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.

• (a) The two magnetic forces are equal and opposite, but they do not cancel out in terms of work done because work depends on the direction of force and displacement, not just the force itself.
• (c) The magnetic forces do not do work on each particle because the force is perpendicular to the direction of motion, hence no work is done.
• (d) The magnetic forces are not necessarily negligible; they are simply perpendicular to the motion, resulting in no work being done.

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.

• (a) The work done is not the same in both experiments because, although the force law is the same, the distance moved by the positron is larger due to its smaller mass, resulting in more work done on the positron.
• (b) The work done is not less for the positron. While it is true that the positron moves away more rapidly and the force weakens, the overall work done depends on both the force and the distance moved. Since the positron moves a larger distance, the work done is actually more.
• (d) The work done by the charged particle on the stationary proton is not the same as the work done on the moving particles. The stationary proton does not move, so no work is done on it. The work done on the moving particles depends on the distance they travel, which is greater for the positron.

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 $+mg$

$\Rightarrow R>mg$

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

$\Rightarrow$ Reactional force $\approx mg$

• (a) The force of reaction of the ground on the man is not constant and equal to ( mg ) in magnitude because when the man is squatting and then gets up, there is a change in the forces acting on him. Initially, the reaction force must counteract not only his weight but also the additional forces due to his movement, making it variable and not constant.

• (b) The force of reaction of the ground on the man is not constant and greater than ( mg ) in magnitude because the reaction force changes during the process. It is greater than ( mg ) when the man is accelerating upwards but eventually becomes equal to ( mg ) when he stands still.

• (c) The force of reaction of the ground on the man is not always greater than ( mg ) because once the man stands up and is no longer accelerating, the reaction force equals ( mg ). Therefore, it is not always greater than ( mg ); it is only greater during the upward acceleration phase.

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 $200\times 10=-2000J$.

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.

• Option (a) +2000 J: This option is incorrect because the work done by the frictional force is negative, as it opposes the motion. Therefore, the work done cannot be positive.

• Option (b) -200 J: This option is incorrect because the work done by the frictional force is calculated as the force multiplied by the distance, which is $200N\times 10m=2000J$. The correct value should be $-2000J$, not $-200J$.

• Option (d) $-20,000$ J: This option is incorrect because it overestimates the work done by the frictional force. The correct calculation is $200N\times 10m=2000J$, so the work done should be $-2000J$, not $-20,000J$.

Answer (c) As the body is falling freely under gravity, the potential energy decreases and kinetic energy increases but total mechanical energy ( $PE+KE$ ) of the body and earth system will be constant as external force on the system is zero.

• Kinetic energy: The kinetic energy of the body increases as it falls because its velocity increases due to the acceleration caused by gravity.

• Potential energy: The potential energy of the body decreases as it falls because its height above the ground decreases.

• Total linear momentum: The total linear momentum of the body increases as it falls because its velocity increases due to the acceleration caused by gravity.

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.

• (a) Total kinetic energy: In an inelastic collision, some of the kinetic energy is converted into other forms of energy such as heat, sound, or deformation energy. Therefore, the total kinetic energy is not conserved.

• (b) Total mechanical energy: Total mechanical energy, which includes both kinetic and potential energy, is not conserved in an inelastic collision because some of the mechanical energy is transformed into other forms of energy like heat or sound.

• (d) Speed of each body: The speed of each body changes during an inelastic collision due to the transfer of momentum and energy between the colliding bodies. Therefore, the speed of each body is not 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,

$$\begin{array}{rl}\frac{1}{2}m{v}^2& =mgh\text{ (for both tracks I and II) }\\ v& =\sqrt{2gh}\end{array}$$

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.

• Option (a): This option is incorrect because although both stones will reach the bottom with the same speed due to conservation of mechanical energy, they will not reach the bottom at the same time. The stone on the steeper track (stone II) will have a greater acceleration and a shorter path, causing it to reach the bottom earlier than the stone on the gradual track (stone I).

• Option (b): This option is incorrect because while it correctly states that both stones will reach the bottom with the same speed, it incorrectly states that stone I will reach the bottom earlier than stone II. In reality, stone II, which is on the steeper track, will reach the bottom earlier due to its greater acceleration and shorter path.

• Option (d): This option is incorrect because it states that both stones will reach the bottom at different times and with different speeds. However, due to the conservation of mechanical energy, both stones will reach the bottom with the same speed. The difference lies only in the time taken to reach the bottom, with stone II reaching 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) $E=PE+0=PE=V(x_m)=\frac{1}{2}k{x}_m^2$

• Option (a) $V=0,K=E$: This option is incorrect because at the extreme position $x=\pm x_m$, the potential energy $V$ is at its maximum value, not zero. The kinetic energy $K$ is zero at this point because the particle momentarily stops before reversing direction.

• Option (c) $V<E,K=0$: This option is incorrect because at the extreme position $x=\pm x_m$, the potential energy $V$ is equal to the total energy $E$, not less than $E$. The kinetic energy $K$ is indeed zero, but the potential energy should be equal to the total energy.

• Option (d) $V=0,K<E$: This option is incorrect because at the extreme position $x=\pm x_m$, the potential energy $V$ is not zero; it is equal to the total energy $E$. The kinetic energy $K$ is zero at this point, not less than the total energy.

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 $3v_2=0,v_3=v$

• Option (a): This option shows all three balls moving with the same velocity after the collision. This is incorrect because, in an elastic collision involving identical masses, the velocities are exchanged, not shared equally among all the balls.

• Option (c): This option shows the incoming ball stopping and the two initially stationary balls moving apart with equal and opposite velocities. This is incorrect because, in an elastic collision, the incoming ball’s velocity would be transferred to one of the stationary balls, not split between them.

• Option (d): This option shows the incoming ball continuing to move with the same velocity while the two initially stationary balls remain stationary. This is incorrect because, in an elastic collision, the incoming ball’s velocity would be transferred to one of the stationary balls, causing it to move.

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

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

We know that

Acceleration

$$\begin{array}{rl}{a}_0& =\frac{dv}{dt}=v\frac{dv}{dx}=a{x}^{3/2}\frac{d}{dx}(a{x}^{3/2})\\ & =a{x}^{3/2}\times a\times \frac{3}{2}\times x^{1/2}=\frac{3}{2}{a}^2{x}^2\end{array}$$

Now,

$$\text{ Force }=m{a}_0=m\frac{3}{2}{a}^2{x}^2$$

Work done $={\int }_{x=0}^{x=2}Fdx={\int }_0^2\frac{3}{2}m{a}^2{x}^{2}dx$

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

• Option (a) $1.5J$ is incorrect because the calculated work done, based on the given parameters and the integration of the force over the displacement, results in a value of $50J$, not $1.5J$. The mathematical steps and integration clearly show that the work done is significantly higher than $1.5J$.

• Option (c) $10J$ is incorrect because, similar to option (a), the integration of the force over the displacement from $x=0$ to $x=2$ results in $50J$. The value of $10J$ does not match the derived result from the given equations and constants.

• Option (d) $100J$ is incorrect because the work done, as calculated through the integration process, is $50J$. The value of $100J$ is double the correct answer, indicating a possible misunderstanding or miscalculation in the problem-solving process.

Answer (b) Given, power = constant

We know that power $(P)$

$$P=\frac{dW}{dt}=\frac{F\cdot ds}{dt}=\frac{Fds}{dt}(\because \text{ body is moving unidirectionally })$$

Hence,

$$F\cdot ds=Fds\cos 0^{\circ }$$

$$P=\frac{Fds}{dt}=\text{ constant }(\because P=\text{ constant by question })$$

Now, writing dimensions

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

• Option (a): This option shows a linear relationship between displacement and time, which implies a constant velocity. However, since the power is constant and not the velocity, the displacement cannot be linear with time.

• Option (c): This option shows a parabolic relationship between displacement and time, which implies a constant acceleration. However, with constant power, the acceleration is not constant; instead, the velocity increases as the square root of time, leading to a displacement-time relationship of ( t^{3/2} ).

• Option (d): This option shows an exponential relationship between displacement and time, which implies an exponentially increasing velocity. However, with constant power, the velocity increases as the square root of time, leading to a displacement-time relationship of ( t^{3/2} ), not an exponential one.

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).

• Option (a): This option shows a constant kinetic energy, which is incorrect because the kinetic energy of the Earth varies as it moves in its elliptical orbit around the Sun. The speed of the Earth changes, leading to changes in kinetic energy.

• Option (b): This option shows a linear increase and decrease in kinetic energy, which is incorrect because the variation in kinetic energy is not linear. The kinetic energy changes more rapidly when the Earth is closer to the Sun and more slowly when it is farther away.

• Option (c): This option shows the kinetic energy dropping to zero, which is incorrect because the Earth’s speed never becomes zero in its orbit around the Sun. The kinetic energy is always positive and never reaches zero.

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).

• Option (a): This diagram shows a constant total mechanical energy over time, which is incorrect because a pendulum oscillating in air loses energy due to air resistance, causing its total mechanical energy to decrease continuously.

• Option (b): This diagram shows an increasing total mechanical energy over time, which is incorrect because a pendulum oscillating in air cannot gain energy on its own; it loses energy due to air resistance.

• Option (d): This diagram shows a periodic variation in total mechanical energy, which is incorrect because the total mechanical energy of a pendulum oscillating in air should decrease continuously due to energy loss from air resistance, not oscillate periodically.

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

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

• Option (b) $100{\pi }^2$: This option is incorrect because it underestimates the kinetic energy. The correct calculation shows that the kinetic energy is $250{\pi }^2$ J, not $100{\pi }^2$ J. The error likely comes from an incorrect application of the kinetic energy formula or a miscalculation of the linear speed.

• Option (c) $5{\pi }^2$: This option is incorrect because it significantly underestimates the kinetic energy. The correct kinetic energy is $250{\pi }^2$ J. This large discrepancy suggests a fundamental error in the calculation process, possibly in the conversion of angular velocity to linear speed or in the application of the kinetic energy formula.

• Option (d) 0: This option is incorrect because it implies that the mass has no kinetic energy. Since the mass is moving with a non-zero velocity along a circular path, it must have kinetic energy. The correct kinetic energy is $250{\pi }^2$ J, not zero.

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 $PE$ and $KE$ is best represented by (b).

• Option (a): This option shows the kinetic energy (KE) increasing linearly throughout the fall, which is incorrect because the raindrop reaches terminal velocity after falling through a height of ((3/4)h). At terminal velocity, KE should become constant, not continue to increase. Additionally, the potential energy (PE) should decrease continuously, but the graph does not show this correctly.

• Option (c): This option shows the kinetic energy (KE) increasing and then decreasing, which is incorrect. Once the raindrop reaches terminal velocity, KE should remain constant, not decrease. The potential energy (PE) is shown as decreasing correctly, but the incorrect representation of KE makes this option invalid.

• Option (d): This option shows the kinetic energy (KE) increasing and then decreasing, similar to option (c), which is incorrect. KE should remain constant after reaching terminal velocity. Additionally, the potential energy (PE) is shown as decreasing correctly, but the incorrect representation of KE makes this option invalid.

Thinking Process

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

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

From conservation of mechanical energy.

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

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

• Option (a) $2.5J$: This value is too low because it only considers a small fraction of the total energy. The initial potential energy due to the height and the initial kinetic energy due to the speed are both significant and add up to much more than $2.5J$.

• Option (b) $5.0J$: This value is also too low. It only accounts for the initial kinetic energy of the shotput ($\frac{1}{2}m{v}^2=5J$) and ignores the potential energy due to the height from which the shotput is thrown.

• Option (c) $52.5J$: This value is incorrect because it underestimates the total energy. It might be considering only part of the potential energy or a miscalculation. The correct total energy should include both the potential energy from the height and the initial kinetic energy, which sums up to $155J$.

Answer (b) First velocity of the iron sphere increases and after sometime becomes constant, called terminal velocity. Hence, accordingly first KE increases and then becomes constant which is best represented by (b).

• Option (a): This diagram shows a linear increase in kinetic energy without any indication of it becoming constant. This does not accurately represent the scenario where the iron sphere reaches terminal velocity, at which point the kinetic energy should level off.

• Option (c): This diagram shows a decrease in kinetic energy after an initial increase. This is incorrect because once the iron sphere reaches terminal velocity, its kinetic energy should remain constant, not decrease.

• Option (d): This diagram shows a constant kinetic energy from the beginning, which is incorrect. Initially, the kinetic energy of the iron sphere should increase as it accelerates, and only after reaching terminal velocity should it become constant.

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

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

Change in momentum of the ball

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

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

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

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

• Option (a) $10.5N$: This option is incorrect because it significantly underestimates the force required. The calculation of force involves a change in momentum divided by the time of contact. Given the mass and velocity of the ball, the resulting force is much higher than 10.5 N.

• Option (b) $21N$: This option is incorrect because it also underestimates the force required. The change in momentum for the given mass and velocity, when divided by the very short contact time of 0.001 seconds, results in a force that is much greater than 21 N.

• Option (d) $2.1\times {10}^{4}N$: This option is incorrect because it overestimates the force required. The correct calculation of the force, based on the given mass, velocity, and contact time, results in a force of $1.05\times {10}^{4}N$, which is half of this value.

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 $mgL$ as the change in kinetic energy is almost zero.

Hence, total work done $=-mgL+mgL=0$

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.

• (a) Work done by all forces on man is equal to the rise in potential energy mgL: This is incorrect because the work done by all forces on the man is zero. The work done by the gravitational force is (-mgL) and the work done by the internal muscular forces is (mgL), which cancel each other out, resulting in a net work of zero.

• (c) Work done by the gravitational force on man is (mgL): This is incorrect because the work done by the gravitational force on the man is actually (-mgL). The gravitational force acts downward while the man moves upward, resulting in negative work done by gravity.

Answer $(b,d,f)$

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

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

$U_i+K_i=U_f+K_f$

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

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

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

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,

$$\begin{array}{rl}& \text{ By question, }=\frac{1}{2}(m{v}^{\prime \prime }{)}^2=\frac{1}{2}[\frac{1}{2}m(v^{\prime }{)}^2]\\ & \frac{1}{2}m(v^{\prime \prime }{)}^2=\frac{1}{4}m(v^{\prime }{)}^2=\frac{1}{4}m[v^2-2gh]\\ & \Rightarrow (v^{\prime \prime }{)}^2=\frac{v^2-2gh}{2}=\frac{v^2}{2}-gh\\ & \Rightarrow v^{\prime \prime }=\sqrt{\frac{v^2}{2}-gh}\end{array}$$

From Eqs. (i) and (ii)

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

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

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

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

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.

• (a) The velocity of the bullet will be reduced to half its initial value: This statement is incorrect because the kinetic energy of the bullet is reduced to half after emerging from the target, not the velocity. Since kinetic energy is proportional to the square of the velocity, reducing the kinetic energy to half results in the velocity being reduced by a factor of (\sqrt{2}), not 2.

• (c) The bullet will continue to move along the same parabolic path: This statement is incorrect because the velocity of the bullet changes after it emerges from the target. Since the velocity is different, the trajectory of the bullet will also change, resulting in a different parabolic path.

• (e) The bullet will fall vertically downward after hitting the target: This statement is incorrect because the bullet still retains a horizontal component of velocity after emerging from the target. Therefore, it will continue to follow a parabolic trajectory rather than falling straight down vertically.

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.

• (a) While spring is fully compressed all the KE of $M_1$ is stored as PE of spring: This statement is incorrect because, at the point of maximum compression, both blocks $M_1$ and $M_2$ will have the same velocity. Therefore, not all the kinetic energy of $M_1$ is converted into the potential energy of the spring; some of it is also converted into the kinetic energy of $M_2$.

• (b) While spring is fully compressed the system momentum is not conserved, though final momentum is equal to initial momentum: This statement is incorrect because the momentum of the system is conserved at all times, including when the spring is fully compressed. The conservation of momentum is a fundamental principle that holds true regardless of the state of compression of the spring.

• (d) If the surface on which blocks are moving has friction, then collision cannot be elastic: This statement is incorrect because the presence of friction does not necessarily make the collision inelastic. The collision can still be elastic if the kinetic energy is conserved during the collision. Friction would affect the motion of the blocks after the collision, but it does not inherently change the nature of the collision itself.

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

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

$W=Fs\cos \theta =Fs\cos 0^{\circ }=Fs(\because \cos 0^{\circ }=1)$

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

(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

$$W=Fs\cos {180}^{\circ }=-Fs$$

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

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

$$W=FS\cos {90}^{\circ }=0(\because \cos {90}^{\circ }=0)$$

Answer Given,

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

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

Total work done during 72 beats

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

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 $\mathrm{\Delta }K=0$.

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 $D,KE$ is maximum. Therefore, $v$ is $\pm$ max.

$KE>0$ (positive)

At $B,KE$ 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

$2m{v}_0=m{v}_1+m{v}_2$

$2v_0=v_1+v_2$

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

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

$\Rightarrow v_2=v_1+2v_{0}e$

$\therefore 2v_1=2v_0-2e{v}_0$

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

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 $\frac{p^2}{2m}>\frac{p_1^2}{2m}+\frac{p_2^2}{2m}$

$\therefore p^2>p_1^2+p_2^2$

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 $KE<0$.

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

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 $PE$ 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

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

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 $(m)=1.00g$

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

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

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

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

(a) Loss of PE of the drop $=mgh$

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

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

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

(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 $t=2T/4,B$ reaches its extreme right position when $KE$ of $A$ is converted into $PE=E_2=E$. Energy of $B,E_1=0$.

At $t=3T/4,A$ 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 $0\le t\le 2T$ 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 $=(V)=9m/s$.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$\begin{array}{rlr}\mathrm{\Delta }E& =\frac{1}{2}k{x}^2=10& =\frac{10}{100}\times 2.5\times {10}^{6}J\text{ or }k=\frac{2\times 2.5\times {10}^6}{10\times (1{)}^2}\\ & =5.0\times {10}^{5}N/m\end{array}$$

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 $(w)=mg=600N$

Height of each step $=h=0.25m$

Length of each step $=1m$

Total distance travelled $=6km=6000m$

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

Total energy utilised in jogging $=n\times mgh$

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

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 $=0.5$

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

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

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

If $f$ is the force of friction then,

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

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

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

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

$$\begin{array}{rl}& =mg\times \text{ Vertical distance travelled }\\ & =mg\times s(\sin \theta )=(mgs)\sin \theta \\ & =1\times 10\times 10\times \sin {30}^{\circ }=50J\end{array}$$

(b) Work done against friction

$$\begin{array}{rl}wf& =f\times s=\mu N\times s=\mu mg\cos \theta \times s\\ & =0.1\times 1\times 10\times \cos {30}^{\circ }\times 10\\ & =10\times 0.866=8.66J\end{array}$$

(c) Increase in $PE=mgh=mg(s\sin \theta )$

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

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

$$\Rightarrow \begin{array}{rl}(W)& =\mathrm{\Delta }K\\ \mathrm{\Delta }k& =W_g+W_f+W_f\\ & =-mgh-fs+FS\\ & =-50-8.66+10\times 10\\ & =50-8.66=41.34J\end{array}$$

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

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

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 $\mathrm{\Delta }m=$ mass of gas ejected in time interval $\mathrm{\Delta }t$.

Relative speed of gas ejected $=u$.

Consider at time $t+\mathrm{\Delta }t$

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

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

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

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

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

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

Since, action-reaction forces are equal.

Hence,

$$\begin{array}{rl}M\frac{dv}{dt}& =\frac{dm}{dt}|u|\\ M\mathrm{\Delta }v& =\mathrm{\Delta }mu\\ \Rightarrow \mathrm{\Delta }K& =\frac{1}{2}\mathrm{\Delta }m{u}^2\end{array}$$

Now, by work-energy theorem,

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

Answer Let

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

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

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

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

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

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

$$\begin{array}{rl}V& =\text{ Volume of balloon }\\ {\rho }_{\text{He }}& =\text{ Density of helium }\\ {\rho }_{\text{air }}& =\text{ Density of air }\end{array}$$

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

So,

$$V({\rho }_{\text{air }}-{\rho }_{He})g=ma=m\frac{dv}{dt}=\text{ up thrust }$$

Integrating with respect to $t$,

$$\begin{array}{rl}V({\rho }_{\text{air }}-{\rho }_{He})gt& =mv\\ \frac{1}{2}m{v}^2& =\frac{1}{2}m\frac{V^2}{m^2}({\rho }_{\text{air }}-{\rho }_{He}{)}^2{g}^{2}t\\ & =\frac{1}{2m}{V}^2({\rho }_{\text{air }}-{\rho }_{He}{)}^2{g}^2{t}^2\end{array}$$

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

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

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

From Eqs. (iii) and (ii),

Rearranging the terms,

$$\begin{array}{rl}\frac{1}{2}m{v}^2& =[V({\rho }_a-{\rho }_{He})g][\frac{1}{2m}V({\rho }_{\text{air }}-{\rho }_{He})g{t}^2]\\ & =V({\rho }_a-{\rho }_{He})gh\end{array}$$

$\Rightarrow \frac{1}{2}m{v}^2+V{\rho }_{He}gh=V_{{\rho }_{\text{air }}}hg$

$K{E}_{\text{balloon }}+P{E}_{\text{balloon }}=$ 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].