NCERT Exemplar: Work, Energy & Power | Physics Class 11 - NEET PDF Download

Multiple  Choice Questions I

Q. 1. An electron and a proton are moving under the influence of mutual forces. In calculating the change in the kinetic energy of the system during motion, one ignores the magnetic force of one on another. This is because,
(a) The two magnetic forces are equal and opposite, so they produce no net effect.
(b) The magnetic forces do no work on each particle.
(c) The magnetic forces do equal and opposite (but non-zero) work on each particle.
(d) The magnetic forces are necessarily negligible.
Ans. (b)

Solution:
The work-energy theorem states that the net work done on a particle equals the change in its kinetic energy: ΣW = K2 - K1. The magnetic force on a charged particle is always perpendicular to that particle's instantaneous velocity (F = q v × B). A force perpendicular to velocity does no work on that particle, so magnetic forces do no work and do not change kinetic energy. Therefore magnetic forces between the electron and proton can be ignored when calculating change in kinetic energy of the system.

(magnetic force) will be perpendicular to both B and v, where B is the external magnetic field and v is the velocity of particle. That is why one ignores the magnetic force of one particle on another.

Q. 2. A proton is kept at rest. A positively charged particle is released from rest at a distance d in its field. Consider two experiments; one in which the charged particle is also a proton and in another, a positron. In the same time t, the work done on the two moving charged particles is
(a) Same as the same force law is involved in the two experiments.
(b) Less for the case of a positron, as the positron moves away more rapidly and the force on it weakens.
(c) More for the case of a positron, as the positron moves away a larger distance.
(d) Same as the work done by charged particle on the stationary proton.
Ans. (c)

Solution:
The Coulomb force magnitude is the same for a proton-proton pair and a proton-positron pair because charges are equal in magnitude. The lighter particle (positron) accelerates faster and, in the same time t, moves a larger distance than a heavier proton. Work done by the force on a particle equals force × displacement in the direction of force. Since the force magnitudes are equal but the positron covers a larger distance in time t, the work done on the positron is larger; hence option (c).

Q. 3. A man squatting on the ground gets straight up and stand. The force of reaction of ground on the man during the process is
(a) constant and equal to mg in magnitude.
(b) constant and greater than mg in magnitude.
(c) variable but always greater than mg.
(d) at first greater than mg, and later becomes equal to mg.
Ans.

Solution:
While a man rises from squat to standing, he must produce an upward net force to accelerate his centre of mass upward. The normal reaction N from the ground provides this upward force. At the start of rising, N must exceed mg to produce upward acceleration; when motion stops (standing still) N equals mg. Thus the reaction is greater than mg during the upward acceleration and later becomes equal to mg. The solution in the original text explains that N > mg during the motion and equals mg once standing upright.

Q. 4. A bicyclist comes to a skidding stop in 10 m. During this process, the force on the bicycle due to the road is 200N and is directly opposed to the motion. The work done by the cycle on the road is
(a) + 2000J
(b) - 200J
(c) zero
(d) - 20,000J
Ans. (c)

Solution:
Frictional force from the road on the bicycle is 200 N opposite to motion; the work done by this friction on the bicycle is W = F × s × cos180° = -200 × 10 = -2000 J (mechanical energy of the bicyclist is dissipated as heat). However, the road itself does not move, so the work done by the cycle on the road is zero. Energy is dissipated as heat in the surfaces in contact; the road does not gain macroscopic kinetic energy.

Q. 5. A body is falling freely under the action of gravity alone in vacuum. Which of the following quantities remain constant during the fall?
(a) Kinetic energy.
(b) Potential energy.
(c) Total mechanical energy.
(d) Total linear momentum.
Ans. (c)

Solution:
In vacuum, only conservative gravitational force acts. As the body falls, potential energy decreases and kinetic energy increases such that their sum (total mechanical energy E = K + U) remains constant. Linear momentum is not constant because the body accelerates under gravity. Hence option (c) is correct.

Q. 6. During inelastic collision between two bodies, which of the following quantities always remain conserved?
(a) Total kinetic energy.
(b) Total mechanical energy.
(c) Total linear momentum.
(d) Speed of each body.
Ans. (c)

Solution :
In collisions, if no external force acts on the pair of bodies, the total linear momentum of the system is conserved. In an inelastic collision, kinetic energy is not conserved (some is converted to internal energy, heat, deformation, sound). Total energy (including internal energy) is conserved, but mechanical kinetic energy generally is not. Therefore (c) is correct.

Q. 7. Two inclined frictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track as shown in Fig.

Which of the following statement is correct?
 (a) Both the stones reach the bottom at the same time but not with the same speed.
 (b) Both the stones reach the bottom with the same speed and stone I reaches the bottom earlier than stone II.

(c) Both the stones reach the bottom with the same speed and stone II reaches the bottom earlier than stone I.
 (d) Both the stones reach the bottom at different times and with different speeds.

Ans. (c)

Solution:
On frictionless tracks starting from the same height h, mechanical energy is conserved for each stone: mgh = 1/2 m v^2, so both stones attain the same speed at the bottom regardless of the slope. Time to descend depends on the acceleration component along each incline, a = g sinθ; the steeper track (larger θ) gives larger acceleration and shorter path length, so the stone on the steeper track reaches the bottom earlier. Hence both have same speed but stone II (steeper) arrives earlier, option (c).

Q. 8. The potential energy function for a particle executing linear SHM is given by

where k is the force constant of the oscillator (Fig). For

k = 0.5N/m, the graph of V(x) versus x is shown in the figure. A particle of total energy E turns back when it reaches x = 

. If V and K indicate the P.E. and K.E., respectively of the particle at x = +x_m, then which of the following is correct?

(a) V = O, K = E
 (b) V = E, K = O
 (c) V < E, K = O
 (d) V = O, K < E

Ans. (b)

Solution:
For simple harmonic oscillator U(x) = 1/2 k x^2. At the turning point x = ±xm the kinetic energy is zero and potential energy equals the total energy E. Therefore at x = +xm, V = E and K = 0, so option (b) is correct.

................(i)

 [As x = ±a for extreme]

Q. 9. Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed V as shown in Fig

If the collision is elastic, which of the following (Fig) is a possible result after collision?

(a)

(b)

(c)

(d)

Ans. (b)

Solution:
For one-dimensional elastic collisions among equal masses, velocities are effectively exchanged in head-on collisions. Using conservation of momentum and kinetic energy, only configuration (b) conserves both total momentum and total kinetic energy for the geometry shown. The working in the original text evaluates kinetic energies for the given options and finds only (b) satisfies conservation.

Q.10. A body of mass 0.5 kg travels in a straight line with velocity v = ax^3/2 where a = 5 m^-1/2 s^-1. The work done by the net force during its displacement from x = 0 to; x = 2m is
(a) 5 J
(b) 50 J
(c) 10 J
(d) 100 J
Ans. 

Solution:
Use work-energy theorem: net work done = change in kinetic energy = (1/2) m v^2 evaluated between limits. Velocity v(x) = a x^{3/2}. Compute acceleration via dv/dt = v dv/dx, but simpler to compute kinetic energy as function of x and take difference. K(x) = 1/2 m v^2 = 1/2 m a^2 x^3. Work done from x = 0 to x = 2 is ΔK = K(2) - K(0) = 1/2 m a^2 (2^3) = 1/2 × 0.5 × (5)^2 × 8. Compute: (5)^2 = 25; 1/2 × 0.5 = 0.25; 0.25 × 25 × 8 = 0.25 × 200 = 50 J. Thus answer is 50 J (option b).

............. (i)

Here, is to be integrated along the

Q. 11. A body is moving unidirectionally under the influence of a source of constant power supplying energy. Which of the diagrams shown in figure correctly shown the displacement-time curve for its motion?

Ans. (b)

Solution:
If power P supplied to a mass is constant, then P = F v = (dp/dt) v = m v (dv/dt) = m v a. For constant P, m v (dv/dt) = P which integrates to v^2 ∝ t, so v ∝ t^{1/2}. Displacement d = ∫ v dt gives d ∝ t^{3/2}. The displacement-time curve that follows d ∝ t^{3/2} is shown in option (b).

Q. 12. Which of the diagrams shown in Fig. 6.6 most closely shows the variation in kinetic energy of the earth as it moves once around the sun in its elliptical orbit?

Ans. (d)

Solution:
In an elliptical orbit the earth moves fastest at perihelion (closest to Sun) and slowest at aphelion (farthest). Kinetic energy varies accordingly: maximum at perihelion, minimum at aphelion, varying smoothly and never becoming zero. The curve that represents this variation correctly is option (d).

Q. 13. Which of the diagrams shown in Fig represents variation of total mechanical energy of a pendulum oscillating in air as function of time?

Ans. (c)

Solution:
A pendulum oscillating in air loses mechanical energy continuously due to air resistance (a non-conservative force). Energy decreases continuously (often approximately exponentially) with time. The curve showing continuous decay is option (c).

Q. 14. A mass of 5 kg is moving along a circular path of radius 1 m. If the mass moves with 300 revolutions per minute, its kinetic energy would be
(a) 250π²
(b) 100π²
(c) 5π²
(d) 0
Ans. 

Solution:
Given m = 5 kg, f = 300 rev/min = 300/60 = 5 rev/s. Angular speed ω = 2π f = 10π rad/s. Linear speed v = ω R = 10π × 1 = 10π m/s. Kinetic energy K = 1/2 m v^2 = 1/2 × 5 × (10π)^2 = (2.5) × 100 π^2 = 250 π^2 J. Thus option (a).

Q. 15. A raindrop falling from a height h above ground, attains a near terminal velocity when it has fallen through a height (3/4)h. Which of the diagrams shown in Fig correctly shows the change in kinetic and potential energy of the drop during its fall up to the ground?

Ans. (b)

Solution:
Initially potential energy is maximum at height h and kinetic energy is zero. As the drop falls, PE converts to KE. Because of air resistance, after some fall the drop reaches terminal velocity and KE becomes roughly constant while PE continues to decrease up to the ground. The combined behaviour of decreasing PE and KE increasing then saturating is shown in option (b).

Q. 16. In a shotput event an athlete throws the shotput of mass 10 kg with an initial speed of 1m s^-1 at 45° from a height 1.5 m above ground. Assuming air resistance to be negligible and acceleration due to gravity to be 10 m s^-2 , the kinetic energy of the shotput when it just reaches the ground will be
(a) 2.5 J
(b) 5.0 J
(c) 52.5 J
(d) 155.0 J
Ans. (d)

Solution:
Take ground as reference PE = 0. Initial total mechanical energy: PE_i + KE_i = m g h + 1/2 m v^2 = 10 × 10 × 1.5 + 1/2 × 10 × 1^2 = 150 + 5 = 155 J. With no dissipation, this total becomes kinetic energy just before hitting ground: K_f = 155 J. Hence option (d).

Q. 17. Which of the diagrams in Fig correctly shows the change in kinetic energy of an iron sphere falling freely in a lake having sufficient depth to impart it a terminal velocity?

Ans. (b)

Solution:
When an iron sphere falls through a fluid it initially accelerates (KE increases) until viscous drag (and buoyant forces) balance gravity, producing terminal velocity; beyond that KE is constant. The diagram that shows KE rising then leveling off is (b).

Q. 18. A cricket ball of mass 150 g moving with a speed of 126 km/h hits at the middle of the bat, held firmly at its position by the batsman.
The ball moves straight back to the bowler after hitting the bat.
Assuming that collision between ball and bat is completely elastic and the two remain in contact for 0.001s, the force that the batsman had to apply to hold the bat firmly at its place would be
(a) 10.5 N
(b) 21 N
(c) 1.05 ×10⁴ N
(d) 2.1 × 10⁴ N
Ans. 

Solution:
Mass m = 150 g = 0.150 kg. Initial speed u = 126 km/h = 35 m/s toward the bat. After elastic head-on rebound from a rigid bat held fixed, speed reverses to v = -35 m/s (same magnitude opposite direction). Change in momentum Δp = m(v - u) = 0.150 (-35 - 35) = 0.150 (-70) = -10.5 kg·m/s. Magnitude Δp = 10.5 kg·m/s. Contact time Δt = 0.001 s. Average force magnitude F = Δp / Δt = 10.5 / 0.001 = 1.05 × 10^4 N. So option (c).

Multiple Choice Questions - II

Q. 19. A man, of mass m, standing at the bottom of the staircase, of height L climbs it and stands at its top.
(a) Work done by all forces on man is equal to the rise in potential energy mgL.
(b) Work done by all forces on man is zero.
(c ) Work done by the gravitational force on man is mgL.
(d) The reaction force from a step does not do work because the point of application of the force does not move while the force exists.
Ans. (b,d)

Solution:
Work done by gravity: W_g = -mgL (it does negative work raising the man). Internal (muscular) forces do positive work mgL to raise the body. Sum of all works = mgL - mgL = 0. Reaction forces from the steps do no work because the point of contact on the step does not move relative to the step while the force exists. Friction does no net work in ideal step-climbing without slipping. Hence (b) and (d).

Q. 20 A bullet of mass m fired at 30° to the horizontal leaves the barrel of the gun with a velocity v. The bullet hits a soft target at a height h above the ground while it is moving downward and emerges out with half the kinetic energy it had before hitting the target.
Which of the following statements are correct in respect of bullet after it emerges out of the target?
 (a) The velocity of the bullet will be reduced to half its initial value.
 (b) The velocity of the bullet will be more than half of its earlier velocity.
 (c ) The bullet will continue to move along the same parabolic path.
 (d) The bullet will move in a different parabolic path.
 (e) The bullet will fall vertically downward after hitting the target.
 (f) The internal energy of the particles of the target will increase.
Ans. (b, d, f)

Solution:
If the bullet emerges with half its kinetic energy, then 1/2 m v_out^2 = 1/2 (1/2 m v_in^2) ⇒ v_out = v_in / √2, which is greater than v_in/2. Thus (a) is false, (b) true. Since the speed and therefore velocity change at impact, its subsequent trajectory will be a different parabola (d). It will not fall vertically (e is false). Energy lost by the bullet is transferred into the target (internal energy, heating, deformation), so (f) is true.

Q. 21. Two blocks M₁ and M₂ having equal mass are free to move on a horizontal frictionless surface. M₂ is attached to a massless spring as shown in Fig. Initially M₂ is at rest and M₁ is moving toward M₂ with speed v and collides head-on with M₂.
(a) While spring is fully compressed all the KE of M₁ is stored as PE of spring.
 (b) While spring is fully compressed the system momentum is not conserved, though final momentum is equal to initial momentum.
(c) If spring is massless, the final state of the M₁ is state of rest.
 (d) If the surface on which blocks are moving has friction, then collision cannot be elastic

Ans.(c, d)

Solution:
During compression part of M1's kinetic energy is stored as spring potential energy and part is transferred to M2's kinetic energy; therefore (a) is false. Momentum of the isolated two-body system is conserved at all times (no external horizontal force), so (b) is false. For equal masses in a head-on elastic collision, velocities exchange; if the spring is massless (ideal), M1 eventually comes to rest and M2 moves with initial speed v, so (c) is correct. If the surface has friction, some kinetic energy is dissipated and the collision cannot be perfectly elastic, so (d) is correct.

Very Short Answer Type Questions

Q. 22. A rough inclined plane is placed on a cart moving with a constant velocity u on horizontal ground. A block of mass M rests on the incline. Is any work done by force of friction between the block and incline? Is there then a dissipation of energy?
Ans. 

Because the cart moves with constant velocity, the block is at rest relative to the incline (no relative motion at contact). Static friction acts to prevent sliding, but the point of application of static friction does not move on the surface of the incline. Therefore the work done by friction on the block is zero and there is no dissipation of energy by that friction.

Q. 23. Why is electrical power required at all when the elevator is descending? Why should there be a limit on the number of passengers in this case?
Ans.

When an elevator descends under control, its motor (often working as a brake or generator or providing controlled torque) must prevent free fall and regulate speed. The motor therefore does (or absorbs) power to balance forces and control motion; electrical power may be required for active braking, control electronics, or to lift counterweights. Cables and motor system have a maximum safe tension; too many passengers increase load and maximum tension required, so a passenger limit ensures the motor and cable operate within safe limits.

Q. 24. A body is being raised to a height h from the surface of earth.
What is the sign of work done by
(a) Applied force
(b) Gravitational force?
Ans.

(a) Work done by the applied (upward) force: positive. W = F s cos0° = +Fs.
(b) Work done by gravitational force: negative. W = (-mg) s cos0° (force opposite to displacement) = -mg h.

Q. 25. Calculate the work done by a car against gravity in moving along a straight horizontal road. The mass of the car is 400 kg and the distance moved is 2m.
Ans.

The displacement is horizontal while weight acts vertically. Angle between weight and displacement is 90°, so work done by weight = mg s cos90° = 0. Therefore the work done by the car against gravity in moving horizontally is zero.

Q. 26. A body falls towards earth in air. Will its total mechanical energy be conserved during the fall? Justify.
Ans. 

If air resistance is negligible, mechanical energy (K + U) is conserved. If air resistance is present, it does negative work (non-conservative), converting some mechanical energy into internal energy (heat), so total mechanical energy of the body is not conserved during the fall. In that case gain in KE < loss in PE; the difference appears as heat and internal energy of the air and body.

Q. 27. A body is moved along a closed loop. Is the work done in moving the body necessarily zero? If not, state the condition under which work done over a closed path is always zero.
Ans. 

Work done over a closed loop is not necessarily zero. It is zero if the net force doing work is conservative (i.e., work depends only on endpoints). For conservative forces (gravitational, electrostatic, ideal spring), the work over any closed path is zero. For non-conservative forces (friction, viscous drag) the work over a closed path is not zero.

Q. 28. In an elastic collision of two billiard balls, which of the following quantities remain conserved during the short time of collision of the balls (i.e., when they are in contact)?
(a) Kinetic energy
(b) Total linear momentum
Give reason for your answer in each case.
Ans. 

Total linear momentum of the two-ball system is conserved during the brief collision because external forces are negligible compared to internal impulsive forces. Kinetic energy is conserved only if the collision is elastic; in general, during deformation some kinetic energy may be temporarily stored as elastic potential energy and then returned-if the collision is elastic the kinetic energy is conserved; if not (inelastic) kinetic energy is not conserved (part becomes heat, permanent deformation). The important point: total energy (including internal) is always conserved.

Q. 29. Calculate the power of a crane, in watts, which lifts a mass of 100 kg to a height of 10 m in 20 s.
Ans. 

Work done = m g h = 100 × 9.8 × 10 ≈ 9.8 × 10^3 J (if g = 9.8 m/s^2). Power = Work / time = 9800 / 20 = 490 W. (If g = 10 m/s^2 is used, W = 100 × 10 × 10 = 10^4 J and power = 500 W.)

Q. 30. The average work done by a human heart while it beats once is 0.5 J. Calculate the power used by heart if it beats 72 times in a minute.
Ans. 

Total work per minute = 72 × 0.5 = 36 J. Power = 36 J / 60 s = 0.6 W.

Q. 31. Give example of a situation in which an applied force does not result in a change in kinetic energy.
Ans.

Example: A mass on a string moving in a vertical or horizontal circle experiences tension directed toward the centre while instantaneous displacement is tangential. Tension is always perpendicular to instantaneous displacement, so work done by tension = 0 and it does not change kinetic energy (it provides centripetal force only).

Q. 32. Two bodies of unequal mass are moving in the same direction with equal kinetic energy. The two bodies are brought to rest by applying retarding force of same magnitude. How would the distance moved by them before coming to rest compare?
Ans. 

By work-energy theorem, initial kinetic energy = work done by retarding force = F d. Since the two bodies have equal initial kinetic energy and the same retarding force is applied, the displacement d before stopping is the same for both bodies. Thus both travel equal distances before coming to rest.

Q. 33. A bob of mass m suspended by a light string of length L is whirled into a vertical circle as shown in Fig. What will be the trajectory of the particle if the string is cut at
(a) Point B?
(b) Point C?
(c) Point X?

Ans. 

When the string is cut, no centripetal force acts and the bob moves under gravity with the instantaneous tangential velocity at the release point:

(a) If cut at B where the tangential velocity is vertically downward, the bob moves straight down (vertical motion under gravity).
(b) If cut at C where the tangential velocity is horizontal, it moves along a parabolic trajectory (projectile motion).
(c) If cut at X where the velocity makes an angle θ to the horizontal, it follows the corresponding projectile (parabolic) path determined by that initial speed and angle; it rises or falls according to the vertical component and then follows a parabola.

Short Answer Type Questions

Q. 34. A graph of potential energy V (x) verses x is shown in Fig. A particle of energy E₀ is executing motion in it. Draw graph of velocity and kinetic energy versus x for one complete cycle AFA.

Ans. 

(i) Kinetic energy versus x: Using energy conservation E0 = K(x) + V(x) ⇒ K(x) = E0 - V(x). At turning points where V(x) = E0, K = 0. At minima of V, K is maximum equal to E0 - Vmin. The KE graph is the vertical translation (downwards) of the V(x) curve by E0, clipped at zero at turning points.

At A and F (turning points): K = 0.
At C and D at potential minimum: K = E0 (maximum).
Graph of KE vs x is drawn accordingly.

(ii) Velocity v versus x: v(x) = ±√(2K/m) = ±√(2(E0 - V(x))/m). Velocity is zero at turning points and maximum where V is minimum. The v-x graph mirrors the KE behaviour with square-root scaling and sign indicating direction of motion in different halves of the cycle.

Q. 35. A ball of mass m, moving with a speed 2v₀, collides inelastically (e > 0) with an identical ball at rest. Show that
(a) For head-on collision, both the balls move forward.
(b) For a general collision, the angle between the two velocities of scattered balls is less than 90°.
Ans.

(a) For head on collision:
Conservation of momentum: 2 m v0 = m v1 + m v2 ⇒ 2 v0 = v1 + v2.
Using coefficient of restitution e, relations give v1 = v0 (1 - e) and v2 = v0 (1 + e). Since e < 1, v1 remains positive (same direction as initial v0). Hence both balls move forward after collision.

(b)
Vector momentum before collision p = 2 m v0 is equal to vector sum p1 + p2 after collision. For inelastic collisions some kinetic energy is lost, so |p|^2 > |p1|^2 + |p2|^2, which implies the angle θ between p1 and p2 is acute (cosθ > 0). Hence the angle between the two outgoing velocities is less than 90°.

Q. 36. Consider a one-dimensional motion of a particle with total energy E. There are four regions A, B, C and D in which the relation between potential energy V, kinetic energy (K) and total energy E is as given below:
Region A : V > E
Region B : V < E
Region C : K > E
Region D : V > K
State with reason in each case whether a particle can be found in the given region or not.
Ans. 

Region A: V > E - Forbidden region for classical motion because K = E - V would be negative. Particle cannot be found classically in A.
Region B: V < E - Allowed region; K = E - V > 0 so particle can be present.
Region C: K > E - Possible only if potential energy V is negative enough so that K = E - V > E; classically allowed if V < 0 and magnitude makes K > E. So yes, particle can be found if V is negative.
Region D: V > K - This is possible since V and K are not directly constrained except by E = K + V; particle can be found if E is large enough. More directly, V > K means V > E/2 in some cases - allowed if K = E - V ≥ 0.

Q. 37. The bob A of a pendulum released from horizontal to the vertical hits another bob B of the same mass at rest on a table as shown in Fig.

If the length of the pendulum is 1m, calculate
(a) the height to which bob A will rise after collision.
(b) the speed with which bob B starts moving.

Neglect the size of the bobs and assume the collision to be elastic.

Ans. 

Solution:
Initial potential energy of bob A released from horizontal (height h = L - L cos90° = L) relative to lowest point is m g L. At bottom just before collision, this converts to kinetic energy: 1/2 m v^2 = m g L ⇒ v = √(2 g L). With two identical masses and elastic head-on collision, velocities are exchanged: A comes to rest and B moves with speed v. Thus:

(a) Bob A rises to zero height after collision (it stops at bottom, v = 0), so it does not rise back up beyond bottom. (b) Speed of bob B after collision is v = √(2 g L). For L = 1 m and g ≈ 9.8 m/s^2, v = √(19.6) ≈ 4.43 m/s.

Q. 38. A raindrop of mass 1.00 g falling from a height of 1 km hits the ground with a speed of 50 m s^-1. Calculate
(a) the loss of P.E. of the drop.
(b) the gain in K.E. of the drop.
(c) Is the gain in K.E. equal to loss of P.E.? If not why.
Take g = 10 ms^-2
Ans. 

(a) Loss of potential energy = m g h = 1.00×10^-3 kg × 10 × 1000 m = 10 J.
(b) Kinetic energy at impact = 1/2 m v^2 = 0.5 × 1.00×10^-3 × (50)^2 = 0.5 × 10^-3 × 2500 = 1.25 J.
(c) The gain in KE (1.25 J) is not equal to loss of PE (10 J). The difference (~8.75 J) has been dissipated as work done against air resistance (viscous drag), heating the air and the drop-non-conservative forces remove mechanical energy during fall.

Q. 39. Two pendulums with identical bobs and lengths are suspended from a common support such that in rest position the two bobs are in contact (Fig). One of the bobs is released after being displaced by 10^o so that it collides elastically head-on with the other bob.

(a) Describe the motion of two bobs.
(b) Draw a graph showing variation in energy of either pendulum with time, for 0 ≤ t≤2T , where T is the period of each pendulum.
Ans. 

Solution:
On elastic head-on collision between identical spheres, velocities exchange. The released bob stops after collision and the other bob acquires the velocity and swings to nearly the same amplitude as the first had initially (neglecting losses). Energy of a given bob will show a transfer at the collision instant: its mechanical energy drops suddenly while the other bob's energy increases correspondingly. Over 0 ≤ t ≤ 2T, the energy plots show periodic exchange peaks corresponding to swing and collision instants. (Original figure  illustrates the energy exchange.)

Q. 40. Suppose the average mass of raindrops is 3.0 × 10^-5kg and their average terminal velocity 9 m s^-1. Calculate the energy transferred by rain to each square metre of the surface at a place which receives 100 cm of rain in a year.
Ans. 

Volume of rain per m^2 = 1 m^3 for 100 cm depth. Mass of rain per m^2 = ρ_water × volume = 10^3 kg/m^3 × 1 m^3 = 10^3 kg. Number of drops N = total mass / mass per drop = 10^3 / (3.0 × 10^-5) ≈ 3.333 × 10^7 drops. Energy per drop at terminal velocity: 1/2 m v^2 = 0.5 × 3.0×10^-5 × 9^2 = 0.5 × 3.0×10^-5 × 81 ≈ 1.215×10^-3 J per drop. Total energy ≈ N × 1.215×10^-3 ≈ 3.333×10^7 × 1.215×10^-3 ≈ 4.05×10^4 J per m^2 per year.

Q. 41. An engine is attached to a wagon through a shock absorber of length 1.5m. The system with a total mass of 50,000 kg is moving with a speed of 36 km h^-1 when the brakes are applied to bring it to rest. In the process of the system being brought to rest, the spring of the shock absorber gets compressed by 1.0 m. If 90% of energy of the wagon is lost due to friction, calculate the spring constant.
Ans.

Initial kinetic energy: v = 36 km/h = 10 m/s. K = 1/2 M v^2 = 0.5 × 50,000 × 100 = 2.5 × 10^6 J. 10% stored in spring = 0.1 × 2.5 × 10^6 = 2.5 × 10^5 J. For spring compression x = 1.0 m, stored energy = 1/2 k x^2 = 2.5 × 10^5 ⇒ k = 2 × (2.5 × 10^5) / (1^2) = 5.0 × 10^5 N/m. (Original solution gave k = 5 × 10^4 N/m due to a numeric mismatch; corrected value is 5.0 × 10^5 N/m.)

Q. 42. An adult weighing 600N raises the centre of gravity of his body by 0.25 m while taking each step of 1 m length in jogging. If he jogs for 6 km, calculate the energy utilised by him in jogging assuming that there is no energy loss due to friction of ground and air.
Assuming that the body of the adult is capable of converting 10% of energy intake in the form of food, calculate the energy equivalents of food that would be required to compensate energy utilised for jogging.
Ans. 

Number of steps in 6 km at 1 m per step = 6000 steps. Energy per step = weight × rise = 600 N × 0.25 m = 150 J. Total energy = 6000 × 150 = 9.0 × 10^5 J. If this is 10% of food energy intake, required food energy = 10 × 9.0 × 10^5 = 9.0 × 10^6 J.

Q. 43. On complete combustion a litre of petrol gives off heat equivalent to 3×10^7 J. In a test drive a car weighing 1200 kg. including the mass of driver, runs 15 km per litre while moving with a uniform speed on a straight track. Assuming that friction offered by the road surface and air to be uniform, calculate the force of friction acting on the car during the test drive, if the efficiency of the car engine were 0.5.
Ans. 

Energy available for work per litre = efficiency × chemical energy = 0.5 × 3×10^7 = 1.5×10^7 J used to overcome friction over 15 km = 1.5×10^4 m. Friction force F = Work / distance = 1.5×10^7 / 1.5×10^4 = 1000 N.

Long Answer Type Questions

Q. 44. A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of 30° by a force of 10 N parallel to the inclined surface (Fig).The coefficient of friction between block and the incline is 0.1. If the block is pushed up by 10 m along the incline, calculate

(a) work done against gravity

(b) work done against force of friction

(c) increase in potential energy

(d) increase in kinetic energy

(e) work done by applied force.

Ans. 

Given m = 1 kg, θ = 30°, d = 10 m, F_applied = 10 N, μ = 0.1, g = 10 m/s^2 (use given if needed).

(a) Work against gravity along incline: W_g = m g (vertical rise) = m g d sinθ = 1 × 10 × 10 × 1/2 = 50 J.

(b) Normal reaction N = m g cosθ = 1 × 10 × cos30° = 10 × 0.866 = 8.66 N. Friction force f = μ N = 0.1 × 8.66 = 0.866 N. Work against friction over 10 m: W_f = f d = 0.866 × 10 = 8.66 J.

(c) Increase in potential energy ΔU = m g d sinθ = 50 J (same as (a)).

(d) Net work = W_applied - (W_g + W_f) = m a d = ΔK. Compute net force along incline: F_net = F_applied - m g sinθ - f = 10 - (1 × 10 × 0.5) - 0.866 = 10 - 5 - 0.866 = 4.134 N. Acceleration a = F_net / m = 4.134 m/s^2. Using v^2 = u^2 + 2 a d with u = 0: v^2 = 2 × 4.134 × 10 = 82.68 ⇒ ΔK = 1/2 m v^2 = 1/2 × 1 × 82.68 = 41.34 J (or simply ΔK = F_net × d = 4.134 × 10 = 41.34 J).

(e) Work done by applied force W_applied = F_applied × d = 10 × 10 = 100 J.

Q. 45. A curved surface is shown in Fig. The portion BCD is free of friction. There are three spherical balls of identical radii and masses. Balls are released from rest one by one from A which is at a slightly greater height than C.

With the surface AB, ball 1 has large enough friction to cause rolling down without slipping; ball 2 has a small friction and ball 3 has a negligible friction.
(a) For which balls is total mechanical energy conserved?
 (b) Which ball (s) can reach D?
 (c) For balls which do not reach D, which of the balls can reach back A?
Ans. 

(a) For ball 3 (negligible friction) mechanical energy (translational kinetic + potential) is conserved. For ball 1 (pure rolling without slipping), total mechanical energy including rotational kinetic energy and translational KE is conserved (no dissipation). For ball 2 with small slipping friction, mechanical energy is not conserved (some energy dissipated as heat).

(b) Ball 3 (sliding without friction) can cross C to reach D because it converts maximum fraction of its initial potential to translational KE. Ball 1 loses some translational KE to rotational KE but total mechanical energy is conserved-whether it reaches D depends on numerical heights; with the given statement, ball 3 is the most likely to cross. Ball 2 loses energy to slipping and will not reach D.

(c) Among those that do not reach D, ball 1 (pure rolling) can return back toward A if its mechanical energy (including rotation) permits; ball 2, which dissipates energy, cannot return to original height and thus cannot reach A again. The original text states ball 1 has rotational energy and may not roll back to A due to friction direction; the qualitative conclusion is that dissipative losses prevent return to original height for ball 2.

Q. 46. A rocket accelerates straight up by ejecting gas downwards. In a small time interval ∆t, it ejects a gas of mass ∆m at a relative speed u. Calculate KE of the entire system at t + ∆t and t and show that the device that ejects gas does work = (1/2) ∆m u² in this time interval (neglect gravity).
Ans.

Solution (sketch):
Consider rocket mass M before ejection moving with speed v. After ejecting ∆m downward at relative speed u, the rocket mass becomes M - ∆m and its speed increases to v + ∆v. Use momentum conservation in inertial frame: (M) v = (M - ∆m) (v + ∆v) + ∆m (v - u + higher order), solve for ∆v to first order: (M) v ≈ (M - ∆m) (v + ∆v) + ∆m (v - u) ⇒ M v ≈ M v + M ∆v - ∆m v - (M - ∆m) ∆v + ∆m v - ∆m u ⇒ M ∆v ≈ ∆m u ⇒ ∆v = (∆m u)/M. Compute kinetic energies before and after and expand to first order in small quantities. The increase in kinetic energy of the whole system equals work done by the internal mechanism ejecting mass; algebra gives the work done on the gas equals (1/2) ∆m u^2 while the remainder transfers to rocket kinetic energy. Full derivation is in the figure .

Q. 47. Two identical steel cubes (masses 50 g , side 1cm) collide head-on face to face with a speed of 10cm/s each. Find the maximum compression of each. Young's modulus for steel = Y = 2 × 10¹¹ N/m 2.
Ans. 

Solution (outline):
Treat the faces in contact as springs using elastic deformation and Young's modulus. For a cube face of area A = (0.01 m)^2 = 1×10^-4 m^2 and length L = 0.01 m, an effective spring constant k for compression of each cube face is k = (Y A) / L. Compute k and convert initial kinetic energy (2 × 1/2 m v^2) into elastic potential 1/2 k (ΔL_total)^2, where ΔL_total is combined compression of both cubes. Solve for compression per cube (half of ΔL_total). Carrying out the numbers yields maximum compression on the order of 1.58×10^-7 m (as computed in the original text).

Q. 48. A balloon filled with helium rises against gravity increasing its potential energy. The speed of the balloon also increases as it rises. How do you reconcile this with the law of conservation of mechanical energy? You can neglect viscous drag of air and assume that density of air is constant.
Ans. 

As the balloon rises an equal volume of air descends to replace it. The buoyant force (ρ_air - ρ_He) V g does positive work on the balloon; this increases the balloon's mechanical energy (KE + PE). The energy gained by the balloon comes from the potential energy lost by the descending volume of air (and from the internal energy changes in the displaced air). In a closed account of both balloon and displaced air, total energy is conserved. The original derivation in the text shows that increase in balloon's KE + PE equals decrease in energy of the surrounding air; thus conservation of total energy holds.