Answer (c) In a uniform translatory motion, all parts of the ball have the same velocity in magnitude and direction and this velocity is constant.

The situation is shown in adjacent diagram where a body

$A$ is in uniform translatory motion.

Answer (b) To solve this question we have to apply Newton’s second law of motion, in terms of force and change in momentum.

We known that

$ F=\frac{d p}{d t} $

given that meter scale is moving with uniform velocity, hence, $d p=0$

$ \text { Force }=F=0 \text {. } $

As all part of the scale is moving with uniform velocity and total force is zero, hence, torque will also be zero.

Answer (c) Given,

$ \mathbf{u}=(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) m / s $

and

$ \mathbf{v}=-(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) m / s $

Mass of the ball $=150 g=0.15 kg$

$\Delta p=$ Change in momentum

$ \begin{aligned} & =\text { Final momentum }- \text { Initial momentum } \\ & =m \mathbf{v}-m \mathbf{u} \\ & =m(\mathbf{v}-\mathbf{u})=(0.15)[-(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}})-(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}})] \\ & =(0.15)[-6 \hat{\mathbf{i}}-8 \hat{\mathbf{j}}] \\ & =-[0.15 \times 6 \hat{\mathbf{i}}+0.15 \times 8 \hat{\mathbf{j}}] \\ & =-[0.9 \hat{\mathbf{i}}+1.20 \hat{\mathbf{j}}] \\ \Delta \mathbf{p} & =-[0.9 \hat{\mathbf{i}}+1.2 \hat{\mathbf{j}}] \end{aligned} $

Hence,

Answer (c) By previous solution $\Delta p=-(0.9 \hat{\mathbf{i}}+1.2 \hat{\mathbf{j}})$

$ \begin{aligned} \text { Magnitude } & =|\Delta \mathbf{p}|=\sqrt{(0.9)^{2}+(1.2)^{2}} \\ & =\sqrt{0.81+1.44}=1.5 kg-m s^{-1} \end{aligned} $

Thinking Process

For conservation of momentum we have to see whether net external force is acting on a system or not.

Answer (d) We know that for a system $\quad F_{\text {ext }}=\frac{d p}{d t} \quad$ (from Newton’s second law) If $F_{\text {ext }}=0, d p=0 \Rightarrow p=$ constant

Hence, momentum of a system will remain conserve if external force on the system is zero.

In case of collision’ between particles equal and opposite forces will act on individuel particles by Newtons third law,

Hence total force on the system will be zero.

Note We should not confuse with system and individual particles. As total force on the system of both particles is zero but force acts on individual particles.

Answer (c) Consider the adjacent diagram

Let $\quad O A=p_1$

$=$ Initial momentum of player northward

Clearly

$A B=p_2=$ Final momentum of player towards west.

Change in momentum $=p_2-p_1$

$=AB-OA=AB+(-OA)$

$=$ Clearly resultant $\mathbf{A R}$ will be along south-west.

Thinking Process

We have to apply differentiations to calculate acceleration and then Newton’s second law will be applied.

Answer (a) Given, mass $=2 kg$

$ \begin{aligned} x(t) & =p t+q t^{2}+r t^{3} \\ v & =\frac{d x}{d t}=p+2 q t+3 r t^{2} \\ a & =\frac{d v}{d t}=0+2 q+6 r t \\ \text { at } t & =2 s ; a=2 q+6 \times 2 \times r \\ & =2 q+12 r \\ & =2 \times 4+12 \times 5 \\ & =8+60=68 m / s \\ \text { Force } & =F=m a \\ & =2 \times 68=136 N \end{aligned} $

Answer (b) Given, mass $=m=5 kg$

Acting force $=\mathbf{F}=(-3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) N$

Initial velocity at $t=0, \mathbf{u}=(6 \hat{\mathbf{i}}-12 \hat{\mathbf{j}}) m / s$

Retardation, $\hat{\mathbf{a}}=\frac{\mathbf{F}}{m}=(-\frac{3 \hat{\mathbf{i}}}{\mathbf{5}}+\frac{4 \hat{\mathbf{j}}}{5}) m / s^{2}$

As final velocity is along $Y$-axis only, its $x$-component must be zero.

From $v=u+a t$, for $X$-component only, $0=6 \hat{\mathbf{i}}-\frac{3 \hat{\mathbf{i}}}{5} t$

$ t=\frac{5 \times 6}{3}=10 s $

Answer (b) Given, mass of the car $=m$

As car starts from rest, $u=0$

Velocity acquired along east $=v \hat{\mathbf{i}}$

Duration $=t=2 s$.

We know that

$ \begin{aligned} v & =u+a t \\ \Rightarrow v \hat{\mathbf{i}} & =0+a \times 2 \\ \Rightarrow \mathbf{a} & =\frac{\mathbf{v}}{2} \hat{\mathbf{i}} \end{aligned} $

Force,

$ \mathbf{F}=m \mathbf{a}=\frac{\mathbf{m} \mathbf{v}}{2} \hat{\mathbf{i}} $

Hence, force acting on the car is $\frac{m v}{2}$ towards east. As external force on the system is only friction hence, the force $\frac{m v}{2}$ is by friction. Hence, force by engine is internal force.

Thinking Process

Here, position of the particle is given for different time intervals. Hence, we have to find velocity and acceleration corresponding to the intervals.

Answer $(a, b, d)$

Given,

$ \begin{aligned} x & =0 \text { for } t<0 s . \\ x(t) & =A \sin 4 \pi t ; \text { for } 0<t<\frac{1}{4} s \\ x & =0 ; \text { for } t>\frac{1}{4} s \\ v(t) & =\frac{d x}{d t}=4 \pi \text { Acos } 4 \pi t \\ a(t) & =\text { acceleration } \\ & =\frac{d v(t)}{d t}=-16 \pi^{2} A \sin 4 \pi t \\ \text { At } t & =\frac{1}{8} s, a(t)=-16 \pi^{2} A \sin 4 \pi \times \frac{1}{8}=-16 \pi^{2} A \\ F & =m a(t)=-16 \pi^{2} A \times m=-16 \pi^{2} mA \\ \text { Impulse } & =\text { Change in linear momentum } \\ & =F \times t=(-16 \pi^{2} Am) \times \frac{1}{4} \\ & =-4 \pi^{2} Am \end{aligned} $

$ \text { For, } 0<t<\frac{1}{4} s $

The impulse (Change in linear momentum)

$ \text { at } t=0 \text { is same as, } t=\frac{1}{4} s \text {. } $

Clearly, force depends upon $A$ which is not constant. Hence, force is also not constant.

Note We have to keep in mind that the force is varying for different time intervals. Hence, we should apply differential formulae for each interval separately.

Thinking Process

In this problem we have to find frictional forces on each surface and accordingly we will decide maximum force.

Answer $(a, b, d, e)$

Consider the adjacent diagram. Frictional force on $B(f_1)$ and frictional force on $A(f_2)$ will be as shown.

Let $A$ and $B$ are moving together $a_{\text {common }}=\frac{F-f_1}{m_{A}+m_{B}}=\frac{F-f_1}{(m / 2)+m}=\frac{2(F-f_1)}{3 m}$

Pseudo force on $A=(m_{A}) \times a_{\text {common }}$

$=m_{A} \times \frac{2(F-f_1)}{3 m}=\frac{m}{2} \times \frac{2(F-f_1)}{3 m}=\frac{(F-f_1)}{3}$

The force $(F)$ will be maximum when

Pseudo force on $A=$ Frictional force on $A$

$ \begin{aligned} & \Rightarrow \quad \frac{F_{\max }-f_1}{3}=\mu m_{A} g \\ & =0.2 \times \frac{m}{2} \times g=0.1 mg \\ & \Rightarrow \quad F_{\max }=0.3 mg+f_1 \\ & =0.3 mg+(0.1) \frac{3}{2} mg=0.45 mg \end{aligned} $

$\Rightarrow$ Hence, maximum force upto which bodies will move together is $F_{\max }=0.45 mg$

(a) Hence, for $F=0.25 mg<F_{\max }$ bodies will move together.

(b) For $F=0.5 mg>F_{\max }$, body $A$ will slip with respect to $B$.

(c) For $F=0.5 mg>F_{\text {max }}$, bodies slip.

$ (f_1) _ {\max } $ = $\mu m_{B} g=(0.1) \times \frac{3}{2} m \times g=0.15 mg $

$ (f_2) _ {\max} $= $\mu m_A g=(0.2)(\frac{m}{2})(g)=0.1 mg $

Hence, minimum force required for movement of the system $(A+B)$

$F_{\min} = (f_1) _ {\max } + (f_2)_{\max } $

$=0.15 mg+0.1 mg=0.25 mg$

(d) Given, force $F=0.1 mg<F_{\text {min }}$.

Hence, the bodies will be at rest.

(e) Maximum force for combined movement $F_{\max }=0.45 mg$.

Thinking Process

The friction force always have tendency to oppose the motion. Consider the adjacent diagram.

Answer $(b, d)$

Let $m_1$, moves up the plane. Different forces involved are shown in the diagram.

$ \begin{aligned} N & =\text { Normal reaction } \\ f & =\text { Frictional force } \\ T & =\text { Tension in the string } \\ f & =\mu N=\mu m_1 g \cos \theta \end{aligned} $

For the system $(m_1+m_2)$ to move up

$ \begin{matrix} \Rightarrow & m_2 g-(m_1 g \sin \theta+\mu m_1 g \cos \theta)>0 \\ \Rightarrow & m_2>m_1(\sin \theta+\mu \cos \theta) \end{matrix} $

Hence, option (b) is correct.

Let the body moves down the plane, in this case $f$ acts up the plane.

Hence,

$ \begin{aligned} & m_1 g \sin \theta-f>m_2 g \\ & m_1 g \sin \theta-\mu m_1 g \cos \theta>m_2 g \\ & m_1(\sin \theta-\mu \cos \theta)>m_2 \\ & m_2<m_1(\sin \theta-\mu \cos \theta) \end{aligned} $

Hence, option (d) is correct.

Answer $(b, c)$

Let $A$ moves up the plane frictional force on $A$ will be downward as shown.

When $A$ just starts moving up

$m g \sin \theta_1+f =m g \sin \theta_2 $

$\Rightarrow m g \sin \theta_1+\mu m g \cos \theta_1 =m g \sin \theta_2 $

$\Rightarrow \mu =\frac{\sin \theta_2-\sin \theta_1}{\cos \theta_1}$

When A moves upwards

$ \Rightarrow \quad \sin \theta_2>\sin \theta_1 \Rightarrow \theta_2>\theta_1 $

Answer (c, $d)$

Given,

Initial velocity $(u)=u_1=u_2=5 m / s$

Final velocity $(v)=v_1=v_2=-5 m / s$

Time duration of collision $=10^{-3} s$.

Change in linear momentum $=m(v-u)$

$ \begin{aligned} & =\frac{1}{20}[-5-5]=-0.5 N-s . \\ \text { Force } & =\frac{\text { Impulse }}{\text { Time }}=\frac{\text { Change in momentum }}{10^{-3} s} \\ & =\frac{0.5}{10^{-3}}=500 N \end{aligned} $

Impulse and force are opposite in directions.

Thinking Process

In this problem, we have to use the concept of resultant of two vectors, when they are perpendicular.

Answer (a, c)

Consider the adjacent diagram

Given,

$ \begin{aligned} \text { mass } & =m=10 kg . \\ F_1 & =6 N, F_2=8 N \\ \text { Resultant force } & =F=\sqrt{F_1^{2}+F_2^{2}}=\sqrt{36+64} \\ & =10 N \\ a & =\frac{F}{m}=\frac{10}{10}=1 m / s^{2} ; \text { along } R . \end{aligned} $

Let $\theta_1$ be angle between $R$ and $F_1$

$ \begin{aligned} \tan \theta_1 & =\frac{8}{6}=\frac{4}{3} \\ \theta_1 & =\tan ^{-1}(4 / 3) \text { w.r.t. } F_1=6 N \end{aligned} $

Let $\theta_2$ be angle between $F$ and $F_2$

$ \begin{aligned} \tan \theta_2 & =\frac{6}{8}=\frac{3}{4} \\ \theta_2 & =\tan ^{-1}(\frac{3}{4}) \text { w.r.t. } F_2=8 N \end{aligned} $

Thinking Process

In this problem, we have to apply conservation of linear momentum.

Answer Given, total mass of girl, bicycle and stone $=m_1=(50+0.5) kg=50.5 kg$.

Velocity of bicycle $u_1=5 m / s$, Mass of stone $m_2=0.5 kg$

Velocity of stone $u_2=15 m / s$, Mass of girl and bicycle $m=50 kg$

Yes, the speed of the bicycle changes after the stone is thrown.

Let after throwing the stone the speed of bicycle be $v m / s$.

According to law of conservation of linear momentum,

$ \begin{aligned} m_1 u_1 & =m_2 u_2+m v \\ 50.5 \times 5 & =0.5 \times 15+50 \times v \\ 252.5-7.5 & =50 v \\ \text { or } \quad v & =\frac{245.0}{50} \\ v & =4.9 m / s \\ \text { Change in speed } & =5-4.9=0.1 m / s . \end{aligned} $

Answer When a lift descends with a downward acceleration a the apparent weight of a body of mass $m$ is given by

$ w^{\prime}=R=m(g-a) $

Mass of the person $m=50 kg$

Descending acceleration $a=9 m / s^{2}$

Acceleration due to gravity $g=10 m / s^{2}$

Apparent weight of the person,

$ \begin{aligned} R & =m(g-a) \\ & =50(10-9) \\ & =50 N \\ \therefore \quad \text { Reading of the weighing scale } & =\frac{R}{g}=\frac{50}{10}=5 kg . \end{aligned} $

Answer Given, mass of the body $(m)=2 kg$

From the position-time graph, the body is at $x=0$ when $t=0$, i.e., body is at rest.

$\therefore$ Impulse at $t=0, s=0$, is zero

From $t=0 s$ to $t=4 s$, the position-time graph is a straight line, which shows that body moves with uniform velocity.

Beyond $t=4 s$, the graph is a straight line parallel to time axis, i.e., body is at rest $(v=0)$.

Velocity of the body $=$ slope of position-time graph

$ =\tan \theta=\frac{3}{4} m / s $

Impulse (at $t=4 s)=$ change in momentum

$ \begin{aligned} & =m v-m u \\ & =m(v-u) \\ & =2(0-\frac{3}{4}) \\ & =-\frac{3}{2} kg-m / s=-1.5 kg-m / s \end{aligned} $

Answer When a person driving a car suddenly applies the brakes, the lower part of the body slower down with the car while upper part of the body continues to move forward due to inertia of motion.

If driver is not wearing seat belt, then he falls forward and his head hit against the steering wheel.

Answer Given, mass of the body $m=2 kg$.

Velocity of the body $\mathbf{v}(t)=2 t \hat{\mathbf{i}}+t^{2} \hat{\mathbf{j}}$

$\therefore$ Velocity of the body at $t=2 s$

$ \mathbf{v}=2 \times 2 \hat{\mathbf{i}}+(2)^{2} \hat{\mathbf{j}}=(4 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) $

Momentum of the body $(p)=m \mathbf{v}$

$ =2(4 \hat{\mathbf{i}}+4 \hat{\mathbf{j}})=(8 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}) kg-m / s $

Acceleration of the body $(a)=\frac{d v}{d t}$

$ \begin{aligned} & =\frac{d}{d t}(2 t \hat{\mathbf{i}}+t^{2} \hat{\mathbf{j}}) \\ & =(2 \hat{\mathbf{i}}+2 t \hat{\mathbf{j}}) \\ \text { At } t & =2 s \\ \mathbf{a} & =(2 \hat{\mathbf{i}}+2 \times 2 \hat{\mathbf{j}}) \\ & =(2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) \end{aligned} $

Force acting on the body $(\mathbf{F})=$ ma

$ =2(2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) $

$ =(4 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}) N $

Answer The approximate graph is shown in the diagram

The frictional force $f$ is shown on vertical axis and the applied force $F$ is shown on the horizontal axis. The portion $O A$ of graph represents static friction which is self adjusting. In this portion, $f=F$.

The point $B$ corresponds to force of limiting friction which is the maximum value of static friction. $C D | O X$ represents kinetic friction, when the body actually starts moving. The force of kinetic friction does not increase with applied force, and is slightly less than limiting friction.

Answer When a child falls on a cement floor, her body comes to rest instantly. But $F \times \Delta t=$ change in momentum $=$ constant. As time of stopping $\Delta t$ decreases, therefore $F$ increases and hence, child feel more pain.

When she falls on a soft muddy ground in the garden the time of stopping increases and hence, $F$ decreases and she feels lesser pain.

Answer Mass of the object $(m)=500 g=0.5 kg$

Speed of the object $(v)=25 m / s$

(a) Impulse imparted to the object = change in momentum

$ \begin{aligned} & =m v-m u \\ & =m(v-u) \\ & =0.5(25-0)=12.5 N-s \end{aligned} $

(b)

$ \begin{aligned} & \text { ounding } \\ & =-\frac{25}{2} m / s \\ & v^{\prime}=-12.5 m / s \\ & \therefore \quad \text { Change in momentum }=m(v^{\prime}-v) \\ & =0.5(-12.5-25)=-18.75 N-s \end{aligned} $

Answer While going up a mountain, the force of friction acting on a vehicle of mass $m$ is $f=\mu R=\mu m g \cos \theta$, where $\theta$ is the angle of slope of the road with the horizontal. To avoid skidding force of friction (f) should be large and therefore, $\cos \theta$ should be large and hence, $\theta$ should be small.

That’s why mountain roads are generally made winding upwards rather than going straight upto avoid skidding.

Thinking Process

As the whole system is going upward with an acceleration we have to apply Newton’s laws.

Answer Given, $m_1=5 kg, m_2=3 kg$

$g=9.8 m / s^{2}$ and $a=2 m / s^{2}$

For the upper block

$ \begin{aligned} & \Rightarrow \\ & \text { For the lower block } \\ & \Rightarrow \\ & \text { From Eq. (i) } \end{aligned} $

$ \begin{aligned} T_1-T_2-5 g & =5 a \\ T_1-T_2 & =5(g+a) \\ T_2-3 g & =3 a \\ T_2 & =3(g+a)=3(9.8+2)=35.4 N \\ T_1 & =T_2+5(g+a) \\ & =35.4+5(9.8+2)=94.4 N \end{aligned} $

Answer In equilibrium, the force $m g \sin \theta$ acting on block $A$ parallel to the plane should be balanced by the tension in the string, i.e.,

$ \begin{aligned} & m g \sin \theta=T=F \quad[\because T=\text { Fgiven }] \\ & W=T=F \end{aligned} $

where, $w$ is the weight of block $B$.

From Eqs. (i) and (ii), we get,

$\therefore$

$ \begin{aligned} W & =m g \sin \theta \\ & =100 \times \sin 30^{\circ} \\ & =100 \times \frac{1}{2} N=50 N \end{aligned} $

$ (\because m g=100 N) $

Note While finding normal reaction in such cases, we should be careful it will be $N=m g \cos \theta$, where $\theta$ is angle of inclination.

Answer Given, mass of the block $=M$

Coefficient of friction between the block and the wall $=\mu$

Let a force $F$ be applied on the block to hold the block against the wall. The normal reaction of mass be $N$ and force of friction acting upward be $f$. In equilibrium, vertical and horizontal forces should be balanced separately.

$\therefore$

$f=M g$

and

$F=N$

But force of friction $(f)=\mu N$

$$ \mu F \quad \quad \quad [\text{using Eq.} (ii)]…(iii)$$

From Eqs. (i) and (iii), we get $$ \mu F = Mg$$ $$F= \frac{Mg}{\mu}$$

Answer Given, mass of the gun $(m_1)=100 kg$

Mass of the ball $(m_2)=1 kg$

Height of the cliff $(h)=500 m$

Horizontal distance travelled by the ball $(x)=400 m$

From

$ h=\frac{1}{2} g t^{2} \quad( \because \text {Initial velocity in downward direction is zero) }$

From

$ \begin{aligned} 500 & =\frac{1}{2} \times 10 t^{2} \\ t & =\sqrt{100}=10 s \\ x=u t, u & =\frac{x}{t}=\frac{400}{10}=40 m / s \end{aligned} $

If $v$ is recoil velocity of gun, then according to principle of conservation of linear momentum,

$ \begin{aligned} m_1 v & =m_2 u \\ v & =\frac{m_2 u}{m_1}=\frac{1}{100} \times 40=0.4 m / s \end{aligned} $

Thinking Process

To solve this question, we have to find the relation for $x$ and time $(t), y$ and time (t) from the given diagram.

Answer Clearly from diagram (a), the variation can be related as

$ \begin{aligned} & x=t \Rightarrow \frac{d x}{d t}=1 m / s \\ & a_{x}=0 \end{aligned} $

From diagram (b)

$\Rightarrow$

Hence,

Hence, net force,

$ \begin{matrix} y & =t^{2} \\ \frac{d y}{d t} & =2 t \quad \text { or } a_{y}=\frac{d^{2} y}{d t^{2}}=2 m / s^{2} \\ I_{y} & =m a_{y}=500 \times 10^{-3} \times 2=1 N & \\ F_{x} & =m a_{x}=0 \\ F & =\sqrt{F_x^{2}+F_y^{2}}=F_{y}=1 N & \quad(\because m=500 g) \\ & \text { (along } y \text {-axis) } \end{matrix} $

Answer Here, initial speed of the coin $(u)=20 m / s$

Acceleration of the elevator $(a)=2 m / s^{2}$

(upwards)

Acceleration due to gravity $(g)=10 m / s^{2}$

$\therefore$ Effective acceleration $a^{\prime}=g+a=10+2=12 m / s^{2} \quad$ (here, acceleration is w.r.t. the lift)

If the time of ascent of the coin is $t$, then

or

$ \begin{aligned} & v=u+a t \\ & 0=20+(-12) \times t \\ & t=\frac{20}{12}=\frac{5}{3} s \end{aligned} $

Time of ascent $=$ Time of desent

$\therefore$ Total time after which the coin fall back into hand $=(\frac{5}{3}+\frac{5}{3}) s=\frac{10}{3} s=3.33 s$

Note While calculating net acceleration we should be aware that if lift is going upward net acceleration is $(g+a)$ and for downward net acceleration is $(g-a)$.

Thinking Process

As the body is found to move with uniform velocity hence, we can say that total force acting will be zero.

Answer As the body is moving with uniform speed (velocity) its acceleration $a=0$.

$\therefore$ The sum of the forces is zero, $F_1+F_2+F_3=0$

(a) Let $F_1, F_2, F_3$ be the three forces passing through a point. Let $F_1$ and $F_2$ be in the plane $A$ (one can always draw a plane having two intersecting lines such that the two lines lie on the plane). Then $F_1+F_2$ must be in the plane $A$.

Since, $F_3=-(F_1+F_2), F_3$ is also in the plane $A$.

(b) Consider the torque of the forces about $P$. Since, all the forces pass through $P$, the torque is zero. Now, consider torque about another point $O$. Then torque about $O$ is

Torque $=OP \times(F_1+F_2+F_3)$

Since, $F_1+F_2+F_3=0$, torque $=0$

Answer Consider the diagram where a body slides down from along an inclined plane of inclination $\theta(=45^{\circ})$.

On smooth inclined plane Acceleration of a body sliding down a smooth inclined plane Here,

$ \begin{aligned} & a=g \sin \theta \\ & \theta=45^{\circ} \\ & a=g \sin 45^{\circ}=\frac{g}{\sqrt{2}} \end{aligned} $

$ \therefore $

Let the travelled distance be $s$

Using equation of motion, $s=u t+\frac{1}{2} a t^{2}$, we get

$ \begin{aligned} s & =0 . t+\frac{1}{2} \frac{g}{\sqrt{2}} T^{2} \\ \text { or } \quad s & =\frac{g T^{2}}{2 \sqrt{2}} \end{aligned} $

On rough inclined plane Acceleration of the body $a=g(\sin \theta-\mu \cos \theta)$

$ \begin{aligned} & =g(\sin 45^{\circ}-\mu \cos 45^{\circ}) \\ & =\frac{g(1-\mu)}{\sqrt{2}} \end{aligned} \quad(\text { As, } \sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}) $

Again using equation of motion, $s=u t+\frac{1}{2} a t^{2}$, we get

or

$ \begin{aligned} & s=0(p T)+\frac{1}{2} \frac{g(1-\mu)}{\sqrt{2}}(p T)^{2} \\ & s=\frac{g(1-\mu) p^{2} T^{2}}{2 \sqrt{2}} \end{aligned} $

From Eqs. (i) and (ii), we get

or

$ \begin{aligned} \frac{g T^{2}}{2 \sqrt{2}} & =\frac{g(1-\mu) p^{2} T^{2}}{2 \sqrt{2}} \\ (1-\mu) p^{2} & =1 \\ 1-\mu & =\frac{1}{p^{2}} \\ \mu & =(1-\frac{1}{p^{2}}) \end{aligned} $

Answer Consider figure (a)

$ \begin{aligned} v_{x} & =2 t \text { for } 0<t<1 s \\ & =2(2-t) \text { for } 1<t<2 s \\ & =0 \text { for } t>2 s . \\ v_{y} & =t \text { for } 0<t<1 s \\ & =1 \text { for } t>1 s \\ F_{x} & =m a_{x}=m \frac{d v_{x}}{d t} \\ & =1 \times 2 \\ & =1(-2) \end{aligned} $

From figure (b)

for $0<t<1$ s

for $1<t<2 s$ for $2<t$

(a)

(b)

Hence,

$ \begin{aligned} F_{y} & =m a_{y}=m \frac{d v_{y}}{d t} \\ & =1 \times 1 \text { for } 0<t<1 s=0 \text { for } 1<t \\ F & =F_{x} \hat{\mathbf{i}}+F_{y} \hat{\mathbf{j}} \\ & =2 \hat{\mathbf{i}}+\hat{\mathbf{j}} \\ & =-2 \hat{\mathbf{i}} \\ & =0 \end{aligned} $

$ \text { for } 0<t<1 s $

for $1<t<2 s$

for $t>2 s$.

Thinking Process

The necessary centripetal force required for the circular motion will be provided by the frictional force.

Answer Balancing frictional force for centripetal force $\frac{m v^{2}}{r}=f=\mu N=\mu m g$ where, $N$ is normal reaction.

$\therefore \quad V=\sqrt{\mu r g} \quad$ (where, $r$ is radius of the circular track)

For path $A B C$

$ \text { Path length }=\frac{3}{4}(2 \pi 2 R)=3 \pi R=3 \pi \times 100 $

$ \begin{aligned} & =300 \pi m \\ v_1=\sqrt{\mu 2 R g} & =\sqrt{0.1 \times 2 \times 100 \times 10} \\ & =14.14 m / s \end{aligned} $

$\therefore \quad t_1=\frac{300 \pi}{14.14}=66.6 s$

For path $D E F$

$ \text { Path length }=\frac{1}{4}(2 \pi R)=\frac{\pi \times 100}{2}=50 \pi $

$ \begin{aligned} & v_2=\sqrt{\mu R g}=\sqrt{0.1 \times 100 \times 10}=10 m / s \\ & t_2=\frac{50 \pi}{10}=5 \pi s=15.7 s \end{aligned} $

For paths, $C D$ and $F A$

$ \text { Path length }=R+R=2 R=200 m $

$ t_3=\frac{200}{50}=4.0 s . $

$\therefore$ Total time for completing one round

$ t=t_1+t_2+t_3=66.6+15.7+4.0=86.3 s $

Thinking Process

To find trajectory, we will relate $x$ and $y$ in terms of constants $A$ and $B$.

Answer (a) Displacement vector of the particle of mass $m$ is given by

$ \mathbf{r}(t)=\hat{\mathbf{i}} A \cos \omega t+\hat{\mathbf{j}} B \sin \omega t $

$\therefore$ Displacement along $x$-axis is,

$ \begin{aligned} x & =A \cos \omega t \\ \frac{x}{A} & =\cos \omega t \end{aligned} $

$ \text { or } $

Displacement along $y$-axis is,

and

$ \begin{aligned} y & =B \sin \omega t \\ \frac{y}{B} & =\sin \omega t \end{aligned} $

Squaring and then adding Eqs. (i) and (ii), we get

$ \frac{x^{2}}{A^{2}}+\frac{y^{2}}{B^{2}}=\cos ^{2} \omega t+\sin ^{2} \omega t=1 $

This is an equation of ellipse.

Therefore, trajectory of the particle is an ellipse. (b) Velocity of the particle

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

or

$\therefore$ Force acting on the particle,

$ F=m \mathbf{a}=-\mathbf{m} \omega^{2} \mathbf{r} $

Hence proved.

Thinking Process

The horizontal component of velocity will remain unaffected by gravity.

Answer (a) When ball is given only horizontal velocity Horizontal velocity at the time of release $(u_{x})=v_{s}$

During projectile motion, horizontal velocity remains unchanged,

Therefore,

In vertical direction,

$ \begin{aligned} & v_{x}=u_{x}=v_{s} \\ & v_y^{2}=u_y^{2}+2 g H \end{aligned} $

$ v_{y}=\sqrt{2 g H} $

$\therefore$ Resultant speed of the ball at bottom,

$ \begin{aligned} v & =\sqrt{v_x^{2}+v_y^{2}} \\ & =\sqrt{v_s^{2}+2 g H} \end{aligned} $

(b) When ball is given horizontal velocity and a small downward velocity

Let the ball be given a small downward velocity $u$.

In horizontal direction

In vertical direction

$ \begin{aligned} & v_x^{\prime}=u_{x}=v_{s} \\ & v_y^{\prime 2}=u^{2}+2 g H \\ & v_y^{\prime}=\sqrt{u^{2}+2 g H} \end{aligned} $

or

$\therefore$ Resultant speed of the ball at the bottom

$ v^{\prime}=\sqrt{v_x^{\prime 2}+v_y^{\prime 2}}=\sqrt{v_s^{2}+u^{2}+2 g H} $

From Eqs. (i) and (ii), we get $\quad v^{\prime}>v$

Thinking Process

To balance the forces, we have to resolve them along two mutually perpendicular directions.

Answer Consider the adjacent diagram, in which forces are resolved.

On resolving forces into rectangular components, in equilibrium forces $(F_1+\frac{1}{\sqrt{2}}) N$ are equal to $\sqrt{2} N$ and $F_2$ is equal to $(\sqrt{2}+\frac{1}{\sqrt{2}}) N$.

$\therefore \quad F_1+\frac{1}{\sqrt{2}}=\sqrt{2}$

and

$ F _1=\sqrt{2}-\frac{1}{\sqrt{2}}=\frac{2-1}{\sqrt{2}}=\frac{1}{\sqrt{2}}=0.707 N $

and $\quad F_2=\sqrt{2}+\frac{1}{\sqrt{2}}=\frac{2+1}{\sqrt{2}} N=\frac{3}{\sqrt{2}} N=2.121 N$

Answer (a) Consider the adjacent diagram, force of friction on the box will act up the plane. For the box to just starts sliding down $m g$

or $\quad \tan \theta=\mu \Rightarrow \theta=\tan ^{-1}(\mu)$

(b) When angle of inclination is increased to $\alpha>\theta$, then net force acting on the box, down the plane is

$ \begin{aligned} F_1 & =m g \sin \alpha-f=m g \sin \alpha-\mu N \\ & =m g(\sin \alpha-\mu \cos \alpha) \end{aligned} $

(c) To keep the box either stationary or just move it up with uniform speed, upward force needed, $F_2=m g \sin \alpha+f=m g(\sin \alpha+\mu \cos \alpha)$ (In this case, friction would act down the plane).

(d) If the box is to be moved with an upward acceleration a, then upward force needed, $F_3=m g(\sin \alpha+\mu \cos \alpha)+m a$.

Answer Given, mass of helicopter $(m_1)=2000 kg$

Mass of the crew and passengeres $m_2=500 kg$

Acceleration in vertical direction $a=15 m / s^{2}(\uparrow)$ and $g=10 m / s^{2}(\downarrow)$

(a) Force on the floor of the helicopter by the crew and passengers

$ \begin{aligned} & m_2(g+a)=500(10+15) N \\ & 500 \times 25 N=12500 N \end{aligned} $

(b) Action of the rotor of the helicopter on the surrounding air $=(m_1+m_2)(g+a)$

$ \begin{aligned} & =(2000+500) \times(10+15)=2500 \times 25 \\ & =62500 N(\text { downward }) \end{aligned} $

(c) Force on the helicopter due to the surrounding air

$ \begin{aligned} & =\text { reaction of force applied by helicopter. } \\ & =62500 N \text { (upward) } \end{aligned} $

Note We should be very clear when we are balancing action and reaction forces. We must know that which part is action and which part is reaction due to the action.