Exercises — Work, Energy and Power

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5.13 marksNCERT Cl-11 Physics Part I, Ch5 Exercises, Q5.1
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative: (a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket, (b) work done by gravitational force in the above case, (c) work done by friction on a body sliding down an inclined plane, (d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity, (e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
Hint. Work is positive when force and displacement point the same way, and negative when they oppose each other — check the direction of each named force against the direction the body actually moves.
Step 1 — (a) Man lifting the bucket. The rope pulls the bucket upward, and the bucket moves upward — force and displacement agree, so this is positive.

Step 2 — (b) Gravity in the same case. Gravity pulls down while the bucket moves up — opposite directions, so this is negative.

Step 3 — (c) Friction on a body sliding down an incline. Friction always opposes relative motion, so it acts up the incline while the body slides down — opposite directions, negative.

Step 4 — (d) Applied force at uniform velocity on a rough surface. The applied force must point along the direction of motion to balance friction and keep the velocity constant — same direction, positive.

Step 5 — (e) Air resistance damping a pendulum. Resistive forces always oppose the instantaneous velocity, whichever way the pendulum happens to be swinging at that moment — negative.

✦ Answer: (a) positive (b) negative (c) negative (d) positive (e) negative.
Where students slip. Assuming (d) must be negative just because friction is present on the surface — the question asks about the work done by the applied force, not friction; the applied force itself still points along the motion, making its own work positive.
5.25 marksNCERT Cl-11 Physics Part I, Ch5 Exercises, Q5.2
A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the (a) work done by the applied force in 10 s, (b) work done by friction in 10 s, (c) work done by the net force on the body in 10 s, (d) change in kinetic energy of the body in 10 s, and interpret your results.
Hint. Find the friction force and the resulting net acceleration first, then use s = half a t squared to get the distance travelled in 10 s before computing any of the four work values.
Step 1 — Friction force and net acceleration. Friction = μmg = 0.1 × 2 × 9.8 = 1.96 N. Net force = 7 − 1.96 = 5.04 N. Acceleration a = 5.04/2 = 2.52 m/s².

Step 2 — Distance travelled in 10 s. Starting from rest: s = ½at² = ½(2.52)(100) = 126 m.

Step 3 — (a) Work by the applied force. W = 7 × 126 = 882 J.

Step 4 — (b) Work by friction. W = −1.96 × 126 ≈ −246.96 J, negative since friction opposes the motion.

Step 5 — (c) Work by the net force. W = 5.04 × 126 ≈ 635.04 J — this also equals (a) + (b), since the net force's work is simply the sum of the individual forces' work.

Step 6 — (d) Change in kinetic energy. By the work-energy theorem, ΔKE equals the work done by the net force: ΔKE ≈ 635.04 J, since the body started from rest.

✦ Answer: (a) 882 J (b) ≈ −246.96 J (c) ≈ 635.04 J (d) ≈ 635.04 J — the work-energy theorem confirms (c) and (d) must match exactly.
Where students slip. Computing (c) as simply the applied force's work minus the friction force's magnitude without tracking signs — friction's work is already negative, so the net force's work is (a) PLUS the (already negative) value from (b), not a fresh subtraction of magnitudes.
5.35 marksNCERT Cl-11 Physics Part I, Ch5 Exercises, Q5.3
Given in Fig. 5.11 are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.
Hint. A particle can only exist where its kinetic energy, E minus V(x), is non-negative — scan each panel for exactly where V(x) rises above the marked energy level E.
Step 1 — (i) A single step up to V0 at x = a. Since E is drawn below V0, KE = E − V0 < 0 for x ≥ a — forbidden there. For x < a, V = 0 < E, so KE = E > 0 — allowed. Minimum energy needed: 0 (matching the lowest value of V(x) in the allowed region). Physical context: a ball rolling toward a step or wall it lacks the energy to climb over.

Step 2 — (ii) A rising staircase of steps. E sits at the lowest plateau (the flat region for x < 0), with every higher step (V0 and beyond, up through the tallest step and even the lower step after it) sitting above E. So the particle is confined to the lowest plateau and forbidden from climbing any step. Minimum energy needed: the height of that lowest plateau. Physical context: a ball on level ground facing a flight of stairs too high for it to hop up onto even the first step.

Step 3 — (iii) A finite square well between a and b. For x < a and x > b, V = V0 > E — forbidden. Between a and b, V = −V1, and since E is above zero (hence above −V1), KE = E − (−V1) > 0 — allowed. The particle is trapped in the well. Minimum energy needed: −V1 (the bottom of the well). Physical context: a ball resting in a valley between two hills it cannot climb out of.

Step 4 — (iv) A well surrounded by decaying barriers. The central well (|x| < a/2) is allowed for the same reason as (iii). Between a/2 and b/2, the decaying curve stays above E throughout — forbidden. Beyond b/2, the curve has decayed below E — allowed again. So the particle is either trapped in the central well or free far away, but cannot sit in the barrier region between. Minimum energy needed: −V1. Physical context: the interaction between two nuclei — a short-range attractive well surrounded by a repulsive (Coulomb) barrier, relevant to nuclear fusion and alpha decay.

✦ Answer: (i) forbidden x≥a, min energy 0 (ii) confined to the lowest plateau (x<0), min energy = that plateau's height (iii) confined to a≤x≤b, min energy −V1 (iv) allowed only in |x|<a/2 or |x|>b/2, min energy −V1.
Where students slip. Forgetting that a flat region of V(x) exactly at the energy level E is a valid turning point, not automatically forbidden — the particle can still exist there (with momentarily zero kinetic energy at the very edge), which is exactly what marks the boundary of each allowed region.
5.44 marksNCERT Cl-11 Physics Part I, Ch5 Exercises, Q5.4
The potential energy function for a particle executing linear simple harmonic motion is given by V(x) = kx^2/2, where k is the force constant of the oscillator. For k = 0.5 N/m, the graph of V(x) versus x is shown in Fig. 5.12. Show that a particle of total energy 1 J moving under this potential must 'turn back' when it reaches x = +/- 2 m.
Hint. A particle 'turns back' exactly where all of its total energy has become potential energy, leaving zero kinetic energy — find where V(x) equals the given total energy.
Step 1 — Write out V(x) with the given k. V(x) = kx²/2 = 0.5x²/2 = 0.25x².

Step 2 — Evaluate V(x) at x = 2 m. V(2) = 0.25 × 2² = 0.25 × 4 = 1 J.

Step 3 — Compare with the total energy. Since the total energy is exactly 1 J, at x = 2 m the entire energy is potential, leaving KE = E − V(2) = 1 − 1 = 0 — the particle is momentarily at rest there. Beyond x = 2 m, V(x) exceeds 1 J, which would require negative kinetic energy — impossible — so the particle cannot go any further and must turn back. The same reasoning applies by symmetry at x = −2 m.

✦ Answer: Since V(±2) = 1 J exactly equals the total energy, the particle has zero kinetic energy there and must turn back — it can never be found beyond x = ±2 m for this energy.
Where students slip. Trying to find the turning point by setting some derivative to zero — the turning point of a particle's actual motion is found from where V(x) equals the total energy E, not from any calculus feature of the V(x) curve itself.
5.56 marksNCERT Cl-11 Physics Part I, Ch5 Exercises, Q5.5
Answer the following: (a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere? (b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet's velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why? (c) An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth? (d) In Fig. 5.13(i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig. 5.13(ii), he walks the same distance pulling the rope behind him, with a 15 kg mass hanging at its other end over a pulley. In which case is the work done greater?
Hint. For (c), remember that total mechanical energy and kinetic energy are not the same thing — a shrinking orbit can lose total energy while its kinetic energy alone still rises.
Step 1 — (a) Where the heat comes from. The heat is generated by friction acting on the rocket casing, which does negative work on the rocket — so the heat energy is obtained at the expense of the rocket's own kinetic energy, not the atmosphere's.

Step 2 — (b) Zero work over a full orbit. Gravity is a conservative force, and the defining property of any conservative force is that the work it does over a closed path — returning to the same starting point — is always exactly zero, regardless of how the force's direction varies with the comet's velocity along the way.

Step 3 — (c) Speeding up while losing energy. As drag removes total mechanical energy, the orbit shrinks and gravitational potential energy (which is negative and grows more negative as radius decreases) drops faster than the total energy itself decreases — so kinetic energy, being the difference between the two, must actually increase even as total energy falls.

Step 4 — (d) Comparing the two ways of moving 15 kg. In (i), the man's supporting force on the mass is vertical (upward), perpendicular to his horizontal walking displacement, so the work done on the mass is zero. In (ii), pulling the rope 2 m over the pulley lifts the hanging 15 kg mass by 2 m vertically, doing work W = mgh = 15 × 9.8 × 2 = 294 J.

✦ Answer: (a) the rocket's kinetic energy (b) gravity is conservative, and conservative forces always do zero net work over a closed loop (c) potential energy falls faster than total energy as the orbit shrinks, so kinetic energy rises (d) case (ii), about 294 J, versus 0 J for carrying the mass horizontally in (i).
Where students slip. In (d), assuming carrying the box in (i) must also do some positive work since the man is clearly exerting effort — the physics definition of work only counts the force component along the direction of displacement, and a purely vertical supporting force does zero work on a purely horizontal displacement, however tiring it feels.
5.64 marksNCERT Cl-11 Physics Part I, Ch5 Exercises, Q5.6
Underline the correct alternative: (a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered. (b) Work done by a body against friction always results in a loss of its kinetic/potential energy. (c) The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces on the system. (d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies.
Hint. For (c) and (d), remember that internal forces between the particles of a system always cancel out in pairs by Newton's third law, leaving only external effects and conserved totals to account for.
Step 1 — (a) Conservative force doing positive work. Since W = −ΔPE for a conservative force, positive work means potential energy decreases.

Step 2 — (b) Work against friction. Friction converts kinetic energy into heat, so working against it costs kinetic energy, not potential energy.

Step 3 — (c) Rate of change of total momentum. Internal forces between particles of the system cancel out in equal-and-opposite pairs, so only the external force can change the system's total momentum.

Step 4 — (d) What survives an inelastic collision. Kinetic energy is generally lost in an inelastic collision, but total linear momentum of the two-body system is always conserved, regardless of how elastic or inelastic the collision is.

✦ Answer: (a) decreases (b) kinetic (c) external force (d) total linear momentum.
Where students slip. Picking 'total energy' in (d) instead of 'total linear momentum' — while total energy (including heat and deformation) is technically always conserved too, the question is contrasting mechanical quantities, and momentum is the one that stays unchanged even as kinetic energy visibly drops.
5.75 marksNCERT Cl-11 Physics Part I, Ch5 Exercises, Q5.7
State if each of the following statements is true or false. Give reasons for your answer. (a) In an elastic collision of two bodies, the momentum and energy of each body is conserved. (b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present. (c) Work done in the motion of a body over a closed loop is zero for every force in nature. (d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.
Hint. For (a), check carefully whether the claim is about each individual body or about the system as a whole — these are two very different claims.
Step 1 — (a) Momentum and energy of each body. False. It is the TOTAL momentum and TOTAL kinetic energy of the two-body system that are conserved in an elastic collision — each individual body's own momentum and energy generally change during the collision.

Step 2 — (b) Total energy always conserved. True. Accounting for every form of energy (heat, sound, deformation, and so on), total energy is always conserved — it is only mechanical energy (KE + PE) alone that fails to be conserved when non-conservative forces like friction are present.

Step 3 — (c) Zero work over a closed loop for every force. False. This only holds for conservative forces. A non-conservative force like friction does negative work regardless of the direction of motion, so its net work over a closed loop is generally negative, not zero.

Step 4 — (d) Final KE always less in an inelastic collision. False. While KE typically decreases in an inelastic collision, there exist cases — such as an explosive separation or a spring-loaded release during the collision — where stored internal energy converts into kinetic energy, making the final KE greater than the initial KE.

✦ Answer: (a) False (b) True (c) False (d) False.
Where students slip. Assuming (d) must be true because 'inelastic' sounds like it always means energy is lost — inelastic strictly only means kinetic energy is not conserved (it changes), which includes the less commonly discussed case of it increasing, not only decreasing.
5.85 marksNCERT Cl-11 Physics Part I, Ch5 Exercises, Q5.8
Answer carefully, with reasons: (a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact)? (b) Is the total linear momentum conserved during the short time of an elastic collision of two balls? (c) What are the answers to (a) and (b) for an inelastic collision? (d) If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic?
Hint. Momentum conservation follows purely from Newton's third law holding at every instant of contact, while kinetic energy conservation is only guaranteed by comparing the states well before and well after the collision.
Step 1 — (a) KE during contact, elastic case. No. During the brief moment of contact, the balls compress slightly, and some kinetic energy is momentarily stored as elastic potential energy of deformation — it is only conserved when comparing the initial and final states, not at every instant during contact.

Step 2 — (b) Momentum during contact, elastic case. Yes. Momentum conservation follows directly from Newton's third law (equal and opposite mutual forces between the balls at every instant), so it holds throughout the contact, not just before and after.

Step 3 — (c) Same questions for an inelastic collision. For KE: still no, and even more so, since some KE is permanently lost to heat/deformation by the end as well. For momentum: still yes, since momentum conservation depends only on there being no external force, not on whether the collision is elastic.

Step 4 — (d) Potential energy depending only on separation. Elastic. If the interaction can be described entirely by a potential energy function of separation distance, the force is conservative — meaning all the energy stored as the balls approach and compress is fully returned as they separate again, with none lost, which is exactly the condition for an elastic collision.

✦ Answer: (a) No (b) Yes (c) same as (a) and (b) (d) elastic.
Where students slip. Assuming momentum can't be conserved during contact just because the balls are pushing on each other — those mutual pushes are exactly what Newton's third law guarantees are equal and opposite, which is precisely why the system's total momentum doesn't change even while the individual balls' momenta are actively changing.
5.92 marksNCERT Cl-11 Physics Part I, Ch5 Exercises, Q5.9
A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to (i) t^1/2 (ii) t (iii) t^3/2 (iv) t^2.
Hint. Write power as force times velocity, then substitute how velocity grows with time under constant acceleration starting from rest.
Step 1 — Express power. P = Fv = (ma)v, where both m and a are constant.

Step 2 — Substitute v(t) for constant acceleration from rest. v = at, so P = ma × at = ma²t.

Step 3 — Read off the proportionality. Since m and a are constants, P ∝ t.

✦ Answer: (ii) t.
Where students slip. Confusing this with the constant-power case (Q5.10) and answering t^1/2 — here acceleration is constant, not power; velocity grows linearly with time, which makes power grow linearly with time too.
5.103 marksNCERT Cl-11 Physics Part I, Ch5 Exercises, Q5.10
A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to (i) t^1/2 (ii) t (iii) t^3/2 (iv) t^2.
Hint. Since power P = mv(dv/dt) is now the constant, integrate this relation for v(t) first, then integrate again to get displacement.
Step 1 — Set up the constant-power relation. P = Fv = m(dv/dt)v = constant.

Step 2 — Integrate to find v(t). Rewriting as v dv = (P/m) dt and integrating from rest: v²/2 = (P/m)t, so v = √(2Pt/m) ∝ t^(1/2).

Step 3 — Integrate v(t) to find displacement. x = ∫v dt ∝ ∫t^(1/2) dt ∝ t^(3/2).

✦ Answer: (iii) t^3/2.
Where students slip. Reusing the Q5.9 result (power ∝ t) here — that was for constant acceleration with varying power; this question flips the setup to constant power with varying acceleration, which changes how v and x depend on t.
5.113 marksNCERT Cl-11 Physics Part I, Ch5 Exercises, Q5.11
A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by F = -i + 2j + 3k N, where i, j, k are unit vectors along the x-, y- and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis?
Hint. Only the component of a force along the actual direction of displacement does any work — the other two components are irrelevant here since the body is constrained to the z-axis.
Step 1 — Identify the relevant force component. The body only moves along z, so only F_z = 3 N contributes to the work done; the x- and y-components (−1 N and 2 N) do no work since there is no displacement in those directions.

Step 2 — Compute the work. W = F_z × d = 3 × 4 = 12 J.

✦ Answer: The work done is 12 J.
Where students slip. Computing the full magnitude of F and multiplying by 4 m — since the body is constrained to move only along z, the x- and y-components of the force are irrelevant to the work done, however large they are.
5.124 marksNCERT Cl-11 Physics Part I, Ch5 Exercises, Q5.12
An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton? Obtain the ratio of their speeds. (electron mass = 9.11x10^-31 kg, proton mass = 1.67x10^-27 kg, 1 eV = 1.60x10^-19 J).
Hint. Rearrange KE = half m v squared to solve for v, then take the ratio directly in terms of the given kinetic energies and masses rather than computing each speed from scratch.
Step 1 — General expression for speed from kinetic energy. v = √(2·KE/m).

Step 2 — Set up the ratio of speeds. v_e/v_p = √[(KE_e/KE_p) × (m_p/m_e)] = √[(10/100) × (1.67×10⁻²⁷/9.11×10⁻³¹)].

Step 3 — Compute the ratio. m_p/m_e ≈ 1833, so v_e/v_p = √(0.1 × 1833) = √183.3 ≈ 13.5.

✦ Answer: The electron is faster, moving about 13.5 times as fast as the proton, even though its kinetic energy is ten times smaller.
Where students slip. Assuming the proton must be faster since it carries ten times the kinetic energy — kinetic energy depends on mass too, and the proton's mass is about 1833 times larger, which more than cancels its tenfold energy advantage.
5.135 marksNCERT Cl-11 Physics Part I, Ch5 Exercises, Q5.13
A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 m/s?
Hint. Gravity's work only depends on mass and the vertical distance fallen, never on speed — so both halves of a 500 m fall (250 m each) get exactly the same gravitational work, regardless of what the resistive force is doing.
Step 1 — Find the raindrop's mass. Volume = (4/3)πr³ = (4/3)π(2×10⁻³)³ ≈ 3.351×10⁻⁸ m³. Mass = density × volume ≈ 1000 × 3.351×10⁻⁸ ≈ 3.351×10⁻⁵ kg.

Step 2 — Work by gravity in each half. Each half is 250 m, so W = mgh = 3.351×10⁻⁵ × 9.8 × 250 ≈ 0.082 J — the same value in both halves, since gravity's work depends only on the vertical drop, not on the speed profile.

Step 3 — Apply the work-energy theorem over the whole 500 m journey. Total work by gravity = mg × 500 ≈ 0.164 J. Final KE = ½mv² = ½ × 3.351×10⁻⁵ × 10² ≈ 1.676×10⁻³ J. Initial KE = 0 (drop starts at rest).

Step 4 — Solve for the work done by the resistive force. W_gravity + W_resistive = ΔKE, so W_resistive = 1.676×10⁻³ − 0.164 ≈ −0.163 J.

✦ Answer: Work by gravity ≈ 0.082 J in each half of the journey; work by the resistive force over the entire journey ≈ −0.163 J.
Where students slip. Assuming the work done by gravity must differ between the two halves because the drop's speed behaves so differently in each — gravity's work formula (mgh) never involves speed at all, only the vertical distance fallen, which is identical (250 m) in both halves.
5.143 marksNCERT Cl-11 Physics Part I, Ch5 Exercises, Q5.14
A molecule in a gas container hits a horizontal wall with speed 200 m/s and angle 30 degrees with the normal, and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?
Hint. Think about which momentum is really being asked about — the molecule's alone, or the whole molecule-plus-wall system.
Step 1 — Momentum conservation. The molecule's own momentum does change (its normal component reverses on rebound), but for the system as a whole (molecule and wall together), momentum is conserved — the wall absorbs an equal and opposite change in momentum, imperceptibly, owing to its enormous relative mass.

Step 2 — Elastic or inelastic? Since the molecule rebounds with exactly the same speed, its kinetic energy is unchanged — no kinetic energy is lost, which is precisely the condition for an elastic collision.

✦ Answer: Momentum is conserved for the molecule-wall system as a whole; the collision is elastic, since speed (and hence kinetic energy) is unchanged.
Where students slip. Concluding momentum is not conserved just because the molecule's own momentum visibly changes — conservation of momentum is a statement about the whole interacting system, not about any one participant in isolation.
5.154 marksNCERT Cl-11 Physics Part I, Ch5 Exercises, Q5.15
A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m^3 in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump?
Hint. Find the useful mechanical power delivered to the water first, then divide by the efficiency to get the actual electrical power drawn — efficiency always makes the required input larger than the useful output.
Step 1 — Find the mass of water and the useful work done. Mass = 30 m³ × 1000 kg/m³ = 30,000 kg. Useful work = mgh = 30,000 × 9.8 × 40 = 1.176×10⁷ J.

Step 2 — Find the useful power output. Time = 15 min = 900 s. Useful power = 1.176×10⁷/900 ≈ 13,067 W.

Step 3 — Account for the pump's efficiency. Since only 30% of the electrical input becomes useful work, electric power consumed = useful power / 0.30 ≈ 13,067/0.30 ≈ 43,556 W.

✦ Answer: The pump consumes about 4.36×10⁴ W (≈ 43.6 kW) of electric power.
Where students slip. Multiplying the useful power by 0.30 instead of dividing by it — an inefficient pump needs to draw MORE electrical power than the useful work it delivers, not less, since some of what it draws is wasted.
5.164 marksNCERT Cl-11 Physics Part I, Ch5 Exercises, Q5.16
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. If the collision is elastic, which of the following (Fig. 5.14) is a possible result after collision?
Hint. Check each of the three pictured outcomes against BOTH conservation laws that an elastic collision must satisfy — momentum and kinetic energy — not just one of them.
Step 1 — Set up the check for each option. Initial momentum = mV, initial KE = ½mV². Every option must match both values exactly.

Step 2 — Option (i): balls 2 and 3 move together at V/2, ball 1 at rest. Momentum = 2m(V/2) = mV ✓. KE = ½(2m)(V/2)² = mV²/4 — only half the initial KE, so this fails energy conservation.

Step 3 — Option (ii): balls 1 and 2 at rest together, ball 3 moves at V. Momentum = m(V) = mV ✓. KE = ½mV² — exactly matches the initial KE ✓. Both conservation laws hold.

Step 4 — Option (iii): all three balls move together at V/3. Momentum = 3m(V/3) = mV ✓. KE = ½(3m)(V/3)² = mV²/6 — far less than the initial KE, so this fails energy conservation too.

✦ Answer: Only option (ii) is possible for an elastic collision — balls 1 and 2 end up at rest while ball 3 flies off with the full original speed V, exactly like a Newton's cradle.
Where students slip. Accepting an option just because momentum checks out — momentum is conserved in EVERY one of the three options here (any shared final velocity does that automatically), so it's the kinetic-energy check that actually distinguishes the genuinely elastic outcome from the other two.
5.173 marksNCERT Cl-11 Physics Part I, Ch5 Exercises, Q5.17
The bob A of a pendulum released from 30 degrees to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 5.15. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.
Hint. Recall what happens in a one-dimensional elastic collision between two equal masses where one starts at rest.
Step 1 — Apply the equal-mass elastic collision result. For a head-on elastic collision between two bodies of equal mass, where one is initially at rest, the incoming body comes to a complete stop, and the one at rest picks up the full initial velocity — this is a standard, general result of solving the momentum and energy equations together for equal masses.

Step 2 — Apply it here. Bob A (moving) transfers all its velocity to bob B (at rest) and itself ends up with zero velocity immediately after the collision.

Step 3 — What that means for bob A's rise. With zero velocity right after the collision, bob A has no kinetic energy left to convert into height, so it simply stays at the bottom, since there's nothing left to lift it back up.

✦ Answer: Bob A does not rise at all after the collision — it comes to a complete stop.
Where students slip. Assuming some height must be regained since the collision is elastic (no energy 'lost') — the energy isn't lost, but for equal masses it is transferred entirely to bob B, leaving bob A with none of its own to rise on.
5.184 marksNCERT Cl-11 Physics Part I, Ch5 Exercises, Q5.18
The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance?
Hint. Released from horizontal, the bob's initial height above the lowest point equals the pendulum's full length — then account for the 5% lost to air resistance before converting the rest to kinetic energy.
Step 1 — Height dropped. Released from horizontal, the bob starts level with the pivot, so it falls a height equal to the pendulum's length: h = 1.5 m.

Step 2 — Available energy at the bottom. Only 95% of the initial potential energy mgh converts to kinetic energy at the lowest point, the rest being dissipated: KE = 0.95mgh.

Step 3 — Solve for speed. ½mv² = 0.95mgh, so v² = 2(0.95)(9.8)(1.5) = 27.93, giving v ≈ 5.28 m/s.

✦ Answer: The bob arrives at the lowest point with a speed of about 5.28 m/s.
Where students slip. Using the full height drop with no energy loss (the familiar v = √(2gh) formula) — that formula assumes no dissipation at all, but 5% of the energy here is explicitly lost to air resistance before the bob reaches the bottom.
5.194 marksNCERT Cl-11 Physics Part I, Ch5 Exercises, Q5.19
A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.05 kg/s. What is the speed of the trolley after the entire sand bag is empty?
Hint. Ask what horizontal velocity the leaking sand carries with it as it falls out — that determines whether the trolley experiences any horizontal recoil from the leak at all.
Step 1 — What happens to the leaking sand's momentum. As sand leaks out through a hole in the floor, it falls out carrying the SAME horizontal velocity the trolley had at that instant — it isn't ejected sideways or backward, so it removes no horizontal momentum relative to what the trolley system already had per unit mass.

Step 2 — Why this means no horizontal force acts on the trolley. Since the escaping sand exerts no horizontal reaction force on the trolley (unlike, say, a rocket ejecting fuel backward at high relative speed), there is nothing to change the trolley's horizontal velocity, even as its total mass decreases.

✦ Answer: The trolley's speed remains unchanged at 27 km/h even after the entire sandbag has leaked out.
Where students slip. Treating this like a rocket-recoil problem and expecting the trolley to speed up as it loses mass — that effect only happens when the departing mass is forcefully ejected at a different velocity than the main body; here the sand simply falls out at the trolley's own current speed, causing no recoil at all.
5.204 marksNCERT Cl-11 Physics Part I, Ch5 Exercises, Q5.20
A body of mass 0.5 kg travels in a straight line with velocity v = a x^3/2 where a = 5 m^-1/2 s^-1. What is the work done by the net force during its displacement from x = 0 to x = 2 m?
Hint. Rather than integrating force over distance directly, use the work-energy theorem: work done by the net force always equals the change in kinetic energy, which only needs the velocities at the two endpoints.
Step 1 — Find v² directly, avoiding an extra square-root step. v = ax^(3/2), so v² = a²x³ = 25x³ (this form is more convenient than computing v itself).

Step 2 — Evaluate at the endpoints. At x = 0: v² = 0. At x = 2: v² = 25 × 2³ = 25 × 8 = 200 m²/s².

Step 3 — Apply the work-energy theorem. W = ΔKE = ½m(v_f² − v_i²) = ½ × 0.5 × (200 − 0) = 0.25 × 200 = 50 J.

✦ Answer: The net force does 50 J of work.
Where students slip. Computing v itself (taking a square root) before squaring it again for the kinetic energy formula — since only v² is ever needed, it's simpler and less error-prone to compute v² = a²x³ directly and skip the detour through v.
5.215 marksNCERT Cl-11 Physics Part I, Ch5 Exercises, Q5.21
The blades of a windmill sweep out a circle of area A. (a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t? (b) What is the kinetic energy of the air? (c) Assume that the windmill converts 25% of the wind's energy into electrical energy, and that A = 30 m^2, v = 36 km/h and the density of air is 1.2 kg/m^3. What is the electrical power produced?
Hint. Picture the air that passes through the circle in time t as a cylinder of length v times t and cross-sectional area A — its volume gives the mass once multiplied by air density.
Step 1 — (a) Mass of air passing through in time t. The air sweeps out a cylinder of length vt and cross-section A, so volume = Avt, and mass = ρAvt.

Step 2 — (b) Kinetic energy of that air. KE = ½(mass)v² = ½(ρAvt)v² = ½ρAv³t.

Step 3 — (c) Convert to power and apply the given numbers. Power (before efficiency) = KE/t = ½ρAv³. With v = 36 km/h = 10 m/s: Power = ½ × 1.2 × 30 × 10³ = ½ × 1.2 × 30 × 1000 = 18,000 W.

Step 4 — Apply the 25% conversion efficiency. Electrical power = 0.25 × 18,000 = 4500 W.

✦ Answer: (a) ρAvt (b) ½ρAv³t (c) 4500 W (4.5 kW) of electrical power.
Where students slip. Forgetting to convert 36 km/h to 10 m/s before plugging into the power formula — mixing km/h with the SI units used for A (m²) and ρ (kg/m³) would give a power value off by a large, wrong factor.
5.225 marksNCERT Cl-11 Physics Part I, Ch5 Exercises, Q5.22
A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated. (a) How much work does she do against the gravitational force? (b) Fat supplies 3.8x10^7 J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?
Hint. Multiply the work done in a single lift by the number of repetitions first, then work backward through the 20% efficiency to find how much fat-energy (and hence fat mass) that work actually cost her.
Step 1 — (a) Total work against gravity. Each lift does mgh = 10 × 9.8 × 0.5 = 49 J of work. Over 1000 lifts: W = 1000 × 49 = 49,000 J = 4.9×10⁴ J.

Step 2 — (b) Energy required from fat. Since only 20% of the fat's chemical energy converts to usable mechanical work, the energy that must be drawn from fat is larger than the mechanical work itself: Energy from fat = 4.9×10⁴ / 0.20 = 2.45×10⁵ J.

Step 3 — Convert that energy to a mass of fat. Fat used = 2.45×10⁵ / 3.8×10⁷ ≈ 6.45×10⁻³ kg (about 6.45 g).

✦ Answer: (a) 4.9×10⁴ J of work against gravity (b) about 6.45×10⁻³ kg (≈ 6.45 g) of fat used up.
Where students slip. Dividing by 3.8×10⁷ before accounting for the 20% efficiency — the mechanical work alone understates how much fat-energy was actually drawn on, since 80% of whatever energy is pulled from fat never becomes useful mechanical work at all.
5.234 marksNCERT Cl-11 Physics Part I, Ch5 Exercises, Q5.23
A family uses 8 kW of power. (a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW? (b) Compare this area to that of the roof of a typical house.
Hint. Find how much usable electrical power a single square metre of solar collection actually provides before dividing that into the family's total requirement.
Step 1 — Usable power per square metre. Only 20% of the incident 200 W/m² becomes electrical power: 0.20 × 200 = 40 W/m².

Step 2 — (a) Required area. Area = total power needed / power per m² = 8000 / 40 = 200 m².

Step 3 — (b) Comparing to a typical house roof. 200 m² is comparable to, and in many cases larger than, the entire roof area of a typical house, since a modest single-family house roof is often itself only on the order of 150-250 m². This shows that meeting a household's full power demand from rooftop solar at this efficiency and insolation would need close to (or more than) the whole available roof, not just a small corner of it.

✦ Answer: About 200 m² of collecting area is needed — roughly comparable to an entire typical house roof, illustrating how much surface solar power realistically demands at these efficiencies.
Where students slip. Assuming 200 m² is a trivially small area compared to a house roof — it is actually on the same order as, or larger than, a typical roof's total area, which is the whole point of the comparison in part (b).

Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Physics Part I textbook, Reprint 2026-27 (keph105.pdf) — one end-of-chapter Exercises set (23 questions, 5.1-5.23); all five figures (5.11's four potential-energy panels, 5.12, 5.13, 5.14, 5.15) were rendered directly from the PDF at up to 900dpi and read visually before answering, including a careful pixel-level check of exactly where the marked total-energy level E sits relative to each potential-energy step in Fig 5.11. Questions are referenced from the NCERT textbook for identification.

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