Thinking Process

In this problem as the electron and proton are moving under the influence of mutual forces, they will perform circular motion about their centre (i.e., about middle point of the line joining them).

Answer (b) When electron and proton are moving under influence of their mutual forces, the magnetic forces will be perpendicular to their motion hence no work is done by these forces.

Answer (c) Force between two protons is same as that of between proton and a positron.

As positron is much lighter than proton, it moves away through much larger distance compared to proton.

We know that work done $=$ force $\times$ distance. As forces are same in case of proton and positron but distance moved by positron is larger, hence, work done will be more.

Answer (d) When the man is squatting on the ground he is tilted somewhat, hence he also has to balance frictional force besides his weight in this case.

$R=$ reactional force $=$ friction $+m g$

$\Rightarrow \quad R>m g$

When the man gets straight up in that case friction $\approx 0$

$\Rightarrow \quad$ Reactional force $\approx m g$

Thinking Process

In this problem energy will be lost due to dissipation by friction.

Answer (c) Here, work is done by the frictional force on the cycle and is equal to $200 \times 10=-2000 J$.

As the road is not moving, hence, work done by the cycle on the road = zero.

Note We should be aware that here the energy of bicyclist is lost during the motion, but it is lost due to friction in the form of heat.

Thinking Process

In an inelastic collision between two bodies due to some deformation, energy may be lost in the form of heat and sound etc.

Answer (c) When we are considering the two bodies as system the total external force on the system will be zero.

Hence, total linear momentum of the system remain conserved.

Answer (c) As the given tracks are frictionless, hence, mechanical energy will be conserved. As both the tracks having common height, $h$.

From conservation of mechanical energy,

$$ \begin{aligned} \frac{1}{2} m v^{2} & =m g h \quad \text { (for both tracks I and II) } \\ v & =\sqrt{2 g h} \end{aligned} $$

Hence, speed is same for both stones. For stone $I, a_1=$ acceleration along inclined plane $=g \sin \theta_1$

Similarly, for stone $| a_2=g \sin \theta_2$ as $\theta_2>\theta_1$ hence, $a_2>a_1$.

And both length for track II is also less hence, stone II reaches earlier than stone I.

Answer (b) Total energy is $E=PE+KE$

When particle is at $x=x_{m}$ i.e., at extreme position, returns back. Hence, at $x=x_{m}$; $x=0 ; KE=0$

From Eq. (i) $\quad E=P E+0=P E=V(x_{m})=\frac{1}{2} k x_m^{2}$

Answer (b) When two bodies of equal masses collides elastically, their velocities are interchanged.

When ball 1 collides with ball-2, then velocity of ball-1, $v_1$ becomes zero and velocity of ball-2, $v_2$ becomes $v$, i.e., similarly.

when ball 2 collides will ball $3 \quad v_2=0, v_3=v$

Answer Given, $v=a x^{3 / 2}$

$$ m=0.5 kg, a=5 m^{-1 / 2} s^{-1}, \text { work done }(W)=? $$

We know that

Acceleration

$$ \begin{aligned} a_0 & =\frac{d v}{d t}=v \frac{d v}{d x}=a x^{3 / 2} \frac{d}{d x}(a x^{3 / 2}) \\ & =a x^{3 / 2} \times a \times \frac{3}{2} \times x^{1 / 2}=\frac{3}{2} a^{2} x^{2} \end{aligned} $$

Now,

$$ \text { Force }=m a_0=m \frac{3}{2} a^{2} x^{2} $$

Work done $=\int_{x=0}^{x=2} F d x=\int_0^{2} \frac{3}{2} m a^{2} x^{2} d x$

$$ \begin{aligned} & =\frac{3}{2} m a^{2} \times(x^{3} / 3)_0^{2} \\ & =\frac{1}{2} m a^{2} \times 8=\frac{1}{2} \times(0.5) \times(25) \times 8=50 J \end{aligned} $$

Answer (b) Given, power = constant

We know that power $(P)$

$$ P=\frac{d W}{d t}=\frac{F \cdot d s}{d t}=\frac{F d s}{d t} \quad(\because \text { body is moving unidirectionally }) $$

Hence,

$$ F \cdot d s=F d s \cos 0^{\circ} $$

$$ P=\frac{F d s}{d t}=\text { constant } \quad(\because P=\text { constant by question }) $$

Now, writing dimensions

$ {[F][v]} =\text{ constant } $ $\Rightarrow {[MLT^{-2}][LT^{-1}]} =\text{ constant } $ $\Rightarrow L^{2} T^{-3} =\text{ constant } (\because \text{ mass is constant })$ $\Rightarrow L \propto T^{3 / 2} \Rightarrow \text{ Displacement }(d) \propto t^{3 / 2}$

Answer (d) When the earth is closest to the sun, speed of the earth is maximum, hence, KE is maximum. When the earth is farthest from the sun speed is minimum hence, KE is minimum but never zero and negative.

This variation is correctly represented by option(d).

Answer (c) When a pendulum oscillates in air, it will lose energy continuously in overcoming resistance due to air. Therefore, total mechanical energy of the pendulum decreases continuously with time.

The variation is correctly represented by curve (c).

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

$$ \begin{aligned} \omega & =300 rev / min \\ & =(300 \times 2 \pi) rad / min \\ & =(300 \times 2 \times 3.14) rad / 60 s \\ & =\frac{300 \times 2 \times 3.14}{60} rad / s=10 \pi rad / s \\ \Rightarrow \quad \text { linear speed } & =v=\omega R \\ & =(\frac{300 \times 2 \pi}{60})(1) \\ & =10 \pi m / s \\ KE & =\frac{1}{2} m v^{2} \\ & =\frac{1}{2} \times 5 \times(10 \pi)^{2} \\ & =100 \pi^{2} \times 5 \times \frac{1}{2} \\ & =250 \pi^{2} J \end{aligned} $$

Thinking Process

During fall of a raindrop first velocity of the drop increases and then become constant after sometime.

Answer (b) When drop falls first velocity increases, hence, first KE also increases. After sometime speed (velocity) is constant this is called terminal velocity, hence, KE also become constant. PE decreases continuously as the drop is falling continuously.

The variation in $P E$ and $K E$ is best represented by (b).

Thinking Process

As air resistance is negligible, total mechanical energy of the system will remain constant.

Answer (d) Given, $h=1.5 m, v=1 m / s, m=10 kg, g=10 ms^{-2}$

From conservation of mechanical energy.

$$ \begin{aligned} & (PE) i+(KE) i=(PE) f+(KE) f \\ & \Rightarrow \quad m g h+\frac{1}{2} m v^{2}=0+(KE) f \\ & \Rightarrow \quad(KE) f=m g h+\frac{1}{2} m v^{2} \\ & \Rightarrow \quad(KE) f=10 \times 10 \times 1.5+\frac{1}{2} \times 10 \times(1)^{2} \\ & =150+5=155 J \end{aligned} $$

Note We should be careful about the reference taken for $P E$, it may or may not be the ground.

Answer (c) Given, $m=150 g=\frac{150}{1000} kg=\frac{3}{20} kg$

$$ \begin{aligned} \Delta t & =\text { time of contact }=0.001 s \\ u=126 km / h & =\frac{126 \times 1000}{60 \times 60} m / s=35 m / s \\ v & =-126 km / h=-35 m / s \end{aligned} $$

Change in momentum of the ball

$$ \begin{aligned} \Delta p=m(v-u) & =\frac{3}{20}(-35-35) kg-m / s \\ & =\frac{3}{20}(-70)=-\frac{21}{2} \end{aligned} $$

We know that force $F=\frac{\Delta p}{\Delta t}$

$$ =\frac{-21 / 2}{0.001} N=-1.05 \times 10^{4} N $$

Here, - ve sign shown that force will be opposite to the direction of movement of the ball before hitting.

Answer $(b, d)$

When a man of mass $m$ climbs up the staircase of height $L$, work done by the gravitational force on the man is-mgl work done by internal muscular forces will be $m g L$ as the change in kinetic energy is almost zero.

Hence, total work done $=-m g L+m g L=0$

As the point of application of the contact forces does not move hence work done by reaction forces will be zero.

Note Here work done by friction will also be zero as there is no dissipation or rubbing is involved.

Answer $(b, d, f)$

Consider the adjacent diagram for the given situation in the question.

(b) Conserving energy between " $O$ " and " $A$ "

$ U_{i}+K_{i} =U_{f}+K_{f} $

$\Rightarrow 0+\frac{1}{2} m v^{2} =m g h+\frac{1}{2} m v^{\prime} $

$\Rightarrow \frac{(v^{\prime})^{2}}{2} =\frac{v^{2}}{2}=-g h $

$\Rightarrow (v^{\prime})^{2} =v^{2}-2 g h \Rightarrow v^{\prime}=\sqrt{v^{2}-2 g h}$

where $v^{\prime}$ is speed of the bullet just before hitting the target. Let speed after emerging from the target is $v^{\prime \prime}$ then,

$$ \begin{aligned} & \text { By question, } \quad=\frac{1}{2}(m v^{\prime \prime})^{2}=\frac{1}{2}[\frac{1}{2} m(v^{\prime})^{2}] \\ & \frac{1}{2} m(v^{\prime \prime})^{2}=\frac{1}{4} m(v^{\prime})^{2}=\frac{1}{4} m[v^{2}-2 g h] \\ & \Rightarrow \quad(v^{\prime \prime})^{2}=\frac{v^{2}-2 g h}{2}=\frac{v^{2}}{2}-g h \\ & \Rightarrow \quad v^{\prime \prime}=\sqrt{\frac{v^{2}}{2}-g h} \end{aligned} $$

From Eqs. (i) and (ii)

$ \frac{v^{\prime}}{v^{\prime \prime}}=\frac{\sqrt{v^{2}-2 g h}}{\frac{\sqrt{v^{2}-2 g h}}{\sqrt{2}}}=\sqrt{2} $

$\Rightarrow v^{\prime \prime}=\frac{v^{\prime}}{\sqrt{2}}=v^{2}(\frac{v^{\prime}}{2}) $

$\Rightarrow \frac{v^{\prime \prime}}{v^{\prime}}=\sqrt{2}=1.414>1 $

$\Rightarrow \frac{v^{\prime \prime}}{2}=\frac{v^{\prime}}{2}$

Hence, after emerging from the target velocity of the bullet $(v^{\prime \prime})$ is more than half of its earlier velocity $v^{\prime}$ (velocity before emerging into the target).

(d) As the velocity of the bullet changes to $v^{\prime}$ which is less than $v^{1}$ hence, path, followed will change and the bullet reaches at point $B$ instead of $A^{\prime}$, as shown in the figure.

(f) As the bullet is passing through the target the loss in energy of the bullet is transferred to particles of the target. Therefore, their internal energy increases.

Answer (c) Consider the adjacent diagram when $M_1$ comes in contact with the spring, $M_1$ is retarded by the spring force and $M_2$ is accelerated by the spring force.

(a) The spring will continue to compress until the two blocks acquire common velocity.

(b) As surfaces are frictionalless momentum of the system will be conserved. (c) If spring is massless whole energy of $M_1$ will be imparted to $M_2$ and $M_1$ will be at rest, then

(d) Collision is inelastic, even if friction is not involved.

Answer Consider the adjacent diagram. As the block $M$ is at rest. Hence, $f=$ frictional force $=M g \sin \theta$

The force of friction acting between the block and incline opposes the tendency of sliding of the block. Since, block is not in motion, therefore, no work is done by the force of friction. Hence, no dissipation of energy takes place.

Answer When the elevator is descending, then electric power is required to prevent it from falling freely under gravity.

Also, as the weight inside the elevator increases, its speed of descending increases, therefore, there should be a limit on the number of passengers in the elevator to prevent the elevator from descending with large velocity.

Answer (a) Force is applied on the body to lift it in upward direction and displacement of the body is also in upward direction, therefore, angle between the applied force and displacement is $\theta=0^{\circ}$

$\therefore$ Work done by the applied force

$ W=F s \cos \theta=F s \cos 0^{\circ}=F s \quad(\because \cos 0^{\circ}=1) $

$\text { i.e., } \quad W=\text { Positive }$

(b) The gravitational force acts in downward direction and displacement in upward direction, therefore, angle between them is $\theta=180^{\circ}$.

$\therefore$ Work done by the gravitational force

$$ W=F s \cos 180^{\circ}=-F s $$

$(\because \cos 180^{\circ}=1)$

Answer Force of gravity acts on the car vertically downward while car is moving along horizontal road, i.e., angle between them is $90^{\circ}$.

Work done by the car against gravity

$$ W=F S \cos 90^{\circ}=0 \quad(\because \cos 90^{\circ}=0) $$

Answer Given,

$$ \begin{aligned} \text { mass } & =m=100 kg \\ \text { height } & =h=10 m \text { time duration } t=20 s \\ \text { power } & =\text { Rate of work done } \\ & =\frac{\text { change of PE }}{\text { time }}=\frac{m g h}{t} \\ & =\frac{100 \times 9.8 \times 10}{20} \\ & =5 \times 98=490 W \end{aligned} $$

Answer Given, average work done by a human heart per beat $=0.5 J$

Total work done during 72 beats

$$ \begin{aligned} & =72 \times 0.5 J=36 J \\ \text { Power } & =\frac{\text { Work done }}{\text { Time }}=\frac{36 J}{60 s}=0.6 W \end{aligned} $$

Answer When a charged particle moves in a uniform normal magnetic field, the path of the particle is circular, as given field is uniform hence, radius of the circular path is also constant.

As the force is central and movement is tangential work done by the force is zero. As speed is also constant we can say that $\Delta K=0$.

Answer According to work-energy theorem,

Change in $KE=$ Work done by the retarding force

$KE$ of the body $=$ Retarding force $\times$ Displacement

As KE of the bodies and retarding forces applied on them are same, therefore, both bodies will travel equal distances before coming to rest.

Thinking Process

In a uniform circular motion, velocity is always tangential in the direction of motion at any point.

Answer When bob is whirled into a vertical circle, the required centripetal force is obtained from the tension in the string. When string is cut, tension in string becomes zero and centripetal force is not provided, hence, bob start to move in a straight line path along the direction of its velocity

(a) At point $B$, the velocity of $B$ is vertically downward, therefore, when string is cut at $B$, bob moves vertically downward.

(b) At point $C$, the velocity is along the horizontal towards right, therefore, when string is cut at $C$, bob moves horizontally towards right.

Also, the bob moves under gravity simultaneously with horizontal uniform speed. So, it traversed on a parabolic path with vertex at $C$.

(c) At point $X$, the velocity of the bob is along the tangent drawn at point $X$, therefore when string is cut at point $C$, bob moves along the tangent at that point $X$.

Also, the bob move under gravity simultaneously with horizontal uniform speed. So, it traversed on a parabolic path with vertex higher than $C$.

Thinking Process

We will assume total mechanical energy of the system to be constant.

Answer KE versus $x$ graph

We know that Total ME = KE + PE

$\Rightarrow$ $E_o = KE + V(x)$

$\Rightarrow$ $ KE = E_o - V(x)$

at $A_1 x=0, V(x)=E_0$

$\Rightarrow$ KE = $E_o - E_o$

at $B_1 V(x)<E_0$

$\Rightarrow$ KE > 0

at $C$ and $D_1 V(x)=0$

$\Rightarrow KE$ is maximum at $F_1 V(x)=E_0$

Hence, $KE=0$

The variation is shown in adjacent diagram.

Velocity versus $x$ graph

As

$$ KE=\frac{1}{2} m v^{2} $$

$\therefore$ At $A$ and $F$, where $KE=0, v=0$.

At $C$ and $D, K E$ is maximum. Therefore, $v$ is $\pm$ max.

$KE>0$ (positive)

At $B, K E$ is positive but not maximum.

Therefore,

$v$ is $\pm$ some value

$(<\max$.

The variation is shown in the diagram.

Answer (a) Let $v_1$ and $v_2$ are velocities of the two balls after collision.

Now, by the principle of conservation of linear momentum,

or

$2 m v_0 =m v_1+m v_2 $

$2 v_0 =v_1+v_2 $

$e =\frac{v_2-v_1}{2 v_0}$

$ \text { and } e =\frac{v_2-v_1}{2 v_0} $

$\Rightarrow v_2 =v_1+2 v_0 e $

$\therefore 2 v_1 =2 v_0-2 e v_0 $

$\therefore v_1 =v_0(1-e)$

Since, $e<1 \Rightarrow v_1$ has the same sign as $v_0$, therefore, the ball moves on after collision.

(b) Consider the diagram below for a general collision.

By principle of conservation of linear momentum,

$$ P=P_1+P_2 $$

For inelastic collision some KE is lost, hence $\frac{p^{2}}{2 m}>\frac{p_1^{2}}{2 m}+\frac{p_2^{2}}{2 m}$

$\therefore \quad p^{2}>p_1^{2}+p_2^{2}$

Thus, $p, p_1$ and $p_2$ are related as shown in the figure.

$\theta$ is acute (less than $.90^{\circ})(p^{2}=p_1^{2}+p_2^{2}.$ would given $.\theta=90^{\circ})$

Thinking Process

A particle cannot be found in the given region when $K E<0$.

Answer We know that

$$ \text { Total energy } E=PE+KE $$

$E=V+K$

For region $A$ Given, $V>E$, From Eq. (i)

$$ K=E-V $$

as $\quad V>E \Rightarrow E-V<0$

Hence, $K<0$, this is not possible.

For region $B$ Given, $V<E \Rightarrow E-V>0$

This is possible because total energy can be greater than PE (V).

For region $C$ Given, $K>E \Rightarrow K-E>0$

from Eq. (i) $PE=V=E-K<0$

Which is possible, because $P E$ can be negative.

For region $D$ Given, $V>K$

This is possible because for a system PE $(V)$ may be greater than $KE(K)$.

Thinking Process

When two bodies of equal masses collides elastically momentum is interchanged. At the bottom point bob $A$ is having almost horizontal velocity.

Answer When ball $A$ reaches bottom point its velocity is horizontal, hence, we can apply conservation of linear momentum in the horizontal direction. (a) Two balls have same mass and the collision between them is elastic, therefore, ball $A$ transfers its entire linear momentum to ball $B$. Hence, ball $A$ will come to at rest after collision and does not rise at all.

(b) Speed with which bob B starts moving

$$ \begin{aligned} & =\text { Speed with which bob } A \text { hits bob } B \\ & =\sqrt{2 g h} \\ & =\sqrt{2 \times 9.8 \times 1} \\ & =\sqrt{19.6} \\ & =4.42 m / s \end{aligned} $$

Note When the bob $A$ is at the bottommost point, its velocity is horizontal and tension is the external force on the bob but still momentum can be considered to be conserved in horizontal direction, because the tension has no effect in horizontal direction at the bottommost point.

Answer Given, mass of the rain drop $(m)=1.00 g$

$$ =1 \times 10^{-3} kg $$

Height of falling $(h)=1 km=10^{3} m$

$$ g=10 m / s^{2} $$

Speed of the rain $drop(v)=50 m / s$

(a) Loss of PE of the drop $=m g h$

$$ =1 \times 10^{-3} \times 10 \times 10^{3}=10 J $$

(b) Gain in KE of the drop $=\frac{1}{2} m v^{2}$

$$ \begin{aligned} & =\frac{1}{2} \times 1 \times 10^{-3} \times(50)^{2} \\ & =\frac{1}{2} \times 10^{-3} \times 2500 \\ & =1.250 J \end{aligned} $$

(c) No, gain in $KE$ is not equal to the loss in its $PE$, because a part of $PE$ is utilised in doing work against the viscous drag of air.

Thinking Process

As collision is elastic, mechanical energy of the system is conserved. We have to apply energy conservation principle to describe the motion of the two bobs.

Answer (a) Consider the adjacent diagram in which the bob $B$ is displaced through an angle $\theta$ and released.

At $t=0$, suppose bob $B$ is displaced by $\theta=10^{\circ}$ to the right. It is given potential energy $E_1=E$. Energy of $A, E_2=0$.

When $B$ is released, it strikes $A$ at $t=T / 4$. In the head-on elastic collision between $B$ and $A$ comes to rest and $A$ gets velocity of $B$. Therefore, $E_1=0$ and $E_2=E$. At $t=2 T / 4, B$ reaches its extreme right position when $KE$ of $A$ is converted into $PE=E_2=E$. Energy of $B, E_1=0$.

At $t=3 T / 4, A$ reaches its mean position, when its $PE$ is converted into $KE=E_2=E$. It collides elastically with $B$ and transfers whole of its energy to $B$. Thus, $E_2=0$ and $E_1=E$. The entire process is repeated.

(b) The values of energies of $B$ and $A$ at different time intervals are tabulated here. The plot of energy with time $0 \leq t \leq 2 T$ is shown separately for $B$ and $A$ in the figure below.

Answer Given, average mass of rain drop

$$ (m)=3.0 \times 10^{-5} kg $$

Average terminal velocity $=(V)=9 m / s$.

$$ \text { Height }(h)=100 cm=1 m $$

$$ \text { Density of water }(\rho)=10^{3} kg / m^{3} $$

Area of the surface $(A)=1 m^{2}$

Volume of the water due to rain $(V)=$ Area $\times$ height

$$ \begin{aligned} & =A \times h \\ & =1 \times 1=1 m^{3} \end{aligned} $$

Mass of the water due to rain $(M)=$ Volume $\times$ density

$$ \begin{aligned} & =V \times \rho \\ & =1 \times 10^{3} \\ & =10^{3} kg \end{aligned} $$

$\therefore$ Energy transferred to the surface $=\frac{1}{2} m v^{2}$

$$ \begin{aligned} & =\frac{1}{2} \times 10^{3} \times(9)^{2} \\ & =40.5 \times 10^{3} J=4.05 \times 10^{4} J \end{aligned} $$

Answer Given, mass of the system $(m)=50,000 kg$

Speed of the system $(v)=36 km / h$

$$ =\frac{36 \times 1000}{60 \times 60}=10 m / s $$

Compression of the spring $(x)=1.0 m$

$$ \begin{aligned} KE \text { of the system } & =\frac{1}{2} m v^{2} \\ & =\frac{1}{2} \times 50000 \times(10)^{2} \\ & =25000 \times 100 J=2.5 \times 10^{6} J \end{aligned} $$

Since, $90 %$ of KE of the system is lost due to friction, therefore, energy transferred to shock absorber, is given by

$$ \begin{aligned} \Delta E & =\frac{1}{2} k x^{2}=10 % \text { of total } KE \text { of the system } \\ & =\frac{10}{100} \times 2.5 \times 10^{6} J \text { or } k=\frac{2 \times 2.5 \times 10^{6}}{10 \times(1)^{2}} \\ & =5.0 \times 10^{5} N / m \end{aligned} $$

Thinking Process

Here, shift in centre of gravity of his body is equal to the height of each step.

Answer Given, weight of the adult $(w)=m g=600 N$

Height of each step $=h=0.25 m$

Length of each step $=1 m$

Total distance travelled $=6 km=6000 m$

$\therefore \quad$ Total number of steps $=\frac{6000}{1}=6000$

Total energy utilised in jogging $=n \times m g h$

$$ =6000 \times 600 \times 0.25 J=9 \times 10^{5} J $$

Since, $10 %$ of intake energy is utilised in jogging.

$\therefore$ Total intake energy $=10 \times 9 \times 10^{5} J=9 \times 10^{6} J$.

Answer Energy is given by the petrol in the form of heat of combustion.

Thus, by question,

Energy given by 1 litre of petrol $=3 \times 10^{7} J$

Efficiency of the car engine $=0.5$

$\therefore \quad$ Energy used by the car $=0.5 \times 3 \times 10^{7} J$

$E=1.5 \times 10^{7} J$

Total distance travelled $(s)=15 km=15 \times 10^{3} m$

If $f$ is the force of friction then,

$$ \begin{aligned} & E=f \times s \quad(\because \text { Energy is utilised in working against friction }) \\ & 1.5 \times 10^{7}=f \times 15 \times 10^{3} \\ & \Rightarrow \quad f=\frac{1.5 \times 10^{7}}{15 \times 10^{3}}=10^{3} N \\ & f=1000 N \end{aligned} $$

Answer Consider the adjacent diagram the block is pushed up by applying a force $F$.

Normal reaction $(N)$ and frictional force $(f)$ is shown.

Given, mass $=m=1 kg, \theta=30^{\circ}$

$F=10 N, \mu=0.1$ and $s=$ distance moved by the block along the inclined plane $=10 m$

(a) Work done against gravity = Increase in PE of the block

$$ \begin{aligned} & =m g \times \text { Vertical distance travelled } \\ & =m g \times s(\sin \theta)=(m g s) \sin \theta \\ & =1 \times 10 \times 10 \times \sin 30^{\circ}=50 J \end{aligned} $$

(b) Work done against friction

$$ \begin{aligned} w f & =f \times s=\mu N \times s=\mu m g \cos \theta \times s \\ & =0.1 \times 1 \times 10 \times \cos 30^{\circ} \times 10 \\ & =10 \times 0.866=8.66 J \end{aligned} $$

(c) Increase in $PE=m g h=m g(s \sin \theta)$

$$ \begin{aligned} & =1 \times 10 \times 10 \times \sin 30^{\circ} \\ & =100 \times \frac{1}{2}=50 J \end{aligned} $$

(d) By work-energy theorem, we know that work done by all the forces = change in KE

$$ \Rightarrow \quad \begin{aligned} (W) & =\Delta K \\ \Delta k & =W_{g}+W_{f}+W_{f} \\ & =-m g h-f s+F S \\ & =-50-8.66+10 \times 10 \\ & =50-8.66=41.34 J \end{aligned} $$

(e) Work done by applied force, $F=F S$

$$ =(10)(10)=100 J $$

Answer (a) As ball 1 is rolling down without slipping there is no dissipation of energy hence, total mechanical energy is conserved.

Ball 3 is having negligible friction hence, there is no loss of energy.

(b) Ball 1 acquires rotational energy, ball 2 loses energy by friction. They cannot cross at $C$. Ball 3 can cross over.

(c) Ball 1,2 turn back before reaching $C$. Because of loss of energy, ball 2 cannot reach back to $A$. Ball 1 has a rotational motion in “wrong” sense when it reaches $B$. It cannot roll back to $A$, because of kinetic friction.

Thinking Process

As the gas is ejected, the rocket gets propelled in forward direction due to upward thrust.

Answer Let $M$ be the mass of rocket at any time $t$ and $v_1$ the velocity of rocket at the same time $t$. Let $\Delta m=$ mass of gas ejected in time interval $\Delta t$.

Relative speed of gas ejected $=u$.

Consider at time $t+\Delta t$

$(KE)_{t}+\Delta t =KE \text { of rocket }+KE \text { of gas } $

$ =\frac{1}{2}(M-\Delta m)(v+\Delta v)^{2}+\frac{1}{2} \Delta m(v-u)^{2} $

$ =\frac{1}{2} M v^{2}+M v \Delta v-\Delta m v u+\frac{1}{2} \Delta m u^{2} $

$(KE)_{t} =KE \text { of the rocket at time } t=\frac{1}{2} M v^{2} $

$\Delta K =(KE) _{t} + \Delta t-(KE) _{t} $

$ =(M \Delta v-\Delta m u) v+\frac{1}{2} \Delta m u^{2}$

Since, action-reaction forces are equal.

Hence,

$$ \begin{aligned} M \frac{d v}{d t} & =\frac{d m}{d t}|u| \\ M \Delta v & =\Delta m u \\ \Rightarrow \Delta K & =\frac{1}{2} \Delta m u^{2} \end{aligned} $$

Now, by work-energy theorem,

$$ \begin{aligned} & \Delta K = \Delta W \\ \Rightarrow & \Delta W=\frac{1}{2} \Delta m u^{2} \end{aligned} $$

Answer Let

$$ \begin{aligned} & m=50 g=50 \times 10^{-3} kg \\ & \text { Side }=L=1 cm=0.01 m \\ & \text { Speed }=V=10 cm / s=0.1 m / s \\ & \text { Young’s modulus }=Y=2 \times 10^{11} N / m^{2} \\ & \text { Maximum compression } \Delta L=? \end{aligned} $$

In this case, all $KE$ will be converted to $PE$ By Hooke’s law,

$$ \frac{F}{A}=Y \frac{\Delta L}{L} $$

where $A$ is the surface area and $L$ is length of the side of the cube. If $k$ is spring or compression constant, then

$$ \begin{aligned} \text { force } F & =K \Delta L \\ K & =Y \frac{A}{L}=Y L \\ \text { Initial } K E & =2 \times \frac{1}{2} m v^{2}=5 \times 10^{-4} J \\ \text { Final } P E & =2 \times \frac{1}{2} k(\Delta L)^{2} \\ \Delta L=\sqrt{\frac{K E}{K}}=\sqrt{\frac{K E}{Y L}} & =\sqrt{\frac{5 \times 10^{-4}}{2 \times 10^{11} \times 0.1}}=1.58 \times 10^{-7} m \quad[\because P E=K E] \end{aligned} $$

Thinking Process

In this problem, as viscous drag of air is neglected, hence there is no dissipation of energy.

Answer Let $m=$ Mass of balloon

$$ \begin{aligned} V & =\text { Volume of balloon } \\ \rho_{\text {He }} & =\text { Density of helium } \\ \rho_{\text {air }} & =\text { Density of air } \end{aligned} $$

Volume $V$ of balloon displaces volume $V$ of air.

So,

$$ V(\rho_{\text {air }}-\rho_{He}) g=m a=m \frac{d v}{d t}=\text { up thrust } $$

Integrating with respect to $t$,

$$ \begin{aligned} V(\rho_{\text {air }}-\rho_{He}) g t & =m v \\ \frac{1}{2} m v^{2} & =\frac{1}{2} m \frac{V^{2}}{m^{2}}(\rho_{\text {air }}-\rho_{He})^{2} g^{2} t \\ & =\frac{1}{2 m} V^{2}(\rho_{\text {air }}-\rho_{He})^{2} g^{2} t^{2} \end{aligned} $$

$$ \Rightarrow \quad \frac{1}{2} m v^{2}=\frac{1}{2} m \frac{V^{2}}{m^{2}}(\rho_{\text {air }}-\rho_{He})^{2} g^{2} t^{2} $$

If the balloon rises to a height $h$, from $s=u t+\frac{1}{2} a t^{2}$,

$$ \text { We get } h=\frac{1}{2} a t^{2}=\frac{1}{2} \frac{V(\rho_{\text {air }}-\rho_{He})}{m} g t^{2} $$

From Eqs. (iii) and (ii),

Rearranging the terms,

$$ \begin{aligned} \frac{1}{2} m v^{2} & =[V(\rho_{a}-\rho_{He}) g][\frac{1}{2 m} V(\rho_{\text {air }}-\rho_{He}) g t^{2}] \\ & =V(\rho_{a}-\rho_{He}) g h \end{aligned} $$

$\Rightarrow \quad \frac{1}{2} m v^{2}+V \rho_{He} g h=V_{\rho_{\text {air }}} h g$

$KE_{\text {balloon }}+PE_{\text {balloon }}=$ Change in $PE$ of air.

So, as the balloon goes up, an equal volume of air comes down, increase in PE and KE of the balloon is at the cost of PE of air [which comes down].