$(1.2)kg(7.5m/s)=(1.2kg)(-3.8m/s)+(12kg)(v)$

$1.1m/s=v$

Illustration 25

Two particles of mass $200g$ and $500g$ are released from rest at some mutual separation. If the velocity of the smaller mass be $2.5m{s}^{-1}$ at any instant, then what is the velocity of the larger mass.

Solution : Since the particles are released from rest, so initial momentum of the system of particles is zero. Further, since the interaction between the particles is mutual, so the forces acting on them are purely internal, and hence the momentum of the system remains constant (and is zero) with time.

$\begin{array}{cc}\therefore & \overrightarrow{p}=m_1{\overrightarrow{v}}_1+m_2{\overrightarrow{v}}_2=0\\ \Rightarrow & 200\times {10}^{-3}\times 2.5+500\times {10}^{-3}{\overrightarrow{v}}_2=0\\ & {\overrightarrow{v}}_2=-1.0m{s}^{-1}\end{array}$

The negative sign indicates that the direction of velocity of larger mass is opposite to that of the smaller mass.

More About Friction

We studied back that friction is a resistance to the relative motion between two objects in contact (in case of solid objects) or the body and its surroundings (in case object is moving in a fluid). Actually, when two objects are kept in contact, a reaction force $R$ acts between the two objects as shown in the figure. This reaction

force $R$ has two components $-f$, along the surface and $N$, perpendicular to the surface. The force $f$ which acts along the surface is called the force of friction.

It is not appreciated that the existence of friction between surfaces does not depend on their roughness or smoothness in the everyday sense of these words. In fact there can be very large frictional force between two highly polished flat metal surfaces. Consequently, it is important, in mechanics, to interpret smooth as frictionless rather than free from projections.

The results of experimental investigation into the behaviour of frictional forces confirm that:

(a) frictional forces opposes the movement of an object across the surface of another with which it is in rough contact.

(b) the direction of the frictional force is opposite to the potential direction of motion.

(c) the magnitude of the frictional force is only just sufficient to prevent movement and increases as the tendency to move increases, up to a limiting value. When the limiting value is reached, the frictional force cannot increase any further and motion is about to begin (limiting equilibrium). When the frictional force $F$ reaches its limit, its value then is related to the normal reaction $N$ in the following way:

$F=\mu N$

The constant $\mu$ is called the coefficient of friction and each pair of surfaces has its own value for this constant. In limiting equilibrium $F=\mu N$

In general $F\le \mu N$

Do you know

The atomic and molecular forces of attraction between the two surfaces at the point of contact give rise to friction between the surfaces.

Static Frictional Force

When there is no relative motion between the contact surfaces, frictional force is called static frictional force. It is a self-adjusting force, it adjusts its value according to requirement (of no relative motion).

In the taken example static frictional force is equal to applied force. Hence one can say that the portion of graph $ab$ will have a slope of ${45}^{\circ }$. $(f_s\le {\mu }_{s}N)$

Limiting Frictional Force

This frictional force acts when body is about to move. This is the maximum frictional force that can exist at the contact surface. We calculate its value using laws of friction.

Kinetic Frictional Force

Once relative motion starts between the surfaces in contact, the frictional force is called as kinetic frictional force. The magnitudeof kinetic frictional force is also proportional to normal force.

$f_k={\mu }_{k}N$

From the previous observation we can say that ${\mu }_k<{\mu }_s$. Although the coefficient of kinetic friction varies with speed, we shall neglect any variation i.e., one relative motion starts a constant frictional force starts opposing its motion.

Angle of friction

At a point of rough contact, where slipping is about to occur, the two forces acting on each object are the normal reaction $N$ andfrictional force $\mu N$.

IDEA BOX

The friction between two surfaces increases (rather than to decrease), when the surfaces are made highly smooth.

The resultant of these two forces is $S$, and $S$ makes an angle $\lambda$ with the normal, where

$\tan \lambda =\frac{\mu N}{N}=\mu$

The angle $\lambda$ is called the angle of friction.

At a point of rough contact when slipping is about to occur we can, therefore, use either.

components $N$ and $\mu N$ at right angles to each other

$S$ at angle $\lambda$ to the normal where $S$ is the resultant contact force or total reaction and $\lambda$ is the angle of friction.

The use of $S$ instead of $N$ and $\mu N$ reduces the number of forces in a problem and can often lead to a three force problem.

1. When the surfaces of two objects in rough contact tend to move relative to each other, equal and opposite frictional forces act on the objects, opposing the potential movement.
2. Up to a limiting value, the magnitude of a frictional force, $F$, is just sufficient to prevent motion.
3. When the limit is reached $F=\mu N$ where $N$ is the normal reaction and $\mu$ is the coefficient of friction for the two surfaces in contact.
4. At all times $F\le \mu N$
5. The resultant of $N$ and $\mu N$ makes an angle $\lambda$ with the normal, where $\tan \lambda =\mu$ and $\lambda$ is the angle of friction.

Angle of Repose

Suppose a body is kept above a rough inclined surface, whose angle of inclination is slowly increased such that at angle ’ $\lambda$ ’ it is about to slide down, then this angle is called angle of repose.

$f=mg\sin \lambda Butf=\mu N,N=mg\cos \lambda$

So, $\mu mg\cos \lambda =mg\sin \lambda$

$\mu =\tan \lambda$ Angle of Repose $=$ Angle of friction

In case $\theta >\lambda$, body will slide down and its acceleration can be found as under

$\begin{array}{rl}& N=mg\cos \theta \\ & f=\mu N=\mu mg\cos \theta \text{ but }mg\sin \theta -f=ma\end{array}$

so $a=g\sin \theta -\mu g\cos \theta$

If body is projected upward on the inclined plane, then $a=-(g\sin \theta +\mu g\cos \theta )$

Illustration 26

Find acceleration and friction force for the body as shown in figure ( $g=10m/s$ )

Solution : Applied force, $F=20N$

Limiting force of friction $=\mu N=0.5\times 10\times 10=50N$

$(N=mg)$

As the applied force is less than the limiting force of friction the body will not accelerate and hence, force of friction will be equal to the applied force

So, $a=0f=20N$

Illustration 27

In Ilustration-26 if the applied force is $100N$, find acceleration of the block. $(g=10m/s^2)$

Solution : $\mu =0.5,F=\mu N:N=mg$

so, $f=0.5\times 10\times 10=50N$

$100-f=10a(\mathrm{\Sigma }F=ma)$

$100-50=10a\Rightarrow a=5m/s^2$

Illustration 28

When a force of $50N$ is applied on a body of mass $10kg$ it is about to move and once the body is set into motion the same force of $50N$ produces an acceleration of $0.5m/s^2$. Find coefficient of static and kinetic friction. $(g=10m/s^2)$

Solution : As on the application of $50N$ force, body does not move the applied force is equal to the limiting force of friction

$\begin{array}{rl}& f=50N\\ & {\mu }_{s}N=50,N=10\times 10=100N\\ & {\mu }_s=0.5\end{array}$

When the body is set into motion then kinetic friction will come into play, that is why the same force will produce accelerativil

$\begin{array}{rl}& 50-{\mu }_k\times 100=10\times 0.1\\ & 50-1={\mu }_k\times 100\\ & {\mu }_k=0.49\end{array}$

Illustration 29

Find acceleration of masses and tension in the string for the system shown $(g=10m/s^2)$.

Solution :

$N=10g;f=\mu N$

So, $T-f=10a\text{ or }T-0.4\times 100=10a$

$T-40=10a$ …(i)

and $50g-T=50a$ …(ii)

$(i)+(ii)500-40=60a\Rightarrow a=7.67m/s^2$

$T=116.67$

Illustration 30

Find acceleration of masses and tension in the string for the system shown in fig $(g=10m/s^2)$.

Solution : Drawing F.B.D of two masses

$N=mg\cos \theta$ & $f=\mu N$

$100g\sin {30}^{\circ }-T-f=100a$

$100\times 10\times \frac{1}{2}-T-0.2\times 100\times 10\times \frac{\sqrt{3}}{2}=100a$

$10g-T=10a$

$\text{ (i) }+\text{ (ii) }$

$500+100-173=110a\Rightarrow a=3.88m/s^2$

$\text{ and }T=10g-10\times 3.88$

$T=61.2N$

Illustrotion 31

Find maximum force applied on the lower mass so that the two masses move together, for the system of masses shown.

Solution : In order to find the direction of friction on the mass $m$ let us initially assume that friction is absent, when lower mass is pulled by $F$ the upper mass will fall backwards hence friction the upper mass has to be in forward direction.

(As friction on the upper body is in forward direction so on the lower body it will be in backward direction i.e. action -reaction pair)

$R=$ is reaction force of the ground on $M$

$N=$ Normal reaction on $m$ by $M$ and on $M$ by $m$ i.e action-reaction pair $f=\mu N:N=mg$ $f=\mu mg\dots$ (1) and $f=ma$

$\begin{array}{rl}& \text{ Also }F-f=Ma\Rightarrow F-ma=Ma\Rightarrow a=\frac{F}{M+m}\\ & F=f+ma\\ & F=\mu mg+M\frac{F}{M+m}\Rightarrow F-\frac{MF}{m+M}=\mu mg\Rightarrow F=\mu g(M+m)\end{array}$

Illustration 32

In illustration-31 if force is applied on the upper mass then what will be your answer.

Solution :

As the two masses are moving together so

$a=\frac{F}{M+m}$

(As the upper mass is likely to slide over the lower mass in forward direction, friction will be in backward direction)

$\begin{array}{rl}& f=\mu N=\mu mg\\ & F-f=ma\Rightarrow F-\mu mg=m\cdot \frac{F}{M+m}\\ & F(1-\frac{m}{M+m})=\mu mgF=\frac{\mu mg}{M}(M+m)\end{array}$

Illustration 33

Explain why it is easier to pull than push a box on a rough surface

Solution : Pushing a box

Along $y$-axis

$N=mg+F\cos \theta$

$f=\mu N=\mu (mg+F\cos \theta )$

Pulling a box

Along $y$-axis, $N+F\cos \theta =mg;N=mg-F\cos \theta$

So, $\mathit{f}=\mathit{\mu }(\mathit{m}\mathit{g}-F\cos \theta )$…(ii) It is quite evident from equation (i) and (ii) that in pushing friction force is greater due to larger normal reaction that is why pulling is easier than pushing. But if friction is absent pulling and pushing would require the same effort.

Illustration 34

A mass of $5kg$ having initial velocity of $20m/s$ comes to rest after travelling a distance of $200m$. Find coefficient of friction and friction force acting on the body:

Solution :

$u=20m/s:v=0;s=200m$

Using $v^2=u^2+2as$

$0=(20{)}^2+2a\times 200\Rightarrow a=-1m/s^2$

But $f=ma;f=\mu N=\mu mg$

$\mu mg=ma,a=\mu g$

So, $\mu \times 10=1\Rightarrow \mu =0.1$

Friction force, $f=0.1\times 5\times 10=5N$

Illustration 35

A chain of mass ’ $m$ ’ and length ’ $r$ ’ is kept on a rough table such that $1/n^{\text{th }}$ part is overhanging. If the chain is at rest find coefficient of friction.

Solution :

From F.B.D

$f=T;T=\frac{m}{n}g$ (As chain is equilibrium)

$\mu N=T$

$\mu m(1-\frac{1}{n})g=\frac{m}{n}g$

$\mu =\frac{1}{n-1}$

Illustration 36

Find acceleration and contact force between two masses in the given fig $(g=10m/s^2)$.

Solution :

Drawing F.B.D (Two mass will slide down together as ${\mu }_1<{\mu }_2$ ) $R$ - contact force

$f_1=0.5\times 10\times g\cos {30}^{\circ }$

$f_2=0.1\times 2\times g\times \cos {30}^{\circ }$

$10g\sin {30}^{\circ }+R-f_1=10a$

and $2g\sin {30}^{\circ }-R-f_2=2a$

Adding (i) and (ii)

$100\times \frac{1}{2}-0.5\times 10\times 10\times \frac{\sqrt{3}}{2}+2\times 10\times \frac{1}{2}-0.1\times 2\times 10\times \frac{\sqrt{3}}{2}=12a$

$60-26\sqrt{3}=12a\Rightarrow a=1.25m/s^2$ and

(from (i)) $R=-50+25\sqrt{3}+10\times 1.25\Rightarrow R=5.75$

EXERCISE 1

Fill in the blanks

DIRECTIONS : Complete the following statements with an appropriate word / term to be filled in the blank space(s).

1. Application of a force changes the ____ object.
2. An object moving at constant speed is in a state of ____
3. Impulse is the product of force and ____
4. The change in the momentum of an object is equal to the ____ applied to it.
5. The SI unit of force is the ____
6. It takes about 4.45 newtons to ____
7. If a force of 200 newtons is required to move a wagon up a frictionless hill at constant speed, the force needed to lef the wagon roll downhill at constant speed is ____
8. In any interaction between two or more isolated objects, the total ____ does not change.
9. An object will move in a circular path at constant speed if a force is applied to it in a direction that is kept ____ to its velocity.
10. The change in the velocity of an object is proportional to the ____ applied to it.
11. The ratio of net force applied to an object to the acceleration it produces is the ____ of the object.
12. If there are several forces on an object, its acceleration depends on its mass and the ____ force.
13. Magnitude of kinetic friction ____ with increase in inclination of the inclined plane.
14. Conical shape of rockets reduce ____
15. ____ is equal to change in momentum.

True/False

DIRECTIONS : Read the following statements and write your answer as true or false.

1. When we push our foot against the ground backwards (action), the ground exerts an equal and opposite force (reaction) on our foot which causes us to more forward.
2. It is easier to start motion in a lighter body than a heavier body.
3. A rocket can propel itself in a vacuum.
4. Action and reaction force acts on the same object.
5. Particle of different masses falls with different acceleration on earth.
6. Particle is at rest, if force is zero.
7. Particle moves in the direction of force.
8. If particle is initially at rest then it moves in direction of net force.
9. No net force acts on a rain drop falling vertically with a constant speed.
10. If net force acting on the body is zero, momentum of the body remains constant.
11. Friction is electromagnetic in nature.
12. Static friction is self-adjusting.
13. Magnitude of kinetic friction is constant.
14. Forces of action and reaction cancel each other
15. A body can be in equilibrium under the action of three coplaner forces.

Match the Columns

DIRECTIONS : Each question contains statements given in two columns which have to be matched. Statements $(A,B,C,D\dots .$. in Column I have to be matched with statements $(p,q,r,s...$. in Column II.

Column I	Column II
(A) To every action there is	(p) Momentum
(B) During collision, there is conservation of	(q) Force
(C) Rate of change of velocity is	(r) Reaction
(D) Rate of change of momentum is	(s) Acceleration
(E) Force that opposes motion	(t) Force of Friction

2. A block of mass $m$ is placed on a weighing machine which is placed on the floor of a lift. Match the different situations (Column I) with readings of the weighing machine (Column II)

Column I	Column II
(A) Lift moving upwards with some acceleration	(p) Equal to $m$
(B) Lift moving downwards with some acceleration	(q) Greater than $m$
(C) Lift moving upwards with constant velocity	(r) Less than $m$
(D) Lift falling freely under gravity	(s) Zero

3. Consider the arrangement shown in the figure. There is no friction between block $A$ and ground, coefficients of static and kinetic frictions between blocks $A$ and $B$ are 0.5 and 0.4 respectively. If masses of blocks $A$ and $B$ are $3kg$ and $2kg$ respectively, then match the magnitude of applied force $F$ (Column I) with magnitude of friction $(f)$ between $A$ and $B$ (Column II). $(g=10m{s}^{-2})$

Column I	Column II
(A) $F=10N$	(p) $8N$
(B) $F=18N$	(q) $4N$
(C) $F=24N$	(r) $7.2N$
(D) $F=30N$	(s) $9.6N$

Very short answer questions

DIRECTIONS : Give answer in one word or one sentence.

1. A person sitting in the compartment of a train moving with a uniform speed throws a ball in the upward direction. What path of the ball will appear to him? What to a person standing outside?
2. The length of an ideal spring increases by $0.1cm$ when a body of $1kg$ is suspended from it. If this spring is laid on a frictionless horizontal table and bodies of $1kg$ each are suspended from its ends, then what will be the increase in its length?
3. The two ends of a spring-balance are pulled each by a force of $10kg$-wt. What will be the reading of the balance?
4. A retarding force is applied to stop a motorcar. If the speed of the motorcar is doubled, how much more distance will it cover before stopping under the same retarding force?
5. A ball of $0.5kg$ mass moving with a speed of $10m/s$ rebounds after striking normally a perfectly elastic wall. Find the change in momentum of the ball.
6. A body of mass $2kg$ is suspended on a spring balance hung vertically in a lift. If the lift is falling downward under acceleration due to gravity $g$, then what will be the reading of the balance? If going upward with the same acceleration, then?
7. A body is suspended from the ceiling of a transparent cabin falling freely towards the earth. Describe the motion of the body as observed by an observer (a) sitting in the cabin, (b) standing on the earth.
8. A force of $5N$ changes the velocity of a body from $10m{s}^{-1}$ to $20m{s}^{-1}$ in $5s$. How much force is required to bring about the same change in $2s$.
9. What is the angle of friction between two surfaces in contact, if coefficient of friction is $\frac{1}{\sqrt{3}}$ ?
10. Under what circumstances is Newton’s first law same as Newton’s second law.
11. Write the equation of Newton’s second law when the body is moving opposite to the direction in which force is applied.
12. What is friction?
13. Why do we call friction a self adjusting force?
14. Why is friction a non-conservative force ?
15. It is easier to roll a barrel than to pull it along the road. Why?

Short Answer Questions

DIRECTIONS : Give answer in 2-3 sentences.

1. Vehicles stop on applying brakes. Does this phenomenon violate the principle of conservation of momentum?
2. Four blocks of the same mass $m$ connected by cords are pulled by a force $F$ on smooth horizontal surface, as shown in figure.

Determine the tensions $T_1,T_2$, and $T_3$ in the cords.

3. Two bodies of masses $M$ and $m$ are allowed to fall freely from the same height. If air resistance for each body is same, then will both the bodies reach the earth simultaneously?
4. A bird is sitting on the floor of a closed glass cage in the hands of girls. Will the girl experience any change in the weight of the cage when the bird (i) starts flying in the cage with a constant velocity? (ii) flies upwards with acceleration ? (iii) flies downwards with acceleration ?
5. The driver of a truck travelling with a velocity $v$ suddenty notices a brick wall in front of him at a distance $d$. It is better for him to apply brakes or to make a cincular turn without applying brakes in order to just avoid crashing into the wall? Why?
6. Derive a formula for the acceleration of a body moving down rough inclined plane.
7. Derive a relation between angle of friction and angle of repose.
8. What is inertia ? Why do we call the Newton’s first law as the law of inertia? Explain.
9. Show that the Newton’s first law of motion gives the qualitative definition of force and the second law gives the measure (or quantitative definition) of force.
10. State Newton’s second law of motion. With the help of this law, establish a relationship between the force and acceleration produced in a body.
11. State Newton’s second law of motion. Show that it gives the measure of force.
12. Prove that Newton’s second law of motion is the only real law of motion.
13. What is force? What are its absolute and gravitational units ? How are these related to each other?
14. Define the terms momentum and impulse. Obtain the relation between impulse and momentum.
15. State the principle of conservation of linear momentum. Explain, how you will prove this law. Explain one example, where we make use of this law.

Long Answer Question

DIRECTIONS : Give answer in four to five sentences.

1. State Newton’s second law of motion. Hence, derive the equation of motion $F=$ M.a. From it obtain the unit of force in SI. Show that Newton’s second law of motion is the real law of motion.
2. State Newton’s laws of motion and the principle of conservation of momentum. Prove the principle of conservation of momentum from (i) Newton’s second (ii) third law of motion.
3. What is friction ? Why is it called a self adjusting force? Give the old and modern view of the cause of friction.
4. Explain fully the terms ‘friction’ and ’limiting friction’.

State the laws of limiting friction. Justify the statement that friction is a necessity and an evil. Give some familiar methods for reducing friction.

EXERCISE 2

Multiple Choice Questions

DIRECTIONS: This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.

1. A force $F_1$ acting on a body of $2kg$ produces an acceleration of $2.5m/se{c}^2$. An other force $F_2$ acting on the another body of mass $5kg$ produces an acceleration of $2m/se{c}^2$. Find the ratio of $F_2/F_1$

(a) 2

(c) 6

(b) 4

(d) 8

2. A field gun of mass $1.5t$ fires a shell of mass $15kg$ with a velocity of $150m/s$. Calculate the velocity of the recoil of the gun.

(a) $1m/sec$

(b) $1.5m/sec$

(b) $3m/sec$

(d) $5m/sec$

3. A feather of 20 grams is dropped from a height. It is found to fail down arts a constant velocity. The net force acting on it is

(a) $200N$

(b) $0.2N$

(c) $0.02N$

(d) zero

4. A body of mass $1kg$ is kept at rest. A constant force of $6.0N$ acting on it, the time taken by the body to move through a distance of $12m$

(a) 2 sec.

(b) 3 sec.

(c) $4sec$

(d) $5sec$

5. If a constant force acts on a body initially kept at rest the distance moved by the body in time $t$ is proportional to -

(a) $t$

(b) $t^2$

(c) $t^3$

(d) $A$

6. By applying a force of one Newton, one can hold a body of mass

(a) 102 gram

(b) $102kg$

(c) $102mg$

(d) None of these

7. The speed of a falling body increases continuously, this is because

(a) no force acts on it

(b) it is very light

(c) the air exert the frictional force

(d) the earth attract it

8. A gun of mass $4.5kg$ fires a bullet of mass $20g$. with a velocity of $108km/hr$. the recoil velocity of the gun is

(a) $1.33m/sec$

(b) $0.133m/sec$

(c) $13.3m/s$

(d) $133m/sec$

9. If an object is in a state of equilibrium

(a) it is at rest

(b) it is in motion at constant velocity

(c) it is in free fall

(d) may be more than one of the above

10. A fish is swimming upward at an angle of ${30}^{\circ }$ with the horizontal. The direction of the force of gravity acting on it is

(a) upward

(b) downward

(c) horizontal

(d) at an angle upward

11. If a boat is moving along a constant speed, it may be assumed that

(a) a net force is pushing it forward

(b) the sum of only vertical forces is zero

(c) the buoyant force is greater than gravity

(d) the sum of all forces is zero

12. An astronaut with all her equipment has a mass of 95 kilograms. How much will she weigh on the moon, where the acceleration due to gravity is 1.67 meters per second squared?

(a) $159N$

(b) $169N$

(c) $149N$

(d) $100N$

13. How much is the viscous drag acting on a rocket-driven sled that is going at constant speed against a frictional force of 22000 newtons when the thrust of the engine is 31000 newtons-

(a) $8500N$

(b) $9000N$

(c) $9500N$

(d) $7500N$

14. To slow down a car, a braking force of 1200 newtons is applied for 10 seconds. How much force would be needed to produce the same change in velocity in 6 seconds -

(a) $2000N$

(b) $3000N$

(c) $2500N$

(d) $1500N$

15. A frictionless wagon is pushed from rest, with a force of 60 newtons for 14 seconds. If it then strikes a wall and comes to rest in 0.15 second, how much average force does the wall exert on it ?

(a) $6000N$

(b) $5600N$

(c) $4500N$

(d) $4000N$

16. What force is needed to a accelerate a 60 -kilogram wagon from rest to 5.0 meters per second in 2.0 seconds

(a) $100N$

(b) $120N$

(c) $150N$

(d) $130N$

17. A frictionless wagon going at 2.5 meters per second is pushed with a force of $380N$, and its speed increases to 6.2 meters per second in 4.0 seconds. What is its mass -

(a) $410kg$

(b) $420kg$

(c) $480kg$

(d) $310kg$

18. A 2.0 kilogram weather balloon is released and begins to nse against 6.5 newtons of viscous drag. If its buoyancy is 32 newtons, what is its acceleration

(a) $1.0m/s^2$

(c) $2.0m/s^2$

(b) $3.0m/s^2$

(d) $4.0m/s^2$

19. What braking force is needed to bring a 2200 kilogram car going 18 meters per second to rest in 6.0 seconds?

(a) $C6O0N$

(b) $6500N$

(c) $6000N$

(d) $6200N$

20. The 35 kilogram girl is standing on a 20 kilogram wagon and jumps off, giving the wagon a kick that sends it off at 3.8 meters per second. How fast is the girl moving

(a) $1.2m/s$

(b) $3m/s$

(c) $4m/s$

(d) $2.2m/s$

21. A $500kg$ rocket is fired straight up from the earth, the engines providing 7500 newtons of thrust. Its acceleration is

(a) $4.5m/s$ square

(c) $9.8m/s$ square

(b) $5.2m/s$ square

(d) $15m/s$ square

22. In outer space, a $250kg$ rocket is to be speeded up from 60 meters per second to 75 meters per second in 5.0 seconds. The thrust needed is -

(a) 17 newton

(c) 3000 newton

(b) 750 newton

(d) 3200 newton

23. Two stars of masses $m$ and $5m$ are 3000 parsec apart. If the force on the large one is $F$, the force on the small star is

(a) $F/25$

(b) $F/5$

(c) $F$

(d) $5F$

24. A moving object can come to rest only if it

(a) has a frictional force acting on it

(b) has no net force acting on it

(c) is completely isolated

(d) applies an impulse to something else

25. If a jet engine provides a thrust of 45000 newtons, how long must it fire to produce 1 million newton-seconds of impulse?

(a) $22s$

(b) $18s$

(c) $25s$

(d) $15s$

26. A rocket-driven sled speeds up from 40 meters per second to 55 meters per second in 5.0 seconds, using an engine that produces 3500 newtons of thrust. How much thrust would be needed to get the same increase in speed in 2.0 seconds

(a) 8550

(b) 8750

(c) 8700

(d) 8500

27. What force is needed to speed up a frictionless $60kg$ cart from 4.0 meters per second to 6.5 meters per second in 3.0 seconds?

(a) $50N$

(b) $100N$

(c) $5N$

(d) $20N$

28. What force must the brakes and tires apply to a $2800kg$ truck going 30 meters per second to bring it to rest in 8.0 seconds

(a) $12000N$

(b) $13000N$

(c) $11000N$

(d) $12500N$

29. A $680kg$ rocket is to be lifted off the surface of the moon, where $g=1.67$ meters per second squared. What force is needed to give it an upward acceleration of 2.0 meters per second squared

(a) $2000N$

(b) $2100N$

(c) $2200N$

(d) $2400N$

30. A $35kg$ girl on roller skates, standing still, throws a $6kg$ medicine ball forward at 3.5 metes per second. How much is her recoil velocity (the backward speed she acquires as a result of the throw)

(a) $-0.6m/s$

(b) $-1.6m/s$

(c) $-2.6m/s$

(d) $-5.6m/s$

31. The recoil velocity of a $7.5kg$ rifle if it fires an $8.0gram$ bullet with a muzzle velocity of 640 meters per second is

(a) $0.12m/s$

(b) $0.68m/s$

(c) $2.68m/s$

(d) $6.8m/s$

32. A $750kg$. rocket, at rest in outer space, propels itself forward by ejecting $45kg$. of hot gas, which leaves the nozzle at 85 meters per second. The change in the momentum of the fuel is

(a) $3825kgm/s$

(b) $3025kgm/s$

(c) $3725kgm/s$

(d) $3125kgm/s$

33. When a body is stationary

(a) There is no force acting on it

(b) The force acting on it not in contact with it

(c) The combination of forces acting on it balances each other

(d) The body is in vacuum

34. A rider on horse falls back when horse starts running, all of a sudden because

(a) rider is taken back

(b) rider is suddenly afraid of falling

(c) inertia of rest keeps the upper part of body at rest while lower part of the body moves forward with the horse

(d) none of the above

35. A man getting down a running bus, falls forward because

(a) due to inertia of rest, road is left behind and man reaches forward

(b) due to inertia of motion upper part of body continues to be in motion in forward direction while feet come to rest as soon as they touch the road

(c) he leans forward as a matter of habit

(d) of the combined effect of all the three factors stated in (a), (b) and (c)

36. The heart is pumping blood at $xkg$ per unit time, with constant velocity $v$. The force needed is

(a) $xv$

(c) $x\frac{dv}{dt}$

(b) $v\frac{dx}{dt}$

(d) zero

37. A force of 50 dynes is acted on a body of mass $5g$ which is at rest for an interval of $3sec$., then impulse is

(a) $0.15\times {10}^{-13}Ns$

(b) $0.98\times {10}^{-3}Ns$

(c) $1.5\times {10}^{-3}Ns$

(d) $2.5\times {10}^{-3}Ns$

38. A force $10N$ acts on a body of mass $20kg$ for $10sec$. Change in its momentum is

(a) $5kgm/s$

(b) $100kgm/s$

(c) $200kgm/s$

(d) $1000kgm/s$

39. Swimming is possible on account of

(a) First law of motion

(b) Second law of motion

(c) Third law of motion

(d) Newton’s law of gravitation

40. The distance $x$ covered in time $t$ by a body having velocity $v_0$ and having a constant acceleration a is given by $x=v_{0}t+\frac{1}{2}a{t}^2$. This result follows from

(a) Newton’s first law

(b) Newton’s second law

(c) Newton’s third law

(d) none of these

41. Gravels are dropped on a conveyor belt at the rate of $0.5kg/sec$. The extra force required in newtons to keep the belt moving at $2m/sec$. is

(a) 1

(b) 2

(c) 4

(d) 0.5

42. A cricket ball of mass $150gm$ is moving with a velocity of $12m/sec$. and is hit by a bat so that the ball is turned back with a velocity of $20m/sec$. The force of bat acts for $0.01s$ on the ball then the average force exerted by the bat on the ball.

(a) $840N$

(b) $48N$

(c) $84N$

(d) $480N$

43. The average force necessary to stop a hammer having momentum $25N$-s in 0.05 second is

(a) $25N$

(b) $50N$

(c) $1.25N$

(d) $500N$

44. A force of 100 dynes acts on mass of $5gm$ for $10sec$. The velocity produced is

(a) $2cm/sec$

(b) $20cm/sec$

(c) $200cm/sec$

(d) $2000cm/sec$

45. The Newton’s laws of motion are valid in

(a) inertial frames

(b) non-inertial frames

(c) rotating frames

(d) accelerated frames

46. Newton’s third law is equivalent to the

(a) law of conservation of linear momentum

(b) law of conservation of angular momentum

(c) law of conservation of energy

(d) law of conservation of energy and mass

47. When we jump out a boat standing in water it moves

(a) forward

(b) backward

(c) side ways

(d) none of the above

48. A man is at rest in the middle of a pond on perfectly smooth ice. He can get himself to the shore by making use of Newton’s

(a) first law

(b) second law

(c) third law

(d) all the laws

49. A parrot is sitting on the floor of a closed glass cage which is in a boy’s hand. If the parrot starts flying with a constant speed, the boy will feel the weight of the cage as

(a) unchanged

(b) reduced

(c) increased

(d) nothing can be said

50. A cannon after firing recoils due to

(a) conservation of energy

(b) backward thrust of gases produced

(c) Newton’s third law of motion

(d) Newton’s first law of motion

51. Newton’s third law of motion leads to the law of conservation of

(a) angular momentum

(b) energy

(c) mass

(d) momentum

More than one correct

DIRECTIONS : This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d) out of which ONE OR MORE may be correct.

1. Acceleration of a body can be calculated using

(a) $a=F/m$

(b) $a=\frac{v-u}{t}$

(c) $a=P/m$

(d) $a=s/t$

2. Which physical quantity has its unit as ‘Newton’?

(a) Friction

(b) Acceleration

(c) Force

(d) Momentum

3. The acceleration produced by a force of $5N$ on a mass of $10kg$ is

(a) $5m/s^2$

(c) $0.5m{s}^{-2}$

(b) $2m/s^2$

(d) $50cm{s}^{-2}$

4. What force would be required to produce an acceleration of $4m/s^2$ in a ball of mass $6kg$ ?

(a) $24N$

(c) $42kgm/s^2$

(b) $42N$

(d) $24kgm/s^2$

5. Which of the following is/are a vector quantity ?

(a) Speed

(b) Acceleration

(c) Momentum

(d) Velocity

6. A body is in translatory equilibrium if

(a) it is in uniform motion

(b) resultant force on it is zero

(c) it is at rest

(d) it is in accelerated motion

7. If a particle under the action of a force $\overrightarrow{F}$ has potential energy $U$, then in equilibrium

(a) $\overrightarrow{F}=0$ but $U\ne 0$

(c) $\overrightarrow{F}\ne 0$ but $U=0$

(b) $\overrightarrow{F}\ne 0$ and $U\ne 0$

(d) $\overline{F}=0$ but $U=0$

8. If the tension in the cable supporting a lift is equal to the weight of the lift, the lift may be

(a) going up with uniform speed

(b) going up with increasing speed

(c) going down with uniform speed

(d) going down with increasing speed

9. In which of the following cases, the net force is zero?

(a) A car moving with constant speed of $30km/hr$ on a rough road.

(b) A drop of train falling down with terminal velocity

(c) A cork of mass $10g$ floating on water. (d) A charged particle moving parallel to the magnetic field.

10. A metal sphere is hung by a string fixed on a wall. The force acting on the sphere are shown in the figure. Which of the following statements are correct?

(a) $T^2=R^2+W^2$

(b) $T=R+W$

(c) $\overline{R}+\overline{T}+\overline{W}=0$

(d) $R=W\tan \theta$

11. A block of mass $1kg$ is stationary with respect to a conveyor belt that is accelerating with $1m/s^2$ upwards at an angle of ${30}^{\circ }$ as shown in figure. Which of the following statements are correct? $(g=10m/s^2)$.

(a) Force of friction on the block is $1.5N$ upwards

(b) Contact iorce between the block and the belt is $5/\sqrt{3}N$.

(c) Force of friction on the block is $6N$ upwards

(d) Contact force between the block and the belt is $10.5N$.

12. A monkey of mass $mkg$ slides down a light rope attached to a fixed spring balance with an acceleration $a$. The reading of the spring is $Wkg$ ( $g=$ acceleration due to gravity). Then:

(a) the tension in the slope is $W_{g}N$

(b) the force of friction exerted by the rope on the monkey is $m(g-a)N$.

(c) $m=\frac{W_g}{g-a}$

(d) $m=W(1+\frac{a}{g})$

13. A man pulls a block heavier than himself with a light rope.

The coefficient of friction is the same between the man and the ground and between the block and the ground. Then

(a) if both move, the acceleration of the man is greater than the acceleration of the block.

(b) the block will not move unless the man also moves

(c) the man can move even when the block is stationary.

(d) none of these assertions is correct.

14. A man tries to remain in equilibrium by pushing with his hands and feet against two parallel walls. For equilibrium:

(a) the coefficient of friction must be the same between both walls and the man

(b) he must exert equal forces on the two walls

(c) the forces of friction at the two walls must be equal

(d) friction must be present on both walls

15. The pulley arrangements of following figures (a) and (b) are identical. The mass of the rope is negligible. In figure (a), the mass $m$ is lifted up by attaching a mass $2m$ to the other end of the rope. In figure (b) the mass $m$ is lifted up by pulling the other end of the rope with a constant downward force of $2mg$. Which of the following are correct?

(a)

(b)

(a) The acceleration of mass $m$ in figure (b) is $g$.

(b) The acceleration of mass $m$ in figure (a) is $g/3$

(c) The tension in the string in figure (a) is $4mg/3$

(d) The tension is the string in figure (b) is $2mg$.

16. System shown in figure consists of two blocks of masses $m_1$ and $m_2$ placed on a smooth horizontal surface. The system is being pushed with a force $F$. Which of the following are correct?

(a) The reaction force of $m_2$ on $m_1$ is $F{m}_2/(m_1+m_2)$

(b) The acceleration of the system is $F(m_1+m_2)$

(c) The acceleration of mass $m_1$ is the same as the acceleration of mass $m_2$

(d) None of the above

17. Refer to the figure, the pulleys and the string are light. $T$ is the tension in the string. Take $g=10m{s}^{-2}$. Which of the following are correct?

(a) The acceleration of the system is $5m{s}^{-2}$

(b) $T=\mathbf{0}\mathbf{N}$

(c) The acceleration of the system is $10m{s}^{-2}$

(d) $T=5N$

Multiple matching questions

DIRECTIONS : Following question has four statements $(A,B$, $C$ and D) given in Column I and four statements ( $p,q,r,s..$. ) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II. Match the entries in Column I with entries in Column II.

Column I	Column II
(A) Newton’s first law	(p) Dust coming out of carpet beaten with stick
(B) Newton’s second law	(q) Recoil of gun
(C) Newton’s third law	(r) Law of inertia
(D) Law of conservation of momentum	(s) Catching a cricket ball
	(t) Use of seat belts in car
	(u) Flying of rockets and aeroplanes

2. For the system shown in the figure, the incline is frictionless and the string is massless and inextensible pulley is light and frictionless. As the system is released from rest.

Column I	Column II
(A) $M_1>M_2$	(p) $M_2$ accelerates down
(B) $M_2>M_1$	(q) $M_2$ accelerates up
(C) $M_1=M_2$	(r) $M_1,M_2$ in equilibrium
(D) $M_1»M_2$	(p) Tension in string equals the weight of either block

3. If ${\mu }_1$ and ${\mu }_2$ represent coefficients of friction between blocks $A$ and inclined plane and block $B$ and inclined plane respectively, then match the following:

Fill in the blanks

DIRECTIONS : Complete the following passage(s) with an appropriate word/term to be filled in the blank spaces.

I. force, simultaneously, opposite, equal, Opposite, different, equal

To every action there is an ____ (1) ____ and ____ (2) ____ reaction. Action and reaction act on two ____ (3) ____ bodies, but they act ____ (4) ____. Whenever one body exerts a ____ (5) ____ on another body, the second body exerts an ____ (6) ____and ____ (7) ____ force on the first body.

II.

Motion, force, second, direction, Newton, momentum, directly directly

According to Newton’s ____ (1) ____ law of ____ (2) ____, the rate of change of ____ (3) ____ of a body is ____ (4) ____ proportional to the ____ (5) ____. It takes place in the ____ (6) ____ in which the force acts. The S.I. unit of force is ____ (7) ____ or $kgm/s^2$.

III. electromagnetic, Contact, tend to move, chemical bonding, kinetic, constant, static, depends, relative motion, does not depend

Friction is an opposing force which act between two surfaces in ____ (1) ____ whenever the surfaces move or ____ (2) ____ w.r.t. each other. The force of friction is ____ (3) ____ in nature and its cause is ____ (4) ____ between the surfaces contact. When there is no relative motion between the surface, friction is called ____ (5) ____ and when there is relative motion between the surfaces, friction is called ____ (6) ____ and its magnitude is ____ (7) ____ The magnitude of friction ____ (8) ____ on nature of surfaces in contact and ____ (9) ____ on area of contact. Friction always opposes ____ (10) ____ between the surfaces.

Passage Based Questions

DIRECTIONS : Study the given paragraph(s) and answer the following questions.

PASSAGE-I

A block of mass $4kg$ is placed on a rough horizontal surface with coefficient of friction $\frac{1}{\sqrt{3}}$. A force $F$ is applied on the block making an $\theta$ with the horizontal as shown in the figure.

1. For what value of $\theta$, force required to move the block is minimum?

(a) $0^{\circ }$

(b) ${30}^{\circ }$

(c) ${45}^{\circ }$

(d) ${80}^{\circ }$

2. What is the minimum value of force $F$ required to move the block?

(a) $\frac{20}{\sqrt{3}}N$

(b) $20N$

(c) $\frac{40}{\sqrt{3}}N$

(d) $40N$

3. If the value of $F$ is $4N$ more than its value calculated in previous problem, the acceleration of the block will be

(a) $\frac{2}{\sqrt{3}}m{s}^{-2}$

(b) $2\sqrt{3}m{s}^{-2}$

(c) $\frac{4}{\sqrt{3}}m{s}^{-2}$

(d) $1m{s}^{-2}$

PASSAGE-II

Two blocks $A$ and $B$ of mass $2kg$ and $3kg$ respectively are connected with the help of a massless, inextensible string passing over a smooth pulley as shown, The system is released from rest at $t=0$, then: (take $g=10m{s}^{-2}$ )

4. Acceleration of blocks is

(a) $5m{s}^{-2}$

(b) $2m{s}^{-2}$

(c) $20/3m{s}^{-2}$

(d) $6m{s}^{-2}$

5. If at $t=1s$, block $B$ is stopped momentarily and released, after how much time will the string become taut again?

(a) $0.2s$

(b) $0.4s$

(c) $0.5s$

(d) $1s$

6. What will be the velocity of the blocks, just after the string becomes taut?

(a) $0.2m{s}^{-1}$

(b) $0.4m{s}^{-1}$

(c) $1m{s}^{-1}$

(d) $2m{s}^{-1}$

PASSAGE-III

Consider the situation shown in the adjoining figure, strings are massless and inextensible, pulleys are massless and frictionless, the inclined surface is rough having coefficient of friction $\frac{1}{2\sqrt{3}}$. The mass of block $A$ is $M$ and those of blocks $B$ and $C$ is $m$ each. Using the above information answer the following.

7. The minimum value of ratio $\frac{M}{m}$ so that block $C$ remains at rest is

(a) 1

(b) $\frac{1}{2}$

(c) 2

(d) $\frac{1}{15}$

8. If the ratio $\frac{M}{m}$ is equal to what calculated in previous problem, acceleration of blocks will be

(a) $g$

(b) $\frac{7g}{8}$

(c) 3

(d) zero

9. The maximum value of ratio $\frac{M}{m}$ so that block $C$ remains at rest is

(a) 2

(b) 3

(c) $\frac{3}{13}$

(d) $\frac{5}{2}$

Assertion & Reasan

DIRECTIONS : Each of these questions contains an Assertion followed by reason. Read them carefully and answer the question on the basis of following options. You have to select the one that best describes the two statements.

(a) If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

(b) If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

(c) If Assertion is correct but Reason is incorrect.

(d) If Assertion is incorrect but Reason is correct.

1. Assertion: When a bullet is fired from a gun, there is a forward force on the bullet and recoil of gun.

Reason: Every action has an equal and opposite reaction.

2. Assertion: When astronauts throw something in space, that object would continue moving in the same direction and with the same speed.

Reason: The acceleration of an object produced by a net applied force is directly related to the magnitude of the force, and inversely related to the mass of the object.

3. When we sit on a chair, our body exerts a force downward and

Assertion: That chair needs to exert an equal force upward or the chair will collapse.

Reason: The third law says that for every action there is an equal and opposite reaction.

4. Assertion: The wings of a bird push air upwards and the air must be pushing the bird downwards.

Reason: For every action there is an equal and opposite reaction.

5. Assertion: When a firefly hits a bus, each of them exerts the same force

Reason: Firefly has more mass as compared to the windshield.

6. Assertion: Force exerted by the ground on the man moves him forward.

Reason: It is a reactional force.

7. Assertion: A quick collision between two bodies is more violent than a slow collision, even when the initial and the final velocities are identical.

Reason: Because the rate of change of momentum which determines the force is greater in the first case.

8. Assertion: Change in momentum is impulse.

Reason: Impulse is the area between $(F-t)$ graph and time axis.

9. Assertion: A body is momentarily at rest when it reverses the direction.

Reason: A body cannot have acceleration if its velocity is zero at a given instant of time.

10. Assertion: For stable equilibrium force has to be zero and potential energy should be minimum.

Reason: For equilibrium, it is not necessary that the force is not zero.

11. Assertion: While walking on ice, one should take small steps to avoid slipping.

Reason: This is because smaller steps ensure smaller friction.

12. Assertion: Force required to accelerate a mass in two perpendicular directions is never same.

Reason: The presence of $g$ will not influence the acceleration.

Hots Subjective Questions

DIRECTIONS : Answer the following questions.

1. On which of these hills does the ball roll down with increasing speed and decreasing acceleration along the path? (Use this example if you wish to explain to someone the difference between speed and acceleration.)

2. If you drop an object, its acceleration toward the ground is $10m/s^2$. If you throw it down instead, would its acceleration after throwing be greater than $10m/s^2$ ? Why or why not?
3. Two $100-N$ weights are attached to a spring scale as shown Does the scale read $0N,100N$, or $200N$, or does it give some other reading? (Hint: Would it read any differently if one of the ropes were tied to the wall instead of to the hanging $100-N$ weight?)

4. Suppose two carts, one twice as massive as the other, fly apart when the compressed spring that joins them is released. How fast does the heavier cart roll compared with the lighter cart?

5. If you exert a horizontal force of $200N$ to slide a crate across a factory floor at a constant velocity, how much friction is exerted by the floor on the crate? Is the force of friction equal and oppositely directed to your $200-N$ push? Does the force of friction make up the reaction force to your push? Why not?
6. Two people of equal mass attempt a tug-of-war with a 12-m rope while standing on frictionless ice. When they pull on the rope, each person slides toward the other. How do their accelerations compare, and how far does each person slide before they meet?
7. When your hand turns the handle of a faucet, water comes out. Do your push on the handle and the water coming out comprise an action-reaction pair? Defend your answer.
8. A stone is shown at rest on the ground.

(a) The vector shows the weight of the stone. Complete the vector diagram showing another vector that results in zero net force on the stone.

(b) What is the conventional name of the vector you have drawn?

9. Here a stone at rest suspended by a stringe.

(a) Draw force vectors for all the forces that act on the stone.

(b) Should your vectors have a zero resultant? Why, or why not?

10. Suppose that the string in the preceding exercise breaks and that the stone slows in its upward motion. Draw a force diagram of the stone when it reaches the top of its path.
11. What is the net force on the stone in the preceding exercise when it is at the top of its path? What is its instantaneous velocity? Its acceleration?
12. Here is the stone sliding down a friction-free incline (a) Idenitify the forces that act on it, and draw appropriate force vectors. (b) By the parallelogram rule, construct the resultant force on the stone.

(carefully showing that it has a direction parallel to the incline- the same direction as the stone’s acceleration).

13. Here is the stone at rest, interacting with both the surface of the incline and the block. (a) Identify all the forces that act on the stone and draw appropriate force vectors. (b) Show that the net force on the stone is zero. (b) Show that the net force on the stone is zero. (Hint 1 : There are two normal forces on the stone. Hint $2:$ Be sure the vectors you draw are for forces that act on the stone, not on the surfaces by the stone.)

14. A force produces an acceleration of $16m/s^2$ in a body of mass $0.5kg$, and an acceleration of $4m/s^2$ in another body. If both the bodies are fastened together, then what is the acceleration produced by that force?
15. A $20gm$ bullet moving at $300m/s$ stops after penetrating $3cm$ of bone. Calculate the average force exerted by the bullet.
16. A force of $50N$ is inclined to the vertical at an angle of ${30}^{\circ }$. Find the aceeleration it produces in a body of mass 2 $kg$ which moves in the horizontal direction.
17. A gun weighing $10kg$ fires a bullet of $30g$ with a velocity of $330m/s$. With what velocity does the gun recoil? What is the resultant momentum of the gun and the bullet before and after firing?

SOLUTIONS

EXERCISE - I

Fill in the blanks

1. velocity
2. equilibrium
3. time
4. impulse
5. newton
6. pound
7. 200 newtons
8. momentum
9. perpendicular
10. impulse
11. mass
12. net unbalanced
13. decrease
14. atmospheric friction
15. impulse

True/False

1. True	2. True	3. True	4. False
5. False	6. False	7. False	8. True
9. True	10. True	11. True	12. True
13. True	14. False	15. True

Match the columns

1. (A) $\to$ (r) ;(B) $\to$ (p) ;(C) $\to$ (s) ;(D) $\to$ (t)
2. (A) $\to$ (q) ;(B) $\to$ (r) ;(C) $\to$ (p) ;(D) $\to$ (s)
3. (A) $\to$ (q) ;(B) $\to$ (r) ;(C) $\to$ (s) ;(D) $\to$ (p)

$FBD$ ’s of $B$ and system $(A+B)$ are shown below For motion of $B$,

$N_{BA}=m_{B}g$

d $f=m_B{a}_B$

but $f\le {\mu }_s{N}_{AB}$

$\therefore a_B\le {\mu }_{s}g$

For system $(A+B)$

$F=(m_P+m_B)a$

Both $A$ and $B$ will move together if

$F\le {\mu }_s(m_A+m_B)g\text{ or }F\le 25N$

Now, if $F=10N,a=2m{s}^{-2}$ & $f=m_B{a}_B=uN$,

if $F=18N,a=3.6m{s}^{-2}$ & $f=7.2N$,

if $F=24N,a=4.8m{s}^{-2}$ & $f=9.6N$

if $F=30N$, block $B$ will slip on block $A$,

hence $f={\mu }_k{N}_{BA}=8N$

Very Short Answer Questions

1. It will appear to fall in vertically downward direction to the person in the compartment and parabolic to a person standing outside.

2. It would increase in length by $0.1cm$ only.

3. $10kg-wt$.

4. Since distance $\propto (.$ velocity) ${}^2$, motorcar will cover the distance four times longer than before.

5. Change in linear momentum $=-mv-mv=-2mv$

$=-2\times 0.5\times 10=-10kgm{s}^{-1}\text{. }$

6. Physical balance will show no change whereas, spring balance will show higher reading.

7. (a) The body will appear stationary in air, (b) The body will appear falling freely under the acceleration in air.

8. $F_1=dp/d{t}_1$ and $F_2=dp/d{t}_2$

$\therefore F_2=F_{1}d{t}_1/d{t}_2=5\times 5/2=12.5N$

9. $\mu =\frac{1}{\sqrt{3}},\tan \theta =\frac{1}{\sqrt{3}}$

$\therefore \theta ={30}^{\circ }$.

10. When the force applied on a body is zero, Newton’s first law becomes a special case of Newton’s second law.

11. $F=-m\times a$ or $-F=m\times a$

12. The opposing force that is set up between the surfaces of contact, when one body slides or rolls or tends to do so on the surface of another body is called friction.

13. When applied force is zero, friction is zero. As the applied force is increased, friction also increases and becomes equal to the applied force. It happens so, till the body does not start moving. For this reason, friction is called self adjusting force.

14. It is because, work done against friction along a closed path is non-zero.

15. The rolling friction is lesser as compared to the sliding friction.

Short Answer Questions

1. Vehicle stops on applying brakes, which is in accordance with the law of conservation of momentum. When brakes are applied, opposition force acts on the vehicle. In pursuit, car will be at rest. Loss of momentum of the vehicle is exactly equal to the impulse of the applied force.

2.

Let a be the common acceleration of the whole system.

$\therefore F=(m+m+m+m)a=4ma$, or, $a=F/4m$

Force equation, $F-T_1=ma,T_1-T_2=ma,T_2-T_3=ma$, $T_3=ma$, solving these equations, we get,

$T_1=\frac{3}{4}F,T_2=\frac{1}{2}F$ and $T_3=\frac{1}{4}F$

3. No. Net force on the body of mass $M$ is $(Mg-F)$

$\therefore a=(Mg-F)/M=(g-F/M)$, thus acceleration would be greater if the mass will be higher and it will reach the earth earlier.

4. In closed glass cage, air inside is bound with the cage. Therefore, (i) no change in weight of the cage, when bird flies with constant velocity, (ii) cage will be heavier, when bird flies with upward acceleration, (iii) cage will be lighter, when bird flies with downward acceleration.

5. While applying brakes, let $F_B$ be the force required to stop the truck in distance $d$.

$\therefore F_B\times d=1/2m{v}^2$ or, $F_B=m{v}^2/2d$

For taking a turn radius $d$, the force required, $F_T=m{v}^2/d$ $=2F_B$ or, $F_B=1/2F_T$, which means it is better to apply brakes.

6.

$R=mg\cos \theta$ and,

$f=mg\cos \theta -F,\mu =F/R$

i.e. $f=ma=mg\sin \theta -\mu R$

$ma=mg\sin \theta -\mu mg\cos \theta =mg(\sin \theta -\mu \cos \theta )$

$\therefore a=g(\sin \theta -\mu \cos \theta )$

7.

If $\alpha$ - angle of repose,

$R=mg\sin \alpha$

$F=mg\cos \alpha$

$F/R=mg\cos \alpha /mg\sin \alpha =\tan \alpha =\mu$

By definition, coefficient of friction,

$\mu =\tan \theta$

$\therefore \theta =\alpha$, which means angle of repose is equal to the angle of friction.

Multiple Choice Questions

1. (a)	2. (b)	3. (b)	4. (a)	5. (b)
6. (a)	7. (d)	8. (a)	9. (d)	10. (b)
11. (d)	12. (a)	13. (b)	14. (a)	15. (b)
16. (c)	17. (a)	18. (b)	19. (a)	20. (d)
21. (b)	22. (b)	23. (c)	24. (d)	25. (a)
26. (b)	27. (a)	28. (c)	29. (d)	30. (a)
31. (b)	32. (a)	33. (c)	34. (c)	35. (b)
36. (a)	37. (c)	38. (b)	39. (c)	40. (b)
41. (a)	42. (d)	43. (d)	44. (c)	46. (a)
47. (b)	48. (c)	49. (a)	50. (c)	51. (d)

More than one correct

1. (a, b)
2. (a, c)
3. (c, d)
4. (a, d)
5. (b, c, d)
6. (a, b, c)

For translatory equilibrium $F_{\text{net }}=0$

$\therefore a=0$ or $v=$ const

7. (a, d) For equilibrium, $f=0$ while $U$ may or may not be zero.
8. (a, c) If $T=mg,a=0$
9. $(a,b,c,d)$
10. (a, c, d)
11. (c, d) $f=mg\sin \theta +ma=5+1=6N$

$\begin{array}{rl}& R=\sqrt{f^2+N^2}=\sqrt{6^2+(5\sqrt{3}{)}^2}\approx 10.5N\\ & N=mg\cos \theta =5\sqrt{3}\end{array}$

12. $(a,b,c)$
13. $(a,b,c)$
14. (b, c, d)
15. $(a,b,c,d)$
16. $(a,b,c)$
17. $(a,c)$

For motion of block $A$

$mg-T=ma$ … (1)

For motion of block $B$

$T=ma$ …(2)

On solving

$a=\frac{g}{2}=5m{s}^{-2}$ & $T=\frac{1}{2}mg=5N$

Multiple Matching Questions

1. (A) $\to$ (p, r) ;(B) $\to$ (s, r) ;(C) $\to$ (q, u) ;(D) $\to$ (q, u)
2. (A) $\to$ (p, q, r, s) ;(B) $\to$ (p, s) ;(C) $\to$ (p) ;(D) $\to$ (q)$
3. (A) $\to$ (p, q) ;(B) $\to$ (p, s) ;(C) $\to$ (q, r) ;(D) $\to$ (p, q, r, s)

Fill In The Passage

I. (1) equal

(2) opposite

(3) different

(4) simultaneously

(5) force

(6) equal

(7) opposite

II. (1) second

(2) motion

(3) momentum

(4) directly

(5) force

(6) direction

(7) Newton

III. (1) Contact

(2) tend to move

(3) electromagnetic

(4) chemical bonding

(5) static

(6) kinetic

(7) constant

(8) depends

(9) does not depend

(10) relative motion,

Passage Based Questions

1. (b) FBD of the block is shown below, Here,

$F\sin \theta +N=mg$

or $N=mg-F\sin \theta$

Block will move if,

$F\cos \theta \ge f_1$

or $F\cos \theta \ge \mu N$ or $F\cos \theta \ge \mu (mg-F\sin \theta )$

or $F\ge \frac{\mu mg}{\cos \theta +\mu \sin \theta }$

Now $F$ is minimum, when

$\cos \theta +\mu \sin \theta =x$ (say) is maximum

Now, maximum value of $x$ is

$x=\sqrt{1+{\mu }^2}$ at $\theta ={\tan }^{-1}\mu ={30}^{\circ }$

2. (b) Here $f_{min}=\frac{\mu mg}{\sqrt{1+{\mu }^2}}=\frac{\frac{1}{\sqrt{3}}\times 4\times 10}{\sqrt{1+\frac{1}{3}}}=20N$.

3. (a) If $F=20+4=24N$

$F\cos \theta -f=ma$

$F\cos \theta -\mu (mg-F\sin \theta )=ma$ or $24\times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{3}}(40-24\times \frac{1}{2})=4a$

or $a=\frac{2}{\sqrt{3}}m{s}^{-2}$

4. (b) $FBD$ is of $A$ and $B$ are shown in figure, as $m_B>m_A$, $A$ moves up and $B$ moves down with same acceleration $Q$.

Here,

$\begin{array}{rl}{m}_{B}g-T& =m_{B}a\\ & T-m_{A}g& =m_{A}a\end{array}$

On solving

$a=\frac{m_B-m_A}{m_B+m_A}g=\frac{g}{5}=2m{s}^{-2}$

5. (a) Speed of $A$ at $t=1s$,

$v=u+at=2m{s}^{-1}$

The string will become taut again when distance covered by both the blocks is equal, i.e.

$s_A=s_B$

$2t-\frac{1}{2}g{t}^2=\frac{1}{2}g{t}^2$

or $t=0.2s$.

6. (b) Let $u_A$ and $u_B$ be their velocities, just before the string become taut, then

$u_A=2-gt=0,u_B=gt=2m{s}^{-1}$

Let $v$ be their common speed just after the string becomes taut, then

impulse of $A$ on $B=-$ impulse of $B$ on $A$

$m_A(v-0)=-m_B(v-2)$

$2v=-3(v-2)$

or $v=0.4m{s}^{-1}$

7. (d) For $C$ to remain at rest,

$2T=mg\sin \theta -f$

$=mg\sin {30}^{\circ }-\mu mg\cos {30}^{\circ }$

or $T=\frac{1}{8}mg$

As $C$ is at rest, acceleration of $A$ and $B$ must equal,

$\therefore T=\frac{2mM}{m+M}g$

or $\frac{2mM}{m+M}g=\frac{1}{8}mg$ or $16M=M+m$

or $\frac{M}{m}=\frac{1}{15}$

8. (b) $a=\frac{m-M}{m+M}g=\frac{1-\frac{M}{m}}{1+\frac{M}{m}}g=\frac{7}{8}g$

9. (c) As the block $C$ tends to slip upwards, friction will at downwards, hence,

$2T=mg\sin \theta +\mu mg\cos \theta$

or $T=\frac{3}{8}mg$

$\therefore T=\frac{2mM}{m+M}g=\frac{3}{8}mg$ or $\frac{M}{m}=\frac{3}{13}$

Assertion and Reasons

1. (a)
2. (b)
3. (a)
4. (d)
5. (c)
6. (b)
7. (a)
8. (b)
9. (c)
10. (c)
11. (a)
12. (c)

Hots Subjective Questions

1. When the ball rolls down the hill as shown in the figure, a component of weight of the ball acts downwards along the inclined plane. It is in the opposite direction of the force of friction. The net force along the inclined plane remains constant throughout the motion and hence the acceleration is also constant. Since a constant acceleration is acting on the ball, therefore the ball rolls down with increasing speed.

When the ball rolls down the hill-as shown in the figure, its acceleration starts decreasing after a certain height because the slope of the hill is not constant. When the ball rolls down from the top of the hill its acceleration increases but as soon as it reaches the point from where the slope is lesser, the acceleration starts decreasing.

2. If an object is dropped from rest i.e. its initial speed is zero, it falls under gravity with an acceleration $10m{s}^{-2}$. On the other hand if the object is thrown down instead then we impart some initial velocity (non zero) to it. As soon as it leaves our hand it moves under the force of gravity and hence its acceleration would be the same as $10m{s}^{-2}$. This is because the acceleration due to gravity remains constant in both the cases whether an object is dropped from rest or thrown downwards with some initial velocity:
3. When two weights are attached to a spring scale then the forcas acting on the strings of the scale are as shown in fig.

The spring balance will show $100N$

The two carts have masses say $m$ and $2m$ respectively when the compressed spring is released then the carts are pushed apart. The two carts are connected by the same spring so the action force of the spring on one mass would produce an equal and opposite reaction force on the other. The two carts therefore move away in opposite direction under equal force. We know that acceleration is inversely proportional to mass i.e, $a\propto \frac{1}{m}$

Let the accelerations of the two carts be $a_1$ and $a_2$ and their masses be $m_1$ and $m_2$ respectively where

$\begin{array}{rl}& m_1=m\text{ and }\\ & m_2=2m\text{ (given) }\\ & \Rightarrow a_1\propto \frac{1}{m_1}\text{ and }\\ & \Rightarrow a_2\propto \frac{1}{m_2}\\ & \Rightarrow \frac{a_1}{a_2}=\frac{m_2}{m_1}\\ & =\frac{2m}{m}\\ & a_1=\frac{2a_2}{2}\\ & \text{ or, }{a}_2=\frac{a_1}{2}\end{array}$

This means that cart twice as massive has half the acceleration as compared with the other.

If the crate slides on the floor with a constant velocity then it means that there is no net force acting on the crate. It means that the force with which we are pushing the crate is exactly equal in magnitude to the force of friction that is opposing the motion of the crate so that there is not net acceleration of the crate but only constant speed. This implies that the force of friction is equal to $200N$.

Yes, the force of friction is equal and opposite to the $200N$ push that we apply on the crate. The force of friction does not make up the reaction force to our push. This is because we are pushing the crate (action) and in turn the crate is pushing back our hands. The force of the crate on our hands is the reaction force here and not frictional force. Friction is a contact force that comes into play when one body moves (slides or rolls) over another. It acts on the two surfaces in contact only.

Since the two people pulling each other in a tug-of-war are of equal mass, and the ground is frictionless, this means that they both exert equal and opposite force on each other. Assuming that the rope does not extend or slag in length then the action-reaction pair of forces which are formed by the pull of both persons would accelerate both of them towards each other with equal magnitude no friction is present to retard their motion. This means that both persons slide towards each other with same speed and same distance

(acceleration $\propto \frac{1}{m}$ is same for both).

Hence they would meet each other after sliding equal distances from their initial positions.

This means they would meet at the centre of the rope length i.e, they meet after sliding $6m$ from each side at the centre of the rope.

7. When our hand turns the handle of a faucet, the faucet in turn applies a pressure on our fingers. These two forces i.e, the push of our hand on the handle of the tap and the push of the tap on our hand make the action-reaction pair.

When force is applied by our hand turns the faucet which moves the screw inside the tap and releases the water which flows out of the nozzle of the tap. The force of screw of the tap on the water and force of water on it make another action-reaction pair. But this means that our push on the handle and the water coming out of the tap do not make an action-reaction pair since our hand and water is not touching each other.

8. (a)

The figure shows a stone resting on the ground. The net force on the stone is zero as its acceleration is zero. One of the force that acts on stone is the weight force of the stone, due to earth’s gravitational pull, acting in the downward direction. This force acts on the ground and pushes it down. As a reaction to this force the ground also exerts an equal and opposite force on the stone in the upward direction. As a result the two forces cancel and no net force acts on the stone. The force vectors can be represented as shown in figure 1

(b) The downward weight force is balanced by an upward force. This upward force is conventionally called reaction force of the ground that prevents the ground from sinking and keeps the stone at rest. It acts normal to the surface of the ground

9. The various force vectors that act on the stone are shown in the figure.

$T=$ tension in the string

$mg=$ weight of the stone

$T=mg$

Net force $=0$

(a) There is a downward force of weight of the stone given as $mg$ where $\mathit{m}$ is mass of the stone and $g$ is acceleration due to gravity. This weight force pulls the stone down and so the sting attached to it also experiences a pull in the downward direction. The string in turn pulls the fixed wall to which it is attached in downward direction. the fixed wall as a reaction to this pull (weight of stone) pulls the string back in upward direction. This pull is expressed in form of tension in the string. This tension or stretching force acts in upward direction and is said to be a force exerted by the fixed point of suspension on the stone.

(b) Since the stone is at rest and shows no acceleration this means that there is no net force on the stone. This happens because the tension and weight force are equal and opposite. They both cancel each other and net force on stone is zero. That is, $T=mg$ and $F_{\text{net }}=$ $T-\mathit{m}\mathit{g}=\mathbf{0}$ acceleration of the stone $a=0$.

Therefore resultant of two vectors is zero.

(c) The stone suspended by the string shows no motion which means its acceleration is zero and net force is also zero.

This happens because downward weight force is balanced by upward tension in the string. Since these two forces are action and reaction forces hence they cancel each other and give the resultant force as zero.

10. The string attached to the stone puils it upward. This means that an external force is applied from the fixed point to pull the stone up. The stone accelerates in upward direction because the upward force (of pull) is greater than the downward force of weight

When the string breaks from the fixed point then this upward force (of pull) on the stone stops acting on it and the stone is under the influence of its weight force only when this happens then there is no upward force on the stone. At the top of its path the stone is momentarily at rest because its upward velocity becomes zero due to deceleration. The only force acting on it at that point is the force of gravity which is represented by weight of the stone as shown in the figure.

11. At the top of the path the stone is at rest. This happens because the upward acceleration of the stone is balanced by the downward acceleration due to gravity. The net force on the stone at the top is the force due to gravity

At that moment, its instantaneous velocity Vinst is zero (rest point). The acceleration of the stone is not zero but it is equal to ’ $g$ ’ i.e. acceleration due to gravity. This is because the upward moving stone slows down under the effect of gravity which acts at a retarding force and brings it to rest after which the stone falls down. At all times gravitational force is acting on the stone and it changes its velocity by decreasing it to zero and then increasing it in reverse direction

(a) The various forces that act on the stone sliding down are

(i) Weight of the stone given as $mg$ in the downward direction.

(ii) reaction force of the incline acting in a normal direction to the incline.

When a stone slides on an frictionless incline then the force of gravity acts on it at all points a vertically downward direction. This is due to the fact that acceleration due to gravity always acts forwards the centre of the earth and hence its direction is always given by a downward vertical \to. Secondly, the motion of the stone on the incline pushes the ground which produces a reaction force by the ground which is always perpendicular to the surface and acts on the moving object in a direction away from it this force of reaction is represented by vector $R$ in the fig. 1

Since it is a frictionless incline so we do not consider frictional force.

(b) The two forces acting on the stone form a parallelogram as shown in fig 2. and their resultant is given by the diagonal of the parallelogram. As seen in the figure the diagonal vector $F_R$ represents resultant of the two force vectors and it can be seen that $F_R$ is parallel to the incline. The resultant force is in the same direction as the stone’s acceleration.

$R=$ reaction force of surface on stone

$mg=$ weight force of stone

$F_{\text{net }}=$ resultant force of $R$ and $mg$

$F_a=$ force of action of stone on block

$F_r=$ force of reaction of block on stone

(a) Here fig. 1 the stone sliding on a frictionless surface is stopped by a block so that the stone is at rest. The various forces that act on the stone are

(1) weight of the stone acting in a vertically downward direction and represented by ’ $mg$ ‘.

(2) the normal reaction of the surface to its weight force acting perpendicularly upward from the surface on the stone and represented by vector $R$.

These two forces form two adjacent sides of a parallelogram and their resultant is the diagonal of the parallelogram given by vector $F_{\text{net }}$ (fig 3 ). This resultant force acts as the action force ’ $F$ ’ ’ on the block due to the push of the stone. As a reaction to this the block pushes the stone by an equal and opposite force represented by $F_r$

(3) This reaction force of the block on the stone is the third force acting on the stone

(b) As the action-reaction forces cancel i.e, $F_a=F_r$ hence there is no net force acting on the stone and it is therefore in a state of rest.

14. $3.2m/s^2$

Hints: $a_1=16m/s^2,m_1=0.5kg$,

$F=m_1{a}_1=16\times 0.5kg=8N$

$m{a}_2=\frac{F}{a_2}=\frac{4}{8}=2kg;{\text{ddotm}}_1+m_2=0.5+2=2.5kg$,

$F=8N;a=\frac{F}{m_1+m_2}=\frac{8}{2.5}=3.2m/s^2$

15. $-3\times {10}^{4}N$

Hints: $m=\frac{20}{1000}kg,u=300m/s,v=0,s=3cm$

$=3\times {10}^{-2}m$

$\begin{array}{cc}\therefore & a=\frac{v^2-u^2}{2s}=\frac{0-(300{)}^2}{2\times 3\times {10}^{-2}}=\frac{-300\times 300}{2\times 3\times {10}^{-2}}\\ & =-1.5\times {10}^{6}m/s^2\\ \therefore & F=ma=\frac{20}{1000}\times (-1.5\times {10}^6)=-3\times {10}^{4}N\end{array}$

16. $12.5m/s^2$

Hints: Horizontal component of force $=F\sin \theta$

$\Rightarrow a=\frac{F\sin \theta }{m}=\frac{50\sin {30}^{\circ }}{2}=12.5m/s^2$

17. $-0.99m/s$, zero before and after firing

Hints: $M=10kg,m=\frac{30}{1000}kg.,V=?,v=330m/s$

From conservation of momentum, $MV+mv=0$

$\Rightarrow 10\times V=-\frac{30}{1000}\times 330\Rightarrow V=-0.99m/s$

Resultant momentum before and after firing is zero as no external froce is acting on it.

CHAPTER 5

WORK, ENERGY AND POWER

Introduction

In common sense, all physical activities such as reading, writing, carrying a bag, pushing a wall etc. are considered to be work. Let, for example, you employ a worker to push and move a high wall of a building. He tries his best whole day but he doesn’t become able to move the wall a little bit. But he asks for his labouring charge. Would you pay him the charge or not? Definitely, your answer would be ‘yes’because he has performed his task or work. But in language of physics he hasn’t done any work.

The meaning of work in physics is a big different from its meaning in common language. Actually, in physics work has a meaning only when a displacement is caused on a body by the applied force on it. If there is no displacement in a body by an applied force, no work is said to be done on the body by the force.

Suppose Pari gets tired after four hours of study but Dolly doesn’t tire even after eight hours of study. Who does more work-Pari or Dolly? We can’t answer it. But we can surely say that Dolly has more capacity of doing work than Pari. The capacity of doing work is defined as the ‘Energy’. Also, the rate of doing work is termed as the ‘Power’.

This chapter is all about Work, Energy and Power.

Work

In physics work is defined if force applied on object displaces the object in direction of force. Here all three terms force, displacement and direction of force are important. If force is zero, work is zero; if force is non-zero but displacement is zero (like pushing the wall) work is zero and if force is non-zero, displacement is non-zero but no part of force is in the direction of displacement, work in zero. Hence, we define the work as the product of the force and displacement in the direction of applied force or product of displacement and force in the direction of displacement.

Note that the force in above formula is the component of force is the direction of displacement in the diretion fo force

Unit of Work

$\begin{array}{rl}W& =\text{ Force }\times \text{ displacement in the direction of force }\\ & =FS\end{array}$

The $Sl$ unit of force is newton and the unit of length is metre ( $m$ ). So, the $SI$ unit of work is newton-metre which is written as $Nm$. This unit $(Nm)$ is also called joule $(J)$, i.e.

$1\text{ joule }=1\text{ newton }1\text{ metre }$

Abbreviated as $1J=1Nm$

When a force of $t$ newton moves a body through a distance of 1 metre in its own direction the work done is 1 joule.

Other units of work

In c.g.s. system of measurement force is measured in dyne, displacement in $cm$, and work is measured in erg.

From $W=Fs$

if $F=1$ dyne, $s=1cm,=1$ then $W=1$ erg.

Thus, if a force of 1 dyne, displaces the point of application, by $1cm$, in the direction of force, then work done by the force is said to be 1 dyne is as follows.

Relation between 1 joule and erg is as follows.

$1\text{ joule }=1N\times 1m={10}^5\text{ dyne }\times {10}^{2}cm={10}^{7}erg$

Higher units of work are $kJ={10}^3$ joule

$\text{ and }1MJ={10}^6\text{ joule }$

Do you know !!

If nothing is actually moving, no work is done-no matter how great the force involved!

Work Done By A Force Applied At An Angle

Let Rishabh pulls a toy car through a string, then as shown in figure the force applied by him is along the string (direction $OA$ whereas the toy car moves horizontally (direction $OX$ ). The pulling force $(F)$ makes an angle $\theta$ with the displacement of car. In this case only a part of the force, say $F_1$, which acts along the horizontal direction is being actually used for the motion of the car. In such a case the amount of work done by the force on the car is defined as the product of its component along the motion and the magnitude of displacement.

$W=$ component of force in the direction of displacement $\times$ magnitude of displacement $W=F\cos \theta s$

$\therefore$ Work done by a force can be positive or negative according as the value of $\cos \theta$ is positive or negative. ( $\because F$ and $s$, being magnitudes, are always positive)

Values of $\cos \theta$

Thus, if the displacement (of the point of application) has a component along the direction of applied force, then/work done is positive. On the other hand, if the displacement has a component opposite to the direction of applied force, then work done will be negative.

If the applied force and particle’s displacement be mutually perpendicular, then work done by the force on the particle is zero. ( $\cos {90}^{\circ }$ being zero).

It is important to understand that work done by the force does not depend on the time taken in the displacement of point of action. For example, one porter takes 5 minutes to put a box on the roof of a bus while other put the same box on the roof in 10 minutes, work done by both the porter is same.

Do you know !!

Work is a scalar quantity but you can have positive and negative work. Positive work is where the force pulls in the same direction as the movement. Negative work is where the force is in the opposite direction.

Example:

(i) If the wall doesn’t move, the person does no work.

(ii) A body attached to a string revolves in a horizontal circle (figure). The tension $T$ in the string does no work on the body, because it has no component in the direction of displacement. In general, for a body moving with uniform speed the centripetal force is always perpendicular to displacement, hence no work will be done by this force.

(iii) A coolie with a luggage on his head, moving on a horizontal platform, does no work, since the direction of force is vertically up and displacement horizontal (even though he might feel physically tired).

(iv) If a boy tries to push a heavy boulder, by applying a force, but unable to displace it, then work done by the boy is zero.

(v) When a horse pulls a cart, the applied force and the displacement are in the same direction. So, work done by the horse is positive.

(vi) When brakes are applied to a moving vehicle, the work done by the braking force is negative. This is because the braking force and the displacement act in opposite directions.

(vii) When a spring is compressed then the force applied by the spring and the displacement will be in opposite to cach other, so work done by the spring will be negative.

Work Done-Against Gravity

Consider a body of mass $m$ which is raised a vertical distance $h$.

The work done by the weight is $-mgh$.

$mgh$ is called the work done against gravity.

If an agent, such as crane, is responsible for lifting the body, then $mgh$ is referred to as the work done by the crane against gravity. Similarly if a vehicle of mass $m$ climbs a hill, and in doing so raises itself a vertical distance $h$, then $mgh$ is called the work done by the vehicle against gravity,

Work Done By A Moving Vehicle

The diagram shows the forces that commonly act on a moving vehicle. $R$ is the resistance to motion (this is always in the direction opposite to the direction of motion) and $F$ is the driving force of the engine.

The work done by $F$ is referred to as the work done by the vehicle.

If the vehicle is not accelerating, the forces acting on it are in equilibrium.

KNOWLEDG NHANCER

(ii) Internal work:

Suppose that a man sets himself in motion backward by pushing against a wall. The forces acting on the man are his weight ’ $W$ ‘, the upward force $N$ exerted by the ground and the horizontal force $N$ ’ exerted by the wall. The works of ’ $W$ ’ and of $N$ are zero because they are perpendicular to the motion. The force $N$ is the unbalanced horizontal force that imparts to the system a horizontal acceleration. The work of $N$, however, is zero because there is no motion of its point of application. We are, therefore, confronted with a curious situation in which a force is responsible for acceleration, but its work, being zero, is not equal to the increase in kinetic energy of the system.

The new feature in this situation is that the man is a composite system with several parts that can move in relation to each other and thus can do work on each other, even in the absence of any interaction with externally applied forces. Such work is called internal work. Although internal forces play no role in acceleration of the composite system, their points of application can move so that work is done; thus the man’s kinetic energy can change even though the external forces do no work

$✓$ CHECK POINT

Figure shows four situations in which a force acts on a box while the box slides rightward a distance $d$ across a frictionless floor. The magnitudes of the forces are identical, their orientations are as shown. Rank the situations according to the work done on the box during the displacement, from most positive to most negative.

(A)

(B)

(C)

(D)

Check Your Answer

We know that the positivity or negativity of work depends upon the angle between the force and the displacement caused by the force. The work done becomes less positive as $\theta$ increases from 0 to 90 and becomes negative afterwards upto ${270}^{\circ }$. Hence, the correct order is

D, C, B, A

Illustration 1

How much work is done by a force of $250N$ in moving an object through a distance of $100m$ in the direction of force?

Solution : The work done is calculated by the formula

$W=Fs$

$=250\times 100$

$=25\times {10}^3$ joule

Illustration 2

A body of mass $2kg$ is raised to a height of $1m$. Find the work done by the force of gravity.

Solution : The force of gravity on the body, is the force exerted by earth on it and is $mg$ (weight) acting vertically down. However, the displacement $s=1m$ vertically up.

Now from $W=Fs\cos \theta$

$\begin{array}{rl}W& =Fs\cos \theta \\ W& =2\times 9.8\times 1\times \cos {180}^{\circ }[\because \cos {180}^{\circ }=-1]\\ & =-19.6\text{ joule }\end{array}$

Illustration 3

A child pulls a toy car through a distance of 10 meters on a horizontal floor. The string held in child’s hand makes an angle of ${60}^{\circ }$ with the horizontal surface. If the force applied by the child be $10N$, calculate the work done by the child in pulling the toy car.

Solution : In this case, as the applied force and the displacement are not in same direction, we will calculate the work done by the formula.

$W=Fs\cos \theta$

Here, Force $F=10N$

Magnitude of the displacement $s=10m$

$\theta ={60}^{\circ }$

So substituting these values in the above formula, we get

$W=(10)\times (10)\cos {60}^{\circ }$

From the table of natural cosines $\cos {60}^{\circ }=1/2$

$W=10\times 10\times \frac{1}{2}=50\text{ joule }$

Illustration 4

A man lifts 20 boxes each of mass $15kg$ to a height of $1.5m$. Find the work done by the man against gravity.

Solution :

The work done against gravity in lifting one box $=15g\times 1.5J=22.5gJ$ The work done against gravity in lifting 20 boxes $=20\times 22.5gJ=450gJ$

Illustration 5

From the graph (figure given below) of force versus displacement of a particle, find the total work done by the force.

Solution :

The total work done $=$ Total area under the $(F-x)$ curve

$=\frac{1}{2}\times 6\times 30=90\text{ joule }$

Illustrotion 6

A force of $10N$ displaces a body by $5m$, the angle between force and displacement is ${60}^{\circ }$, then find the work done.

Solution : Force, $F=10N$, Displacement, $d=5m$

Angle between force and displacement, $\theta ={60}^{\circ }$

Work done, $W=Fd\cos \theta =10\times 5\times \cos {60}^{\circ }\therefore \cos {60}^{\circ }=1/2$

then, $W=10\times 5\times \frac{1}{2}=25$ Joule

Energy

Some people have a lot of energy when they get up in the morning. Carbohydrates are high-energy foods. Oil is the main source of the energy that keeps incustry and cars going. The centre in a football formation has to put a lot of energy into his attack.

The word energy has a different meaning in each of those four sentences. As used in physics, the word has a very precise meaning, although it is a little difficult to define because energy takes many different forms. We can approach a definition by noting the relationship between energy and work. Energy is defined as the capacity to do work.

In other words, anything which has the capacity to do work is said to possess energy. The implies that work can be done only at the expense (cost) of energy i.e., to do work, we need to spend energy, whatsoever be its form.

Unit of Energy: Its unit is same as that of work i.e., joule (J).

IDEA BOX

Let us understand energy-work equivalence with some examples

(i) When a fast moving cricket ball hits a stationary stump, the stump is thrown away. Here the work is done on the stump by the ball and the ball has the capacity to do this work because of its motion (kinetic energy).

(ii) A body can acquire the ability to do work when it is deformed temporily. For example, a compressed watch spring is able to drive the wheels of a watch.

(iii) If a boy (mass $=m$ ) climbs upstairs to a height $(h$ ) then work done by him would be $mgh$ and consequently he would have lost $mgh$ joule of energy.

Do you know !!

Energy is a promise of work to be done in future. It is the stored ability to do work.

Mechanical Energy

The energy in a body may be by virtue of its motion (kinetic energy) or by virtue of its position (potential energy). Energy in a body due to these conditions is called mechanical energy. For example, energy of water in a water tank on the roof, energy of moving bullet, energy of small spring in ball-pen, energy of moving air etc. are the forms of mechanical energy.

Kinetic Energy

Energy possessed by a body by virtue of its state of motion is called kinetic energy. Kinetic energy is always positive and is a scalar. The fact, that moving bodies carry energy with them is proved by some of the several happenings in day to day life.

Examples :

(i) A stone thrown with some velocity, breaks the window pane.

(ii) A moving vehicle, when accidently happens to collide with another vehicle at rest or motion, leads to destruction.

Do you know !!

The kinetic energy of a body always positive.

Expression of kinetic energy

Suppose that a constant force $F$ is applied on a body of mass $m$. Its velocity becomes $v$ in a displacement $s$, then according to Newton’s 3rd equation of motion .

$v^2=0+2(\frac{F}{m})s\text{ or }(F)=m(\frac{v^2}{2s})$

Work done by force $F$ in displacing the body by a distance $s$ in the direction of force

$W=F_s.=m(\frac{v^2}{2s})s\text{ or }W=\frac{1}{2}m{v}^2$

This work done by the force which makes a stationary body to move with a velocity $v$, is measured as its kinetic energy i.e.

Kinetic energy $K=\frac{1}{2}m{v}^2$

From this expression it is clear that the kinetic energy possessed by a moving body is directly proportional to its mass and to the square of its velocity, if velocity is doubled KE becomes 4 times.

KNOWLEDG NHANCER

Relation between kinetic energy and momentum :

A body of mass $m$ is moving with velocity $v$, then the momentum of the body, $p=mv$

The kinetic energy of the body, $K=\frac{1}{2}m{v}^2$

or

$K=\frac{1}{2}m{v}^2\times \frac{m}{m}$ or $K=\frac{1}{2}\frac{m^2{v}^2}{m}$

$(.$ Here $.m^2{v}^2=p^2)$

$K=\frac{p^2}{2m}$

or $p=\sqrt{2mK}$

(i) For same momentum :

K-energy varies inversely as the mass $K\propto \frac{1}{m}$

(ii) For same K-energy :

Momentum varies directly as the square root of mass of the body. $p\propto \sqrt{m}$

CHECK POINT

(i) If a golf ball and a Ping-Pong ball both move with the same kinetic energy, can you say which has the greater speed? Explain in terms of K.E. Similarly, in a gaseous mixture of massive molecules and light molecules with the same average $KE$, can you say which has the greater speed?

(ii) How is it possible that a flock of birds in flight can have a momentum of zero but not have zero kinetic energy?

Check Your Answer

(i) The kinetic energy of an object depends on the mass and square of the velocity . As the kinetic enengy of a golf ball and a ping-pong ball is the same therefore a ping-pong ball has greater speed because its mass is less than a golf-ball. Similarly, in a gaseous mixture of massive molecules and light molecules with the same average K.E, the light molecules have the greater speed.

(ii) Momentum is a vector quantity having both magnitude and direction whereas kinetic energy is a scalar quantity having only magnitude. Momentum that is directional is capable of being cancelled entirely.

The vector sum of the momenta of a flock of birds in flight can be zero because of birds flying in different directions in the flock.

Each flying bird has some kinetic energy and the algebraic addition of the kinetic energies of all birds in the flock cannot be zero.

Illustration 7

If a stone of mass $3kg$ be thrown with a kinetic energy of 37.5 joule, find its velocity.

Solution: From K.E. $=\frac{1}{2}m{v}^2$

$\Rightarrow 37.5=\frac{1}{2}\times 3v^2\Rightarrow v^2=\frac{75}{3}=25\Rightarrow v=5m/s$

Illustration 8

A scooter is moving with a velocity of $15m/sec$. Calculate its kinetic energy if its mass along with the rider is $150kg$.

Solution: Kinetic energy is given by

$=\frac{1}{2}m{v}^2=\frac{1}{2}\times 150\times (15{)}^2=75\times 225=16875J$

Illustration 9

A bullet is fired from a gun. What will be the ratio of kinetic energy of bullet and gun ?

Solution: When a bullet is fired from a gun, the gun has same momentum backward, which the bullet has a forward momentum

$\because K\propto \frac{1}{m}$

$\frac{\text{ K.E. of bullet }}{\text{ K.E. of gun }}=\frac{\text{ mass of gun }}{\text{ mass of bullet }}=\frac{M}{m}$

So bullet has more K-energy than the gun.

Conservative And Non Conservative Forces

Nature has been gifted by several types of forces, coming into operation, now and then, as and when possible. To name a few, the wind force, the force of friction, the viscous drag, the gravitational force, the electric force, etc.

The forces which always tend to oppose the motion e.g., frictional forces, viscous forces, etc. always tend to dissipate energy when a body moves relative to another, in their presence i.e., work has always to be done against them, to move a body from one position to another, which cannot be recovered, after regaining the original position. These are hence known as non-conservative forces. At the same time, there are several forces which are conservative in nature and in which energy can be recovered, provided, the body is restored to its original configuration. Commonly encountered such forces are gravitational, elastic forces.

Hence we can define conservative and non-conservative forces as follows.

$1^{\text{st }}$ definition : If the work done by a force in moving a body from an initial location to a final location is independent of the path taken between the two points, then the force is conservative, otherwise it is non-conservative.

$W_1=W_2=W_3$ for a conservative force.

Example : Both blocks acquire the same gravitational potential energy, $mgh$. The same work is done on each block. What matters is the 2 definition: If a body moves under the action of a force that does no net work during any round trip, then the force is conservative, otherwise it is non-conservative.

Conservative Force and Work done

As we have already discussed in the previous chapter that work done by a conservative force doesn’t depend upon the path taken i.e. it only depends upon the initial and final positions of the body. There is another important very useful relation between work done by conservative force and potential energy. The work done by a conservative force is equal to the negative of change in the potential energy.

$W_c=-(U_f-U_i)=-\mathrm{\Delta }U$

$U_f=$ final potential energy

$U_i=$ initial potential energy

$F_c\mathrm{\Delta }x=-\mathrm{\Delta }U$

$F_c-\frac{\mathrm{\Delta }U}{\mathrm{\Delta }x}$ or $F_c=-\frac{dU}{dx}$ or, $U_f-U_i=-\int \overrightarrow{F}\cdot \overline{dr}$

POTENTIAL ENERGY

Potential energy is energy due to position. If a body is in a position such that if it were released it would begin to move, it has potential energy. There are two common forms of potential energy, gravitational and elastic.

(i) Gravitational Potential Energy

When an object is allowed to fall from one level to a lower level it gains speed due to gravitational pull, i.e. it gains kinetic energy. Therefore, in possessing height, a body has the ability to convert its height into kinetic energy, i.e. it possesses potential energy.

The magnitude of its gravitional potential energy is equivalen to the amount of work done by the weight of the body in causing the descent.

If a mass $m$ is at a height $h$ above a lower level, the P.E. possessed by the mass is $(mg)(h)$.

Since $h$ is the height of an object above a specified level, an object below the specified level has negative potential energy.

Note: The chosen level from which height is measured has no absolute position. It is, therefore, important to indicate clearly the zero P.E. level in any problem in which P.E. is to be calculated.

Do you know !!

It does not matter, what path is taken; work done to move a body between any two points in a vertical plane is equal to work done against gravity through the vertical distance between the points!

(ii) Elastic Potential Energy

This is a kind of potential energy which is due to a change in the shape of a body. The change in shape of a body can be brought about by stretching, compressing, bending and twisting the body. Some work has to be done to change the shape of a body. This work gets stored in the deformed body in the form of elastic potential energy. For example, the energy stored in a stretched rubber band or a spring is elastic potential energy and is equal to the work done in stretching the rubber band or spring. When this deformed body is released, it attains its original shape and the potential energy is converted into some other form, usually in kinetic energy. Elastic potential energy is never negative whether due to extension or to compression.

Consider a massless spring whose length is $\ell$ on its natural state. Now it is elongated

by ’ $x$ ’ then the workdone by external force will be $\frac{1}{2}k{x}^2$. Similarly, if the spring is compressed by ’ $x$ ’ then also the work will be $\frac{1}{2}k{x}^2$ only.

So, the energy associated with the state of a elongation or compression of a spring is called elastic potential energy of spring or simply potential energy $(U)$.

$U=\frac{1}{2}k{x}^2$

Where ’ $k$ ’ is force constant of spring. It is also called spring factor, its unit is $N/m$ or dyne $/cm$. Remember spring contant $(k)$ is inversely proportional to the lengths of spring $(\ell )$ i.e.,

$k\propto \frac{1}{\ell }\text{ or }k\ell =\text{ constant }$

In which car will you be moving the fastest at the very bottom of the incline?

Check your answer

Of course, at any moment all cars have the same speed but the gravitational potential energy of the system of cars is least when the middle car is at the bottom - when the centre of mass of the system is lowest. Lowest potential energy means highest kinetic energy. So sit in the middle car for the finstest ride at the very bottom.

Conservation of Energy

According to this law, energy can only be converted from one form to another, it can neither be created nor destroyed. The total energy before and after the transformation always remains the same.

Conservation of Mechanical Energy

Kinetic and potential energy are both forms of mechanical energy. The total mechanical energy of a body or sywtem of bodies will be changed in value if :

(a) an external force other than weight causes work to be done (work done by weight is potential energy and is therefore already included in the total mechanical energy),

(b) some mechanical energy is converted into another form of energy (e.g. sound, heat, light etc.). Such a conversion of energy usually takes place when a sudden change in the motion of the system occurs. For instance, when two moving objects collide some mechanical energy is converted into sound energy which is heard as bang at impact. Another common example is the conversion of mechanical energy into heat energy when two rough objects rub against each other.

If neither (a) nor (b) occurs then the total mechanical energy of a system remains constant. This is the principle of conservation of Mechanical Energy can be expressed in the form :

The total mechanical energy (K.E. + P.E.) of a system remains constant provided that no external work is done and no mechanical energy is converted into another form of energy.

Examples :

(1) Body falling freely :

Considering earth as a body of zero mechanical energy, if a body of mass $m$ is situated at point $A$ whose height from earth surface is $h$, then the total mechanical energy of the body/system will be potential energy only. i.e. $E_A=m$ gsh Now, as body falls freely from point $A$, its velocity inereases due to acceleration due to gravity. At the point $B$, boty passesses both K.E. and P.E.

So, potential energy at $B:U_B-mg(h-x)$

Kinetic energy at $B:K_B=\frac{1}{2}m{v}_B^2$

Total mechanical energy at $B:E_B=U_B+K_B$

or $E_B=mg(h-x)+\frac{1}{2}m{v}_B^2$

or $E_B=mg(h-x)+\frac{1}{2}m\cdot 2gx$ or $E_B=mgh$

When the body reaches at point $C$ on earth’s surface, $h$ becomes zero. Therefore, the potential energy $(U_c)$ at $C$ is zero. Total mechanical energy at $C$ is K.E. only which equals the work done by gravitational force from $A$ to $C$. Therefore,

$\begin{array}{rl}-E_C& =0+\frac{1}{2}m{v}_C^2\\ \text{ or }{E}_C& =0+\frac{1}{2}m\cdot 2gh\text{ or }{E}_C=mgh\end{array}$

Thus, the total mechanical energy remains conserved at every point on the path of a freely falling body.

(2) When the pendulum is pulled to position $C$, it gains height.

At position $C$, it has maximum potential energy and zero kinetic energy, as the pendulum is held by hand in position $C$.

When the pendulum is released from position $C$, it moves towards position $A$. In doing so, its velocity increases, due to the increase in velocity, its kinetic energy increases, at the expense of potential energy. At position $A$, it has maximum kinetic energy and zero potential energy, as it is at its lowest position. (we can call it as recerence level)

When the pendulum swings from $A$ to $B$, it again gains height and hence, its potential energy increases. However, due to gain in height, its velocity decreases and hence, the kinetic energy decreases. At position $B$, it has maximum potential energy and zero kinetic energy, as pendulum comes to rest at $B$ for a moment, before swinging back to position $A$.

Hence in the system of pendulum and earth, the energy is conserved. It is the potential energy, which changes to the kinetic energy and vice versa.

Illustration 10

A mass of $10kg$ is released from a helght of $40m$ above the surface of the earth. Calculate its mechanical energy when it is at $20m$ above the surface of the earth. (Take $g=10m/s^2$ )

Solution :

(i) The initial energy of the body $(U+K)=mgh+\frac{1}{2}m{v}^2$

Here, initially the body is at rest $v=u=0$

and $m=10kg,h=40m$, and $g=10m/s^2$

$\therefore$ initial energy $=10\times 10\times 40+\frac{1}{2}m\times 0=4000J$

(ii) When the body is at a height of $h_1=20m$ above the ground, it has fallen by a distance of $s_1=(40-20)=20m$ from its initial position, at this position it will have both the potential and kinetic energy.

At this position, kinetic energy $K=\frac{1}{2}m{v}_1^2$

$v_1^2=2g{s}_1$

$K=\frac{1}{2}m\times 2\times g\times s_1=\frac{1}{2}\times 10\times 2\times 2\times 10\times 2$

or $K=\frac{1}{2}\times 10\times 400=2000J$

Here the body is at height $h_1$ above the surface of earth hence the (gravitational) potential energy

$U=mg{h}_1=10\times 10\times 20=2000J$

$\therefore$ Total energy $=K+U=2000+2000=\mathbf{4}\mathbf{0}\mathbf{0}\mathbf{0}J$

Illustration 11

A body of mass $10kg$ is kept at a height $10m$ from the ground, when it is released, after sometime its kinetic energy becomes 450 joule. What will be the potential energy of the body at that instant.

Solution : At a height of $10m$ the mechanical energy of the body,

$E=\text{ Kinetic energy }+\text{ Potential energy }$

$E=m(0{)}^2+mgh(\because \text{ initial velocity of the body is zero })$

$E=10\times 10\times 10=1000\text{ Joule }$

After sometime the kinetic. energy is 450 joule, suppose at that instant potential energy is $u$, then by the law of conservation of mechanical energy

$E=450+u$

$1000=450+u$

or $u=1000-450$

or $u=550$ joule

Illustration 12

Starting at rest, the cart of figure slides frictionlessly to point $P$, which is 4.5 meters below the top of the hill. How fast is it going at $P$ ?

Solution : Considering its gravitational energy to have dropped to zero at $P$, its gravitational energy when it is at the top of the hill is $(mgh{)}_{\text{start }}$ Its kinetic energy starts at zero and increases to $(\frac{1}{2}m{v}^2{)}_{\text{end }}$. Since its total energy does not change,

$(mgh{)}_{\text{start }}=(\frac{1}{2}m{v}^2{)}_{\text{end }}$

The m’s drop out, and the equation becomes

$v=\sqrt{2gh}=\sqrt{2(9.8m/s^2)(4.5m)}=9.4m/s$

WORK-ENERGY THEOREM

According to the work-energy theorem, total work done on a system by forces (external, internal, conservative or non-conservative) equals the change in kinetic energy.

$W_c+W_{nc}+W_{ext}=K_f-K_i$

Where, $W_c=$ Work done by conservative forces

$W_{nc}=$ Work done by non-conservative forces

$W_{ext}=$ Work done by external forces

$K_f=$ Final kinetic energy

$K_i=$ Initial kinetic energy.

But work done by conservative forces equals the negative of change in potential energy,

$W_c=-(U_f-U_i)$

So, $W_{nc}+W_{ext}=(K_f+U_f)-(K_i+U_i)$

$W_{nc}+W_{ext}=M{E}_f-M{E}_i$ (ME - Mechanical Energy)

If $W_{nc}=0;$ then $W_{ext}=M{E}_f-M{E}_i$

So, if non-conservative forces do not act then work done by external forces equals the change in mechanical energy and if there is no external force acting then mechanical energy of the system remains conserved.

Do you know !!

Work-energy theorem is particularly useful in calculation of minimum stopping force or minimum stopping distance. If a body is brought to a halt, the work done to do so is equal to the kinetic energy lost.

Illustrotion 13

A bullet leaving the nuzzle of a rifle barrel with a velocity $v$ penetrates a plank and loses one fifth of its velocity. It then strikes second plank, which it just penetrates through. Find the ratio of the thickness of the planks supposing average resistance to the penetration is same in both the cases.

Solution : Let $R=$ resistance force offered by the planks, $t_1=$ thickness of first plank, $t_2=$ thickness of second plank.

For first plank :

Loss in $KE=$ Work against resistance

$\frac{1}{2}m{v}^2-\frac{1}{2}m(\frac{4}{5}v{)}^2=R{t}_1\Rightarrow \frac{1}{2}m{v}^2(\frac{9}{25})=R{t}_1$ … (1)

For second plank

$\frac{1}{2}m(\frac{4}{5}v{)}^2-0=R{t}_2\Rightarrow \frac{1}{2}m{v}^2(\frac{16}{25})=R{t}_2$ … (2)

$\text{ Dividing (1) \& (2) }\Rightarrow \frac{t_1}{t_2}=\frac{9}{16}$

Illustrotion 14

A particle of mass $0.5kg$ travels in a straight line with velocity $v=a{x}^{3/2}$ where $a=5m^{-1/2}{s}^{-1}$. What is the work done by the net force during its displacement from $x=0$ to $x=2m$ ?

Solution : $m=0.5kg,v=a{x}^{3/2},a=5m^{-1/2}{s}^{-1},W=$ ?

Initial velocity at $x=0,v_0=a\times 0=0$

Final velocity at $x=2,v_2=a\times 2^{3/2}=5\times 2^{3/2}$

Work done $=$ increase in kinetic energy $=\frac{1}{2}m(v_2^2-v_0^2)=\frac{1}{2}\times 0.5[(5\times 2^{3/2}{)}^2-0]=50J$

Illustration 15

A force acts on a body of mass $3kg$ causing its speed to increase from 4 metre per second to $5m/s$. How much work has the force done?

Solution : Initial K.E. $=\frac{1}{2}m{v}_1^2=\frac{1}{2}(3)(4{)}^{2}J=24J$

Final K.E. $=\frac{1}{2}m{v}_2^2=\frac{1}{2}(3)(5{)}^{2}J=37.5J$

Work done $=$ Change in energy

Hence, work done by force $=(37.5-24)J=13.5J$

Illustration 16

A block of mass $2.0kg$ is given an initital velocity of $3m/s$ on a horizontal table where the coefficient of friction is 0.2 . Find the distance covered by the block before it stops. Take $g=10m/s^2$.

Solution : Alternatively, you can apply work-energy methods in a very precise way as follows :

By using Work-Energy theorem

For the block, in the system under consideration, the statement of work-energy theorem can be written as

$W_{mg}=W_N=K_{fk}=K_2-K_1$

$\Rightarrow 0+0+(-{\mu }_{k}mg\times s)=0-\frac{1}{2}m{v}_0^2$

$\Rightarrow s=\frac{v_0^2}{2{\mu }_{k}g}=\frac{(3{)}^2}{2\times 0.2\times 10}=2.25m$

Notice the difference between the “work done against friction” and the “work done by the frictional force”. In this example, work done against friction is ${\mu }_{k}mgs$, and the work done by the frictional force is $(-{\mu }_{k}mgs)$.

Motion In a Vertical Circle

Consider a particle of mass ’ $m$ ’ tied to an inelastic, massless string of length $R$ moving in a vertical circle as shown in fig.

At $\mathit{A}$ :

$\begin{array}{rl}& .T_A-mg=\frac{m{v}_A^2}{R}\text{ (Newton’s Second law }\sum f=ma,\text{ here }a=\frac{v^2}{R})\\ & T_A=mg+\frac{m{v}_A^2}{R}\end{array}$

At $B$ :

$\begin{array}{rl}& T_B-mg\cos \theta =\frac{m{v}_B{}^2}{R}\\ & T_B=mg\cos \theta +\frac{m{v}_B{}^2}{R}\end{array}$

Applying conservation of mechanical energy between $A$ and $B$

$\frac{1}{2}m{v}_A^2=\frac{1}{2}m{v}_B^2+mgR(1-\cos \theta )$

$v_A^2=v_B^2+2gR(1-\cos \theta )$

(Taking ’ $A$ ’ as reference line of zero $PE$ }

Substituting in (ii), we get

$T_B=\frac{m{v}_A{}^2}{R}-2mg+3mg\cos \theta$

Using equation (iii), tension at any point can be found by simply substituting the value of angle ’ $\theta$ ‘.

At $C$ :

$\begin{array}{rl}& \theta ={90}^{\circ },T_c=\frac{m{v}_A{}^2}{R}-2mg(v_c^2=v_A^2-2gR)\\ & T_c=\frac{m{v}_c{}^2}{R}\end{array}$

At $D$ :

$\begin{array}{rl}\theta ={180}^{\circ },T_D& =\frac{m{v}_A^2}{R}-5mg(v_c^2=v_A^2-4gR)\\ T_D& =\frac{m{v}_D{}^2}{R}-mg\end{array}$

Condition for just completing vertical circle :

Using $T_D\ge 0$; we get $v_D=\sqrt{Rg}$

$v_A=\sqrt{5Rg}\text{ and }{v}_c=\sqrt{3Rg}$

Similarly, $T_A=6mg;T_c=3mg$

Condition for leaving circle :

$\sqrt{2Rg}<v_A<\sqrt{5Rg}\text{ when }{90}^{\circ }<\theta <{180}^{\circ }$

At this point tension will be zero but velocity will not be zero.

Condition for oscillation :

When $0<v_A<\sqrt{2Rg}$ i.e. $0<\theta <{90}^{\circ }$, the particle will oscillate.

Motion of a body on a smooth hemisphere : Consider a small body of mass ’ $m$ ’ initially at rest kept over a smooth hemisphere of radius ’ $R$ ‘. It is given a little push and it starts sliding over the surface of hemisphere and at a point ’ $B$ ’ it leaves off the contact with the hemisphere.

Using Newton’s $2^{\text{nd }}$ law at $B$

$mg\cos \theta -N=\frac{m{v}^2}{R}$ …(i)

Applying conservation of energy between $A$ and $B$

$mgR=mgR\cos \theta +\frac{1}{2}m{v}^2$ …(ii)

When body leaves the contact, $N=0$

$\begin{array}{rl}& mg\cos \theta =\frac{m{v}^2}{R},\text{ using (ii) }\\ & \cos \theta =\frac{2}{3}\text{ or, }\theta ={48.2}^{\circ }\text{ and }v=\sqrt{\frac{2gR}{3}}\text{ and }AC=\frac{R}{3}\end{array}$

Looping the loop : Consider a smooth inclined plane of vertical height $H$ which is connected to a smooth vertical circular strack of radius ’ $R$ ‘. A mass ’ $m$ ’ slides down from the point $A$ and moves along the circular track and at point ’ $CD$ ’ leaves the contact from the track with speed $v$.

Using Newton’s $2^{\text{nd }}$ Law at $D$,

$N+mg\cos \theta =m(\frac{v^2}{R})$

As it leaves the contact, $N=0$

So, $mg\cos \theta =\frac{m{v}^2}{R}$ …(i)

Applying conservation of mechanical energy at $A$ and $D$,

$mgH=mg(R+R\cos \theta )+\frac{1}{2}m{v}^2$ …(ii)

Solving (i) and (ii),

$v=\sqrt{\frac{2g(H-R)}{3}}$

Also $\cos \theta =\frac{2(H-R)}{3R}$

Illustration 17

You may have seen in a circus a motorcyclist driving in vertical loops inside a ‘death-well’ (a hollow apherical chamber with holes, so the spectators can watch from outside). Explain clearly why the motoreyclist does not drop down when he is at the uppermost point, with no support from below. What is the minimum speed required to perform a vertical loop, if the radius of the chamber is $25m$ ?

Solution : The motorcyclist is expected to drop down under the action of his weight, when he is at the uppermost point. However, it does not happen so. The reason is that at the uppermost point, the weight of the motorcyclist provides the necessary centripetal force in order to go round the vertical circle.

The minimum speed, the motor cyclist is required to possess at the uppermost point is given by

$v_2=\sqrt{gr}$

Here, $g=9.8m{s}^{-2};r=25m$

$v_2=\sqrt{9.8\times 25}=15.65m{s}^{-1}$

Illustration 18

A body weighing $0.4kg$ is whirled in a vertical circle maling 2 revolutions per second. If the radius of the circle $1s1.2m$, find the tension in the string, when body is (a) at the bottom of the circle, (b) at the top of the circle.

Solution : Here, $M=0.4kg;r=1.2m;v=2$ r.p.s.

Now, angular speed, $\omega =2\pi v=2\pi \times 2=4\pi rad{s}^{-1}$

(a) When body is at the bottom of the circle : Let $T_1$ be tension in the string, when the body is at the bottom of the circle. Then,

$\begin{array}{rl}{T}_1-Mg& =\text{ centripetal force }=Mr{\omega }^2\\ \text{ or }{T}_1& =M(g+r{\omega }^2)=0.49.8+1.2\times (4\pi {)}^2\\ & =0.4(9.8+1.2\times 16\times 9.87)=0.4(9.8+189.5)\\ & =79.72N\end{array}$

(b) When body is at the top of the circle : Let $T_2$ be tension in the string, when the body is at the top of the circle. Then,

$T_2+Mg=$ centripetal force $=Mr{\omega }^2$

or $T_1=M(r{\omega }^2-g)=0.41.2\times (4\pi {)}^2-9.8$

$=0.4(189.5-9.8)=71.88N$

Illustration 19

A pendulum consisting of a small heavy bob suspended from a rigid support oscillates in a vertical plane. When the beb passes through the position of equilibrium, the rod is subjected to a tension equal to twice the weight of the hob, ’ fhrouthi what maximum angle from the vertical will the pendulum be deflected? Disregard the weight of the rod and the resiotanet of the air.

Solution : Suppose thatasmallbob of weight $Mg$ issuspended fromarigidsupport $S$ withastring of length $l$. In equilibrium position $O$, the bob is under the action of two forces ; its weight $Mg$ (acting vertically downwards) and tension $T$ in the string (acting vertically upwards) as shown in fig.

The resultant of these two forces provides the necessary centripetal force to the bob, so as to make it move along circular path of radius $l$ i.e.

$T-Mg=\frac{M{v}^2}{l}$

Since $T=2Mg$, we have

$2Mg-Mg=\frac{M{v}^2}{l}\text{ or }v=\sqrt{gl}$

As the bob gets deflected from its equilibrium position, its velocity decreases. It goes up to the point $A$, where its velocity becomes zero. Suppose that in this position, the string makes an angle $\theta$ with the vertical. If $h$ is the height of the bob above its equilibrium position, then

$h=SO-SN=l-l\cos \theta$

From the principle of conservation of energy, we have kinetic energy of the bob at the point $O=$ potential energy of the bob at the point $A$

or

$Mgh=\frac{1}{2}M{v}^2$

or $Mg(l-l\cos \theta )=\frac{1}{2}M(\sqrt{gl}{)}^2$

$\begin{array}{rl}& \text{ or }\cos \theta =\frac{1}{2}\\ & \text{ or }\theta ={60}^{\circ }\end{array}$

Illustrotion 20

A small sphere tied to the string of length $0.8m$ is describing a vertical circle so that the maximum and minimum tensions in the string are in the ratio $3:1$. The fixed end of the string is at a height of $5.8m$ above ground.

(a) Find the velocity of the sphere at the lowest position.

(b) If the string suddenly breaks at the lowest position, when and where will the sphere hit the ground? (take $g=10m/s^2$ )

Solution : (a) Let $u$ and $v$ be the speeds of sphere at the bottom and the top positions and $m$ be the mass.

Radius of circle $=$ length of string $=r=0.8m$ $T_b=$ tension at the bottom or the maximum tension

$T_t=$ tension at the top or the minimum tension

$T_b-\tilde{m}g=\frac{m{u}^2}{r}$

$T_t+mg=\frac{m{v}^2}{r}$

$\Rightarrow T_b=3T_t$

$(\frac{m{u}^2}{r}+mg)=3(\frac{m{v}^2}{r}-mg)$

$(3v^2-u^2)=4rg$ …(i)

Using conservation of energy,

loss in $KE$ from bottom to the top = gain in GPE

$\frac{1}{2}m{u}^2-\frac{1}{2}m{v}^2=mg(2r)$

$\Rightarrow v^2=u^2-4rg$ …(ii)

Using Eqs. (i) and (ii), we get

$3(u^2-4gr)-u^2=4rg$

$\Rightarrow 2u^2=16rg$

$\Rightarrow u=\sqrt{8rg}=\sqrt{8(0.8)10}=8$

(b) After breaking away from the string, the sphere moves along a parabolic path, and strikes the ground at $G$. Vertical displacement of the sphere,

$s_y=5.8-0.8=5m$

Let $t=$ time after which the sphere hits the ground.

$\Rightarrow s_y=0t+\frac{1}{2}g{t}^2$

$\Rightarrow t=\sqrt{\frac{2s_y}{g}}=1s$

The horizontal displacement $=x=ut=8\times 1=8m$

Hence the sphere hits the ground $1s$ after breaking off the string and at point $G$.

Collisons

Collision is an event in which two or more than two bodies interact with one another for a short time and exchange momentum and kinetic energy. Collisions are of two types :

(i) Elastic collision

(ii) Inelastic collision

Do you know !!

Linear momentum is always conserved, whereas kinetic energy is conserved only in elastic collision.

Elastic Collision

A.collision in which there is no loss of kinetic energy is called elastic or perfectly elastic collision. The basic oharacteristics of perfectly elastic collision are

(i) linear momentum is conserved

(ii) kinetic energy is conserved

(iii) total energy is conserved

(iv) coefficient of restitution is unity $(e=1)$

Elastic Collision in one Dimension

Let us discuss elastic collision is one dimension. Let two balls $A$ and $B$ of mass $m_1$ and $m_2$ move in the same direction along a straight line such that $u_1>u_2$. After collision their velocities are $v_1$ and $v_2$ such that $v_1<v_2$.

Conserving linear momentum,

$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$ …(i)

(Note : this is a vector equation so take care of sign)

Conserving kinetic energy,

$\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2$ …(ii)

From (i) and (ii), we get

$v_1=\frac{(m_1-m_2)u_1+2m_2{u}_2}{(m_1+m_2)};v_2=\frac{(m_2-m_1)u_2+2m_1{u}_1}{(m_1+m_2)}$

Special Cases

(a) When $m_1=m_2$ then $v_1=u_2$ and $v_2=u_1$ i.e. velocities of two colliding bodies are interchanged.

(b) $m_1\gg m_2$ and $u_2=0$ (Ball $A$ is very heavy and lighter ball $B$ is at rest)

$v_1=u_1$ and $v_2=2u_1$ i.e. heavier ball $A$ keeps moving with the same speed and lighter ball $B$ moves with double the speed of $A$.

(c) $m_2\gg m_1$ and $u_2=0$ (Ball $B$ is very heavy and it is at rest)

$v_1=-u_1,v_2=0$ i.e. (lighter ball $A$ rebounds with the same speed whereas heavy ball $B$ remains at rest)

Coefficiant of Restitution

It is defined as the ratio of the velocity of separation to the velocity of approach. That is,

$e=\frac{\text{ velocity of separation }}{\text{ velocity of approach }}$

or, $e=\frac{{\nu }_2-v_1}{u_1-u_2}$

Do you know !!

The collision in one dimension is also known as head-on collision.

Elastic Collision in two Dimensions

Suppose two balls $A$ and $B$ moving with velocities $u_1$ and $y_2$ initially in a straight line collide and after collision move at angles $\theta$ and $\varphi$ from the line of action ( $x$-axis) with velocities $v_4$ and $v_2$ respectively :

Conserving kinetic energy,

$\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2$ …(i)

Conserving linear momentum along $x$-axis

$m_1{u}_1+m_2{u}_2=m_1{v}_1\cos \theta +m_2{v}_2\cos \varphi$ …(ii)

Conserving linear momentum along $y$-axis

$0=m_1{v}_1\sin \theta -m_2{v}_2\sin \varphi$ (iii)

Inelastic Collision

In an inelastic collision kinetic energy is lost during collision. The basic characteristic of an inelastic collision are : ab? lalsog?

(i) linear momentum is conserved

(ii) kinetic energy is not conserved

(iii) total energy is conserved

(iv) coefficient of restitution is $0<e<1$

For solving problems of inelastic collision we can use

$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$ and $e=\frac{v_2-v_1}{u_1-u_2}$

And the loss in kinetic energy is given by

$\mathrm{\Delta }KE=\frac{1}{2}\frac{m_1{m}_2}{(m_1+m_2)}(u_1-u_2{)}^2(1-e{)}^2$

In completely inelastic collision the two bodies get stuck together and move with common velocity, that is why for perfectly inelastic collision;

$e=0$ and common velocity, $v=\frac{m_1{u}_1+m_2{u}_2}{m_1+m_2}$

Loss in K.E. is completely inelastic collision

$\mathrm{\Delta }K.E=\frac{1}{2}\frac{m_1{m}_2}{(m_1+m_2)}\cdot (u_1-u_2{)}^2$

Also, the fraction of K.E. lost is given by (when $m_2$ is at rest) $=\frac{\text{ Loss of KE }}{\text{ Initial KE }}=\frac{m_2}{m_1+m_2}$

Do you know !!

A collision in which colliding bodies stick together, is always an inelastic collision. It is because, Kinetic energy is never conserved in such collisions.

$✓$ CHECK POINT

The bob $A$ of a pendulum released from ${30}^{\circ }$ to the vertical hits another bob $B$ of the same mass at rest on a table as shown in fig. How high does the bob $A$ rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.

Check Your Answer

The positions of the bob of the pendulum and the target bob are shown in the fig. The bob of the pendulum will not rise as explained below:

If the bob of the pendulum acquires a velocity, say $v$, on reaching the lower most position, then the target bob will start moving with the velocity $v$ and the bob of the pendulum will comes to rest. It is because of the fact that in a perfectly elastic collision, when a moving object collides against a target object of the same mass, the two exchange their velocities.

Illustration 21

A mass is released from a height $H$ and after hitting the ground it rebounds to a height $h$, find the coefincient of restitition between the ground and mass.

Solution : Take the mass as body 1 and the earth as body 2 . So

$u_1=\sqrt{2gH},v_1=-\sqrt{2gh}$ and $v_1=v_2=0$

Using, $e=\frac{v_2-v_1}{u_1-u_2}\Rightarrow e=\sqrt{\frac{h}{H}}$

Illustration 22

Two balls each of mass $M$ moving in opposite directions with equal speeds $v$ undergo a head-on collision. Calkabate the velucity of the two balls after collision.

Solution : Here, $M_1=M_2=M;u_1=v$ and $u_2=-v$

Now,

$\begin{array}{rl}{v}_1& =\frac{(M_1-M_2)u_1+2M_2{u}_2}{M_1+M_2}\\ & =\frac{(M-M)v+2M(-v)}{M+M}=\frac{0-2Mv}{2M}\end{array}$

or, $v_1=-v$

Also, $\begin{array}{rl}{v}_2& =\frac{(M_2-M_1)u_2+2M_1{u}_1}{M_1+M_2}\\ & =\frac{(M-M)(-v)+2Mv}{M+M}=\frac{0+2Mv}{2M}\end{array}$

or, $v_2=v$

Thus, after the collision, the two balls bounce back with equal speeds $v$.

Illustration 23

A neutron having mass of $1.67\times {10}^{-27}kg$ and moving at ${10}^{8}m{s}^{-1}$ collides with a deutron at rest and sticks to it. Glven that the mass of neutron is $3.34\times {10}^{-27}kg$, calculate the speed of the combination. The composite particle is called triton.

Solution : Here, mass of the neutron, $M_1=1.67\times {10}^{-27}kg$

Initial velocity of the neutron, $u_1={10}^{8}m{s}^{-1}$;

Mass of the deutron, $M_2=3.34\times {10}^{-27}kg$

and initial velocity of the deutron, $u_2=0$ (at rest)

The mass of the composite particle i.e. triton,

$M_1+M_2=1.67\times {10}^{-27}+3.34\times {10}^{-27}=5.01\times {10}^{-27}kg$

As the neutron sticks to the deutron after the collision, the collision is inelastic in nature. For an inelastic collision, only the law of conservation of linear momentum holds. If $v$ is the velocity of the composite particle, then according to the law of conservation of linear momentum,

$\begin{array}{rl}& M_1{u}_1+M_2{u}_2=(M_1+M_2)v\\ & \text{ or }v=\frac{M_1{u}_1+M_2{u}_2}{M_1+M_2}\\ & =\frac{1.67\times {10}^{-27}\times {10}^8+3.34\times {10}^{-27}\times 0}{5.01\times {10}^{-27}}\\ & =\frac{1.67\times {10}^8}{5.01}=3.33\times {10}^{7}m{s}^{-1}\end{array}$

Illustration 24

A ball is dropped from rest at a height of $12m$. If it loses 25 percent of its kinetic energy on striking the ground, what is the height to which it bounces? How do you account for the loss in K.E. ?

Solution : Here, $u=0;s=12m;a=g=9.8m{s}^{-2}$;

Let $v$ be the velocity of the ball on reaching the ground.

From the relation : $v^2-u^2=2$ as, we have

$\begin{array}{rl}& v^2-(0{)}^2=2\times 9.8\times 12\\ & \text{ or }v=\sqrt{2\times 9.8\times 12}=\sqrt{235.2}=15.34m{s}^{-1}\end{array}$

Let $M$ be the mass of the ball. Then,

K.E. of the ball on reaching the ground,

$\frac{1}{2}M{v}^2=\frac{1}{2}M\times (15.34{)}^2=117.66M\text{ joule }$

On striking the ground, the ball loses $25$ of its K.E.

Therefore, K.E. possessed by the ball on bouncing from the ground

$=\frac{117.66M\times 75}{100}=88.245M$

If $u^{\prime }$ is the velocity of the ball after bouncing from the ground, then,

$\frac{1}{2}M{u}^1=88.245M$

or

$u^{\prime }=\sqrt{88.245\times 2}=13.285m{s}^{-1}$

Suppose that after striking the ground, the ball bounces up to a height $h$.

Then, for the vertical motion of the ball after bouncing :

initial velocity, $u^{\prime }=13.285m{s}^{-1}$; final velocity, $v^{\prime }=0$

and acceleration, $a^{\prime }=-g=-9.8m{s}^{-2}$

Now, $v^{\prime 2}-u^{\prime 2}=2a^{\prime }s$

$(0{)}^2-(13.285{)}^2=2(-29.8)h$

or $h=\frac{(13.285{)}^2}{2\times 9.8}=9m$

The loss in kinetic energy of the ball occurs, because of the fact that the collision between the ball and the ground is not perfectly elastic.

Power

In several situations, it is not enough only to know that how much work is done but it is also required that how quickly it is done i.e, it is also important to know the rate of work done by the force.

The time rate of doing work is defined as power $(P)$. If equal works are done in different times, power will be different. More quickly work is done, power will be more.

Power $=\frac{\text{ work }}{\text{ time }}$

Unit of Power

The unit of power is the joule per second and this is called the watt $(W)$. When large amounts of power are involved, a more convenient unit is the kilowatt $(kW)$ where $1kW=1000W$.

1 Megawatt $={10}^6$ watt

Power was also measured earlier in a unit called horse power. Even these days, the unit of horse power is in common use.

1 horse power $=746$ watt

The unit kilowatt-hour means one kilowatt of power supplied for one hour. It is, therefore, the unit of energy.

$1KWh=(1000J/s)\times 60\times 60s=3.6\times {10}^{6}J$

IDEA BOX

The Power of a moving vehicle

The power of a vehicle is defined as the rate at which the driving force is working.

Consider a vehicle moving at a constant speed $v$ meters per second. The driving force is $F$ newtons.

The distance moved in 1 second is $v$ meters.

The work done by the driving force in 1 second is $Fv$ joules.

Hence the power of the vehicle is $Fv$ watts.

So, if $P$ is the power, $P=Fv$

i.e. the power of a vehicle is given by multiplying the driving force by the velocity.

When the velocity is not constant this relationship gives the power at the instant when the velocity is $v$.

Do you know !!

Instantaneous power (P) of a body is defined as the dot product of force $(\overrightarrow{F})$ and the instantaneous velocity $(\overrightarrow{v})$ of the body i.e. $P=\overrightarrow{F}\cdot \overrightarrow{v}$

KNOWLEDG NHANCER

Area under power-time curve gives the work done.

Illustration 25

What is the power of an engine which can lift 20 metric ton of coal per hour from a 20 metre deep mine ?

Solution : Mass, $m=20$ metric ton $=20\times 1000kg$, Distance, $s=20m$, Time, $t=1$ hour $=3600s$

$\text{ Power }=\frac{\text{ work }}{\text{ time }}=\frac{mg\times s}{t}=\frac{20\times 1000\times 9.8\times 20}{3600}\text{ watt }=1.09\times {10}^{3}W$

Illustration 26

A student pulls a bucket of water in 1 minute from a $30m$ deep well. If the mass of bucket with water is $25kg$, then calculate the power of this student.

Solution : Work done by the student $=mgh$

$=25\times 10\times 30=7500J$

Time duration of the work is $=1min.=60sec$.

Hence, the power of student $=\frac{\text{ work }}{\text{ time }}=\frac{7500J}{60s}=125$ watt

Illustration 27

One coolie takes one minute to ralse a box through a height of 2 metre. Another one takes 30 second for the same job and does the same amount of work. Which one of the two has greater power and which one uses greater energy ?

Solution : Power of first coolie $=\frac{\text{ Work }}{\text{ time }}=\frac{M\times g\times s}{t}=\frac{M\times 9.8\times 2}{60}J{s}^{-1}$

Power of second coolie $=\frac{M\times 9.8\times 2}{30}J{s}^{-1}=2(\frac{M\times 9.8\times 2}{60})J{s}^{-1}=2\times$ Power of first coolie

So, the power of the second coolie is double that of the first.

Both the coolies spend the same amount of energy.

Aliter, we know that $W=Pt$

For the same work, $W=P_1{t}_1=P_2{t}_2$

or $\frac{P_2}{P_1}=\frac{t_1}{t_2}=\frac{1\text{ minute }}{30s}=2$ or $P_2=2P_1$

Illustrotion 28

A 100W bulb operates for 5 hours. How much electric energy will it consume?

Solution : Work $=$ Power $\times$ Time

$=100$ watt $\times 5$ hour

$=500$ watt hour

$=0.5$ kilowatt hour

EXERCISE 1

Fill in the Blanks

DIRECTIONS: Complete the following statements with an appropriate word/term to be filled in the blank space(s).

1. The work done by the external force on a system equals the change in ____ energy.
2. One horse-power $=$ ____ watt.
3. An electric motor exerts a force of $40N$ on a cable and pulls it through a distance of $30m$ in one minute. The power supplied by the motor in watts is ____
4. A one kilogram mass has a kinetic energy of one joule when its speed is ____ meter/sec.
5. A truck and a car moving with the same kinetic energy are brought to rest by the application of brakes which provide equal retarding forces. The distance moved by the truck, in coming to rest, will then be ____ the distance moved by car.
6. The work done in holding $15kg$ suitcase while waiting for a bus for 15 minute is ____
7. Two bodies of masses $m_1$ and $m_2$ have equal moments. Their kinetic energies are in the ratio ____
8. The energy possessed by the body due to its ____ called kinetic energy.
9. Work done by a force is maximum when angle between force and displacement is ____
10. Energy is a ____ quantity.
11. Energy stored in an elongated rubber is ____
12. Watt second is a unit of ____
13. ____ energy can never be negative.
14. When angle between force and displacement is obtuse, work done by the force is ____

True/False

DIRECTIONS: Read the following statements and write your answer as true or false.

1. Work is always done on a body when it experiences an increase in energy through a mechanical influence.
2. Work done by the resultant force is always equal to change in kinetic energy.
3. A light and a heavy body, having equal momenta, have equal kinetic energies.
4. Work done in the motion of a body over a closed loop is zero for every force in nature.
5. More work is done in compressing a litre of air than a litre of water from a pressure of one atmosphere to three atmospheres.
6. No work is done on a particle which remains at rest.
7. A man rowing a boat up stream is at rest with respect to the shore, is doing no work.
8. The total energy of a body in motion is equal to the work it can do in being brought to rest.
9. A body cannot have momentum when its energy is zero.
10. Kinetic energy of a body depends upon the direction of motion.
11. Work done by friction can-never be positive.
12. Work done by conservative force in round trip is zero.
13. Gravitational force is non-conservative.
14. Work done by conservative forces is equal to increase in potential energy.
15. A man carrying a bucket of water, walking on a level road with a uniform velocity does no work.

Match the columns

DIRECTIONS: Each question contains statements given in two columns which have to be matched. Statements $(A,B,C,D)$ in Column I have to be matched with statements $(p,q,r,s)$ in Column II.

1. A boy of mass $55kg$ runs up a flight of 40 stairs, each measuring $0.15m$ in $5s$. Column II given numerical value for quantities describe in Column I, match them correctly.

Column I	Column II
(A) Force acting on the boy	(p) 550
(B) Work done by the boy	(q) 3300
(C) Gain of potential energy by the boy	(r) 33000
(D) Power developed by the boy	(s) 220

2. Column I shows some devices and column II shows transformation of energy for which they are used. Then match the following.

Column I	Column II
(A) Electric motor	(p) Electrical energy to heat energy
(B) Engine on an automobile	(q) Electrical energy to mechanical energy
(C) Electric heater	(r) Light energy to electrical energy
(D) Photocell	(s) Heat energy to mechanical energy.

3. In the situation shown in the figure, coefficient of friction between the block and ground is $0.2,m=2kg,F_1=5N$ and $F_2=4N$. If the block is displaced by $10m$, match the following. $(g=10m{s}^{-2})$

Column I	Column II
(A) Magnitude of work done by $F_1$	(p) 20 J
(B) Magnitude of work done by $F_2$	(q) 30 J
(C) Magnitude of work done by friction	(r) 40 J
(D) Magnitude of increase in K.E of the block	(s) 50 J

Very Short Answer Questions

DIRECTIONS: Give answer in one word or one sentence.

1. A man raises a mass ’ $m$ ’ to a height ’ $h$ ’ and then shifts it horizontally by a length ’ $x$ ‘. What is the work done against the force of gravity?
2. A mass $m$ collides with another mass $2m$ and sticks to it. What is the nature of collision?
3. Two bodies $m_1$ and $m_2(m_1>m_2)$ have equal kinetic energies. Which will have more momentum?
4. Ten identical balls are placed in contact on a smooth surface. If an eleventh identical ball moving with a speed ’ $u$ ’ collides on the first, what will be the resultant motion of the system?
5. Can a constant velocity be maintained in a body moving on a rough source without doing any work on it?
6. Does the work done in raising a box on to a platform depends upon how fast it is raised up? If not, why?
7. A man rowing a boat upstream is at rest with respect to the shore, is he doing work?
8. A light body and a heavy body have the same momentum. Which one will have greater kinetic energy?
9. A spring is cut into two equal halves. How is spring constant of each half affected?
10. A truck and a car are moving with the same kinetic energy on a straight road. Their engines are simultaneously switched off. Which one will stop at a lesser distance?

Short Answer Questions

DIRECTIONS: Give answer in 2-3 sentences.

1. Two springs $A$ and $B$ with constant $k_A$ and $k_B$ $(k_A>k_B)$ are given. In which of the springs more work is to be done, if

(a) they are stretched by the same amount,

(b) they are stretched by same force?

2. Calculate the work done to stretch a spring from $x$ to $2x$. Given the spring constant is $k$.

3. Two inclined frictionless tracks are gradual and the other steep meet at $A$ from where two stones are allowed to slide down from rest on each track. Will the stones reach the bottom at the same time and at same speed. Why?

4. What is an elastic collision? What will happen when

(i) a heavy body collides on a light mass at rest.

(ii) a light body collides on a heavy mass at rest.

5. Show that in case of a freely falling body the total energy remains constant at every point in its paths.
6. A motor pump is used to deliver water at a certain rate from a given pipe. To obtain ’ $n$ ’ times water from the same pipe in the same time by what amount (a) the force and (b) power of the motor should be increased.
7. Two masses, one $n$ times heavier than the other are dropped from same height. How do their momentum compare just before they hit the ground?
8. Two masses, one $n$ times heavier than the other have equal kinetic energy. Find the ratio of their momentum.
9. The momentum of a body is increased by $50$. What is the percentage change in its $K.E$ ?

Long Answer Questions

DIRECTIONS: Give answer in four to five sentences.

1. Define work. Write its unit. How will you define 1 joule.
2. What do you mean by energy? Explain different kinds of the mechanical energy.
3. Define kinetic energy. Derive an expression for it.
4. How will you define a collision. Define different kind of collision.
5. Find expressions for the velocities of the colliding objects after an elastic collision.
6. Explain conservative and non-conservative forces.
7. Derive an expression for the velocity of an object undergoing a vertical circular motion at the highest point of its path.

EXERCISE 2

Multiple Choice Questions

DIRECTIONS : This section contains multiple choice questions. Each questions has 4 choices (a), (b), (c) and (d), out of which only one is correct.

1. Which is not a unit of energy

(a) Watt second

(b) kilo watt hour

(c) watt

(d) joule

2. kilowatt hour is the unit of

(a) time

(b) power

(c) energy

(d) force

3. 1 kilowatt hour is equal to

(a) 1 joule

(c) 36 joule

(b) 100 joule

(d) $3.6\times {10}^3$ kilo joule

4. A stone of mass $1kg$ is raised through $1m$ height

(a) The loss of gravitational potential energy by the stone is 1 joule

(b) The gain of gravitational potential energy by the stone is 1 joule

(c) The loss of gravitational potential energy is 9.8 joule

(d) The gain of gravitational potential energy is 9.8 joule

5. The energy of $4900J$ was expanded in lifting a $50kg$ mass. The mass was raised to a height of

(a) $10m$

(b) $98m$

(c) $960m$

(d) $245000m$

6. A body of mass $1kg$ has kinetic energy $1J$ when its speed is

(a) $0.45m/s$

(b) $1m/s$

(c) $1.4m/s$

(d) $4.4m/s$

7. The kinetic energy of a body will become eight times if

(a) its mass is made four times

(b) its velocity is made four times

(c) both the mass and velocity are doubled

(d) both the mass and velocity are made four times

8. For a body falling freely under gravity from a height

(a) only the potential energy goes on increasing

(b) only the kinetic energy goes on increasing

(c) both kinetic energy as well as potential energy go on increasing

(d) the kinetic energy goes on increasing while potential energy goes on decreasing

9. A ball is dropped from a height of $10m$. If the energy of the ball reduces by $40$ after striking the ground, the ball will rebound to

(a) $4m$

(b) $6m$

(c) $10m$

(d) $9.8m$

10. A $2kg$ mass kept on a horizontal table is acted upon by a force of $5N$ at an angle of ${60}^{\circ }$ from horizontal and moved through a distance of $3m$ on the table. The work done on the body is

(a) $5J$

(b) $7.5J$

(c) $10J$

(d) $15J$

11. The kinetic energy acquired by a body of mass ’ $m$ ’ after travelling a fixed distance from rest under the action of a constant force is

(a) directly proportional to mass $m$

(b) inversely proportional to mass $m$

(c) inversely proportional to mass $m^{1/2}$

(d) independent of mass $m$

12. If a force $F$ is applied on a body and it moves with velocity $v$, the power will be

(a) $Fv$

(b) $F/v$

(c) $F{v}^2$

(d) $F/v^2$

13. The kinetic energy of a body becomes twice its initial value. The new momentum of the body will be

(a) 2 times

(b) $\sqrt{2}$ times

(c) 4 times

(d) unchanged

14. A car weighing $500kg$. working against a resistance of $500N$, accelerates from rest to $20m/s$ in $100m$. $(g=10m/s^2)$. The work done by the engine of car is

(a) $1.0\times {10}^{5}J$

(b) $1.5\times {10}^{5}J$

(c) $1.05\times {10}^{5}J$

(d) The information given is insufficient

15. A uniform force of $4N$ acts on a body of mass $8kg$ for a distance of $2.0m$. The K.E. acquired by the body is

(a) $8J$

(b) $64J$

(c) $4J$

(d) $16J$

16. With what speed must a ball be thrown down for it to bounce $10m$ higher than its original level ? Neglect any loss in energy against air resistance and in collision with the ground

(a) $5m/s$

(b) $14m/s$

(c) $20m/s$

(d) The information given is incomplete

17. The heart of a man pumps $4000c{m}^3$ of blood through the arterise per minute at a pressure of $130mm$ of $Hg$. If the density of $Hg$ is $13.6\times {10}^{3}kg/m^3$, what is the power of heart?

(a) $1.15W$

(c) $0.115W$

(b) $2.30W$

(d) $0.15W$

18. A cord is used to lower vertically a block of mass $M$ a distance $d$ at a constant downward acceleration of $g/4$. Then the work done by the cord on the block is

(a) $Mgd/4$

(b) $3Mgd/4$

(c) $-3Mgd/4$

(d) Mgd

19. Figure shows the frictional force versus displacement for a particle in motion. The loss of kinetic energy (work done against frcition) in travelling over $s=0$ to $s=20m$ will

(a) $80J$

(b) $160J$

(c) $240J$

(d) $24J$

20. Work done in time $t$ on a body of mass $m$ which is accelerated from rest to a speed of $v$ in time $t_1$ as a function of time $t$ is given by

(a) $\frac{1}{2}m\frac{v}{t_1}{t}^2$

(b) $m\frac{v}{t_1}{t}^2$

(c) $\frac{1}{2}m(\frac{mv}{t_1}{)}^2{t}^2$

(d) $\frac{1}{2}m\frac{v^2}{t_1^2}{t}^2$

21. A man $A$ of mass $80kg$ runs up a staircase in 12 seconds. Another man $B$ of mass $60kg$ runs up the same staircase in 11 seconds. The ratio of powers of $A$ and $B$ is

(a) $11:12$

(b) $11:9$

(c) $12:11$

(d) $9:11$

22. A man of weight $60kg$ wt. takes a body of mass $15kg$ at a height $10m$ on a building in 3 minutes. The efficiency of mass is

(a) $10$

(b) $20$

(c) $30$

(d) $40$

23. Kinetic energy of a body moving with speed $10m/s$ is $30J$. If its speed becomes $30m/s$, its kinetic energy will be

(a) $10J$

(b) $90J$

(c) $180J$

(d) $270J$

24. A weight-lifter lifts $200kg$ from the ground to a height of 2 metre in 9 second. The average power generated by the man is

(a) $15680W$

(b) $3920W$

(c) $1960W$

(d) $980W$

25. Two masses $m$ and $9m$ are moving with equal kinetic energies. The ratio of the magnitudes of their momenta is

(a) $1:1$

(b) $1:3$

(c) $3:1$

(d) $1:9$

26. When you compress a coil spring you do work on it. The elastic potential energy

(a) increases

(b) decreases

(c) disappears

(d) remains the same

27. If a stone of mass $m$ falls a vertical distance $d$, the decrease in gravitational potential energy is

(a) $mg/d$

(b) $m{d}^{2/2}$

(c) mgd

(d) $md/g$

28. No work is done when

(a) a nail is plugged into a wooden board

(b) a box is pushed along a horizontal floor

(c) there is no component of force parallel to the direction of motion

(d) there is no component of force normal to the direction of force

29. Potential energy of your body is minimum when you

(a) are standing

(b) are sitting on a chair

(c) are sitting on the ground

(d) lie down on the ground

30. The work done against gravity in moving the block a distance $s$ up the slope is

(a) $mh$

(b) mgs

(c) $ms$

(d) $mgh$

31. A block of weight $W$ is pulled a distance $l$ along a horizontal table. The work done by the weight is

(a) $Wl$

(b) 0

(c) $Wgl$

(d) $Wl/g$

32. A child builds a tower from three blocks. The blocks are uniform cubes of side $2cm$. The blocks are initially all lying on the same horizontal surface and each block has a mass of $0.1kg$. The work done by the child is

(a) $4J$

(b) $0.04J$

(c) $6J$

(d) $0.06J$

33. A car is moving with a constant speed of $20m/s$ against a resistance of $100N$. The power exerted by the car is

(a) $2kW$

(b) $5W$

(c) $200W$

(d) $1kW$

34. A particle of mass $m$ moves from rest under the action of a constant force $F$ which acts for two seconds. The maximum power attained is

(a) $2Fm$

(c) $2F/m$

(b) $F^2/m$

(d) $2F^{2/m}$

35. A body of mass $m$ accelerates uniformly from rest to $v_1$ in time $t_1$. As a function of $t$, the instantaneous power delivered to the body is

(a) $\frac{m{v}_{1}t}{t_2}$

(b) $\frac{m{v}_1{t}^2}{t_1}$

(c) $\frac{m{v}_1{t}^2}{t_1}$

(d) $\frac{m{v}_1^{2}t}{t_1^2}$

36. A particle moves under the effect of a force $F=cx$ from $x=0$ to $x=x_1$, the work done in the process is

(a) $c{x}_1^2$

(b) $\frac{1}{2}c{x}_1^2$

(c) $2c{x}_1^2$

(d) zero

37. A block is acted upon by a force, which is inversely proportional to the distance covered $(x)$. The work done will be proportional to

(a) $x$

(b) $x^{1/2}$

(c) $x^2$

(d) none of the above

38. Two bodies $A$ and $B$ having masses in the ratio of $3:1$ possess the same kinetic energy. The ratio of linear momentum of $B$ to $A$ is

(a) $1:3$

(b) $3:1$

(c) $1:\sqrt{3}$

(d) $\sqrt{3}:1$

39. A small body is projected in a direction inclined at ${45}^{\circ }$ to the horizontal with kinetic energy $K$. At the top of its flight, its kinetic energy will be

(a) zero

(b) $K/2$

(c) $K/4$

(d) $K/\sqrt{2}$

40. If the linear momentum is increased by $5$, the kinetic energy will increase by

(a) $50$

(b) $100$

(c) $125$

(d) $10$

41. A wire suspended vertically from one of its ends is stretched by attaching a weight of $200N$ to the lower end. The weight stretches the wire by $1mm$. Then the elastic energy stored in the wire is

(a) $10J$

(b) $20J$

(c) $0.1J$

(d) $0.2J$

42. Consider the following two statement

I. Linear momentum of a system of particles is zero.

II. Kinetic energy of a system of particles is zero. Then

(a) I implies II but II does not imply I.

(b) I does not imply II but II implies I.

(c) I implies II and II implies I.

(d) I does not imply II and II does not imply I.

43. A bullet of mass ’ $a$ ’ and velocity ’ $b$ ’ is fired into a large block of wood of mass ’ $c$ ‘. The bullet gets embedded into the block of wood. The final velocity of the system is

(a) $\frac{b}{a+b}\times c$

(b) $\frac{a+b}{c}\times a$

(c) $\frac{a}{a+c}\times b$

(d) $\frac{a+c}{a}\times b$

44. A ball is dropped from a height $h$. If the coefficient of restitution be $e$, then to what height will it rise after jumping twice from the ground?

(a) $eh/2$

(c) $eh$

(b) $2eh$

(d) $e^{4}h$

45. A body of mass $5kg$ initially at rest explodes into 3 fragments with mass ratio $3:1:1$. Two of fragments each of mass ’ $m$ ’ are found to move with a speed $60m/s$ in mutually perpendicular directions. The velocity of third fragment is

(a) $60\sqrt{2}$

(b) $20\sqrt{3}$

(c) $10\sqrt{2}$

(d) $20\sqrt{2}$

46. In figure, a carriage $P$ is pulled up from $A$ to $B$. The relevant coefficient of friction is 0.40 . The work done will be

(a) $10kJ$

(b) $23kJ$

(c) $25kJ$

(d) $28kJ$

47. A mass of $0.5kg$ moving with a speed of $1.5m/s$ on a horizontal smooth surface, collides with a nearly weightless spring of force constant $k=50N/m$. The maximum compression of the spring would be

(a) $0.5m$

(b) $0.15m$

(c) $0.12m$

(d) $1.5m$

48. A body of mass $m$ moving with velocity $v$ makes a head on elastic collision with another body of mass $2m$ which in initially at rest. The loss of kinetic energy of the colliding body (mass $m$ ) is

(a) $\frac{1}{2}$ of its initial kinetic energy

(b) $\frac{1}{9}$ of its initial kinetic energy

(c) $\frac{8}{9}$ of its initial kinetic energy

(d) $\frac{1}{4}$ of its initial kinetic energy

More than one correct

DIRECTIONS: This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d), out of which One or More may be correct.

1. The total energy of a swinging pendulum at any instant of time

(a) remains zero

(c) is same

(b) remains conserved

(d) is lost

2. The power of a body doing 20 Joules of work in 5 seconds is

(a) $400J/s$

(b) $4kW$

(c) $4J/s$

(d) $4W$

3. 1 unit of electricity consumed is same as

(a) $1kwh$

(b) $3.6\times {10}^{6}J/s$

(c) $3.6\times {10}^{6}J$

(d) $1m/s^2$

4. Which of the following is/are a scalar quantity?

(a) Acceleration

(b) Velocity

(c) Work done

(d) Energy

5. No work is done by a force on an object if:

(a) the object moves in such a way that the point of application of the force remains fixed

(b) the object is stationary but the point of application of the force moves on the object

(c) the force is always perpendicular to its acceleration

(d) the force is always perpendicular to its velocity

6. When a body is moving up with constant velocity:

(a) work done by force of gravity is positive

(b) work done by lifting force is positive

(c) work done by lifting force is negative

(d) work done by force of gravity is negative

7. When two blocks connected by a spring move towards each other under mutual interaction:

(a) their momentum are equal and opposite

(b) their velocities are equal and opposite

(c) their accelerations are equal and opposite

(d) the force acting on them are equal and opposite.

8. A particle is projected from a point at an angle with the horizontal at $t=0$. At an instant ’ $t$ ‘, if $p$ is linear momentum, $x$ is horizontal displacement, $y$ is vertical displacement and $E$ is kinetic energy of the particle, then which of the following graphs are correct?

9. In which of the following are no work done by the force?

(a) A man walking upon a staircare

(b) A man carrying a bucket of water, walking on a level road with a uniform velocity.

(c) A drop of rain falling vertically with a constant velocity

(d) A man whirling a stone tied to a string in circle with a constant speed.

10. A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that:

(a) the acceleration of the particle is constant

(b) the velocity of the particle is constant

(c) the particle moves in a circular path

(d) the kinetic energy of the particle is constant

11. A body of mass $m$ is moving in a straight line at a constant speed $v$. Its kinetic energy is $K$ and the magnitude of its momentum is $p$. Which of the following relations are correct?

(a) $v=\frac{2K}{p}$

(b) $2K=pv$

(c) $p=\sqrt{2mK}$

(d) $p=\sqrt{\frac{2K}{m}}$

12. A man of mass $m$ is standing on a stationary flat car of mass M. The car can move without friction along horizontal rails. The man starts walking with velocity v relative to the car. Work done by him

(a) can never be less than $1/2m{v}^2$

(b) is equal to $1/2m{v}^2$ if he walks normal wo rails

(c) is less than $1/2m{v}^2$ if he walks along raiks

(d) is greater than $1/2m{v}^2$ if he walks along rails

13. Which of the following are correct?

(a) The power of an agent is $\overrightarrow{F},\overrightarrow{v}$

(b) When a force retards the motion of a body, the work done is negative

(c) A body of weight 1 newton has a potential energy of 1 Joule relative to the ground when it is at a height of 1 metre

(d) A $1kg$ body has a kinetic energy of 1 joule when its veloclity is $1.414m{s}^{-1}$

14. The figure given below shows how the net imteraction force between two particles A and B is related to the distance between them. When the distance between them varies from $x_1$ to $x_4$, then:

(a) kinetic energy increases from $x_1$ to $x_2$ and decreases form $x_2$ to $x_3$

(b) potential enengy of the system increases from $x_1$ to $x_2$

(c) potential energy of the system increases from $x_3$ to $x_4$

(d) potential energy of the system increases form $x_2$ to $x_3$

Multiple Matching Questions

DIRECTIONS : Following question has four statements (A, B. C and D) given in Column I and four statements (a, r, s …. f) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) griven in Column II. Match the entries in Column I with entries in Column II.

1.

Column I	Column II
(A) Gravitational force	(p) Conservative
(B) Electrostatic force	(q) Non-consecutive
(C) Frictional force	(r) Act between surfaces in contact
(D) Viscous force	(s) Opposes relative motion
	(t) Do not dissipate energy

2. A particle of mass $m$ is projected from ground with speed $v$ making an angle $\theta$ with the vertical. Match the following for the motion of the particle.

Column I	Column II
(A) Power of gravity at the highest point	(p) zero
(B) Power of gravity at the point of projection	(q) $\frac{1}{2}m{v}^2{\cos }^2\theta$
(C) Work done by gravity during ascend	(r) $mgu\cos \theta$
(D) Work done by gravity during descend	(s) negative
	(t) positive

Fill in the Blanks

I.

kinetic energy, velocity, mass, halved, one fourth

If the mass of a body is doubled, its (1) is also doubled and if the (2) of the body is halved, its kinetic energy gets (3). However if (4) of a body is doubled its kinetic energy becomes four times and if velocity of a body is halved, its kinetic energy becomes (5).

II. transformation, kinetic, change, potential, another

The (1) of one form of energy into (2) form of energy is known as (3) of energy. Suppose a stone is lying on the roof of a house. In this Position, all the energy of the stone is in the form of (4) energy. When the stone is dropped from the roof, it starts moving downwards towards the ground and the potential energy of the stone starts changing into (5) energy.

III. non-conservative, zero, conservative, non-zero.

The forces, work done by which do not depend on path followed but depend only on the initial and final position are called (1). Work done by these forces in round trip is (2) . On the other hand, the forces work done by which is path dependent are called (3).

Work done by these forces in round trip is (4).

Paragraph Based Questions

DIRECTIONS : Study the given paragraph(s) and answer the following questions.

PASSAGE-I

A block of mass $m=2kg$ strikes an elastic spring of spring constant $k=100N{m}^{-1}$ with $s$ speed $30cm{s}^{-1}$. The surface is rough with coefficient of friction $\mu =0.2$

1. The maximum compression in the spring is

(a) $5cm$

(b) $10cm$

(c) $20cm$

(d) $25cm$

2. The velocity with which the block leaves the spring during return journey is

(a) $10cm{s}^{-1}$

(c) $30cm{s}^{-1}$

(b) $20cm{s}^{-1}$

(d) zero

3. Work done by friction in the round trip is

(a) $0.4J$

(b) $0.8J$

(c) $0.9J$

(d) zero

PASSAGE-II

Consider the situation shown in the figure. All surfaces are smooth, a particle is released from point $A$ as shown. Answer the following for the motion of the particle. (Take $g=10m{s}^{-2}$ )

4. How high will the particle rise above the ground after leaving from point $B$ ?

(a) $45m$

(b) $95m$

(c) $61.25m$

(d) $83.75m$

5. How much time will the particle take to reach the ground after leaving from point $B$ ?

(a) $2s$

(b) $5s$

(c) $8s$

(d) $\sqrt{10s}$

6. What will be the horizontal distance travelled by the particle to reach the ground after leaving from point $B$ ?

(a) $30\sqrt{3}m$

(b) $75\sqrt{3}m$

(c) $50m$

(d) $30\sqrt{10}m$

PASSAGE-III

A pendulum bob of mass $m$ is hanging from a fixed point can oscillate in a vertical plane. A constant horizontal force $F$ is applied on the bob as shown.

7. The maximum angle made by the string with the vertical is

(a) ${\tan }^{-1}(\frac{F}{mg})$

(b) $2{\tan }^{-1}(\frac{F}{mg})$

(c) $2{\tan }^{-1}(\frac{2F}{mg})$

(d) ${\tan }^{-1}(\frac{mg}{F})$

8. The maximum velocity of the $bob$ will be

(a) $[\frac{2l(\sqrt{F^2+m^2{g}^2-mg})}{m}{]}^{1/2}$

(b) $[2l\frac{(\sqrt{F^2+m^2{g}^2}-F)}{m}{]}^{1/2}$

(c) $[\frac{l(\sqrt{F^2+m^2{g}^2}-mg)}{m}{]}^{1/2}$

(d) $[\frac{l(\sqrt{F^2+m^2{g}^2}-F)}{m}{]}^{1/2}$

9. The tension in the bob in the position of problem 7 will be

(a) $\sqrt{F^2+(mg{)}^2}$

(b) $\frac{2Fmg}{F+mg}$

(c) $\frac{F+mg}{2}$

(d) $mg$

Assertions And Reasons

DIRECTIONS: Each of these questions contains an Assertion followed by reason. Read them carefully and answer the question on the basis of following options. You have to select the one that best describes the two statements.

(a) If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

(b) If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

(c) If Assertion is correct but Reason is incorrect.

(d) If Assertion is incorrect but Reason is correct.

1. Assertion : A moving hammer drives a nail into wood.

Reason : A moving hammer has potential energy.

2. Assertion : A man gets completely exhausted in trying to push a stationary wall.

Reason : Work done by the man on the wall is zero.

3. Assertion : The driver increases the speed of his car on approaching a hilly road.

Reason : To give more kinetic energy to the car so that it may go up against gravity.

4. Assertion : Winding up the spring of a toy car gives it energy for moving.

Reason : Work done in winding the spring get stored up as kinetic energý.

5. Assertion : A car engine convert heat energy into chemical energy.

Reason : A car burns fuel to get energy.

6. Assertion : The work done during a round trip is always zero.

7. Assertion : Work done in moving a body over a closed loop is zero for every force in nature.

Reason : Work done depends on nature of force.

8. Assertion : A person walking on horizontal road with a load on his head does no work.

Reason : No work is said to be done, if the directions of force and displacement of load are perpendicular to each other.

Hots Answer Questions

DIRECTIONS: Answer the following questions.

1. It is generally much more difficult to stop a heavy truck than a light car when they move at the same speed. State a case in which the moving car could require more stopping force. (Consider relative times.)
2. Why is a punch more forceful with a bare fist than with a boxing glove?
3. A boxer can punch a heavy bag for more than an hour without tiring, but will tire quickly when boxing with an opponent for a few minutes. Why?
4. Railroad cars are loosely coupled so that there is a noticeable time delay from the time the first car is moved and last cars gre moved from rest by the locomotive. Discuss the advisability of this loose coupling and slack between cars from an impulse-momentum point of view.
5. A high-speed bus and an innocent bug have a head-on collision. The sudden change of momentum for the bug spatters it all over the windshield. Is the change in momentum of the bus greater, less, or the same as the change in momentum of the unfortunate bug?
6. Why is it difficult for a firefighter to hold a hose that ejects large quantities of water at a high speed?
7. You’re on a small raft next to a dock, and you jump from the raft only to fall into the water. What physics principle did you fail to take into account?
8. Your friend says the conservation of momentum is violated when you step off a chair and gain momentum as you fall. What do you say?
9. If a Mack truck and a Honda Civic have a head-on collision, which vehicle will experience the greater force of impact? The greater impulse ? The greater change in its momentum? The greater acceleration?
10. Would a head-on collision between two cars be more damaging to the occupants if the cars stuck together or if the cars rebounded upon impact?
11. Suppose there are three astronauts outside a space-ship, and two of them decide to pay catch using the third man. All the astronauts weigh the same on earth and are equally strong. The first astronaut throws the second one toward the third one and the game begins. Describe the motion of the astronauts as the game proceeds. How long will the game last?
12. When a stationary uranium nucleus undergoes fission, it breaks into two unequal chunks that fly apart. What can you conclude about the momenta of the chunks? What can you conclude about the speed of the chunks?
13. How is it possible that a flock of birds in flight can have a momentum of zero but not have zero kinetic energy?
14. In determining the potential energy of Tenn’s drawn bow (figure), would it be an underestimate or an overestimate to multiply the force with which she holds the \to in its drawn position by the distance the pulled it? Why do we say the work done is the average force $\times$ distance?
15. When a cannon with a long barrel is fired, the force of expanding gases acts on the cannonball for a longer distance. What effect does this have on the velocity of the emerging cannonball?

SOLUTIONS

EXERCISE - I

Fill In The Blanks

1. total	2. 746
3. 20	4. $\sqrt{2}$
5. the same	6. zero
7. $m_2/m_1$	8. motion
9. $0^{\circ }$	10. Scalar
11. potential energy	12. energy
13. kinetic	14. negative

True/False

1. True	2. True	3. False
4. False	5. True	6. True
7. True	8. False	9. True
10. False	11. False	12. True
13. False	14. False	15. True

MATCH THE COLUMNS

1. (A) $\to$ (p); (B) $\to$ (q) ;(C) $\to$ (q) ;(D) $\to$ (s)
2. (A) $\to$ (q) ;(B) $\to$ (s) ; C $\to$ (p) ;(D) $\to$ (r)
3. (A) $\to$ (s) ;(B) $\to$ (p) ;(C) $\to$ (r) ;(D) $\to$ (q)

Work done by $F_1{W}_1=F_{1}s=5\times 10=50J$

Work done by $F_2{W}_2=F_{2}s\cos {60}^{\circ }=4\times 10\times \frac{1}{2}=20J$

Work done by friction, $W_3=-\mu mgs=-0.2\times 2\times 10=-40J$

$\therefore$ Change in K.E $=$ Net work done $=W_1+W_2-W_3=30J$

Very Short Answer Question

1. Work done against gravity is $mgh$, as no work is done in the horizontal displacement.

2. Inelastic, as the two bodies are moving together after collision.

3. Energy $(E)=\frac{p^2}{2m}\therefore p=\sqrt{2mE}$

Heavier the mass, more will be momentum.

4. When the eleventh ball hits the first ball, the tenth ball starts moving with the same speed due to transfer of momentum in equal proportion.

5. No, work has to be done to compensate the energy loss due to friction.

6. Since, the gravitational force is conservative in nature, so work done against it depends only on initial and final points, not on time.

7. Yes, he is doing work to oppose the river current.

8. $K.E=\frac{p^2}{2m}\because p$ is same, $K.E\propto \frac{1}{m}$

$\therefore$ Light body will have greater kinetic energy.

9. $F=kl\therefore k=\frac{F}{l}$

When it is cut into 2 equal halves, length of each half will be $\frac{l}{2}$.

$\therefore$ Spring constant will be doubled.

10. $$ Distance $=\frac{K.E}{\text{ Retarding force }}$

Here the retarding force $=$ frictional force.

Since mass of the truck is more, so its frictional force will be more.

$\therefore$ The truck will come to rest at a lesser distance.

Short Answer Questions

1. $F=-kx$ and energy $=W=\frac{1}{2}k{x}^2$

(a) For same stretch, $\frac{W_A}{W_B}=\frac{k_A}{k_B}\because k_A>k_B$

$\therefore W_A>W_B$

(b) For same force, $\frac{W_A}{W_B}=\frac{F^2}{2k_A}\cdot \frac{2k_B}{F^2}=\frac{k_B}{k_A}$

$\therefore W_A<W_B$

2. $W={\int }_x^{2x}kx\cdot dx=\frac{1}{2}k[(2x{)}^2-x^2]=\frac{1}{2}k3x^2=\frac{3}{2}k{x}^2$

3. Let $AM=h$ and $AB=s,a=$ acceleration of the stone falling through $AB$.

$\therefore a=g\sin \theta =g\cdot \frac{h}{s}$

$\therefore s=ut+\frac{1}{2}a{t}^2=\frac{1}{2}a{t}^2[\because u=0]$

$=\frac{1}{2}\frac{gh}{s}{t}^2$

$\therefore t=s\sqrt{\frac{2}{gh}}\Rightarrow t\propto s$

$\therefore$ The stone falling through $AB$ will take more time than that falling through $AC$.

$v^2-u^2=2gh\because u=0,\therefore v=\sqrt{2gh}$

$\because$ Both the stones are falling through the same vertical height $h$, therefore they will reach the bottom with the same velocity.

4. Elastic collision is defined as that collision in which both momentum and kinetic energy before and after collision are conserved. If two masses $m_1$ and $m_2$ moving with velocities $u_1$ and $u_2$ collides, velocities after collision is

given by $v_1=\frac{(m_1-m_2)u_1+2m_2{u}_2}{(m_1+m_2)}$ and

$v_2=\frac{(m_2-m_1)u_2+2m_1{u}_1}{(m_1+m_2)}$

(i) If $m_1\gg m_2$ and $u_2=0,v_1=\frac{m_1{u}_1}{m_1}=u_1$ and $v_2=2u_1$

$\therefore$ The first body continues to move with the same velocity and the second body acquires double the initial velocity of the first.

(ii) If $m_1\gg m_2$ and $u_1=0$

$v_1=\frac{2m_2{u}_2}{m_1}\approx 0$ and $v_2=-\frac{m_1{u}_2}{m_1}=-u_2$

$\therefore$ After collision, the heavy body remains at rest and the lighter body rebounds with the same velocity.

5.

At $\mathit{A}$ :

K.E. $=\mathbf{0}$ (Since the body is at rest),

$P.E=mgh$

$\therefore$ T.E. $=$ K.E. + P.E $=mgh$

At $B$ :

Let $v_1$ be the velocity of the body.

$\therefore$ Using the relation, $v^2-u^2=2$ as.

$v_1{}^2-0=2gx$ or $v_1{}^2=2gx$

$\therefore K.E.=\frac{1}{2}m{v}_1^2=\frac{1}{2}\times 2mgx=mgx$

P.E. $=mg(h-x)$

$\therefore$ T.E. $=$ K.E. + P.E. $=mgx+mg(h-x)=mgh$ At $C$ :

It touches the ground. $\therefore P.E.=0$

If $v$ is the final velocity, then $v^2=2gh$

$\therefore K.E.=\frac{1}{2}m{v}^2=\frac{1}{2}\times m\times 2gh=mgh\therefore T.E.=mgh$

Total energy remains constant at every point in the path of a freely falling body.

6. Let $A=$ Cross-sectional area of the pipe, $\nu =$ velocity of flow of water, $\rho =$ density of water.

$\therefore$ Mass of water flowing out $/s=\frac{dm}{dt}=Av\rho$

To get $n$ times water in same time,

$(\frac{dm}{dt}{)}^{\prime }=n(\frac{dm}{dt})=nAv\rho$

$A^{\prime }{v}^{\prime }\rho =nAv\rho$

$\because A=A^{\prime }$ and $\rho ={\rho }^{\prime }$

$\therefore v^{\prime }=nv$

(a) $F=v\frac{dm}{dt}$

$\therefore \frac{F^{\prime }}{F}=\frac{v^{\prime }(\frac{dm}{dt}{)}^{\prime }}{v(\frac{dm}{dt})}=\frac{nv(n\frac{dm}{dt})}{v(\frac{dm}{dt})}$

$\therefore F^{\prime }=n^{2}F$

(b) $\frac{P^{\prime }}{p}=\frac{F^{\prime }\times v^{\prime }}{F\times v}=\frac{n^{2}F\times n\nu }{F\times v}=n^3$

$\therefore P^{\prime }=n^{3}P$

7. When dropped from same height, the two masses hit ground with same velocity.

$v=\sqrt{2gh}$

$\therefore \frac{P_2}{P_1}=\frac{m_2}{m_1}=n$

8. $K.E.=\frac{P^2}{2m}=$ constant.

$\therefore P=\sqrt{2m\times K.E}$

$\therefore \frac{P_2}{P_1}=\sqrt{\frac{m_2}{m_1}}=\sqrt{n}$

9. When momentum is increased by $50$, velocity also increases by $50$.

$\therefore$ New velocity $=v^{\prime }=v+\frac{50}{100}v=\frac{3}{2}v$

$\therefore$ New K.E. $=\frac{1}{2}m{v}^{\prime 2}=\frac{9}{4}\times (K.E)-\frac{9}{4}\times 100=225$

$\therefore$ Increase in K.E $=225-100=125$

EXERCISE-2

Multiple Choice Questions

1. (c)

2. (c)

3. (d)

4. (d)

5. (a)

6. (c)

7. (c)

8. (d)

9. (b)

10. (b)

11. (d)

12. (a)

13. (b)

14. (b)

15. (a)

16. (b)

17. (a)

18. (c)

19. (c)

20. (d)

21. (b)

22. (b)

23. (d)

24. (d)

25. (b)

26. (a)

27. (c)

28. (d)

29. (d)

30. (d)

31. (b)

32. (a)

33. (d)

34. (d)

35 (d) From $v=u+at,v_1=0+a{t}_1$

$\therefore a=\frac{v_1}{t_1}$

$F=ma=m{v}_1/t_1$

Velocity acquired in $tec=$ at $=\frac{v_1}{t_1}t$

Power $=F\times v=\frac{m{v}_1}{t_1}\times \frac{v_{1}t}{t_1}=\frac{m{v}_1^{2}t}{t_1^2}$

36. (b) $W={\int }_0^{x_1}Fdx={\int }_0^{x_1}cxdx=[\frac{1}{2}c{x}^2{]}_0^{x_1}$

$=\frac{1}{2}c(x_1^2-0)=\frac{1}{2}c{x}_1^2$

37. (d) $W=F\times s$

$W\propto \frac{1}{x}(x)\therefore W\propto x^0$

38. (c) As $\frac{1}{2}{m}_A{v}_A^2=\frac{1}{2}{m}_B{v}_B^2$

$\frac{v_A}{v_B}=\sqrt{\frac{m_B}{m_A}};$

$=\frac{m_B}{m_A}\sqrt{\frac{m_A}{m_B}}=\sqrt{\frac{m_B}{m_A}}=\frac{1}{\sqrt{3}}$

39. (b) At the top of flight, horizontal component of velocity $=u\cos {45}^{\circ }=u/\sqrt{2}$

$\therefore K.E.=\frac{1}{2}m(\frac{u}{\sqrt{2}}{)}^2=\frac{1}{2}(\frac{m{u}^2}{2})=\frac{1}{2}K$

40. (d) As $E=\frac{p^2}{2m}\therefore \frac{dE}{E}=2(\frac{dp}{p})=2\times 5$

41. (c) Work done $=\frac{1}{2}\times$ Stress $\times$ Volume $\times$ Strain

$=\frac{1}{2}\times$ Force $\times$ Extension $=\frac{1}{2}\times 200\times 1\times {10}^{-3}$

$=0.1J$

42. (b) If $\overrightarrow{L}=0\Rightarrow K$.E may or may not be zero.

If $K.E=0,\overrightarrow{L}=0$.

43. (c) $v=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}=\frac{a\times b+0}{a+c}=\frac{a(b)}{a+c}$.

44. (d) As $e^n=(\frac{h_n}{h_0}{)}^{1/2}\therefore h_n=e^{2n}{h}_0=e^{2\times 2}h=e^{4}h$.

45. (d) Applying the principle of conservation of linear momentum, we get

$3m\times v=\sqrt{(m\times 60{)}^2+(m\times 60{)}^2}$

$v=20\sqrt{2}m/s$

46. (b) Work done against gravity

$Wg=50\times 10\times 30=15kJ$

Work done against friction

$W=\mu mg\cos \theta \times s=0.4\times 50\times 10\times \frac{4}{5}\times 50=8kJ$

Total work done $=W_g+W_f=15kJ+8kJ=23kJ$

47. (b) $\frac{1}{2}m{v}^2=\frac{1}{2}k{x}^2$

$m{v}^2=k{x}^2$ or $0.5\times (1.5)2=50\times x^2$

$\therefore x=0.15m$

48. (c) Fraction of energy transferred $=\frac{4\times 2}{(1+2{)}^2}=\frac{8}{9}$

More Than One Correct

1. (b, c)

2. (c, d)

3. (a, c)

4. (c, d)

5. (a, d)

6. (b, d)

7. (a, d) Net force acting on the system is zero, hence force as well as momentum of both the blocks are equal and opposite.

8. $(a,b,c,d)$

$x=u\cos \theta t$

$y=u\sin \theta t-\frac{1}{2}g{t}^2$

$E=\frac{1}{2}m{u}^2-mgy$

$\begin{array}{rl}& =\frac{1}{2}m{u}^2-mgu\sin \theta t+\frac{1}{2}m{g}^2{t}^2\\ & =\frac{1}{2}m{u}^2-mg\tan \theta x+\frac{m{g}^2}{u^2{\cos }^2\theta }{x}^2\end{array}$

Also $E=\frac{P^2}{2m}$

Hence all the graphs are correct.

9. (b, c, d)

10. (c, d) Acceleration and velocity change in direction, hence are not uniform.

11. $(a,b,c)$

$K=\frac{1}{2}m{v}^2$ and $p=mv$

$\therefore K=\frac{1}{2}pv$ or $2K=pv$

or $v=\frac{2K}{p}$

or $p=\sqrt{2mK}$ 12. $(a,b,d)$

13. (a, b, c, d)

14. (b, d)

Multiple Matching Questions

1. (A) $\to$ (p, t) ;(B) $\to$ (p, t) ;(C) $\to$ (q, r, s) ;(D) $\to$ (q, r, s)
2. (A) $\to$ (p) ;(B) $\to$ (r, s) ;(C) $\to$ (q, s) ;(D) $\to$ (q, t)

Fill In The Passage

I. (1) kinetic energy

(2) mass

(3) halved

(4) velocity

(5) one fourth

II. (1) change

(2) another

(3) transformation

(4) potential

(5) kinetic

III. (1) conservative

(2) zero

(3) non-conservative

(4) non-zero.

Passage Based Questions

1. (b) By work-energy theorem, Work done by all the forces $=$ Change in K.E

or $-\mu mgx-\frac{1}{2}k{x}^2=-\frac{1}{2}m{u}^2(x=$ non. compression $)$ or $\frac{1}{2}k{x}^2+\mu mgx-\frac{1}{2}m{u}^2=0$

or $50x^2+4x-0.9=0$ or $x=0.1m=10cm$

2. (a) $\frac{1}{2}m{v}^2=\frac{1}{2}m{u}^2-2\mu mgx$

or $v=10cm{s}^{-1}$

3. (b) Work done by friction in round trip, $W=-2\mu mgx=-2\times 0.2\times 2\times 10\times 0.1=0.8J$

4. (c) Speed of particle at $B$,

$u=\sqrt{2g(h_p-h_B)}=\sqrt{2\times 10\times 45}=30m{s}^{-1}$

For projectile motion of the particle,

$H=\frac{u^2{\sin }^2{30}^{\circ }}{2g}=\frac{(30{)}^2\times (\frac{1}{2}{)}^2}{2\times 10}=11.25m$

$\therefore$ height above the ground $=50+11.25=61.25m$.

5. (b) Let $x$-and $y$-axis lie along horizontal and vertical respectively, then

$u_x=u\cos {30}^{\circ }=15\sqrt{3}m{s}^{-1},u_y=u\sin {30}^{\circ }$

$=15m{s}^{-1}$

$a_x=0,a_y=-g=-\overline{10}m{s}^{-1}$

at any instant,

$y=y_0+u_{y}t+\frac{1}{2}{a}_y{t}^2=50+15t-5t^2$

When the particle strikes the ground,

$y=0$

or $50+15t-5t^2=0$ or $t^2-3t-10=0$

or

$t=5s$ or $-2s$

but time cannot be negative, hence $t=5s$ is correct choice.

6. (b) Now, $R=u_{x}t=75\sqrt{3}m$

7. (b) For maxiumum angle, work done by $F=$ gain in P.E of bob

or $F_x=mgh$

$Fl\sin {\theta }_0=mgl(1-\cos {\theta }_0)$

$\frac{1-\cos {\theta }_0}{\sin {\theta }_0}=\frac{F}{mg}$ or $\tan {\theta }_0/2=\frac{F}{mg}$

or ${\theta }_0=2{\tan }^{-1}(\frac{F}{mg})$

8. (a) Velocity is maximum when net force on the bob is zero.

Let the angle mode by the string with vertical at that instant be $\theta$, then

$T\cos \theta =mg-\theta$ and $T\sin \theta =F$

or $\theta ={\tan }^{-1}\frac{F}{mg}$

By work energy theorem,

$Fl\sin \theta -mgl(1-\cos \theta )=\frac{1}{2}m{v}^2$

or $v=[\frac{u(\sqrt{F^2{m}^2{g}^2-mg})}{m}{]}^{1/2}$

9. (d) At the position of maximum deflection,

$T=mg\cos {\theta }_0+F\sin {\theta }_0$

using $\cos {\theta }_0=\frac{1-{\tan }^2{\theta }_0/2}{1+{\tan }^2{\theta }_0/2}$ and $\sin {\theta }_0=\frac{2\tan {\theta }_0/2}{W{\tan }^2{\theta }_0/2}$

We get

$T=mg$

Assertion and Reasons

1. (c)
2. (a)
3. (a)
4. (c)
5. (d)
6. (d)
7. (d)
8. (a) $W=\overrightarrow{F}\cdot \overrightarrow{s}=Fs\cos \theta$

if $\theta ={90}^{\circ },W=0$

HOTS Subjective Questions

1. If a moving truck or a car is brought to a stop then the change in momentum is brought to zero.

As both the truck and the car are traveling at the same speed but the truck is more heavier than the car therefore momentum of the truck is more and hence a greater force will be required to change its momentum to zero.

A moving car could require more stopping force if the time during which its momentum is brought to zero is reduced considerably. For example, if the time of impact is reduced by 10 , then force of impact is extended to 10 times.

2. A punch with a bare fist is more forceful than with a boxing glove. A boxing glove allows a longer time of impact thus decreasing the force of impact whereas a punch with a bare fist has smaller time of impact and hence it has greater force of impact.

3. When the boxer is punching a heavy bag, the bag provides impulse to stop the punches and reduce the momentum to zero. The boxer’s punch acts as action to the bag and the bag exerts an equal and opposite force of reaction to the boxer and thus stops the punches. Therefore the boxer can punch the bag for an hour without tiring.

While boxing with an opponent, punches are missed because the opponent bends sideways to protect himself and thus missed punches are in air. In this case air molecules will be providing impulse to stop the punches and that will be very small. There will be no enough impulse to bring the momentum to zero.

As a result the body of boxer will be pulled along with the punches and hence he will tire quickly.

4. Railroad cars are loosely coupled so that there is a noticeable time delay from the time the first car is moved and last cars are moved from rest by the locomotive.

By the impulse-momentum relation

$F.t=\mathrm{\Delta }mv$

where $F$ is the force, $t$ is the time and $\mathrm{\Delta }mv$ is the change in momentum.

The mass of the locomotive and coupled cars is enormous so when locomotive moves from rest it acquires large momentum and this momentum is imparted to the cars through the loose coupling and slack between cars.

Such a large momentum will cause a great force of impact that might hurt the passengers inside the car.

The loose coupling allows a longer time of impact and thereby decreasing the force of impact considerably so that the passengers may be comfortable when cars gain momentum.

5. A high-speed bus and an innocent bug have a head-on collision. The sudden change of momentum of the bus is the same as the change in momentum of the bug. This is because the momentum is conserved for all types of collisions. The mass of the bus is large hence its velocity will change by small amount after collision and as the mass of the bug is very small its velocity will change by large amount but the change in momentum of the bus and the bug will be the same.
6. The water coming out a hose at high speed has a large momentum in the forward direction. In accordance with the conservation of momentum, the same amount of momentum is exerted on the hose pipe in the backward direction therefore a firefighter has to apply a large force to counter the backward momentum of hose and keep the hose in a stable position.
7. A small raft is next to a dock. If someone jumps from the raft, he will fall into the water because he could not take into account the principle of the conservation of momentum. If someone jumps from the raft, he acquires momentum in the forward direction and hence the raft moves in the opposite direction gaining the momentum in backward direction so that the net change in the momentum may be zero as no external force acts here.
8. There is no violation of the conservation of momentum when you step off a chair and gain momentum as you fall. When you are sitting on the chair the total momentum of the system is zero. When you fall you gain in momentum in one direction and the chair moves in the other direction, so that the net change in momentum of the system may be zero.
9. If a Mack truck and a Honda Civic have a head on collision, both of them will experience the same force of impact, the same impulse and the same change in momentum. In any type of collision, the momentum of the system remains conserved. And by the impulse-momentum relation.

$F.t=\mathrm{\Delta }mv$

where $F$ is the force $t$ is time and $\mathrm{\Delta }mv$ is change in momentum. The time of impact during collision is the same for both the Mack truck and Honda Civic therefore the force of impact is also the same because the change in momentum of the two is the same. Now, as acceleration $=\frac{\text{ Force }}{\text{ mass }}$, hence the acceleration of the Honda Civic will be greater because its mass is lesser than the Mack truck.

10. A head-on collision between two cars will be more damaging to the occupants if the cars rebounded upon impact instead of sticking together. Because the impulse required to bring an object to a stop and then to throw it back again is greater than the impulse required merely to bring it to a stop. Thus impulse is greater when an object bounces and hence the damaging force is also greater.
11. When the first astronaut throws the second one towards the third one, the first astronaut will be thrown in the opposite direction. This in accordance with the principle of momentum conservation.

When the second astronaut pushes the third astronaut he comes to rest and the third astronaut moves with the same velocity of the second.

The game will not last long because the astronaut will not meet together further.

12. The forces involved in the fission of a uranium nucleus are internal force which means the net momentum of its flying chunks is the same as the momentum of the uranium nucleus before explosion. As the uranium nucleus is initially at rest therefore its net momentum is zero and hence the chunks fly apart exactly in opposite directions such that the sum of their momenta becomes zero.

Consider an uranium nucleus which is at rest

After fission it breaks into two chunks of masses $m_1$ and $m_2$ which fly off with velocities $v_1$ and $v_2$.

The momentum of each fragment is the same in magnitude but opposite in direction.

Thus, $m_1{\overrightarrow{v}}_1=-m{\overrightarrow{v}}_2$

or, $m_1{\overrightarrow{v}}_1+m{\overline{v}}_2=0$

Now as the momentum of the two chunks is the same in magnitude therefore the speed of the lighter chunk will be greater and the speed of the heavier chunk will be smaller.

13. Momentum is a vector quantity having both magnitude and direction whereas kinetic energy is a scalar quantity having only magnitude. Momentum that is directional is capable of being cancelled entirely.

The vector sum of the momenta of a flock of birds in flight can be zero because of birds flying in different directions in the flock.

Each flying bird has some kinetic energy and the algebraic addition of the kinetic energies of all birds in the flock cannot be zero.

14. In determining the potential energy of Tenny’s drawn bow, it would be accurate to multiply the force with which she holds the \to in its drawn position by the distance she pulled it because work done (average force $\times$ distance) in drawing the \to is the measure of potential energy. When a force is applied on an object it produces a change in the position of the object. The force may be variable thus work done by the force is measured by the product of the average force and the distance that has been covered by the object.
15. The velocity of the emerging cannonball fired from a cannon with a long barrel is greater because the force of expanding gases acts on the cannonball for a longer distance and hence the kinetic energy of the cannonball is more. It is evident from work - energy theorem.

$F.d=\mathrm{\Delta }KE$ where $F$ is the force, $d$ is the distance and $\mathrm{\Delta }KE$ is the charge in the kinetic energy. As the kinetic energy of the emerging cannonball is more therefore its velocity is greater.