Class 11 · Physics · Part 1

Chapter 5: Work, Energy and Power

Revision notes, practice numericals, a question bank, assertion and reason questions and answers to the NCERT questions. Tap a group to open it; opening one closes the others.

5.1Introduction and the scalar productWork, energy and power in physics; A · B

Precise meanings

In everyday speech 'work', 'energy' and 'power' are used loosely. In physics each has a precise meaning. Energy is the capacity to do work. Power is how fast work is done.
A farmer ploughing or a student studying are 'working' in daily life, but in physics work needs a force and a displacement.

The scalar (dot) product

A · B = A B cos θ, where θ is the angle between the vectors. The result is a scalar: A (B cos θ) = magnitude of A times the component of B along A.
It is commutative, A · B = B · A, and distributive, A · (B + C) = A · B + A · C. Also A · (λB) = λ (A · B).
Unit vectors: î · î = ĵ · ĵ = k̂ · k̂ = 1 and î · ĵ = ĵ · k̂ = k̂ · î = 0. In components A · B = Aₓ Bₓ + A_y B_y + A_z B_z and A · A = A².
A · B = 0 if A and B are perpendicular (and non-zero).
Example 5.1: for F = (3 î + 4 ĵ − 5 k̂) and d = (5 î + 4 ĵ + 3 k̂): F · d = 15 + 16 − 15 = 16, F = d = √50, cos θ = 16/50 = 0.32 so θ = cos⁻¹ 0.32.

Work done by a constant force: only the component of the force along the displacement contributes.

5.2–5.4Work, kinetic energy and the work-energy theoremW = F · d, K = ½mv², Kf − Ki = W

From kinematics to work and energy

For constant acceleration v² − u² = 2 a·d. Multiplying by m/2: ½mv² − ½mu² = m a · d = F · d.
Call K = ½ m v² the kinetic energy and W = F · d the work. Then Kf − Ki = W: the work-energy theorem. The change in kinetic energy equals the work done by the net force.

Work

Work of a constant force: W = F d cos θ = F · d. It is done by a force on a body over a displacement. 'Calculate the work done' is incomplete unless we say by which force (or group of forces) and over what displacement.
Work and energy have dimensions [M L² T⁻²]; the SI unit is the joule (1 J = 1 N m). Other units: erg (10⁻⁷ J), kWh (3.6 × 10⁶ J), eV (1.6 × 10⁻¹⁹ J).
Work is a scalar but can be positive (0° ≤ θ < 90°), negative (90° < θ ≤ 180°, e.g. friction) or zero.
No work is done if (i) the displacement is zero (pushing a rigid wall; holding a weight steadily), (ii) the force is zero (a block gliding on a smooth table), or (iii) the force is perpendicular to the displacement (gravity on a block on a smooth table; the earth's gravity on a satellite in a circular orbit).

Example highlights

Example 5.2: a raindrop of 1.00 g falls 1.00 km and reaches 50.0 m s⁻¹. ΔK = ½ × 10⁻³ × 50² = 1.25 J; work by gravity W_g = m g h = 10 J; work by the unknown resistive force W_r = ΔK − W_g = −8.75 J (negative).
Example 5.3: a cyclist skids to a stop in 10 m under a 200 N road force opposite to the motion: work done by the road = 200 × 10 × cos 180° = −2000 J. The work by the cycle on the road is zero because the road does not move. So the work done by the two forces of a third-law pair need not be equal and opposite.
Example 5.4: a 50 g bullet at 200 m s⁻¹ has K = 1000 J. It emerges with 10% of this energy: v_f = √(2 × 100/0.05) = 63.2 m s⁻¹, a drop of only 68% in speed.

Kinetic energy

K is a scalar, always positive (or zero). It measures the work a moving body can do because of its motion (a mill grinding corn with a flowing stream, a sailing ship using the wind).
K and momentum: K = p²/2m. Doubling the speed makes K four times as large.

5.5–5.6Work done by a variable forceArea under the F–x graph, W = ∫ F dx

Area under the curve

For a small displacement Δx, ΔW = F(x) Δx (the area of a thin rectangle). Adding the rectangles and letting Δx → 0: W = ∫ F(x) dx from xᵢ to x_f, the area under the F–x curve.
Areas below the axis count as negative work.

Work done by a variable force: the area under the F–x curve between xᵢ and x_f.

Example 5.5: pushing a trunk

A woman pushes a trunk with 100 N for 10 m, then her force falls linearly to 50 N at 20 m. Friction is 50 N (opposing). Work by the woman = rectangle + trapezium = 100 × 10 + ½ (100 + 50) × 10 = 1750 J. Work by friction = −50 × 20 = −1000 J.

Work-energy theorem for a variable force

In one dimension: dK/dt = d(½mv²)/dt = m v (dv/dt) = F v = F (dx/dt), so dK = F dx. Integrating: Kf − Ki = ∫ F dx = W.
The theorem is the integral form of Newton's second law, in scalar form. It does not contain the time information or the directional information of the vector second law.
Example 5.6: a 1 kg block enters a rough patch (0.1 m to 2.01 m) where F_r = −k/x with k = 0.5 J: K_f = ½mvᵢ² − k ln(2.01/0.1) = 2 − 0.5 ln 20.1 = 0.5 J, so v_f = 1 m s⁻¹.

5.7–5.8Potential energy and conservation of mechanical energyConservative forces, F = −dV/dx, K + V = constant

Potential energy

Potential energy is 'stored' energy by virtue of position or configuration (a stretched bow, a raised ball, a compressed spring). It is defined only for conservative forces.
For gravity near the earth (h ≪ R_E): V(h) = m g h, the negative of the work done by gravity in raising the body. F = −dV/dh = −m g.
In general F(x) = −dV/dx and ∫ F dx = Vᵢ − V_f. Equivalently ΔV = −F Δx.

Conservative forces

Three equivalent definitions: (i) F can be derived from a potential V(x) by F = −dV/dx; (ii) the work done depends only on the end points; (iii) the work done over a closed path is zero.
Gravity, the spring force and electrostatic forces are conservative. Friction is not: the work it does depends on the path and no potential energy can be defined for it.
The zero of potential energy is arbitrary. Once chosen it must be used consistently.

Conservation of mechanical energy

From ΔK = FΔx and −ΔV = FΔx we get Δ(K + V) = 0. So Kᵢ + V(xᵢ) = K_f + V(x_f): the total mechanical energy of a system is conserved if the forces doing work are conservative.
A ball dropped from a height H: E_H = m g H, E_h = m g h + ½ m v_h², E₀ = ½ m v_f². All are equal, giving v_f = √(2 g H) and v_h² = 2 g (H − h).
With a non-conservative force too, E_f − Eᵢ = W_nc, the work done by non-conservative forces (it depends on the path).

Example 5.7: a bob in a vertical circle

A bob of mass m on a string of length L is given a horizontal speed v₀ at the lowest point A so that the string goes slack only at the top C. At C: m g = m v_C²/L so v_C = √(g L). Energy at A equals that at C: ½ m v₀² = ½ m v_C² + 2 m g L so v₀ = √(5 g L).
At B (level of the centre): v_B = √(3 g L). The ratio of kinetic energies K_B : K_C = 3 : 1. After C the bob would move as a horizontal projectile if the string were cut.

Example 5.7: a bob that just completes the vertical circle (the string goes slack only at the top).

5.9Potential energy of a springHooke's law, V = ½kx², energy exchange

The spring force

For an ideal spring F_s = −k x (Hooke's law), where x is the displacement from equilibrium and k the spring constant (N m⁻¹). A stiff spring has large k.
Work done by the spring force in stretching by x_m: W_s = −½ k x_m² (negative); the external force does +½ k x_m². In general W_s = ½ k xᵢ² − ½ k x_f², which depends on the end points only, and is zero for a cyclic process. The spring force is conservative.

Potential energy and conservation

V(x) = ½ k x² (zero at x = 0). Check: −dV/dx = −kx. If the block is released from rest at x_m, ½ k x_m² = ½ k x² + ½ m v² at any x.
The speed is maximum at x = 0: ½ m v_m² = ½ k x_m² so v_m = x_m √(k/m). K and V are complementary parabolas; their sum E is constant.

A block on a spring: V and K are complementary parabolas; E = K + V stays constant.

Examples 5.8 and 5.9

A 1000 kg car at 18 km h⁻¹ (5 m s⁻¹) hits a spring of k = 5.25 × 10³ N m⁻¹ on a smooth road: K = 1.25 × 10⁴ J = ½ k x_m² gives x_m = 2.00 m.
With friction (μ = 0.5) use the work-energy theorem: ½ m v² = ½ k x_m² + μ m g x_m, a quadratic equation; x_m = 1.35 m, less than without friction. The time is not given by energy methods.

5.10PowerRate of doing work, P = F · v, watt, horse power

Definitions

Average power P_av = W/t. Instantaneous power P = dW/dt = F · v. Power is a scalar with dimensions [M L² T⁻³].
SI unit: the watt, 1 W = 1 J s⁻¹. 1 hp = 746 W.
kWh is a unit of energy (not power): 1 kWh = 10³ W × 3600 s = 3.6 × 10⁶ J. A 100 W bulb on for 10 hours uses 1 kWh.

Examples

Example 5.10: an elevator of 1800 kg (with load) rises at 2 m s⁻¹ against a 4000 N friction: F = m g + F_f = 18 000 + 4000 = 22 000 N and P = F v = 44 000 W ≈ 59 hp.
Constant acceleration from rest: P = F v = m a² t, so P ∝ t. Constant power: v ∝ t^(1/2) and x ∝ t^(3/2).

5.11CollisionsElastic, inelastic, one and two dimensions

Conservation laws in collisions

In all collisions of an isolated system the total linear momentum is conserved (the impulses on the two bodies are equal and opposite at every instant).
Elastic collision: the total kinetic energy after the collision equals that before. Completely inelastic: the bodies move together after the collision. Inelastic: some kinetic energy is lost (heat, sound, deformation).
Kinetic energy is conserved (in an elastic collision) only between the beginning and the end; during contact the bodies are deformed and part of the energy is momentarily stored (like a compressed spring).

One dimension

Completely inelastic (m₂ initially at rest): v_f = m₁ v₁ᵢ/(m₁ + m₂) and the loss of kinetic energy is ΔK = ½ [m₁ m₂/(m₁ + m₂)] v₁ᵢ².
Elastic (m₂ initially at rest): v₁f = [(m₁ − m₂)/(m₁ + m₂)] v₁ᵢ and v₂f = [2 m₁/(m₁ + m₂)] v₁ᵢ (also v₂f = v₁ᵢ + v₁f).
Equal masses: v₁f = 0 and v₂f = v₁ᵢ (they exchange velocities). If m₂ ≫ m₁: v₁f ≈ −v₁ᵢ and v₂f ≈ 0 (the light ball rebounds).
Example 5.11 (neutron moderator): fractional KE kept by the neutron f₁ = [(m₁ − m₂)/(m₁ + m₂)]². For deuterium (m₂ = 2m₁) f₁ = 1/9 and the nucleus gets f₂ = 8/9; for carbon f₁ = 71.6%.

One-dimensional collisions: elastic (equal masses exchange velocities) and completely inelastic (bodies move together).

Two dimensions

Momentum conservation gives two equations: m₁v₁ᵢ = m₁v₁f cos θ₁ + m₂v₂f cos θ₂ and 0 = m₁v₁f sin θ₁ − m₂v₂f sin θ₂. With four unknowns (v₁f, v₂f, θ₁, θ₂) and an elastic energy equation, one more quantity (such as θ₁) must be known.
Example 5.12: two equal billiard balls, one at rest, collide elastically and obliquely: after the collision they move at right angles (θ₁ + θ₂ = 90°). For θ₂ = 37°, θ₁ = 53°.
Scattering: when the bodies act on each other at a distance (a comet near the sun, an alpha particle near a nucleus), the final velocities also depend on the type of interaction.

PonderPoints to ponderCommon confusions

Remember

Work is always done by a specific force (or group of forces) on a body over a displacement.
Work can be negative; kinetic energy and mass cannot.
The two forces of a third-law pair sum to zero, but the sum of the work they do need not be zero.
The work-energy theorem is the scalar form of the second law; it also holds in all inertial frames. Conservation of mechanical energy follows for conservative forces.
Potential energy is defined only up to a constant, and only for conservative forces (none for friction).
During a collision, momentum is conserved at each instant; kinetic energy (if the collision is elastic) is conserved only before and after contact.

Quick referenceFormulas at a glanceRevise before the exam

Key results

W = F · d = F d cos θ; W = ∫ F dx; K = ½mv² = p²/2m; Kf − Ki = W
V = mgh; V = ½kx²; F = −dV/dx; K + V = const; E_f − Eᵢ = W_nc
P = W/t; P = F · v; 1 kWh = 3.6 × 10⁶ J; 1 hp = 746 W
Elastic (m₂ at rest): v₁f = (m₁ − m₂)v₁ᵢ/(m₁ + m₂), v₂f = 2m₁v₁ᵢ/(m₁ + m₂); inelastic: v_f = m₁v₁ᵢ/(m₁ + m₂)

Common mistakes

Using W = F d when the force is not along the displacement (forgetting cos θ).
Treating kWh as a unit of power.
Applying conservation of mechanical energy when friction acts.
Saying kinetic energy is conserved in every collision.
Forgetting the sign of the work done by friction or by a spring.

Try each problem first, then open the solution to check your working.

5.1Work and the scalar productConstant forces, signs and zero work · 6 problems

Q1Work by a force at an angle

A boy pulls a sled with a rope that makes an angle of 60° with the ground, with a force of 20 N, over a distance of 5 m. Find the work done by the force.

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Given: F = 20 N, d = 5 m, θ = 60°

Use: W = F d cos θ

W = 20 × 5 × cos 60° = 100 × 0.5 = 50 J.

Answer: 50 J

Q2Work from components

A force F = (2 î + 3 ĵ − k̂) N acts on a body and displaces it by d = (î − 2 ĵ + 4 k̂) m. Find the work done and the angle between F and d.

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Given: F = (2, 3, −1) N, d = (1, −2, 4) m

Use: W = F · d = Fₓ dₓ + F_y d_y + F_z d_z

W = 2 × 1 + 3 × (−2) + (−1) × 4 = 2 − 6 − 4 = −8 J.

F = √(4 + 9 + 1) = √14; d = √(1 + 4 + 16) = √21.

cos θ = −8/(√14 √21) = −0.466, so θ = 117.8° (obtuse, so the work is negative).

Answer: −8 J; 117.8°

Q3Work against friction

A 10 kg block is pushed 5 m along a rough floor at constant speed. μₖ = 0.2 (g = 10 m s⁻²). Find the work done by friction and by the pushing force.

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Given: m = 10 kg, d = 5 m, μₖ = 0.2

Use: Constant speed: pushing force = friction

Friction f = μₖ m g = 0.2 × 100 = 20 N.

Work by friction = −20 × 5 = −100 J.

The pushing force equals 20 N, so the work it does = +100 J (the net work is zero, so K does not change).

Answer: −100 J by friction; +100 J by the push

Q4Gravity on an incline

A 5 kg block slides 4 m down a smooth plane inclined at 30° (g = 10 m s⁻²). Find the work done by gravity and by the normal force.

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Given: m = 5 kg, d = 4 m, θ = 30°

Use: Component of weight along the slope: m g sin θ

Work by gravity = m g sin 30° × d = 5 × 10 × 0.5 × 4 = 100 J.

The normal force is perpendicular to the displacement, so the work done by it is zero.

Answer: 100 J; 0

Q5Lifting a bag

A labourer lifts a 50 kg bag steadily through 3 m (g = 10 m s⁻²). Find the work done by the labourer and by gravity.

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Given: m = 50 kg, h = 3 m

Use: At constant speed the lifting force equals the weight

Work by the labourer = m g h = 50 × 10 × 3 = 1500 J.

Work by gravity = −1500 J (opposite to the displacement). The net work is zero.

Answer: +1500 J; −1500 J

Q6Carrying a suitcase

A porter carries a 20 kg suitcase on his head for 100 m along a level road at constant speed. How much work does his supporting force do on the suitcase?

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Given: m = 20 kg, d = 100 m (horizontal)

Use: The supporting force is vertical; the displacement is horizontal

θ = 90°, cos 90° = 0, so W = 0.

(Energy is still used by his muscles, but no work is done on the suitcase.)

Answer: zero

5.2Kinetic energy and the work-energy theoremSpeeding up and stopping · 6 problems

Q7Kinetic energy of a car

A car of mass 1000 kg moves at 72 km h⁻¹. Find its kinetic energy.

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Given: m = 1000 kg, v = 72 km h⁻¹ = 20 m s⁻¹

Use: K = ½ m v²

K = ½ × 1000 × 20² = 2.0 × 10⁵ J = 200 kJ.

Answer: 2.0 × 10⁵ J

Q8Work needed to speed up

How much work must the engine do to increase the speed of an 800 kg car from 10 m s⁻¹ to 20 m s⁻¹ on a level road (neglect friction)?

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Given: m = 800 kg, u = 10 m s⁻¹, v = 20 m s⁻¹

Use: Work-energy theorem: W = ½ m (v² − u²)

W = ½ × 800 × (400 − 100) = 400 × 300 = 1.2 × 10⁵ J.

Answer: 1.2 × 10⁵ J

Q9Braking distance by energy

A 1500 kg car moving at 20 m s⁻¹ is stopped by a constant braking force of 6000 N. Find the stopping distance.

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Given: m = 1500 kg, u = 20 m s⁻¹, F = 6000 N

Use: Work done by the brake = loss of kinetic energy

K = ½ × 1500 × 400 = 3.0 × 10⁵ J.

F d = K, so d = 3.0 × 10⁵/6000 = 50 m.

Answer: 50 m

Q10Bullet stopped by a block

A 20 g bullet at 300 m s⁻¹ is stopped after penetrating 10 cm into a wooden block. Find the average resistive force.

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Given: m = 0.020 kg, u = 300 m s⁻¹, d = 0.10 m

Use: Work done by the resistive force = −K

K = ½ × 0.020 × 300² = 900 J.

F d = 900, so F = 900/0.10 = 9000 N.

Answer: 9000 N

Q11Equal momentum, different masses

Two bodies of masses 1 kg and 4 kg have the same momentum. Find the ratio of their kinetic energies.

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Given: p₁ = p₂, m₁ = 1 kg, m₂ = 4 kg

Use: K = p²/2m

K₁/K₂ = (p²/2m₁)/(p²/2m₂) = m₂/m₁ = 4/1.

The lighter body has four times the kinetic energy.

Answer: 4 : 1

Q12Kinetic energy and momentum

The kinetic energy of a body is increased by 21%. By what percentage does its momentum increase (mass constant)?

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Given: K′ = 1.21 K

Use: K = p²/2m so p ∝ √K

p′/p = √1.21 = 1.1.

The momentum increases by 10%.

Answer: 10%

5.3Variable forcesIntegrals and graphs · 4 problems

Q13Spring force

A spring exerts a force F = 3x (N) when stretched by x metres. Find the work done in stretching it from x = 0 to x = 2 m.

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Given: F = 3x

Use: W = ∫ F dx

W = ∫₀² 3x dx = (3/2) x² from 0 to 2 = 1.5 × 4 = 6 J.

Answer: 6 J

Q14Work from an F–x graph

A force rises uniformly from 0 to 10 N over the first 4 m and then stays constant at 10 N for the next 6 m. Find the total work done.

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Given: Triangle (0–4 m) and rectangle (4–10 m)

Use: Work = area under the F–x graph

Triangle: ½ × 4 × 10 = 20 J.

Rectangle: 10 × 6 = 60 J.

Total work = 80 J.

Answer: 80 J

Q15Linearly varying force

A force F = (4 + 2x) N acts along the x-axis. Find the work done from x = 1 m to x = 3 m.

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Given: F = 4 + 2x

Use: W = ∫ F dx

W = [4x + x²] from 1 to 3 = (12 + 9) − (4 + 1) = 21 − 5 = 16 J.

Answer: 16 J

Q16Variable force and speed

A force F = 6x² (N) acts on a 2 kg body, initially at rest at x = 0. Find its speed after it has moved to x = 2 m.

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Given: F = 6x², m = 2 kg

Use: Work-energy theorem

W = ∫₀² 6x² dx = 2x³ from 0 to 2 = 16 J.

½ m v² = 16, so v² = 16 and v = 4 m s⁻¹.

Answer: 4 m s⁻¹

5.4Potential energy and conservation of energyFalling bodies, springs, pendulums and circles · 7 problems

Q17Falling body

A stone is dropped from a height of 20 m (g = 10 m s⁻²). Find its speed just before it reaches the ground.

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Given: h = 20 m

Use: Conservation of mechanical energy: m g h = ½ m v²

v = √(2 g h) = √(2 × 10 × 20) = √400 = 20 m s⁻¹.

Answer: 20 m s⁻¹

Q18Energy stored in a spring

A spring of constant 200 N m⁻¹ is compressed by 0.10 m. Find the energy stored and the force at that compression.

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Given: k = 200 N m⁻¹, x = 0.10 m

Use: V = ½ k x², F = kx

V = ½ × 200 × 0.01 = 1.0 J.

F = 200 × 0.10 = 20 N.

Answer: 1.0 J; 20 N

Q19Spring-launched block

A 0.5 kg block is pressed against a spring (k = 800 N m⁻¹) and compressed by 0.10 m on a smooth table. Find its speed when released and leaves the spring.

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Given: m = 0.5 kg, k = 800 N m⁻¹, x = 0.10 m

Use: ½ k x² = ½ m v²

v = x √(k/m) = 0.10 × √(800/0.5) = 0.10 × 40 = 4 m s⁻¹.

Answer: 4 m s⁻¹

Q20Pendulum bob

A pendulum bob on a 1.0 m string is released from rest with the string at 60° to the vertical (g = 10 m s⁻²). Find its speed at the lowest point.

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Given: L = 1.0 m, θ = 60°

Use: Height fallen: h = L (1 − cos θ)

h = 1.0 (1 − 0.5) = 0.5 m.

v = √(2 g h) = √(2 × 10 × 0.5) = √10 = 3.16 m s⁻¹.

Answer: 3.16 m s⁻¹

Q21A rough incline

A 2 kg block slides 5 m from rest down an incline of height 2.5 m. The friction force is 4 N (g = 10 m s⁻²). Find its speed at the bottom.

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Given: m = 2 kg, length 5 m, height 2.5 m, f = 4 N

Use: Energy: Kf = m g h − f d

m g h = 2 × 10 × 2.5 = 50 J.

Work against friction = 4 × 5 = 20 J.

K = 50 − 20 = 30 J, so v = √(2 × 30/2) = √30 = 5.48 m s⁻¹.

Answer: 5.48 m s⁻¹

Q22Just completing a vertical circle

A stone tied to a 0.4 m string is whirled in a vertical circle. Find the minimum speeds at the bottom and at the top for the string to remain taut (g = 10 m s⁻²).

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Given: L = 0.4 m

Use: At the top m g = m v²/L; energy conservation from top to bottom

Top: v_top = √(g L) = √4 = 2 m s⁻¹.

Bottom: v² = v_top² + 4 g L = 4 + 16 = 20, so v_bottom = √(5 g L) = 4.47 m s⁻¹.

Answer: 2 m s⁻¹ at the top; 4.47 m s⁻¹ at the bottom

Q23Where KE equals PE

A ball is thrown vertically upward at 20 m s⁻¹ (g = 10 m s⁻²). At what height are its kinetic and potential energies equal?

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Given: u = 20 m s⁻¹

Use: Total energy is conserved: ½ m u² = K + V = 2 m g h

Total energy = ½ m u². At the required height K = V = m g h, so 2 m g h = ½ m u².

h = u²/(4g) = 400/40 = 10 m.

Answer: 10 m

5.5PowerRate of doing work and units of energy · 3 problems

Q24A pump

A pump lifts 1200 kg of water through 20 m in 60 s (g = 10 m s⁻²). Find the power delivered.

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Given: m = 1200 kg, h = 20 m, t = 60 s

Use: P = m g h/t

Work = 1200 × 10 × 20 = 2.4 × 10⁵ J.

P = 2.4 × 10⁵/60 = 4000 W = 4 kW.

Answer: 4000 W

Q25Power of a car engine

A car moves at a constant speed of 25 m s⁻¹ against a total resistive force of 2000 N. Find the power the engine delivers in watts and in horse power.

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Given: v = 25 m s⁻¹, F = 2000 N

Use: P = F v

At constant speed the driving force equals the resistance: P = 2000 × 25 = 50 000 W = 50 kW.

In horse power: 50 000/746 = 67 hp.

Answer: 50 kW (≈ 67 hp)

Q26The electricity bill

Five 100 W bulbs are used for 6 hours a day. Find the energy used per day in kWh and in joules.

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Given: 5 × 100 W = 500 W; t = 6 h

Use: E = P t

E = 0.5 kW × 6 h = 3 kWh.

In joules: 3 × 3.6 × 10⁶ = 1.08 × 10⁷ J.

Answer: 3 kWh = 1.08 × 10⁷ J

5.6CollisionsElastic and completely inelastic collisions · 2 problems

Q27An elastic collision

A 2 kg ball moving at 6 m s⁻¹ collides head-on and elastically with a 4 kg ball at rest. Find the velocities after the collision and verify that kinetic energy is conserved.

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Given: m₁ = 2 kg, v₁ᵢ = 6 m s⁻¹, m₂ = 4 kg

Use: v₁f = (m₁ − m₂) v₁ᵢ/(m₁ + m₂), v₂f = 2 m₁ v₁ᵢ/(m₁ + m₂)

v₁f = (2 − 4) × 6/6 = −2 m s⁻¹ (rebounds).

v₂f = 2 × 2 × 6/6 = 4 m s⁻¹.

K before = ½ × 2 × 36 = 36 J; K after = ½ × 2 × 4 + ½ × 4 × 16 = 4 + 32 = 36 J. Conserved.

Answer: −2 m s⁻¹ and +4 m s⁻¹

Q28A ballistic pendulum

A 10 g bullet moving at 400 m s⁻¹ embeds itself in a 1.99 kg wooden block suspended by strings. How high does the block rise (g = 10 m s⁻²)?

Show solutionHide solution

Given: m = 0.010 kg, v = 400 m s⁻¹, M = 1.99 kg

Use: Completely inelastic collision, then energy conservation

Momentum: (0.010 + 1.99) V = 0.010 × 400, so V = 4/2.0 = 2 m s⁻¹.

Energy after the collision: h = V²/(2g) = 4/20 = 0.2 m.

(The kinetic energy before the collision, 800 J, is far greater than the 4 J after it: most is lost in the collision.)

Answer: 0.2 m

Answers to every exercise question of the NCERT chapter. Try each one first, then tap Show answer.

5.1Exercises 5.1 to 5.23Work, energy, power, collisions and conservation laws · 23 questions

Q15.1 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.

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(a) Positive. The force on the bucket (upward) and its displacement (upward) are in the same direction.
(b) Negative. Gravity acts downward while the bucket moves up.
(c) Negative. Friction is opposite to the displacement of the body.
(d) Positive. At constant velocity the applied force equals the friction and is in the direction of motion.
(e) Negative. The resistive force of air opposes the motion of the pendulum.

Q25.2 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.

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Take g = 9.8 m s⁻². Friction f = μ m g = 0.1 × 2 × 9.8 = 1.96 N, so the net force is 7 − 1.96 = 5.04 N and a = 5.04/2 = 2.52 m s⁻².

Distance in 10 s: x = ½ a t² = ½ × 2.52 × 100 = 126 m. Speed at 10 s: v = a t = 25.2 m s⁻¹.

(a) Work by the applied force = 7 × 126 = 882 J.
(b) Work by friction = −1.96 × 126 = −247 J (−246.96 J).
(c) Work by the net force = 882 − 247 = 635 J.
(d) ΔK = ½ × 2 × (25.2)² = 635 J.

Interpretation: the work done by the net force (635 J) equals the change in kinetic energy, as the work-energy theorem requires. Friction removes part of the energy supplied by the applied force.

Q35.3 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.

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The kinetic energy K = E − V(x) cannot be negative, so the particle can be found only where V(x) ≤ E. The four panels were read from the printed figure.

(i) V = 0 for x < a and V₀ for x > a, with E < V₀. The particle cannot be found for x > a. The minimum total energy is 0. Example: a ball rolling on a level floor bouncing off a high wall at x = a.
(ii) A staircase rising from V₀ up to V₃ and then dropping again, with E below every level of the curve. The particle cannot be found anywhere for this E. For any motion the total energy must exceed the lowest level of V (the level at x < 0 and x > d, here lying just below V₀). Example: a potential with steps, like a particle meeting a series of barriers.
(iii) V = V₀ for x < a and x > b, and V = −V₁ for a < x < b (a square well), with −V₁ < E < V₀. The particle cannot be found at x < a or at x > b: it is confined to a < x < b. The minimum total energy is −V₁. Example: a ball in a box or a nucleon in a nucleus.
(iv) A well of depth −V₁ between −a/2 and a/2 with potential barriers rising to V₀ at ±a/2 and decaying outwards, E as shown. The particle cannot be found for a/2 < |x| < b/2 (where V > E); it can be in |x| < a/2 or |x| > b/2. The minimum total energy is −V₁. Example: an atomic nucleus: a potential well with a repulsive barrier.

Fig. 5.11 (Ex 5.3): the particle can be found only where V(x) ≤ E.

Q45.4 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⁻¹, 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.

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The particle moves where K = E − V ≥ 0. It turns back where the kinetic energy becomes zero, i.e. where V(x) = E.

½ k x² = E gives x² = 2E/k = 2 × 1/0.5 = 4, so x = ±2 m. For |x| > 2 m, V > E and K would be negative, which is impossible; so the particle turns back at x = ±2 m.

Q55.5 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. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater?

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(a) The heat is obtained at the expense of the rocket: the work done against the frictional force comes from the rocket's kinetic (mechanical) energy, so the rocket loses mechanical energy.
(b) Gravity is a conservative force. The work done by a conservative force over a closed path (a complete orbit) is zero, even though the force is not perpendicular to the velocity at every point.
(c) As the satellite loses energy it spirals to a lower orbit. In a closer orbit the potential energy (negative) decreases by twice as much as the total energy decreases; the kinetic energy equals the magnitude of the total energy, so it increases. The speed therefore increases: the fall in potential energy (work done by gravity) more than makes up for the energy dissipated by drag.
(d) In case (i) the force on the mass (upward, equal to its weight) is perpendicular to the horizontal displacement, so the work done by the man on the mass is zero. In case (ii) the man pulls the rope with a force equal to the weight (15 × 9.8 = 147 N) and the hanging mass rises through 2 m, so the work done is 147 × 2 = 294 J. The work done is greater in case (ii).

Q65.6 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.

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(a) Decreases. W = Vᵢ − V_f > 0 means the potential energy falls.
(b) Kinetic energy. Work done against friction is taken from the body's kinetic energy.
(c) External force. Internal forces cancel in pairs (F = dp/dt for the system).
(d) Total linear momentum and total energy of the system do not change; the total kinetic energy does.

Q75.7 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.

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(a) False. It is the total momentum and the total kinetic energy of the system that are conserved; the momentum and energy of each body generally change.
(b) False. The total energy of an isolated system is conserved. External forces can do work on the system and change its energy.
(c) False. It is zero only for conservative forces. For a non-conservative force such as friction, the work over a closed loop is not zero.
(d) True. In an inelastic collision some kinetic energy is converted into other forms (heat, sound, deformation), so the final kinetic energy is less than the initial.

Q85.8 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? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).

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(a) No. During contact the balls are deformed, and part of the kinetic energy is stored as elastic potential energy (like a compressed spring). The total kinetic energy is the same only before and after the collision.
(b) Yes. The mutual forces are equal and opposite at every instant, so the total momentum is conserved at every instant, even during contact.
(c) The answers are the same: kinetic energy is not conserved during contact (and is not conserved after the collision either), while momentum is conserved at every instant.
(d) The collision is elastic. If the potential energy depends only on the separation, the force is conservative, so mechanical energy is conserved: the kinetic energy after the collision equals the kinetic energy before it.

Q95.9 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) t².

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The velocity is v = a t and the force is F = m a (constant). The power is P = F v = m a² t, which is proportional to t.

The correct alternative is (ii).

Q105.10 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) t².

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P = F v = m v (dv/dt) = constant, so v dv = (P/m) dt. Integrating: v²/2 = (P/m) t, so v = √(2P/m) t^(1/2).

Then x = ∫ v dt = √(2P/m) (2/3) t^(3/2), so the displacement is proportional to t^(3/2). The correct alternative is (iii).

Q115.11 A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by F = (−î + 2ĵ + 3k̂) N where î, ĵ, k̂ 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?

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The displacement is d = 4 k̂ m. The work done is W = F · d = (−î + 2ĵ + 3k̂) · (4 k̂) = 3 × 4 = 12 J. (Only the z-component of the force contributes; the x- and y-components are perpendicular to the displacement.)

Q125.12 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 = 9.11 × 10⁻³¹ kg, proton mass = 1.67 × 10⁻²⁷ kg, 1 eV = 1.60 × 10⁻¹⁹ J).

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Use K = ½ m v², so v = √(2K/m).

Electron: K = 10 keV = 10⁴ × 1.60 × 10⁻¹⁹ = 1.60 × 10⁻¹⁵ J; v_e = √(2 × 1.60 × 10⁻¹⁵/9.11 × 10⁻³¹) = 5.93 × 10⁷ m s⁻¹.

Proton: K = 100 keV = 1.60 × 10⁻¹⁴ J; v_p = √(2 × 1.60 × 10⁻¹⁴/1.67 × 10⁻²⁷) = 4.38 × 10⁶ m s⁻¹.

The electron is faster. The ratio v_e/v_p = 5.93 × 10⁷/4.38 × 10⁶ ≈ 13.5. (At 5.9 × 10⁷ m s⁻¹, about 0.2 c, the non-relativistic formula is only approximate.)

Q135.13 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⁻¹?

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Mass of the drop (water, density 1000 kg m⁻³): m = (4/3) π r³ ρ = (4/3) π (2 × 10⁻³)³ × 1000 = 3.35 × 10⁻⁵ kg. Take g = 9.8 m s⁻².

Work by gravity in each half: W = m g h = 3.35 × 10⁻⁵ × 9.8 × 250 = 0.082 J in the first half and 0.082 J in the second half (the same, as the displacement is 250 m in each).

For the whole journey: W_g + W_r = ΔK = ½ m v² − 0 = ½ × 3.35 × 10⁻⁵ × 100 = 1.68 × 10⁻³ J. So W_r = 1.68 × 10⁻³ − 3.35 × 10⁻⁵ × 9.8 × 500 = 0.00168 − 0.1642 = −0.16 J (about −0.163 J).

Q145.14 A molecule in a gas container hits a horizontal wall with speed 200 m s⁻¹ and angle 30° with the normal, and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?

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Momentum: the component of the molecule's momentum along the wall is unchanged, but the normal component reverses, so the momentum of the molecule alone is not conserved. The wall receives the opposite momentum, and since the wall (the container) is attached to the earth it has a huge mass and its velocity is practically unchanged. The total momentum of the molecule + wall (+ earth) system is conserved.

Kinetic energy: the molecule rebounds with the same speed, so its kinetic energy is unchanged and the wall gains negligible energy. The collision is elastic.

Q155.15 A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m³ 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?

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Mass of water = 30 m³ × 1000 kg m⁻³ = 3.0 × 10⁴ kg. Work done in raising it = m g h = 3.0 × 10⁴ × 9.8 × 40 = 1.176 × 10⁷ J.

Useful power (output) = 1.176 × 10⁷/900 s = 1.31 × 10⁴ W.

Efficiency 30%: the input power = output/0.30 = 4.36 × 10⁴ W ≈ 43.6 kW.

Q165.16 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?

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Fig. 5.14 shows three possibilities: (i) ball 1 at rest, balls 2 and 3 moving together with V/2; (ii) balls 1 and 2 at rest, ball 3 moving with V; (iii) all three moving together with V/3.

Initially: momentum = m V, kinetic energy = ½ m V².

(i) Momentum = 2m (V/2) = m V (conserved), but kinetic energy = ½ (2m)(V/2)² = ¼ m V², less than ½ m V². Not elastic.
(ii) Momentum = m V, kinetic energy = ½ m V². Both conserved: possible.
(iii) Momentum = 3m (V/3) = m V, kinetic energy = ½ (3m)(V/3)² = (1/6) m V². Not elastic.

The only possible result for an elastic collision is (ii).

Q175.17 The bob A of a pendulum released from 30° 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.

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In an elastic head-on collision between two bodies of equal mass, with one initially at rest, the moving body stops and the other moves off with the velocity the first had (they exchange velocities).

So after the collision bob A comes to rest at the lowest point, and bob B moves off with the speed of A. Bob A therefore does not rise: it rises to a height of zero (it stays at rest at the lowest point).

Q185.18 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?

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The bob falls through h = L = 1.5 m. Its initial potential energy is m g L. It loses 5%, so the kinetic energy at the lowest point is 95% of m g L:

½ m v² = 0.95 m g L, so v = √(2 × 0.95 × 9.8 × 1.5) = √27.93 = 5.28 m s⁻¹.

Q195.19 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⁻¹. What is the speed of the trolley after the entire sand bag is empty?

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The sand that leaks out falls vertically from the trolley. At the moment it leaves, its horizontal velocity equals that of the trolley, and no horizontal force acts on it (or on the trolley) afterwards. So the sand carries away exactly its share of the horizontal momentum and the trolley's speed does not change.

The speed of the trolley remains 27 km h⁻¹ (7.5 m s⁻¹) after the whole bag is empty.

Q205.20 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?

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By the work-energy theorem, W = ΔK. At x = 0, v = 0, so K = 0. At x = 2 m: v = 5 × 2^(3/2) = 5 × 2.828 = 14.14 m s⁻¹, so v² = 25 × 8 = 200 m² s⁻².

K = ½ m v² = ½ × 0.5 × 200 = 50 J. The work done is W = 50 J.

Q215.21 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 m², v = 36 km/h and the density of air is 1.2 kg m⁻³. What is the electrical power produced?

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(a) In time t the air passing through the circle forms a cylinder of length v t and cross-section A. Volume = A v t, so the mass is m = ρ A v t.

(b) The kinetic energy is K = ½ m v² = ½ ρ A v³ t.

(c) v = 36 km h⁻¹ = 10 m s⁻¹. The power in the wind = K/t = ½ ρ A v³ = ½ × 1.2 × 30 × 1000 = 1.8 × 10⁴ W. The electrical power at 25% = 0.25 × 1.8 × 10⁴ = 4.5 × 10³ W = 4.5 kW.

Q225.22 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 3.8 × 10⁷ J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?

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(a) Work in each lift = m g h = 10 × 9.8 × 0.5 = 49 J. For 1000 lifts: W = 49 × 1000 = 4.9 × 10⁴ J. (The energy lost on lowering is dissipated, so only the lifting counts.)

(b) The chemical energy needed = 4.9 × 10⁴/0.20 = 2.45 × 10⁵ J. The fat used = 2.45 × 10⁵/3.8 × 10⁷ = 6.45 × 10⁻³ kg, about 6.5 g.

Q235.23 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.

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(a) The useful power per square metre = 20% of 200 W = 40 W m⁻². Area needed = 8000/40 = 200 m².

(b) A typical house roof is of the order of 100 m² (say 10 m × 10 m), so the required area is about twice that of a typical roof: comparable, which shows that solar power at this rate can meet a family's needs only with a large collector area.