Thinking Process

To solve such type of questions, we have to use the formula for dot product or cross product.

Answer (b) Given, $\mathbf{A}=\hat{\mathbf{i}}+\hat{\mathbf{j}}$

$\mathbf{B}=\hat{\mathbf{i}}-\hat{\mathbf{j}}$

We know that

$ \begin{gathered} A \cdot B=|A||\mathbf{B}| \cos \theta \\ \Rightarrow \quad(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \cdot(\hat{\mathbf{i}}-\hat{\mathbf{j}})=(\sqrt{1+1})(\sqrt{1+1}) \cos \theta \end{gathered} $

where $\theta$ is the angle between $\mathbf{A}$ and $\mathbf{B}$

$ \begin{aligned} \Rightarrow & & \cos \theta & =\frac{1-0+0-1}{\sqrt{2} \sqrt{2}}=\frac{0}{2}=0 \\ \Rightarrow & & \theta & =90^{\circ} \end{aligned} $

Thinking Process

In this question according to the diagram, we have to decide the components of a given vector.

Answer (b) Clearly from the diagram, $\mathbf{u}=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}$

As $u$ is in the first quadrant, hence both components $a$ and $b$ will be positive.

For $\mathbf{v}=p \hat{\mathbf{i}}+q \hat{\mathbf{j}}$, as it is in positive $x$-direction and located downward hence $x$-component $p$ will be positive and $y$-component $q$ will be negative.

Answer (b) Let $r$ makes an angle $\theta$ with positive $x$-axis component of $r$ along $X$-axis

As

$r_{x} =|\mathbf{r}| \cos \theta $

$(r_{x}) _{\text{maximum }} =|\mathbf{r}|(\cos \theta) _{\text{maximum }} $

$=|r| \cos 0^{\circ}=|r| \quad(\because \cos \theta \text{ is maximum of } \theta=0^{\circ})$

$ \theta = 0^o$ $r$ is along positive $x$-axis.

Answer (c) We know that

where $\theta$ is angle of projection

Given,

$ \theta=15^{\circ} \text { and } R=50 m $

$ \text { Range, } R=\frac{u^{2} \sin 2 \theta}{g} $

Putting all the given values in the formula, we get

$ \begin{matrix} \Rightarrow & & R & =50 m=\frac{u^{2} \sin (2 \times 15^{\circ})}{g} \\ \Rightarrow & & 50 \times g=u^{2} \sin 30^{\circ}=u^{2} \times \frac{1}{2} \end{matrix} $

$ \begin{matrix} \Rightarrow & 50 \times g \times 2 & =u^{2} \\ \Rightarrow & u^{2} & =50 \times 9.8 \times 2=100 \times 9.8=980 & \\ \Rightarrow & u & =\sqrt{980}=\sqrt{49 \times 20}=7 \times 2 \times \sqrt{5} m / s & \\ & & =14 \times 2.23 m / s=31.304 m / s & \\ \text { For } & \theta & =45^{\circ} ; R=\frac{u^{2} \sin 2 \times 45^{\circ}}{g}=\frac{u^{2}}{g} & (\because \sin 90^{\circ}=1) \\ \Rightarrow & R & =\frac{(14 \sqrt{5})^{2}}{g}=\frac{14 \times 14 \times 5}{9.8}=100 m \end{matrix} $

Thinking Process

As speed is a scalar quantity, hence it will be related with path length (scalar quantity) only.

Answer (d) We know that

$ \text { speed, } v_0=\frac{\text { total distance travelled }}{\text { time taken }} $

Hence, total distance travelled $=$ Path length

$ =(\text { speed }) \times \text { time taken } $

Note We should be very careful with the fact, that speed is related with total distance covered not with displacement.

Thinking Process

This question can solved by checking each options one by one.

Answer $(b, d)$

Given $A+B+C=0$

Hence, we can say that $A, B$ and $C$ are in one plane and are represented by the three sides of a triangle taken in one order. Now consider the options one by one.

(a) We can write

$B \times(A+B+C) =B \times 0=0 $

$\Rightarrow B \times A+B \times B+B \times C=0 $

$\Rightarrow B \times A+0 B \times C=0 $

$\Rightarrow B \times A=-B \times C $

$\Rightarrow A A B=B \times C $

$\therefore (A \times B) \times C =(B \times C) \times C$

It cannot be zero.

If $B | C$, then $B \times C=0$, than $(B \times C) \times C=0$.

(b) $(A \times B) \cdot C=(B \times AC) \cdot C=0$ whatever be the positions of $A, B$ and $C$. If $B | C$, then $B \times C=0$, then $(B \times C) \times C=0$.

(c) $(A \times B)=X=AB \sin \theta X$. The direction of $X$ is perpendicular to the plane containing $A$ and $B$. $(A \times B) \times C=X \times C$. Its direction is in the plane of $A, B$ and $C$.

(d) If $C^{2}=A^{2}+B^{2}$, then angle between $A$ and $B$ is $90^{\circ}$

$ \begin{aligned} (\mathbf{A} \times \mathbf{B}) \cdot C & =(A B \sin 90^{\circ} \mathbf{X}) \cdot \mathbf{C}=A B(X \cdot C) \\ & =A B C \cos 90^{\circ}=0 \end{aligned} $

Answer ( $a, b$ )

Given that

$ |\mathbf{A}+\mathbf{B}|=|\mathbf{A}| \text { or }|\mathbf{A}+\mathbf{B}|^{2}=|\mathbf{A}|^{2} $

$ \Rightarrow \quad|\mathbf{A}|^{2}+|\mathbf{B}|^{2}+2|\mathbf{A}||\mathbf{B}| \cos \theta=|\mathbf{A}|^{2} $

where $\theta$ is angle between $\mathbf{A}$ and $\mathbf{B}$.

$ \begin{aligned} \Rightarrow & & |\mathbf{B}|(|\mathbf{B}|+2|\mathbf{A}| \cos \theta) & =0 \\ \Rightarrow & & |\mathbf{B}| & =0 \text { or }|\mathbf{B}|+2|\mathbf{A}| \cos \theta=0 \\ \Rightarrow & & \cos \theta & =-\frac{|\mathbf{B}|}{2|\mathbf{A}|} \end{aligned} $

If $\mathbf{A}$ and $\mathbf{B}$ are antiparallel, then $\theta=180^{\circ}$

Hence, from Eq. (i)

$ -1=-\frac{|\mathbf{B}|}{2|\mathbf{A}|} \Rightarrow|\mathbf{B}|=2|\mathbf{A}| $

Hence, correct answer will be either $|\mathbf{B}|=0$ or $\mathbf{A}$ and $\mathbf{B}$ are antiparallel provided $|\mathbf{B}|=2|\mathbf{A}|$

Thinking Process

In this problem, we have to apply equation for maximum height reached $H=$ $\frac{u^{2} \sin ^{2} \theta}{2 g}$, where $\theta$ is angle of projection and $u$ is speed of projection of a projectile

Answer

$(a, b, c)$

We know that maximum height reached by a projectile,

$ \begin{aligned} & H=\frac{u^{2} \sin ^{2} \theta}{2 g} \\ & H_1=\frac{v_0^{2} \sin ^{2} \theta_1}{2 g} \quad \text { (for first particle) } \\ & H_2=\frac{v_0^{2} \sin ^{2} \theta_2}{2 g} \quad \text { (for second particle) } \end{aligned} $

According to question, we know that

$ \begin{aligned} & H_1>H_2 \\ & \Rightarrow \quad \frac{v_0^{2} \sin ^{2} \theta_1}{2 g}>\frac{v_0^{2} \sin ^{2} \theta_2}{2 g} \\ & \Rightarrow \quad \sin ^{2} \theta_1>\sin ^{2} \theta_2 \\ & \Rightarrow \quad \sin ^{2} \theta_1-\sin ^{2} \theta_2>0 \\ & \Rightarrow \quad(\sin \theta_1-\sin \theta_2)(\sin \theta_1+\sin \theta_2)>0 \\ & \text { Thus, either } \quad \sin \theta_1+\sin \theta_2>0 \\ & \Rightarrow \quad \sin \theta_1-\sin \theta_2>0 \\ & \Rightarrow \quad \sin \theta_1>\sin \theta_2 \text { or } \theta_1>\theta_2 \\ & T=\frac{2 u \sin \theta}{g}=\frac{2 v_0 \sin \theta}{g} \\ & T_1=\frac{2 v_0 \sin \theta_1}{g} \\ & T_2=\frac{2 v_0 \sin \theta_2}{g} \end{aligned} $

As,

Hence, (Here, $T_1=$ Time of flight of first particle and $T_2=$ Time of flight of second particle).

$ \sin \theta_1>\sin \theta_2 $

$ T_1>T_2 $

We know that

Given,

$\Rightarrow$

$\Rightarrow \quad \frac{R_1}{R_2}=\frac{\sin 2 \theta_1}{\sin 2 \theta_2}>1$

$\Rightarrow \quad R_1>R_2$

Total energy for the first particle,

$ \begin{aligned} & U_1=K E+P E=\frac{1}{2} m_1 v_0^{2} \\ & \text { (This value will be constant throughout the journey) } \\ & U_2=K E+P E=\frac{1}{2} m_2 v_0^{2} \quad \text { (Total energy for the second particle) } \end{aligned} $

Total energy for the second particle

If

$ \begin{aligned} & m_1=m_2 \text { then } U_1=U_2 \\ & m_1>m_2 \text { then } U_1>U_2 \\ & m_1<m_2, \text { then } U_1 < U_2 \end{aligned} $

Thinking Process

In this type of question, nature of track is very important of consider, as friction is not in this track, total energy of the particle will remain constant throughout the journey.

Answer (c)

As the given track $y=x^{2}$ is a frictionless track thus, total energy (KE + PE) will be same throughout the journey.

Hence, total energy at $A=$ Total energy at $P$. At $B$, the particle is having only KE but at $P$ some KE is converted to $P$.

Hence, $(KE)_B >(KE)_P$

Total energy at $A=PE=$ Total energy at $B=KE$

$ \begin{aligned} & =\text { Total energy at } P \\ & =PE+KE \end{aligned} $

The potential energy at $A$, is converted to $KE$ and $PE$ at $P$, hence

$ (PE) P<(PE) A $

Hence,

$ \text { (Height) } P<\text { (Height) A } $

As, height of $P<$ Height of $A$

Hence, path length $A B>$ path length $B P$

Hence, time of travel from $A$ to $B \neq$ Time of travel from $B$ to $P$.

Answer (a, c)

If an object undergoes a displacement $\Delta r$ in time $\Delta t$, its average velocity is given by $v=\frac{\Delta r}{\Delta t}=\frac{r_2-r_1}{t_2-t_1}$; where $r_1$ and $r_2$ are position vectors corresponding to time $t_1$ and $t_2$.

It the velocity of an object changes from $v_1$ to $v_2$ in time $\Delta t$. Average acceleration is given by

$ a_{av}=\frac{\Delta v}{\Delta t}=\frac{v_2-v_1}{t_2-t_2} $

But, when acceleration is non-uniform

We can write

$ v_{av} \neq \frac{v_1+v_2}{2} $

$ \text { Hence, } \quad \Delta r=r_2-r_1=(v_2-v_1)(t_2-t_1) $

Answer $(a, b, c)$

For a particle performing uniform circular motion

(i) speed will be constant throughout.

(ii) velocity will be tangential in the direction of motion at a particular point.

(iii) acceleration $a=\frac{v^{2}}{r}$, will always be towards centre of the circular path.

(iv) angular momentum ( $m v r$ ) is constant in magnitude and direction/out of the plane perpendicularly, as well.

Note In uniform circular motion, magnitude of velocity and acceleration is constant but direction changes continuously.

Answer $(b, d)$

Given, $|\mathbf{A}+\mathbf{B}|=|\mathbf{A}-\mathbf{B}|$

$ \begin{matrix} \Rightarrow & \sqrt{|\mathbf{A}|^{2}+|\mathbf{B}|^{2}+2|\mathbf{A}||\mathbf{B}| \cos \theta}=\sqrt{|\mathbf{A}|^{2}+|\mathbf{B}|^{2}-2|\mathbf{A}||\mathbf{B}| \cos \theta} \\ \Rightarrow & |\mathbf{A}|^{2}+|\mathbf{B}|^{2}+2|\mathbf{A}||\mathbf{B}| \cos \theta=|\mathbf{A}|^{2}+|\mathbf{B}|^{2}-2|\mathbf{A}||\mathbf{B}| \cos \theta \\ \Rightarrow & 4|\mathbf{A}||\mathbf{B}| \cos \theta=0 \\ \Rightarrow & |\mathbf{A}||\mathbf{B}| \cos \theta=0 \\ \Rightarrow & |\mathbf{A}|=0 \text { or }|\mathbf{B}|=0 \text { or } \cos \theta=0 \\ \Rightarrow & \theta=90^{\circ} \end{matrix} $

When $\theta=90^{\circ}$, we can say that $\mathbf{A} \perp \mathbf{B}$

Answer As shown in the adjacent figure. The cyclist covers the path $O P R Q O$. As we know whenever an object performing circular motion, acceleration is called centripetal acceleration and is always directed towards the centre.

Hence, acceleration at $R$ is a $=\frac{v^{2}}{r}$

$\Rightarrow \quad a=\frac{(10)^{2}}{1 km}=\frac{100}{10^{3}}=0.1 m / s^{2}$ along $R O$.

Thinking Process

When a particle is under projectile motion, horizontal component of velocity will always be constant and acceleration is always vertically downward and is equal to $g$.

Answer Consider the adjacent diagram in which a particle is projected at an angle $\theta$.

$v_{x}=$ Horizontal component of velocity $=v \cos \theta=$ constant.

$v_{y}=$ Vertical component of velocity $=v \sin \theta$

Velocity will always be tangential to the curve in the direction of motion and acceleration is always vertically downward and is equal to $g$ (acceleration due to gravity).

Answer Consider the adjacent diagram in which a ball is projected from point $O$, and covering the path $O A B C$.

(a) At point $B$, it will gain the same speed $u$ and after that speed increases and will be maximum just before reaching $C$.

(b) During upward journey from $O$ to $A$ speed decreases and will be minimum at point $A$.

(c) Acceleration is always constant throughout the journey and is vertically downward equal to $g$.

Answer (a) Consider the adjacent diagram in which a football is kicked into the air vertically upwards. Acceleration of the football will always be vertical downward and is equal to $g$.

(b) When the football reaches the highest point velocity will be zero as it is continuously retarded by acceleration due to gravity $g$.

Answer The path of the ball observed by a boy standing on the footpath is parabolic. The horizontal speed of the ball is same as that of the car, therefore, ball as well car travells equal horizontal distance. Due to its vertical speed, the ball follows a parabolic path.

Note We must be very clear that we are working with respect to ground. When we observe with respect to the car motion will be along vertical direction only.

Answer Consider the diagram below

(b)

The boy throws the ball at an angle of $60^{\circ}$.

$\therefore$ Horizontal component of velocity $=4 \cos \theta$

$ \begin{aligned} & =(10 m / s) \cos 60^{\circ}=10 \times \frac{1}{2}=5 m / s . \\ \text { Speed of the car } & =18 km / h=5 m / s . \end{aligned} $

As horizontal speed of ball and car is same, hence relative velocity of car and ball in the horizontal direction will be zero.

Only vertical motion of the ball will be seen by the boy in the car, as shown in fig. (b)

Thinking Process

When air resistance is included the horizontal component of velocity will not be constant and obviously trajectory will change.

Answer Due to air resistance, particle energy as well as horizontal component of velocity keep on decreasing making the fall steeper than rise as shown in the figure.

When we are neglecting air resistance path was symmetric parabola $(O A B)$. When air resistance is considered path is asymmetric parabola $(O A C)$.

Thinking Process

When the bomb is dropped from the plane, the bomb will have same velocity as that of plane.

Answer Consider the adjacent diagram. Let a fighter plane, when it be at position $P$, drops a bomb to hit a target $T$.

Let

$ <P^{\prime} P T=\theta $

Speed of the plane $=720 km / h=720 \times \frac{5}{18} m / s=200 m / s$

Altitude of the plane $(P^{\prime} T)=1.5 km=1500 m$

If bomb hits the target after time $t$, then horizontal distance travelled by the bomb,

$ P P^{\prime}=u \times t=200 t $

Vertical distance travelled by the bomb,

$ \begin{aligned} & P^{\prime} T=\frac{1}{2} g t^{2} \Rightarrow 1500=\frac{1}{2} \times 9.8 t^{2} \\ \Rightarrow \quad t^{2} & =\frac{1500}{4.9} \Rightarrow t=\sqrt{\frac{1500}{49}}=17.49 s \end{aligned} $

Using value of $t$ in Eq. (i),

Now,

$ \begin{aligned} P P^{\prime} & =200 \times 17.49 m \\ \tan \theta & =\frac{P^{\prime} T}{P^{\prime} P}=\frac{1500}{200 \times 17.49}=.49287=\tan 23^{\circ} 12^{\prime} \\ \theta & =23^{\circ} 12^{\prime} \end{aligned} $

Note Angle is with respect to target. As seen by observer in the plane motion of the bomb will be vertically downward below the plane.

Answer (a) Radius of the earth $(R)=6400 km=6.4 \times 10^{6} m$

Time period $(T)=1$ day $=24 \times 60 \times 60 s=86400 s$

Centripetal acceleration $(a_{c})=\omega^{2} R=R(\frac{2 \pi}{T})^{2}=\frac{4 \pi^{2} R}{T}$

$ \begin{aligned} & =\frac{4 \times(22 / 7)^{2} \times 6.4 \times 10^{6}}{(24 \times 60 \times 60)^{2}} \\ & =\frac{4 \times 484 \times 64 \times 10^{6}}{49 \times(24 \times 3600)^{2}} \\ & =0.034 m / s^{2} \end{aligned} $

At equator,

latitude $\theta=0^{\circ}$

$\therefore$

$ \frac{a_{c}}{g}=\frac{0.034}{9.8}=\frac{1}{288} $

(b) Orbital radius of the earth around the sun $(R)=1.5 \times 10^{11} m$

Time period $=1 yr=365$ day

$ =365 \times 24 \times 60 \times 60 s=3.15 \times 10^{7} s $

Centripetal acceleration $(a_{c})=R \omega^{2}=\frac{4 \pi^{2} R}{T^{2}}$

$ \begin{aligned} & =\frac{4 \times(22 / 7)^{2} \times 1.5 \times 10^{11}}{(3.15 \times 10^{7})^{2}} \\ & =5.97 \times 10^{-3} m / s^{2} \\ \frac{a_{C}}{g} & =\frac{5.97 \times 10^{-3}}{9.8}=\frac{1}{1642} \end{aligned} $

Thinking Process

In this problem, triangular law of vector addition will be applied.

Answer Consider the adjacent diagram in which vectors $\mathbf{A}$ and $\mathbf{B}$ are corrected by head and tail.

Resultant vector $\mathbf{C}=\mathbf{A}+\mathbf{B}$

(a) from (iv) it is clear that $\mathbf{c}=\mathbf{a}+\mathbf{b}$

(b) from (iii) $\mathbf{c}+\mathbf{b}=\mathbf{a} \Rightarrow \mathbf{a}-\mathbf{c}=\mathbf{b}$

(c) from (i) $\mathbf{b}=\mathbf{a}+\mathbf{c} \Rightarrow \mathbf{b}-\mathbf{a}=\mathbf{c}$

(d) from (ii) $-\mathbf{c}=\mathbf{a}+\mathbf{b} \Rightarrow \mathbf{a}+\mathbf{b}+\mathbf{c}=\mathbf{0}$

Answer Given $|\mathbf{A}|=2$ and $|\mathbf{B}|=4$

(a) $\mathbf{A} \cdot \mathbf{B}=A B \cos \theta=0 \quad \Rightarrow \quad 2 \times 4 \cos \theta=0$ $\Rightarrow \cos \theta=0=\cos 90^{\circ} \Rightarrow \theta=90^{\circ}$

$\therefore$ Option (a) matches with option (ii).

(b) $\mathbf{A} \cdot \mathbf{B}=A B \cos \theta=8 \quad \Rightarrow \quad 2 \times 4 \cos \theta=8$

$\Rightarrow \cos \theta=1=\cos 0^{\circ} \Rightarrow \quad \theta=0^{\circ}$

$\therefore$ Option (b) matches with option (i).

(c) $\mathbf{A} \cdot \mathbf{B}=A B \cos \theta=4 \quad \Rightarrow 2 \times 4 \cos \theta=4$

$\Rightarrow \cos \theta=\frac{1}{2}=\cos 60^{\circ} \Rightarrow \theta=60^{\circ}$

$\therefore$ Option (c) matches with option (iv).

(d) $\mathbf{A} \cdot \mathbf{B}=A B \cos \theta=-8 \quad \Rightarrow \quad 2 \times 4 \cos \theta=-8$

$\Rightarrow \cos \theta=-1=\cos 180^{\circ} \Rightarrow \quad \theta=180^{\circ}$

$\therefore$ Option (d) matches with option (iii).

Answer Given $|\mathbf{A}|=2$ and $|\mathbf{B}|=4$

(a) $|\mathbf{A} \times \mathbf{B}|=A B \sin \theta=0 \quad \Rightarrow 2 \times 4 \times \sin \theta=0$

$\Rightarrow \sin \theta=0=\sin 0^{\circ} \Rightarrow \quad \theta=0^{\circ}$

$\therefore$ Option (a) matches with option (iv).

(b) $|\mathbf{A} \times \mathbf{B}|=A B \sin \theta=8 \quad \Rightarrow \quad 2 \times 4 \sin \theta=8$

$\Rightarrow \sin \theta=1=\sin 90^{\circ} \Rightarrow \theta=90^{\circ}$

$\therefore$ Option (b) matches with option (iii).

(c) $|\mathbf{A} \times \mathbf{B}|=A B \sin \theta=4 \quad \Rightarrow \quad 2 \times 4 \sin \theta=4$

$\Rightarrow \sin \theta=\frac{1}{2}=\sin 30^{\circ} \Rightarrow \quad \theta=30^{\circ}$

$\therefore$ Option (c) matches with option (i).

(d) $|\mathbf{A} \times \mathbf{B}|=A B \sin \theta=4 \sqrt{2} \quad \Rightarrow \quad 2 \times 4 \sin \theta=4 \sqrt{2}$

$\Rightarrow \sin \theta=\frac{1}{\sqrt{2}}=\sin 45^{\circ} \Rightarrow \theta=45^{\circ}$

$\therefore$ Option (d) matches with option (ii).

Answer Given, speed of packets $=125 m / s$

Height of the hill $=500 m$.

To cross the hill, the vertical component of the velocity should be sufficient to cross such height.

But

$ \begin{aligned} u_{y} & \geq \sqrt{2 g h} \\ & \geq \sqrt{2 \times 10 \times 500} \\ & \geq 100 m / s \\ u^{2} & =u_x^{2}+u_y^{2} \end{aligned} $

$\therefore$ Horizontal component of initial velocity,

$ u_{x}=\sqrt{u^{2}-u_y^{2}}=\sqrt{(125)^{2}-(100)^{2}}=75 m / s $

Time taken to reach the top of the hill,

$ t=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 500}{10}}=10 s $

Time taken to reach the ground from the top of the hill $t^{\prime}=t=10 s$. Horizontal distance travelled in $10 s$

$ x=u_{x} \times t=75 \times 10=750 m $

$\therefore$ Distance through which canon has to be moved $=800-750=50 m$

Speed with which canon can move $=2 m / s$

$\therefore \quad$ Time taken by canon $=\frac{50}{2} \Rightarrow t^{\prime \prime}=25 s$

$\therefore$ Total time taken by a packet to reach on the ground $=t^{\prime \prime}+t+t^{\prime}=25+10+10=45 s$

Thinking Process

Horizontal range of a projectile is maximum when it is thrown at an angle $45^{\circ}$ from the horizontal and $R_{\max }=\frac{u^{2}}{g}$, where $u$ is speed of projection of the projectile.

Answer This problem can be approached in two different ways

(i) Refer to the diagram, target $T$ is at horizontal distance $x=R+\Delta x$ and between point of projection $y=-h$.

(ii) From point $P$ in the diagram projection at speed $v_0$ at an angle $\theta$ below horizontal with height $h$ and horizontal range $\Delta x A$ )

Applying method (i)

Maximum horizontal range

$ R=\frac{v_0^{2}}{g}, \text { for } \theta=45^{\circ} $

Let the gun be raised through a height $h$ from the ground so that it can hit the target. Let vertically downward direction is taken as positive

Horizontal component of initial velocity $=v_0 \cos \theta$

Vertical component of initial velocity $=-v_0 \sin \theta$

Taking motion in vertical direction, $h=(-v_0 \sin \theta) t+\frac{1}{2} g t^{2}$

Taking motion in horizontal direction

$(R+\Delta x) =v_0 \cos \theta \times t $

$\Rightarrow t =\frac{(R+\Delta x)}{v_0 \cos \theta}$

Substituting value of $t$ in Eq. (ii), we get

$ \begin{aligned} & h=(-v_0 \sin \theta) \times(\frac{R+\Delta x}{v_0 \cos \theta})+\frac{1}{2} g(\frac{R+\Delta x}{v_0 \cos \theta})^{2} \\ & h=-(R+\Delta x) \tan \theta+\frac{1}{2} g \frac{(R+\Delta x)^{2}}{v_0^{2} \cos ^{2} \theta} \end{aligned} $

As angle of projection is $\theta=45^{\circ}$, therefore

$ \begin{aligned} h & =-(R+\Delta x)+\tan 45^{\circ}+\frac{1}{2} g \frac{(R+\Delta x)^{2}}{v_0^{2} \cos ^{2} 45^{\circ}} \\ h & =-(R+\Delta x) \times 1+\frac{1}{2} g \frac{(R+\Delta x)^{2}}{v_0^{2}(1 / 2)} \\ h & .=-(R+\Delta x)+\frac{(R+\Delta x)^{2}}{R} \quad \quad(\because \tan 45^{\circ}=1 \text { and } \cos 45^{\circ}=\frac{1}{\sqrt{2}}) \quad \text { [Using Eq. (i), } R=v_0^{2} / g] \\ & =-(R+\Delta x)+\frac{1}{R}(R^{2}+\Delta x^{2}+2 R \Delta x) \\ & =-R-\Delta x+(R+\frac{\Delta x^{2}}{R}+2 \Delta x) \\ & =\Delta x+\frac{\Delta x^{2}}{R} \quad \text { Hence proved. } \\ h & =\Delta x \biggl (1+\frac{\Delta x}{R} \biggl) \quad \end{aligned} $

Note We should not confuse with the positive direction of motion. May be vertically upward direction or vertically downward direction is taken as positive according to convenience.

Thinking Process

To solve problems involving projectile motion on an inclined plane, we have to choose two mutually perpendicular axes, one along inclined plane and other perpendicular to the inclined plane.

Answer Consider the adjacent diagram.

Mutually perpendicular $x$ and $y$-axes are shown in the diagram.

Particle is projected from the point $O$.

Let time taken in reaching from point $O$ to point $P$ is $T$.

(b) Considering motion along vertical upward direction perpendicular to $O X$.

For the journey $O$ to $P$.

$ y=0, u_{y}=v_0 \sin \beta, a_{y}=-g \cos \alpha, t=T $

Applying equation,

y = $u_{y} t+\frac{1}{2} a_{y} t^{2} $

$\rightarrow 0 =v_0 \sin \beta T+\frac{1}{2}(-g \cos \alpha) T^{2} $

$\rightarrow \quad T[v _0 \sin \beta-\frac{g \cos \alpha}{2} T] =0 $

$rightarrow T =0, T=\frac{2 v _0 \sin \beta}{g \cos \alpha}$

As $T=0$, corresponding to point $O$

Hence,

$ T=\text { Time of flight }=\frac{2 v_0 \sin \beta}{g \cos \alpha} $

(a) Considering motion along $O X$.

$ \begin{aligned} x & =L, u_{x}=v_0 \cos \beta, a_{x}=-g \sin \alpha \\ t & =T=\frac{2 v_0 \sin \beta}{g \cos \alpha} \\ x & =u_{x} t+\frac{1}{2} a_{x} t^{2} \end{aligned} $

$ \begin{aligned} \Rightarrow \quad L & =v_0 \cos \beta T+\frac{1}{2}(-g \sin \alpha) T^{2} \\ \quad L & =v_0 \cos \beta T-\frac{1}{2} g \sin \alpha T^{2} \\ & =T[v_0 \cos \beta-\frac{1}{2} g \sin \alpha T] \\ & =T[v_0 \cos \beta-\frac{1}{2} g \sin \alpha \times \frac{2 v_0 \sin \beta}{g \cos \alpha}] \\ & =\frac{2 v_0 \sin \beta}{g \cos \alpha}[v_0 \cos \beta-\frac{v_0 \sin \alpha \sin \beta}{\cos \alpha}] \\ & =\frac{2 v_0^{2} \sin \beta}{g \cos ^{2} \alpha}[\cos \beta \cdot \cos \alpha-\sin \alpha \cdot \sin \beta] \\ \Rightarrow \quad L & =\frac{2 v_0^{2} \sin \beta}{g \cos ^{2} \alpha} \cos (\alpha+\beta) \end{aligned} $

(c) For range ( $L$ ) to be maximum,

$\sin \beta \cdot \cos (\alpha+\beta)$ should be maximum.

$ \begin{aligned} & \text { Let, } \quad \begin{aligned} Z & =\sin \beta \cdot \cos (\alpha+\beta) \\ & =\sin \beta[\cos \alpha \cdot \cos \beta-\sin \alpha \cdot \sin \beta) \\ & =\frac{1}{2}[\cos \alpha \cdot \sin 2 \beta-2 \sin \alpha \cdot \sin ^{2} \beta] \\ & =\frac{1}{2}[\sin 2 \beta \cdot \cos \alpha-\sin \alpha(1-\cos 2 \beta)] \\ \Rightarrow \quad Z & =\frac{1}{2}[\sin 2 \beta \cdot \cos \alpha-\sin \alpha+\sin \alpha \cdot \cos 2 \beta] \\ & =\frac{1}{2}[\sin 2 \beta \cdot \cos \alpha+\cos 2 \beta \cdot \sin \alpha-\sin \alpha] \\ & =\frac{1}{2}[\sin (2 \beta+\alpha)-\sin \alpha] \end{aligned} \end{aligned} $

For $z$ to be maximum,

$ \begin{aligned} & \sin (2=\beta+\alpha)=\text { maximum }=1 \\ \Rightarrow \quad 2 \beta+\alpha & =\frac{\pi}{2} \text { or, } \beta=\frac{\pi}{4}-\frac{\alpha}{2} \end{aligned} $

Thinking Process

When particle rebounded elastically speed will remain same.

Answer Considering $x$ and $y$-axes as shown in the diagram. For the motion of the projectile from $O$ to $A$.

$ \begin{aligned} y & =0, u_{y}=v_0 \cos \theta \\ a_{y} & =-g \cos \theta, t=T \end{aligned} $

Applying equation of kinematics,

$ \begin{aligned} y & =u_{y} t+\frac{1}{2} a_{y} t^{2} \\ \Rightarrow \quad 0 & =v_0 \cos \theta T+\frac{1}{2}(-g \cos \theta) T^{2} \\ \Rightarrow \quad T,[v_0 \cos \theta-\frac{g \cos \theta T}{2}] & =0 \\ T & =\frac{2 v_0 \cos \theta}{g \cos \theta} \end{aligned} $

As $T=0$, corresponds to point $O$

Hence, $\quad T=\frac{2 v_0}{g}$

Now considering motion along $O X$.

$ x=L, u_{x}=v_0 \sin \theta, a_{x}=g \sin \theta, t=T=\frac{2 v_0}{g} $

Applying equation of kinematics,

$ \begin{aligned} x & =u_{x} t+\frac{1}{2} a_{x} t^{2} \\ L \quad L & =v_0 \sin \theta t+\frac{1}{2} g \sin \theta t^{2}=(v_0 \sin \theta)(T)+\frac{1}{2} g \sin \theta T^{2} \\ & =(v_0 \sin \theta)(\frac{2 v_0}{g})+\frac{1}{2} g \sin \theta \times(\frac{2 v_0}{g})^{2} \\ & =\frac{2 v_0^{2}}{g} \sin \theta+\frac{1}{2} g \sin \theta \times \frac{4 v_0^{2}}{g^{2}}=\frac{2 v_0^{2}}{g}[\sin \theta+\sin \theta] \\ \quad L & =\frac{4 v_0^{2}}{g} \sin \theta \end{aligned} $

Thinking Process

Draw the vector diagram is for the information given and find $a$ and $b$. We may draw all vectors in the reference frame of ground based observer.

Answer Assume north to be $\hat{i}$ direction and vertically downward to be $-\hat{j}$. Let the rain velocity $v_{r}$ be $a \hat{i}+b \hat{j}$.

$v_r$ = $ a \hat{i} + b \hat{j} $

Case I Given velocity of girl =$ v_{g}=(5 m / s) \hat{i}$

Let $v_{rg}$ = Velocity of rain w.r.t girl

$ =v_{r}-v_{g} = (a \hat{i} + b \hat{j})-5 \hat{i} $

$ =(a-5) \hat{i} + b \hat{j}$

According to question rain, appears to fall vertically downward.

Hence,

$ a-5=0 \Rightarrow a=5 $

Case II Given velocity of the girl, $\mathbf{v} _{g}=(10 m / s) \hat{\mathbf{i}}$

$\therefore \quad v_{rg}=v_{r}-v_{g}$

$ =(a \hat{i}+b \hat{j})-10 \hat{i}=(a-10) \hat{i}+b \hat{j} $

According to question rain appears to fall at $45^{\circ}$ to the vertical hence $\tan 45^{\circ}=\frac{b}{a-10}=1$

$\Rightarrow \quad b=a-10=5-10=-5$

Hence, velocity of rain $=a \hat{i}+b \hat{j}$

$\Rightarrow v_r = 5 \hat{i}-5 \hat{j}$

Speed of rain =$|v_{r}|=\sqrt{(5)^{2}+(-5)^{2}}=\sqrt{50}=5 \sqrt{2} m / s$

Answer Given, Speed of the river $(v_{r})=3 m / s$ (east)

Speed of swimmer $(v_{s})=4 m / s$ (east)

(a) When swimmer starts swimming due north then his resultant velocity

$ \begin{aligned} v & =\sqrt{v_r^{2}+v_s^{2}}=\sqrt{(3)^{2}+(4)^{2}} \\ & =\sqrt{9+16}=\sqrt{25}=5 m / s \\ \tan \theta & =\frac{v_{r}}{v_{s}}=\frac{3}{4} \\ & =0.75=\tan 36^{\circ} 54^{\prime} \\ \theta & =36^{\circ} 54^{\prime} N \end{aligned} $

Hence,

(b) To reach opposite points $B$, the swimmer should swim at an angle $\theta$ of north.

Resultant speed of the swimmer

$ \begin{aligned} v & =\sqrt{v_s^{2}-v_r^{2}}=\sqrt{(4)^{2}-(3)^{2}} \\ & =\sqrt{16-9}=\sqrt{7} m / s \\ \tan \theta & =\frac{v_{r}}{v}=\frac{3}{\sqrt{7}} \\ \theta \quad \theta & =\tan ^{-1}(\frac{3}{\sqrt{7}}) \text { of north } \end{aligned} $

(c) In case (a),

Time taken by the swimmer to cross the river, $t_1=\frac{d}{v_{s}}=\frac{d}{4} s$

In case (b),

Time taken by the swimmer to cross the river

$ \begin{aligned} & t_1=\frac{d}{v}=\frac{d}{\sqrt{7}} \\ & \frac{d}{4}<\frac{d}{\sqrt{7}}, \text { therefore } t_1<t_2 \end{aligned} $

As

Hence, the swimmer will cross the river in shorter time in case (a).

Answer Consider the adjacent diagram.

(a) Initial velocity in

$x -direction, u_x = u+v_0 \cos \theta $

$u_{y} =\text { Initial velocity in } y \text {-direction } $

$ =v_0 \sin \theta$

where angle of projection is $\theta$.

Now, we can write

$ \begin{aligned} \tan \theta & =\frac{u_{y}}{u_{x}}=\frac{u_0 \sin \theta}{u+u_0 \cos \theta} \\ \Rightarrow \quad \theta & =\tan ^{-1}(\frac{v_0 \sin \theta}{u+v_0 \cos \theta}) \end{aligned} $

(b) Let $T$ be the time of flight.

As net displacement is zero over time period $T$.

$ y=0, u_{y}=v_0 \sin \theta, a_{y}=-g, t=T $

$\text { We know that } y =u_{y} t+\frac{1}{2} a_{y} t^{2} $

$\Rightarrow \quad 0 =v_0 \sin \theta T+\frac{1}{2}(-g) T^{2} $

$\Rightarrow \quad T[v_0 \sin \theta-\frac{g}{2} T] =0$

$\Rightarrow T=0, \frac{2 v_0 \sin \theta}{g} $

$T =0_1, \text { corresponds to point } O . $

${ Hence, } T =\frac{2 u_0 \sin \theta}{g}$

(c) Horizontal range, $R=(u+v_0 \cos \theta) T=(u+v_0 \cos \theta) \frac{2 v_0 \sin \theta}{g}$

$ =\frac{v_0}{g}[2 u \sin \theta+v_0 \sin 2 \theta] $

(d) For horizontal range to be maximum, $\frac{d R}{d \theta}=0$

$ \begin{aligned} & \Rightarrow \quad \frac{v_0}{g}[2 u \cos \theta+v_0 \cos 2 \theta \times 2]=0 \\ & \Rightarrow \quad 2 u \cos \theta+2 v_0[2 \cos ^{2} \theta-1]=0 \\ & \Rightarrow \quad 4 v_0 \cos ^{2} \theta+2 u \cos \theta-2 v_0=0 \\ & \Rightarrow \quad 2 v_0 \cos ^{2} \theta+u \cos \theta-v_0=0 \\ & \Rightarrow \quad \cos \theta=\frac{-u \pm \sqrt{u^{2}+8 v_0^{2}}}{4 v_0} \\ & \Rightarrow \quad \theta_{\max }=\cos ^{-1}[\frac{-u \pm \sqrt{u^{2}+8 v_0^{2}}}{4 v_0}] \\ & =\cos ^{-1}[\frac{-u+\sqrt{u^{2}+8 v_0^{2}}}{4 v_0}] \end{aligned} $

(e) If $u=v_0$,

$ \cos \theta=\frac{-v_0 \pm \sqrt{v_0^{2}+8 v_0^{2}}}{4 v_0}=\frac{-1+3}{4}=\frac{1}{2} $

$ \Rightarrow \quad \theta=60^{\circ} $

If $u«v_0$, then $8 v_0{ }^{2}+u^{2} \approx 8 v_0{ }^{2}$

$ \theta_{\max }=\cos ^{-1}[\frac{-u \pm 2 \sqrt{2} v_0}{4 v_0}]=\cos ^{-1}[\frac{1}{\sqrt{2}}-\frac{u}{4 v_0}] $

If $u«v_0$, then $\quad \theta_{\max }=\cos ^{-1}(\frac{1}{\sqrt{2}})=\frac{\pi}{4}$

If $u>u_0$ and

$ u»v_0 $

$ \theta_{\max }=\cos ^{-1}[\frac{-u \pm u}{4 v_0}]=0 \Rightarrow \theta_{\max }=\frac{\pi}{2} $

(f) If $u=0, \theta_{\max }=\cos ^{-1}[\frac{0 \pm \sqrt{8 v^{2} 0}}{4 v_0}]=\cos ^{-1}(\frac{1}{\sqrt{2}})=45^{\circ}$

Answer (a) Given, unit vector

$ \begin{aligned} & \hat{r}=\cos \theta \hat{i}+\sin \theta \hat{j} \\ & \hat{\theta}=-\sin \theta \hat{i}+\cos \theta \hat{j} \end{aligned} $

Multiplying Eq. (i) by $\sin \theta$ and Eq. (ii) with $\cos \theta$ and adding

$\hat{r} \sin \theta+\hat{\theta} \cos \theta=\sin \theta \cdot \cos \theta \hat{i}+\sin ^{2} \theta \hat{j}+\cos ^{2} \theta \hat{j}-\sin \theta \cdot \cos \theta \hat{i}$

$=\hat{j}(\cos ^{2} \theta+\sin ^{2} \theta)=\hat{j}$

$ \Rightarrow \quad \hat{r} \sin \theta+\hat{\theta} \cos \theta=\hat{j} $

By Eq. (i) $\times \cos \theta-$ Eq. (ii) $\times \sin \theta$

$ n(\hat{r} \cos \theta-\hat{\theta} \sin \theta)=\hat{i} $

(b) $\hat{r} \cdot \hat{\theta}=(\cos \theta \hat{\mathbf{i}}+\sin \theta \hat{\mathbf{j}}) \cdot(-\sin \theta \hat{\mathbf{i}}+\cos \theta \hat{\mathbf{j}})=-\cos \theta \cdot \sin \theta+\sin \theta \cdot \cos \theta=0$

$ \Rightarrow \quad \theta=90^{\circ} \text { Angle between } \hat{\mathbf{r}} \text { and } \hat{\theta} \text {. } $

(c) Given, $\hat{r}=\cos \theta \hat{i}+\sin \theta \hat{j}$

$ \begin{aligned} \frac{d \hat{\mathbf{r}}}{d t} & =\frac{d}{d t}(\cos \theta \hat{\mathbf{i}}+\sin \theta \hat{\mathbf{j}})=-\sin \theta \cdot \frac{d \theta}{d t} \hat{\mathbf{i}}+\cos \theta \cdot \frac{d \theta}{d t} \hat{\mathbf{j}} \\ & =\omega[-\sin \theta \hat{\mathbf{i}}+\cos \theta \hat{\mathbf{j}}] \quad[\because \theta=\frac{d \theta}{d t}] \end{aligned} $

(d) Given, $r=a \theta \hat{r}$, here, writing dimensions $[r]=[a][\theta][\hat{r}]$

$ \Rightarrow \quad L=[a] % 1 \Rightarrow[a]=L=[M^{0} L^{1} T^{0}] $

(e) Given, $a=1$ unit

$ \begin{aligned} \mathbf{r} & =\theta \hat{\mathbf{r}}=\theta[\cos \theta \hat{\mathbf{i}}+\sin \theta \hat{\mathbf{j}}] \\ V & =\frac{d r}{d t}=\frac{d \theta}{d t} \hat{\mathbf{r}}+\theta \frac{d}{d t} \hat{\mathbf{r}}=\frac{d \theta}{d t} \hat{\mathbf{r}}+\theta \frac{d}{d t}[(\cos \theta \hat{\mathbf{i}}+\sin \theta \hat{\mathbf{j}})] \\ & =\frac{d \theta}{d t} \hat{\mathbf{r}}+\theta[(-\sin \theta \hat{\mathbf{i}}+\cos \theta \hat{\mathbf{j}}) \frac{d \theta}{d t}] \\ & =\frac{d \theta}{d t} \hat{\mathbf{r}}+\theta \hat{\theta} \omega=\omega \hat{\mathbf{r}}+\omega \theta \hat{\theta} \end{aligned} $

Velocity,

Acceleration,

$ \begin{aligned} a & =\frac{d}{d t}[\omega \hat{r}+\omega \hat{\theta}]=\frac{d}{d t}[\frac{d \theta}{d t} \hat{\mathbf{r}}+\frac{d \theta}{d t}(\theta \hat{\theta})] \\ & =\frac{d^{2} \theta}{d t^{2}} \hat{r}+\frac{d \theta}{d t} \cdot \frac{d \hat{r}}{d t}+\frac{d^{2} \theta}{d t^{2}} \theta \hat{\theta}+\frac{d \theta}{d t} \frac{d}{d t}(\theta \hat{\theta}) \\ & =\frac{d^{2} \theta}{d t^{2}} \hat{r}+\omega[-\sin \theta \hat{\mathbf{i}}+\sin \theta \hat{\mathbf{j}}]+\frac{d^{2} \theta}{d t^{2}} \theta \hat{\theta}+\frac{\omega d}{d t}(\theta \hat{\theta}) \\ & =\frac{d^{2} \theta}{d t^{2}} \hat{\mathbf{r}}+\omega^{2} \hat{\theta}+\frac{d^{2} \theta}{d t^{2}} \times \theta \hat{\theta}+\omega^{2} \hat{\theta}+\omega^{2} \theta(-\hat{\mathbf{r}}) \\ & (\frac{d^{2} \theta}{d t^{2}}-\omega^{2}) \hat{\mathbf{r}}+(2 \omega^{2}+\frac{d^{2} \theta}{d t^{2}} \theta) \theta \end{aligned} $

Answer Consider adjacent diagram.

Time taken to go from $A$ to $C$ via straight line path $A P Q C$ through the $S$ and

$ \begin{aligned} T_{\text {sand }} & =\frac{A P+Q C}{1}+\frac{P Q}{v}=\frac{25 \sqrt{2}+25 \sqrt{2}}{1}+\frac{50 \sqrt{2}}{v} \\ & =50 \sqrt{2}+\frac{50 \sqrt{2}}{v}=50 \sqrt{2}(\frac{1}{v}+1) \end{aligned} $

Clearly from figure the shortest path outside the sand will be $A R C$.

Time taken to go from $A$ to $C$ via this path

Clearly,

$ T_{\text {outside }}=\frac{A R+R C}{1} s $

Clearly

$ A R=\sqrt{75^{2}+25^{2}}=\sqrt{75 \times 75+25 \times 25} $

$ \begin{aligned} & =5 \times 5 \sqrt{9+1}=25 \sqrt{10} m \\ R C & =A R=\sqrt{75^{2}+25^{2}}=25 \sqrt{10} m \end{aligned} $

$\Rightarrow$

$T_{\text {outside }}=2 A R=2 \times 25 \sqrt{10} s=50 \sqrt{10} s$

For

$ T_{\text {sand }}<T_{\text {outside }} $

$\Rightarrow \quad 50 \sqrt{2}(\frac{1}{v}+1)<2 \times 25 \sqrt{10}$

$\Rightarrow \quad \frac{2 \sqrt{2}}{2}(\frac{1}{v}+1)<\sqrt{10}$

$\Rightarrow \quad \frac{1}{v}+1<\frac{2 \sqrt{10}}{2 \sqrt{2}}=\frac{\sqrt{5}}{2} \times 2=\sqrt{5}$

$\Rightarrow \quad \frac{1}{v}<\frac{\sqrt{5}}{2} \times 2-1 \Rightarrow \frac{1}{v}<\sqrt{5}-1$

$\Rightarrow \quad v>\frac{1}{\sqrt{5}-1} \approx 0.81 m / s$

$\Rightarrow \quad v>0.81 m / s$