Class 11 · Physics · Chapter 5
Work, Energy and Power
This chapter builds work, kinetic energy and potential energy from the dot product of force and displacement, proves the work-energy theorem, and uses conservation of mechanical energy to solve motion problems without touching Newton's laws directly. It is not board-examined in Class 11 but is the foundation for Class 12 boards, JEE and NEET, so every derivation here is worth mastering properly.
Before you startWhat the board asks from this chapter
No Class 11 board exam exists, so this chapter carries school-final marks only (commonly 8-12 marks in most school schemes) and is not weighted by CBSE at this stage. Its real weight shows up later: it underlies energy questions in Class 12 boards and is one of the highest-yield chapters in JEE Main/Advanced and NEET.
| Question type | Marks | Usually asked |
|---|---|---|
| MCQ | 1 | Sign of work done, nature of a force (conservative vs non-conservative), units and dimensions of power, graph-based questions on F-x or P-t. |
| Assertion–Reason | 1 | Work done by centripetal/normal/tension force being zero, KE always positive but momentum can be negative, elastic vs inelastic collision statements. |
| Very short answer | 2 | State the work-energy theorem, define conservative force, distinguish elastic from inelastic collision, give SI unit and dimension of power. |
| Short answer | 3 | Derive work-energy theorem, relate power to force and velocity, small numericals combining KE, PE and work done by friction. |
| Long answer | 5 | Full derivation of elastic collision in one dimension (v1′ and v2′), potential energy of a spring with graph, law of conservation of energy applied to a swinging/falling body. |
| Case-based | 4 | A block on an incline or a pendulum/spring scenario with sub-questions testing work, KE-PE conversion and power. |
Hot topics, in order
- Work-energy theorem (derivation)
- Elastic collision in one dimension (derivation)
- Potential energy of a spring
- Conservative vs non-conservative forces
- Law of conservation of energy
- Power = F·v
- Vertical circular motion (minimum speed at top)
- Coefficient of restitution
The whole chapter, in orderNotes
01 5.1 Introduction
The chapter opens by fixing the everyday meaning of 'work' against the physics meaning. In physics, work is done only when a force produces a displacement in the direction of the force (or a component of it). A person holding a heavy suitcase and standing still does zero physics-work, however tired the muscles get, because there is no displacement. The chapter builds three ideas in sequence: work as a scalar defined by a dot product, energy as the capacity to do work (kinetic energy for motion, potential energy for position/configuration), and power as the rate of doing work. These three, along with the law of conservation of energy, are used throughout mechanics, so the definitions must be stated exactly as NCERT gives them, not loosely.
Work, energy and power are all scalar quantities even though force, displacement and velocity (which define them) are vectors. This scalar nature is what makes the work-energy theorem and conservation of energy such powerful problem-solving tools — you can add and subtract energies without worrying about direction.
02 5.2 Notions of Work and Kinetic Energy: The Work-Energy Theorem PYQ CBSE 2020
NCERT introduces work-energy ideas informally first through three familiar situations (a moving cricket ball hitting a stationary wicket, a moving trolley pushed by a mass, a vehicle braking on the road) before formalising them in §5.3–5.4. Each example shows that the change in kinetic energy of a body equals the work done on it by the net force — this is the seed of the work-energy theorem proved rigorously later. The theorem is stated as: work done by the net force on a body equals the change in its kinetic energy, W = ΔKE = KEf − KEi. It holds for both constant and variable forces and is one of the most repeated results in the whole of mechanics.
These qualitative examples are important because they let you reason about direction of force versus motion without a formula: braking force acts opposite to velocity so it does negative work and reduces KE; a falling object has gravity doing positive work and gaining KE.
03 5.3 Work PYQ CBSE 2019 · CBSE 2023
For a constant force F acting on a body while it undergoes a displacement d, work done is defined as the scalar (dot) product: W = F·d = F d cosθ, where θ is the angle between the force and the displacement. Because it is a dot product, work is a scalar — it has magnitude but no direction, even though F and d are vectors.
Three special cases matter for exams. When θ = 0° (force along displacement), W = Fd, maximum positive work. When θ = 90° (force perpendicular to displacement), W = 0 — this is why the work done by centripetal force in circular motion, by the normal reaction on a body moving along a surface, and by tension in a string whirling a stone, are all zero. When θ = 180° (force opposite to displacement), W = −Fd, negative work — as in friction opposing motion or gravity acting on a rising object.
For a variable force, work is computed by summing (integrating) F·dx over small displacements: W = ∫F dx from initial to final position. Graphically, this is the area under the F–x curve, with area below the x-axis counted as negative.
04 5.4 Kinetic Energy PYQ CBSE 2018
Kinetic energy is the energy possessed by a body by virtue of its motion. For a body of mass m moving with speed v, KE = ½mv². It is a scalar and is always positive (or zero), regardless of the direction of velocity — this is a favourite assertion-reason trap since momentum (mv) can be negative but KE (½mv²) cannot.
Using the equation of motion v² = u² + 2as for a body accelerated by a constant force F = ma from speed u to speed v over displacement s, work done by the force is W = Fs = mas = m(v² − u²)/2 s /s... more directly, W = ½mv² − ½mu² = ΔKE. This is the formal proof of the work-energy theorem for a constant force. NCERT extends this proof to a variable force using calculus (W = ∫F dx = ∫m(dv/dt)dx = ∫mv dv = ½mv² − ½mu²), which is the version most often asked as a 3–5 mark derivation.
05 5.5 Work Done by a Variable Force
Most real forces (a spring force, a varying push) change with position, so the simple W = Fd cosθ does not apply directly. NCERT's method: divide the total displacement into a large number of tiny intervals Δxj, treat the force as constant (Fj) over each tiny interval, sum ΣFjΔxj, and take the limit as Δx → 0. This limit is the definite integral W = ∫xixf F(x) dx, and geometrically it is the area under the force–displacement curve between the initial and final positions.
This is the general definition of work; the constant-force formula and the spring's PE formula (§5.9) are both special cases of it. In numericals, when F(x) is given as an algebraic expression (e.g. F = kx or F = a + bx), integrate directly; when F is given as a graph, count the enclosed area (using simple shapes — triangles, rectangles, trapeziums).
06 5.6 The Work-Energy Theorem for a Variable Force PYQ CBSE 2020 · CBSE 2022
This is the rigorous, most-asked derivation of the chapter. Start from Newton's second law F = m(dv/dt). Work done over a small displacement dx is dW = F dx = m(dv/dt) dx = m(dx/dt) dv = mv dv (since dx/dt = v). Integrating both sides from the initial speed u to the final speed v:
W = ∫uv mv dv = m[v²/2]uv = ½mv² − ½mu²
The right-hand side is exactly ΔKE, so W = ΔKE = KEf − KEi. This proves that the work-energy theorem, discovered for a constant force in §5.4, holds equally for a variable force — this generality is why the theorem is used so widely, including for springs and other non-uniform forces later in the chapter. Note that W here means the work done by the net (resultant) force on the body.
07 5.7 The Concept of Potential Energy PYQ CBSE 2021
Potential energy is the energy stored in a body or system by virtue of its position or configuration, associated only with conservative forces (gravity, spring force, electrostatic force) — it cannot be defined for a non-conservative force like friction, because the work done by friction depends on the actual path taken, not just the end points. For a conservative force, the work done in moving a body from one point to another is independent of the path and depends only on the initial and final positions; the negative of this work is defined as the change in potential energy: W = −ΔPE i.e. ΔPE = −W.
The most familiar case is gravitational PE near Earth's surface: raising a body of mass m through height h against gravity, work done by the person = mgh, so PE gained = mgh (taking PE = 0 at the reference/ground level). PE is defined only up to an arbitrary additive constant — only differences (ΔPE) have physical meaning, which is why a reference level must always be chosen and stated.
08 5.8 The Conservation of Mechanical Energy PYQ CBSE 2019 · CBSE 2022 · CBSE 2024
For a body moving under a conservative force only (no friction, no air resistance), the total mechanical energy — kinetic plus potential — remains constant at every point of the motion: KE + PE = constant, i.e. KEi + PEi = KEf + PEf. The proof follows directly from combining the work-energy theorem (W = ΔKE) with the definition of PE (W = −ΔPE): equating the two gives ΔKE = −ΔPE, i.e. ΔKE + ΔPE = 0, so KE + PE stays constant.
The classic application is a body falling freely from height H. At the top, KE = 0, PE = mgH. At height h during the fall, KE = mg(H−h) [from v² = 2g(H−h)], PE = mgh, and their sum is always mgH. At the ground, all PE has converted to KE = mgH. This is tested heavily with a pendulum, a body sliding down a frictionless incline, or a block on a spring.
If a non-conservative force like friction also acts, mechanical energy is not conserved — the 'lost' mechanical energy appears as heat/sound (total energy, including these forms, is still conserved, which is the deeper law in §5.10).
09 5.9 The Potential Energy of a Spring PYQ CBSE 2018 · CBSE 2023
A spring obeys Hooke's law: the restoring force is F = −kx, where x is the displacement of the free end from the natural (unstretched) length and k is the spring (force) constant (unit N m−1), with the negative sign showing the force always opposes the displacement. The applied (external) force needed to stretch or compress it slowly is Fapplied = +kx.
Work done in stretching the spring from x = 0 to x = xm is W = ∫0xm kx dx = ½kxm². Since the spring force is conservative, this work is stored as elastic potential energy: PE(x) = ½kx². This PE is the same whether the spring is stretched or compressed by x (it depends on x², so it's always positive), and graphically PE(x) vs x is a parabola while F(x) vs x is a straight line through the origin with negative slope −k.
At maximum extension xm, if the spring-mass system oscillates, KE = 0 and PE = ½kxm² (all energy is potential); at x = 0 (natural length), PE = 0 and KE is maximum, equal to ½kxm² by conservation.
10 5.10 Various Forms of Energy: The Law of Conservation of Energy PYQ CBSE 2021
Beyond mechanical energy, NCERT surveys other forms — heat energy (from molecular motion, measured via the mechanical equivalent of heat, 4.186 J = 1 calorie), chemical energy (stored in molecular bonds, released or absorbed in reactions, e.g. combustion), electrical energy, and the mass-energy equivalence E = mc² from special relativity (nuclear energy, where a small mass defect releases enormous energy, e.g. nuclear fission and fusion, and the Sun's energy source).
The law of conservation of energy is the deepest generalisation of this chapter: energy can only be transformed from one form to another; it can neither be created nor destroyed. The total energy of an isolated system remains constant. This is far more general than conservation of mechanical energy — it holds even when friction, heat and other dissipative or non-mechanical processes are present, because the 'lost' mechanical energy simply reappears as heat, sound or other forms, keeping the grand total unchanged.
11 5.11 Power PYQ CBSE 2020
Power is the time rate of doing work, i.e. the rate of transfer of energy. Average power over a time interval Δt is Pav = W/Δt. Instantaneous power is the limit as Δt → 0: P = dW/dt. Since dW = F·dr, instantaneous power can also be written as P = F·v, the dot product of the applied force and the instantaneous velocity — a very useful form for numericals involving vehicles, pumps and motors.
SI unit of power is the watt (W), 1 W = 1 J s−1. A common commercial (non-SI) unit is the horsepower, 1 hp = 746 W. Dimensional formula of power is [M L² T−3]. The commercial unit of energy, the kilowatt-hour (kWh, colloquially a 'unit' of electricity), equals 3.6 × 10⁶ J.
12 5.12 Collisions PYQ CBSE 2019 · CBSE 2022 · CBSE 2024
A collision is an event where two (or more) bodies exert strong mutual forces for a short time, changing their velocities. Momentum is always conserved in every collision (no external force acts during the brief collision interval). Collisions are classified by whether kinetic energy is also conserved.
Elastic collision: both momentum and kinetic energy are conserved; bodies separate after collision without any permanent deformation or heat loss (an idealisation, e.g. collisions between atoms/billiard balls approximately). Inelastic collision: momentum is conserved but kinetic energy is not — some KE converts to heat, sound or deformation. In a perfectly inelastic collision, the colliding bodies stick together and move with a common velocity after impact, and the KE loss is maximum for the given momentum.
1-D elastic collision derivation (mass m1 with initial velocity u1 strikes stationary m2, u2=0): conservation of momentum gives m1u1 = m1v1 + m2v2; conservation of KE gives ½m1u1² = ½m1v1² + ½m2v2². Solving these simultaneously (rearranging each as a difference of squares and dividing) yields v1 = [(m1−m2)/(m1+m2)]u1 and v2 = [2m1/(m1+m2)]u1. Special cases: if m1 = m2, v1 = 0 and v2 = u1 (complete transfer of velocity, as in Newton's cradle); if m2 ≫ m1, v1 ≈ −u1 (bounces back) and v2 ≈ 0.
For a perfectly inelastic collision, momentum conservation alone gives the common final velocity: v = m1u1/(m1+m2) when m2 is initially at rest.
VocabularyKey terms
Keep this on one pageFormula sheet
| Name | Expression | Notes |
|---|---|---|
| Work done by a constant force | W = Fd cosθ | θ is the angle between force F and displacement d; θ=90° gives W=0 |
| Work by a variable force | W = ∫F(x) dx | equals the area under the F–x graph between initial and final position |
| Kinetic energy | KE = ½mv2 | always positive; SI unit joule (J) |
| Work-energy theorem | Wnet = ΔKE = ½mv2 − ½mu2 | holds for constant and variable net force |
| Gravitational PE | PE = mgh | h measured from a chosen zero-PE reference level |
| Spring PE | PE(x) = ½kx2 | x is displacement from the spring's natural length; same for stretch or compression |
| Spring restoring force (Hooke's law) | F = −kx | k in N m−1; negative sign shows force opposes displacement |
| Conservation of mechanical energy | KE + PE = constant | valid only when solely conservative forces act (no friction/drag) |
| Average power | Pav = W/Δt | SI unit watt; 1 hp = 746 W |
| Instantaneous power | P = F·v | dot product of applied force and instantaneous velocity |
| 1-D elastic collision final velocities (m<sub>2</sub> at rest) | v1=[(m1−m2)/(m1+m2)]u1, v2=[2m1/(m1+m2)]u1 | derived from simultaneous conservation of momentum and KE |
| Perfectly inelastic collision common velocity | v = m1u1/(m1+m2) | m2 initially at rest; from momentum conservation alone |
| Mass-energy equivalence | E = mc2 | basis of nuclear energy release; c = 3 × 108 m s−1 |
Practise with a pencilDiagrams that earn marks
Force–displacement (F–x) graph for a variable force
Draw x-axis as displacement and y-axis as force, with a curve F(x) that varies (e.g. rising then falling). Shade the region between the curve and the x-axis from xi to xf; label this shaded area as the work done. Show clearly that area below the x-axis (if any) represents negative work and must be subtracted, not added.
Spring force and potential energy vs displacement
Draw two graphs side by side sharing the x-axis (displacement x from natural length, negative for compression, positive for extension). Top graph: F vs x as a straight line through the origin with negative slope −k (Hooke's law). Bottom graph: PE vs x as an upward-opening parabola through the origin, symmetric about x=0, since PE=½kx² depends on x² regardless of sign.
Energy transformation of a freely falling body
Draw a vertical line showing a body falling from height H to the ground. Mark three points: top (h=H: KE=0, PE=mgH), middle (height h: KE=mg(H−h), PE=mgh), and ground (h=0: KE=mgH, PE=0). At each point write both KE and PE values and show their sum is always mgH, illustrating conservation of mechanical energy.
Elastic collision in one dimension, before and after
Draw two horizontal 'before' and 'after' rows. Before: mass m1 moving right with velocity u1 approaching stationary mass m2 (u2=0), with an arrow for u1 only. After: m1 moving with v1 and m2 moving with v2, both arrows shown (direction depends on relative masses). Label all four quantities and note momentum and KE are conserved between the two rows.
TextbookNCERT questions the board keeps asking
5.1 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.3 m
(a) Positive — the applied force (tension in the rope) acts upward, in the same direction as the bucket's upward displacement, so θ=0° and W is positive.
(b) Negative — gravity acts downward while the bucket moves upward, so θ=180° and W is negative.
(c) Negative — friction always opposes relative motion, so it acts up the incline while the body slides down, making W negative.
(d) Positive — the applied force must overcome friction to maintain uniform velocity, and it acts in the direction of motion, so W is positive (this positive work exactly balances the negative work done by friction, keeping KE constant).
(e) Negative — air resistance always opposes the pendulum's velocity, so it continuously removes energy, doing negative work, which is why the pendulum eventually comes to rest.
5.2 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? Give reasons.3 m
Only (c) total mechanical energy remains constant. In vacuum, gravity (a conservative force) is the only force acting, so mechanical energy KE+PE is conserved throughout the fall.
(a) KE increases continuously as speed increases (KE = ½mv²).
(b) PE decreases continuously as height decreases (PE = mgh).
(d) Momentum (mv) keeps increasing in magnitude as v increases under constant g, so it is NOT constant (only in an isolated system with no external force would momentum be conserved; here gravity is an external force acting on the falling body).
5.3 Given below are examples of some potential energy functions in one dimension. State whether the motion for negative and positive values of x is bounded or unbounded, in a general context (using the graph of V(x)).3 m
This exercise (with graphs) tests reading a PE–x curve to decide the region of motion: a particle with total energy E can only exist where PE(x) ≤ E (since KE = E − PE(x) ≥ 0). If PE(x) rises above E on both sides of a minimum, motion is bounded between the two turning points; if PE(x) stays below E as x→∞ on one or both sides, motion is unbounded in that direction. Students should sketch E as a horizontal line on the given V(x) graph and mark the region(s) where V(x) ≤ E as the allowed (bounded/unbounded) region of motion.
5.4 The potential energy function for a particle executing linear simple harmonic motion is given by V(x) = kx²/2, where k is the force constant of the oscillator. For k = 0.5 N/m, the graph of V(x) versus x is shown. Show that a particle of total energy 1 J moving under this potential must 'turn back' when it reaches x = ±2 m.3 m
Total energy E = KE + V(x) = 1 J is constant. At the turning point, the particle momentarily stops, so KE = 0 and all energy is potential: V(x) = E = 1 J.
Using V(x) = kx²/2 = 1 ⇒ (0.5)x²/2 = 1 ⇒ x² = 4 ⇒ x = ±2 m.
For |x| > 2 m, V(x) would exceed 1 J, which would require negative KE — impossible. So the particle cannot go beyond x = ±2 m and must turn back there.
5.5 A block of mass m is placed on top of a spring (compressed) of spring constant k on a frictionless surface. Discuss (through worked reasoning) how work-energy ideas and conservation of energy apply as the spring pushes the block, ending with a standard result for the maximum height/speed the block attains.5 m
This is the standard spring-launch problem style. If a spring is compressed by x0 and released, releasing a block of mass m: energy stored, ½kx0², converts entirely into KE of the block at natural length (frictionless surface, so mechanical energy is conserved): ½kx0² = ½mv², giving v = x0√(k/m). If the block then leaves the spring and rises against gravity (vertical case) or moves along a frictionless horizontal surface, apply KE=PE at the relevant points; e.g. maximum height h reached (if launched vertically) is found from mgh = ½kx0², giving h = kx0²/(2mg).
5.6 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 internal forces on the system. (d) In an inelastic collision of two bodies, the quantities that do not change after the collision are the total kinetic energy/total linear momentum/total energy of the two-body system.4 m
(a) Decreases — positive work by a conservative force reduces PE (since ΔPE = −W).
(b) Kinetic energy — friction converts KE into heat, so it is kinetic energy that is lost.
(c) External force — internal forces cancel in pairs by Newton's third law, so only the net external force changes total momentum.
(d) Total linear momentum and total energy — momentum is always conserved in collisions, and total energy (including heat/sound generated) is always conserved; only the total kinetic energy decreases in an inelastic collision.
5.7 State whether the following statements are true or false, with reasons: (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.4 m
(a) False — only the TOTAL momentum and TOTAL KE of the system of two bodies is conserved; individual bodies can gain or lose momentum/energy through the collision.
(b) True — total energy (in all forms, including heat, sound, etc.) is always conserved regardless of the nature of internal or external forces; this is the universal law of conservation of energy.
(c) False — this is true only for conservative forces (e.g. gravity, spring force); for a non-conservative force like friction, work done over a closed loop is not zero.
(d) True — in any inelastic collision, some KE is converted to heat, sound or deformation, so final KE is always less than initial KE (equal only in the limiting elastic case).
5.8 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 (i.e. when the balls 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? (Note: the answer to this exercise is not obvious and requires match with reality, in particular the elastic collision of billiard balls.)5 m
(a) No — during the brief contact time, the balls deform, so some KE is momentarily stored as elastic PE of deformation; total KE is not conserved AT THAT INSTANT (it is conserved only before and after, when the balls have separated and regained shape).
(b) Yes — momentum is conserved throughout the collision, at every instant, since the mutual forces are internal (action-reaction pairs) and no external force acts.
(c) Same as elastic for momentum (conserved throughout); for KE, it is not conserved at any stage post-collision either, since some KE permanently converts to heat/deformation/sound.
(d) Elastic — if PE depends only on the separation between centres (a conservative interaction with no permanent deformation), then when the balls separate to their original shape, all the temporarily stored PE converts back to KE, so the collision is elastic.
5.13 A molecule in a gas container hits a horizontal wall with speed 200 m/s and angle 30° with the normal, and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?3 m
Momentum is conserved: the wall (attached to a massive container) exerts a normal force on the molecule; the component of the molecule's momentum along the wall (tangential) is unchanged, and the normal component reverses (from +mv cos30° to −mv cos30°), with the wall absorbing the equal and opposite momentum change — so total momentum of (molecule + wall/container) is conserved.
The collision is elastic: the molecule's speed before (200 m/s) equals its speed after (200 m/s), so its kinetic energy is unchanged — KE is conserved, which is the definition of an elastic collision.
5.14 A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m³ in 15 minutes. 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?3 m
Mass of water, m = volume × density = 30 × 1000 = 30,000 kg.
Work done against gravity (useful work), W = mgh = 30,000 × 9.8 × 40 = 1.176 × 107 J.
Time, t = 15 min = 900 s.
Useful power output, Puseful = W/t = 1.176×107/900 = 1.3067 × 104 W.
Since efficiency = (useful power output)/(electric power input) = 30% = 0.3,
Electric power consumed = Puseful/0.3 = 1.3067×104/0.3 ≈ 4.36 × 104 W ≈ 43.6 kW.
5.16 (Exemplar) A bullet of mass 0.012 kg and horizontal speed 70 m/s strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.5 m
Step 1 — Perfectly inelastic collision (bullet embeds in block), momentum conservation: mbulletu = (mbullet+mblock)v
0.012 × 70 = (0.012 + 0.4) v ⇒ 0.84 = 0.412 v ⇒ v = 2.039 m/s.
Step 2 — Height risen (energy conservation after collision, block+bullet rise together): ½(0.412)v² = (0.412)gh
h = v²/(2g) = (2.039)²/(2×9.8) = 4.157/19.6 ≈ 0.212 m ≈ 21.2 cm.
Step 3 — Heat produced = loss in KE during the collision = Initial KE − KE just after collision.
Initial KE = ½ × 0.012 × 70² = ½ × 0.012 × 4900 = 29.4 J.
KE just after = ½ × 0.412 × (2.039)² = ½ × 0.412 × 4.157 ≈ 0.857 J.
Heat produced ≈ 29.4 − 0.857 ≈ 28.5 J.
5.9 / typical 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 speed V. If the collision is elastic, which of the following figures/outcomes is a possible result after collision: (a) one ball bearing moves with speed V, two remain at rest, (b) two ball bearings move with speed V/2 each, third remains at rest, (c) all three move with speed V/3.3 m
(a) is the only possible result. Since all masses are equal, apply both momentum and KE conservation to the system: total momentum before = mV, total KE before = ½mV².
For (a): momentum after = m(V)+0+0 = mV ✓; KE after = ½mV² ✓ — both conserved, valid.
For (b): momentum after = m(V/2)+m(V/2)+0 = mV ✓ but KE after = 2×½m(V/2)² = mV²/4 ≠ ½mV² ✗ — KE not conserved, invalid (extra energy would have to appear from nowhere).
For (c): similarly KE after = 3×½m(V/3)² = mV²/6 ≠ ½mV² ✗ — invalid.
So only outcome (a), where the struck ball bearing that was in direct contact moves off with the full speed V and the front one stays effectively at rest (like Newton's cradle), satisfies both conservation laws.
Previous years and frequently askedBoard questions with model answers
MCQ A body is moving unidirectionally under the influence of a constant power source. Its displacement in time t is proportional to:- t1/2
- t
- t3/2
- t2
Frequently asked1 m
MCQ A body of mass 1 kg is thrown upwards with a velocity 20 m/s. It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction? (g = 10 m/s²)- 10 J
- 20 J
- 30 J
- 40 J
CBSE 20201 m
MCQ Which of the following is a conservative force?- Friction
- Air resistance
- Gravitational force
- Viscous force
Frequently asked1 m
MCQ A spring of spring constant k is cut into two equal halves. The spring constant of each half becomes:- k/2
- k
- 2k
- 4k
CBSE 20191 m
MCQ Two bodies of equal mass moving with equal speed in opposite directions collide and stick together. After collision, the combined body:- moves with double the speed
- moves with half the speed
- remains at rest
- moves with the same speed
Frequently asked1 m
MCQ A 1 kg block situated on a rough incline is connected to a spring of spring constant 100 N/m as shown. The block is released from rest with the spring in the natural length. The block moves 10 cm down the incline before coming to rest. If μ = 0.5, what is the work done against friction?- 0.5 J
- 1.0 J
- 1.5 J
- 2.0 J
CBSE 20231 m
MCQ The dimensional formula for power is:- [M L2 T−2]
- [M L2 T−3]
- [M L T−2]
- [M L2 T−1]
Frequently asked1 m
MCQ A ball is dropped from a height h onto a floor and rebounds to a height h/2. The coefficient of restitution for this collision is:- 1
- 1/2
- 1/√2
- 1/4
Frequently asked1 m
Assertion–Reason Assertion (A): The work done by the centripetal force in uniform circular motion is always zero. Reason (R): The centripetal force acts perpendicular to the velocity of the particle at every instant.- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Frequently asked1 m
Assertion–Reason Assertion (A): Kinetic energy of a body can never be negative. Reason (R): Kinetic energy depends on the square of speed, and mass is always positive.- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Frequently asked1 m
Assertion–Reason Assertion (A): In an elastic collision between two bodies of equal mass, where one is initially at rest, the moving body comes to rest after the collision and the other moves with the initial velocity of the first. Reason (R): In an elastic collision, both momentum and kinetic energy are conserved.- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
CBSE 20221 m
Very short answer State the work-energy theorem. Is it applicable only to constant forces, or also to variable forces? Justify.CBSE 20222 m
Very short answer Define a conservative force. Give one example each of a conservative and a non-conservative force, and state one key difference in the work done by each around a closed path.Frequently asked2 m
Very short answer Derive the SI unit and dimensional formula for power.CBSE 20192 m
Very short answer A light body and a heavy body have the same momentum. Which one has greater kinetic energy?Frequently asked2 m
Short answer A ball of mass 0.2 kg falls from a height of 45 m onto a floor and rebounds to a height of 20 m. Find (a) the velocity of the ball just before it strikes the floor, (b) the velocity just after it rebounds, and (c) the coefficient of restitution. (g = 10 m/s²)CBSE 20213 m
(b) Using v²=2gh for the rebound (20 m): v'=√(2×10×20)=√400=20 m/s (upward, just after impact).
(c) Coefficient of restitution e = (velocity of separation)/(velocity of approach) = 20/30 = 2/3 ≈ 0.67.
Short answer Derive an expression for the potential energy stored in a stretched spring, obeying Hooke's law, when it is stretched by xm from its natural length.Frequently asked3 m
Short answer A pump is required to lift 600 kg of water per minute from a well 25 m deep and eject it with a speed of 50 m/s. Calculate the power required. (g = 10 m/s²)CBSE 20203 m
Power to raise water = (dm/dt)×g×h = 10×10×25 = 2500 W.
Power to give it kinetic energy (ejection speed) = ½(dm/dt)v² = ½×10×50² = ½×10×2500 = 12500 W.
Total power required = 2500 + 12500 = 15000 W = 15 kW.
Short answer Show that in a perfectly inelastic head-on collision between two bodies, the loss in kinetic energy is maximum (i.e. the common velocity result minimises leftover KE compared with any partial-elasticity outcome), using m1u1 with m2 initially at rest.Frequently asked3 m
Initial KE = ½m1u1².
Final KE = ½(m1+m2)v² = ½(m1+m2)×[m1u1/(m1+m2)]² = ½m1²u1²/(m1+m2).
Loss in KE = ½m1u1² − ½m1²u1²/(m1+m2) = ½m1u1²[1 − m1/(m1+m2)] = ½[m1m2/(m1+m2)]u1², which is always positive (maximum possible loss for the given initial momentum, since sticking together removes all relative motion, the maximum KE that can be lost while still conserving momentum).
Long answer State and prove the work-energy theorem for a variable force. Hence derive the relation between kinetic energy and momentum.CBSE 20205 m
Proof (variable force): By Newton's second law, F = m(dv/dt). Work done over small displacement dx: dW = F dx = m(dv/dt)dx = m(dx/dt)dv = mv dv (since dx/dt=v). Total work as the body's speed changes from u to v: W = ∫uv mv dv = m[v²/2]uv = ½mv² − ½mu² = KEf − KEi = ΔKE. This proves W = ΔKE for a variable force.
KE-momentum relation: Momentum p = mv, so v = p/m. KE = ½mv² = ½m(p/m)² = p²/(2m). This shows KE = p²/2m, useful for comparing bodies with equal momentum or equal KE.
Long answer Derive expressions for the final velocities of two bodies undergoing a one-dimensional elastic collision, where a body of mass m1 moving with velocity u1 strikes a stationary body of mass m2. Discuss two special cases: (i) m1=m2, (ii) m2≫m1.CBSE 20185 m
Momentum conservation: m1u1 = m1v1 + m2v2 ⇒ m1(u1−v1) = m2v2 ...(1)
KE conservation: ½m1u1² = ½m1v1² + ½m2v2² ⇒ m1(u1²−v1²) = m2v2² ⇒ m1(u1−v1)(u1+v1) = m2v2² ...(2)
Dividing (2) by (1): u1+v1 = v2 ...(3)
Substituting v2 from (3) into (1): m1(u1−v1) = m2(u1+v1) ⇒ m1u1−m2u1 = m2v1+m1v1 ⇒ v1 = [(m1−m2)/(m1+m2)]u1.
Then from (3): v2 = u1+v1 = [2m1/(m1+m2)]u1.
Case (i) m1=m2=m: v1=0, v2=u1 — the first body stops completely and the second moves off with the first body's initial velocity (as in Newton's cradle).
Case (ii) m2≫m1: v1≈−u1 (the light body rebounds with almost the same speed) and v2≈0 (the heavy body barely moves) — e.g. a ball bouncing off a wall.
Long answer A block of mass 2 kg is released from rest on a frictionless incline of angle 30° and slides down a distance of 4 m before hitting a spring of spring constant 1000 N/m fixed at the bottom of the incline. Using energy conservation, find (a) the speed of the block just before it touches the spring, and (b) the maximum compression of the spring. (g = 10 m/s²)Frequently asked5 m
v² = 2gh = 2×10×2 = 40 ⇒ v = √40 = 2√10 ≈ 6.32 m/s.
(b) After touching the spring, let maximum compression be x. From the point of touching to maximum compression (block momentarily at rest), using energy conservation for the whole journey (release point to max compression), total height dropped = (d+x)sinθ:
mg(d+x)sinθ = ½kx²
2×10×(4+x)×0.5 = ½×1000×x²
10(4+x) = 500x²
40+10x = 500x²
500x² − 10x − 40 = 0 ⇒ 50x² − x − 4 = 0.
Using the quadratic formula: x = [1 ± √(1+800)]/100 = [1 ± 28.3]/100.
Taking the positive root: x = 29.3/100 ≈ 0.293 m ≈ 29.3 cm.
Case-based A pendulum bob of mass 0.5 kg is pulled aside so that the string makes an angle of 60° with the vertical, then released from rest. The length of the string is 1 m. Assume air resistance is negligible. Answer the following: (i) What is the height through which the bob falls before reaching the lowest point? (ii) What is the speed of the bob at the lowest point? (iii) What provides the centripetal force at the lowest point, and is this force doing any work on the bob at that instant? (iv) If air resistance were NOT negligible, would the bob rise back to the same 60° on the other side? Explain.Frequently asked4 m
(ii) By conservation of mechanical energy (only gravity does work, string tension does no work since it's always perpendicular to velocity): mgh = ½mv² ⇒ v = √(2gh) = √(2×10×0.5) = √10 ≈ 3.16 m/s (taking g=10 m/s²).
(iii) The tension in the string provides the centripetal force at the lowest point, directed upward along the string toward the pivot. It does zero work on the bob at every instant because it always acts perpendicular to the bob's velocity (which is tangential to the circular arc).
(iv) No — with air resistance present, some mechanical energy is lost to heat/sound throughout the swing (air resistance does negative work), so by the time the bob reaches the other side, its total mechanical energy is less than at the start, and it would rise to a height/angle less than 60°.
Case-based A car of mass 1200 kg is moving on a level road with a constant speed of 20 m/s against a total resistive force (friction + air drag) of 600 N. Answer the following: (i) What is the power delivered by the car's engine to maintain this constant speed? (ii) If the car now accelerates uniformly from 20 m/s to 30 m/s in 5 s while the resistive force stays the same at 600 N, find the average power delivered by the engine during this interval. (iii) State the work-energy theorem and use it to find the work done by the engine during the acceleration phase in part (ii). (iv) Why does the engine need to deliver more instantaneous power at higher speed, even if the driving force were the same?CBSE 20224 m
(ii) Acceleration a = (30−20)/5 = 2 m/s². Net force needed = ma = 1200×2 = 2400 N. Driving force F = net force + resistive force = 2400+600 = 3000 N. Work done by engine = F×distance; distance s = ut+½at² = 20×5+½×2×25 = 100+25=125 m. W = 3000×125 = 375000 J. Average power = W/t = 375000/5 = 75000 W = 75 kW.
(iii) Work-energy theorem: work done by the net force equals the change in kinetic energy, Wnet=ΔKE. Here ΔKE = ½×1200×30² − ½×1200×20² = ½×1200×(900−400) = 600×500 = 300000 J. This equals the work done by the NET force (2400 N × 125 m = 300000 J ✓), which is less than the total work done by the engine (375000 J) because part of the engine's work (600×125=75000 J) goes into overcoming the resistive force and does not change KE.
(iv) Since P=Fv, for the same driving force F, power required is directly proportional to speed v — so at higher speed, even with unchanged force, the instantaneous power delivered must be greater.
Test yourselfPractice quiz
Marks lost every yearCommon mistakes
- Writing work done as simply 'Force × distance' and forgetting the cosθ factor, leading to wrong sign or magnitude when the force is not along the displacement.
- Treating the work-energy theorem as W(applied force) = ΔKE instead of W(net force) = ΔKE — if friction or other forces also act, only the NET work equals ΔKE.
- Assuming momentum is conserved only in elastic collisions — momentum is conserved in every collision (elastic or inelastic); it is kinetic energy that is conserved only in the elastic case.
- Forgetting to define a reference (zero) level for potential energy before applying conservation of energy, which causes sign errors in the final answer.
- Believing potential energy can be defined for friction or any non-conservative force — PE is defined only for conservative forces.
- Confusing the coefficient of restitution formula direction — writing it as approach/separation instead of separation/approach (e = velocity of separation / velocity of approach), which inverts the value.
- Using g = 9.8 in one part of a numerical and g = 10 in another part of the same answer, giving an internally inconsistent final value.
- Forgetting that spring PE = ½kx² uses x as displacement from the NATURAL length, not from any other reference, and applies equally to compression and extension.
How to prepareStrategy for this chapter
This chapter has no board weightage in Class 11, but treat it as a JEE/NEET and Class 12-board investment: every derivation here (work-energy theorem for constant and variable force, spring PE, elastic collision) reappears almost unchanged in Class 12 topics like circular motion, rotational mechanics and even electrostatics (potential energy of a system of charges uses the same logic).
Start by memorising the three definitions exactly as NCERT states them — work, kinetic energy, potential energy — because assertion-reason and one-mark questions test precise wording (e.g. 'independent of path' for conservative forces). Then master two derivations cold: the work-energy theorem for a variable force (§5.6) and the elastic collision in one dimension (§5.12), including both special cases (equal mass, and one mass much larger). These two account for most of the long-answer weight this chapter carries in school finals.
For numericals, always write down what is conserved before touching numbers — momentum only, or momentum and KE, or mechanical energy — since misidentifying this is the single biggest source of wrong answers. Practise the standard problem types: free fall with air resistance (energy 'lost'), spring-block systems, pendulum swings, and power required by a pump (splitting the answer into gravitational PE and kinetic energy terms). Keep g at a single consistent value (9.8 or 10) throughout one answer as instructed by your school.