PhysicsISC Class 11 · Chapter 5

Work, Energy and Power ISC Class 11: Notes and Collisions

Q: Do I need to derive E = mc²?

No. The syllabus says E = mc² with no derivation. You should know what it means and use it, for example that 1 g of mass is equivalent to 9 × 10¹³ J.

Q: Are two-dimensional collisions in the ISC Class 11 syllabus?

Yes. The syllabus includes elastic and inelastic collisions in one and two dimensions, with oblique collisions named explicitly. Conserve momentum along x and y separately.

Q: Which one-dimensional collision results must I learn?

The velocity equations for unequal masses with the second body at rest, and the special cases: equal masses exchange velocities, a heavy body hitting a light one, and a light body hitting a heavy one.

Q: Is mechanical energy conserved when friction acts?

No. Friction converts mechanical energy into heat. Total energy is still conserved, so the loss of mechanical energy equals the work done against friction.

Q: Why is mgh valid only near the Earth's surface?

It assumes g is constant over the height h. For heights comparable with the Earth's radius, use the gravitational potential energy −GMm/r, studied in Gravitation.

Q: Can kinetic energy be negative?

No. Since K = ½mv² and both m and v² are non-negative, kinetic energy is never negative. Potential energy can be negative, depending on where its zero is chosen.

Updated 15 October 2026 · Fact-checked

Work, Energy and Power is Chapter 5 of ISC Class 11 Physics. It covers work done by constant and variable forces, kinetic energy and the work-energy theorem, potential energy including that of a spring, conservative and non-conservative forces, conservation of mechanical energy, power, and elastic and inelastic collisions in one and two dimensions.

What Work, Energy and Power is about

Newton's laws relate force to acceleration at each instant. The ideas of work and energy relate force to the change of speed over a distance, which often makes problems much simpler.

Work. A constant force does work $W = F \cdot s = F s cos θ$ : positive when the force helps the motion, negative when it opposes it (friction), and zero when it is perpendicular (centripetal force, a porter carrying a load horizontally). A variable force does $W = \int F d x$ , the area under the F-x graph. Stretching a spring by x stores $W = \frac{1}{2} k x^{2}$ .

Energy. Kinetic energy $K = \frac{1}{2} m v^{2} = \frac{p ^{2}}{2 m}$ . The work-energy theorem: the net work done on a body equals its change in kinetic energy. Potential energy is stored energy of configuration: mgh near the Earth (for h much less than R), $\frac{1}{2} k x^{2}$ in a spring. It is defined only for conservative forces (gravity, spring force), whose work does not depend on the path and is zero round a closed path; friction is non-conservative. If only conservative forces act, mechanical energy K + U is conserved, as in an oscillating spring or a swinging pendulum. More generally, energy is never created or destroyed, only converted, and mass itself is a form of energy, $E = m c^{2}$ (no derivation needed).

Power is the rate of doing work: $P = \frac{W}{t} = F \cdot v$ (watt; 1 hp = 746 W). Collisions always conserve momentum. In an elastic collision kinetic energy is conserved too; in an inelastic collision some is lost, and the most is lost when the bodies stick together. You need the velocity equations for one-dimensional collisions with a target at rest, their special cases, and oblique collisions in two dimensions.

Weightage: Work, Energy and Power belongs to unit 4 of the ISC Class XI Physics syllabus. CISCE weights units 4-6 together (Work, Energy and Power; Motion of System of Particles and Rigid Body; Gravitation) at 17 of the 70 marks in Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Physics syllabus, examination year 2028, Class XI unit weightage

Key concepts

Work done by a constant force: Work is done when a force moves its point of application: $W = F \cdot s = F s cos θ$ , a scalar measured in joule (1 J = 1 N m; 1 erg = $1 0^{- 7}$ J). Positive work (0 ≤ θ < 90°): the force helps the motion, e.g. gravity on a falling body. Negative work (90° < θ ≤ 180°): the force opposes the motion, e.g. friction or the brakes. Zero work: no displacement (pushing a wall), or force perpendicular to displacement (the centripetal force, the normal reaction on a horizontal road, a porter carrying a load along level ground). Work depends on the frame of reference, since displacement does.; Worked example
A box is pulled 5 m along a level floor by a 50 N force acting at 37° above the horizontal (cos 37° = 0.8). Find the work done by this force and by the normal reaction.
Try it first, then show the solutionSolution
W = Fs cos θ = 50 × 5 × 0.8 = 200 J
The normal reaction is perpendicular to the displacement, so it does zero work.
Answer: 200 J; zero
Work done by a variable force: If the force changes with position, split the path into tiny steps dx over which F is nearly constant: $d W = F d x$ , so $W = \int_{x_{1}}^{x_{2}} F d x$ . Geometrically, this is the area under the force-displacement graph between $x_{1}$ and $x_{2}$ (area below the x-axis counts as negative work). Spring: the restoring force is $F = - k x$ (Hooke's law; k is the spring constant in N m⁻¹). The work done by an external agent in stretching it from 0 to x is $W = \int_{0}^{x} k x d x = \frac{1}{2} k x^{2}$ ; from $x_{1}$ to $x_{2}$ it is $\frac{1}{2} k (x_{2}^{2} - x_{1}^{2})$ .; Worked example
A force $F = (3 x^{2} + 2)$ N acts along the x-axis. Find the work it does as the body moves from x = 1 m to x = 3 m.
Try it first, then show the solutionSolution
$W = \int_{1}^{3} (3 x^{2} + 2) d x = [x^{3} + 2 x]_{1}^{3}$
= (27 + 6) − (1 + 2) = 30 J
Answer: 30 J
Kinetic energy and the work-energy theorem: Kinetic energy is the energy a body has because of its motion: $K = \frac{1}{2} m v^{2}$ . Since p = mv, $K = \frac{p ^{2}}{2 m}$ . Work-energy theorem: the work done by the net force on a body equals the change in its kinetic energy, $W_{n e t} = K_{f} - K_{i}$ . Proof (variable force): $F = ma = m \frac{d v}{d t} = m v \frac{d v}{d x}$ , so $W = \int F d x = \int_{u}^{v} m v d v = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}$ . It holds for any force, constant or variable, conservative or not. It is especially useful for stopping distances: a resisting force F stops a body in distance d with $F d = \frac{1}{2} m v^{2}$ .; Worked example
A 20 g bullet moving at 400 m s⁻¹ penetrates 10 cm into a wooden block before stopping. Find the average resistive force.
Try it first, then show the solutionSolution
Initial KE = ½ × 0.02 × 400² = 1600 J; final KE = 0
Work by resistance = −1600 J: F × 0.10 = 1600
F = 16 000 N
Answer: $1.6 \times 1 0^{4}$ N
Potential energy; conservative and non-conservative forces: Potential energy U is energy stored because of position or configuration. It is defined only for a conservative force, one whose work between two points is independent of the path, or equivalently zero round any closed path: gravity, the spring force, the electrostatic force. For such a force, $W = - Δ U$ and $F = - \frac{d U}{d x}$ . Gravitational PE near the Earth's surface: ΔU = mgh (valid when h is much less than the Earth's radius). Elastic PE of a spring stretched or compressed by x: $U = \frac{1}{2} k x^{2}$ . Non-conservative forces (friction, air resistance, viscous drag) do work that depends on the path; the energy they take is converted into heat and cannot be recovered as mechanical energy, so no potential energy can be defined for them.; Worked example
A spring of constant 800 N m⁻¹ is compressed by 5 cm. Find the energy stored, and the extra work needed to compress it a further 5 cm.
Try it first, then show the solutionSolution
$U_{1} = \frac{1}{2} \times 800 \times (0.05)^{2} = 1 J$
$U_{2} = \frac{1}{2} \times 800 \times (0.10)^{2} = 4 J$
Extra work = 4 − 1 = 3 J
Answer: 1 J; 3 J (not 1 J, since U grows as x²)
Conservation of mechanical energy and of energy: If only conservative forces do work, the total mechanical energy is constant: $K + U = constant$ , or $\frac{1}{2} m v_{1}^{2} + U_{1} = \frac{1}{2} m v_{2}^{2} + U_{2}$ . Freely falling body: at height h it has mgh and no KE; at the ground, $\frac{1}{2} m v^{2} = m g h$ ; at every point in between, the sum is mgh. If friction acts, $Δ K + Δ U = W_{f r i c t i o n}$ (negative), and the lost mechanical energy appears as heat. The general law of conservation of energy: energy can be converted from one form to another but the total energy of an isolated system is constant. Einstein showed that mass is also a form of energy, $E = m c^{2}$ ; 1 g of mass is equivalent to $9 \times 1 0^{13}$ J.; Worked example
A ball is released from rest 5 m above the ground. Find its speed (a) just before it lands and (b) when it is 2 m above the ground. (g = 9.8 m s⁻², ignore air resistance)
Try it first, then show the solutionSolution
(a) $\frac{1}{2} m v^{2} = m g h$ : $v = 2 \times 9.8 \times 5 = 9.9 m s^{- 1}$
(b) It has fallen 3 m: $v = 2 \times 9.8 \times 3 = 7.67 m s^{- 1}$
Answer: 9.9 m s⁻¹; 7.67 m s⁻¹ (mass does not matter)
Energy of an oscillating spring: A block of mass m on a smooth surface attached to a spring of constant k, pulled to x = A and released, oscillates between +A and −A. Only the spring force does work, so mechanical energy is conserved: $\frac{1}{2} m v^{2} + \frac{1}{2} k x^{2} = \frac{1}{2} k A^{2}$ . At the extreme positions, energy is all potential and v = 0; at x = 0, it is all kinetic and the speed is greatest, $v_{ma x} = A \frac{k}{m}$ . At any x, $v = \frac{k}{m} (A^{2} - x^{2})$ . KE and PE keep exchanging while their sum stays fixed.; Worked example
A 0.5 kg block on a spring of constant 200 N m⁻¹ oscillates with amplitude 0.1 m on a smooth floor. Find its total energy, its maximum speed and its speed at x = 0.06 m.
Try it first, then show the solutionSolution
$E = \frac{1}{2} \times 200 \times 0. 1^{2} = 1 J$
$v_{ma x} = 0.1 \frac{200}{0.5} = 0.1 \times 20 = 2 m s^{- 1}$
$v = 20 0.01 - 0.0036 = 20 \times 0.08 = 1.6 m s^{- 1}$
Answer: 1 J; 2 m s⁻¹; 1.6 m s⁻¹
Power: Power is the rate of doing work: average power $P = \frac{W}{t}$ ; instantaneous power $P = \frac{d W}{d t} = F \cdot v$ . Unit: watt (1 W = 1 J s⁻¹); 1 horsepower = 746 W. The kilowatt hour is a unit of energy, not power: 1 kWh = $3.6 \times 1 0^{6}$ J. A vehicle moving at constant speed v against a resistance F needs power P = Fv; on a slope it must also supply $m g sin θ \times v$ . Efficiency = useful output power ÷ input power.; Worked example
A motor lifts a 200 kg load through 15 m in 20 s at steady speed. Find its output power in watt and horsepower. (g = 9.8 m s⁻²)
Try it first, then show the solutionSolution
W = mgh = 200 × 9.8 × 15 = 29 400 J
P = 29 400/20 = 1470 W
1470/746 = 1.97 hp
Answer: 1470 W, about 2 hp
Collisions in one dimension: In every collision, linear momentum is conserved (internal forces only). Elastic collision: kinetic energy is also conserved (e.g. collisions between atoms, nearly so for steel balls). Inelastic collision: some KE becomes heat, sound or deformation; in a perfectly inelastic collision the bodies stick together. Head-on elastic collision of $m_{1}$ (speed $u_{1}$ ) with $m_{2}$ at rest: from momentum and KE conservation, $v_{1} = \frac{( m _{1} - m _{2} )}{m _{1} + m _{2}} u_{1}$ and $v_{2} = \frac{2 m _{1}}{m _{1} + m _{2}} u_{1}$ ; also the relative speed of separation equals that of approach. Special cases: equal masses exchange velocities ( $v_{1} = 0$ , $v_{2} = u_{1}$ ); a very heavy body hitting a light one carries on almost unchanged and the light one moves off at about $2 u_{1}$ ; a light body hitting a very heavy one rebounds with almost the same speed. Perfectly inelastic with $m_{2}$ at rest: $v = \frac{m _{1} u _{1}}{m _{1} + m _{2}}$ , and the fraction of KE lost is $\frac{m _{2}}{m _{1} + m _{2}}$ .; Worked example
A 2 kg ball moving at 6 m s⁻¹ collides head-on elastically with a 1 kg ball at rest. Find their velocities after the collision and check that KE is conserved.
Try it first, then show the solutionSolution
$v_{1} = \frac{2 - 1}{3} \times 6 = 2 m s^{- 1}$ ; $v_{2} = \frac{2 \times 2}{3} \times 6 = 8 m s^{- 1}$
Momentum: 12 = 2(2) + 1(8) ✓
KE: ½(2)(36) = 36 J; after: ½(2)(4) + ½(1)(64) = 4 + 32 = 36 J ✓
Answer: 2 m s⁻¹ and 8 m s⁻¹, both forward
Collisions in two dimensions (oblique collisions): When the bodies do not move along the line of centres, the collision is oblique. Momentum is conserved separately along x and along y: $m_{1} u_{1} = m_{1} v_{1} cos θ_{1} + m_{2} v_{2} cos θ_{2}$ and $0 = m_{1} v_{1} sin θ_{1} - m_{2} v_{2} sin θ_{2}$ , where $θ_{1}$ and $θ_{2}$ are measured on either side of the original direction. If the collision is elastic, add $\frac{1}{2} m_{1} u_{1}^{2} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$ . Useful result: in an elastic oblique collision between equal masses, one initially at rest, the two move off at right angles to each other ( $θ_{1} + θ_{2} = 9 0^{\circ}$ ), as seen with snooker balls.; Worked example
A ball moving at 4 m s⁻¹ strikes an identical ball at rest elastically and moves off at 30° to its original direction. Find the speeds of both balls after the collision.
Try it first, then show the solutionSolution
Equal masses, elastic: the second ball moves at 90° − 30° = 60° on the other side.
The velocity triangle is right-angled: $v_{1} = 4 cos 3 0^{\circ} = 3.46 m s^{- 1}$ , $v_{2} = 4 sin 3 0^{\circ} = 2 m s^{- 1}$
Check KE: 3.46² + 2² = 12 + 4 = 16 = 4² ✓
Answer: 3.46 m s⁻¹ at 30° and 2 m s⁻¹ at 60° on the other side

Physics in real life: Work, Energy and Power

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Your power when climbing stairs

Suppose a 50 kg student walks up a flight of stairs 3 m high in 6 s, and later runs up it in 3 s.

Find the power developed in each case. (g = 9.8 m s⁻²)

Try it first, then show the solutionSolution

Work done against gravity = mgh = 50 × 9.8 × 3 = 1470 J in both cases
Walking: P = 1470/6 = 245 W; running: P = 1470/3 = 490 W

Answer: 245 W walking and 490 W running: the same work, twice the power.

Memory hook: Same work faster means more power.

Stopping distance and the work-energy theorem

Imagine a car that stops in 10 m from 36 km h⁻¹ with full braking.

With the same braking force, how far would it take to stop from 72 km h⁻¹?

Try it first, then show the solutionSolution

Braking force × distance = initial KE, so d ∝ v²
Doubling the speed: d = 10 × 2² = 40 m

Answer: 40 m: twice the speed needs four times the stopping distance.

Memory hook: Double the speed, four times the skid.

Energy lost on a playground slide

Suppose a child slides from rest down a slide whose top is 3 m above its bottom and reaches the bottom at 6 m s⁻¹.

What fraction of the potential energy is lost to friction? (g = 9.8 m s⁻²)

Try it first, then show the solutionSolution

Per kilogram: PE lost = gh = 29.4 J; KE gained = ½ × 36 = 18 J
Fraction lost = (29.4 − 18)/29.4 = 0.39

Answer: About 39% of the potential energy becomes heat (without friction she would reach 7.7 m s⁻¹).

Memory hook: Missing mechanical energy is heat.

Topics in this chapter

Work, Energy and Power formula sheet

Work by a constant force: $W = F \cdot s = F s cos θ$; 1 J = $1 0^{7}$ erg.
Work by a variable force: $W = \int_{x_{1}}^{x_{2}} F d x$; Area under the F-x graph.
Kinetic energy: $K = \frac{1}{2} m v^{2} = \frac{p ^{2}}{2 m}$; Doubling p quadruples K.
Work-energy theorem: $W_{n e t} = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}$; Holds for any force.
Potential energy: $Δ U = m g h; U_{s p r in g} = \frac{1}{2} k x^{2}; F = - \frac{d U}{d x}$; mgh only for h ≪ R.
Conservation of mechanical energy: $\frac{1}{2} m v_{1}^{2} + U_{1} = \frac{1}{2} m v_{2}^{2} + U_{2}$; Only conservative forces acting.
Oscillating spring: $\frac{1}{2} m v^{2} + \frac{1}{2} k x^{2} = \frac{1}{2} k A^{2}, v_{ma x} = A \frac{k}{m}$; KE and PE exchange.
Power: $P = \frac{W}{t} = F \cdot v$; 1 hp = 746 W; 1 kWh = $3.6 \times 1 0^{6}$ J.
Elastic collision, target at rest: $v_{1} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} u_{1}, v_{2} = \frac{2 m _{1}}{m _{1} + m _{2}} u_{1}$; Equal masses exchange velocities.
Perfectly inelastic collision: $v = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}$; KE lost fraction $\frac{m _{2}}{m _{1} + m _{2}}$ when $u_{2} = 0$ .
Mass-energy: $E = m c^{2}$; No derivation needed.

Work, Energy and Power: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Force-displacement graph and work

Work done by a variable force; work in stretching a spring.

Draw

Axes: force F on the y-axis, displacement x on the x-axis.
For a spring, a straight line through the origin with slope k up to x.
Shade the triangle under the line from 0 to x and label its area ½kx²; for a general force, shade the area under a curve between $x_{1}$ and $x_{2}$ .

Label

F (N)
x (m)
slope = k
area = W
$x_{1}$
$x_{2}$

Watch out: Note that the area between $x_{1}$ and $x_{2}$ for a spring is a trapezium: ½k( $x_{2}^{2} - x_{1}^{2}$ ).

Energy graph of an oscillating spring

Showing conservation of mechanical energy for a spring-block system.

Draw

Axes: energy on the y-axis, position x from −A to +A on the x-axis.
A horizontal line at height ½kA² for the total energy E.
An upward-opening parabola U = ½kx² touching E at ±A, and a downward parabola K = ½k(A² − x²) reaching E at x = 0.

Label

E = ½kA²
U = ½kx²
K
−A
0
+A

Watch out: At any x, the vertical gap between the U curve and the E line is the kinetic energy.

How the school exam asks Work, Energy and Power

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Concepts of work	When is work positive, negative or zero; work by centripetal force or friction; units.	Use W = Fs cos θ and name θ; give an everyday example for each case.
Variable force	Work from F(x) by integration or from the area of an F-x graph; work in stretching a spring.	Integrate with limits, or split the area into triangles and rectangles.
Work-energy theorem	Prove it; stopping distance; bullet in a block; resistive force.	Write net work = ΔK; include every force's work with its sign.
Conservation of energy	Falling bodies, pendulums, springs, slides; energy lost to friction.	Write K + U at two points; any shortfall is the work done against friction.
Power	Motors, pumps, vehicles on level roads and slopes, efficiency.	P = W/t or P = Fv; for pumps include both mgh and ½mv² per second.
Collisions	Derive elastic-collision velocities; special cases; perfectly inelastic collisions and KE loss; oblique collisions.	Momentum always; KE only if elastic; resolve along x and y for oblique collisions.

Tricks to remember Work, Energy and Power

Short memory aids. Say each one aloud once, then use it on a question the same day.

Work signs by angle
Acute angle, positive; right angle, zero; obtuse angle, negative.
Use it when: Classifying work done by a force.
Half k x squared
Spring energy is the triangle under F = kx: half base (x) times height (kx).
Use it when: Spring energy and the work to stretch it.
Momentum always, energy sometimes
Every collision conserves momentum; only elastic ones also conserve KE.
Use it when: Choosing equations for collision problems.
Equal masses swap
Head-on elastic, equal masses: they swap velocities; oblique: they fly apart at 90°.
Use it when: Special-case collisions.
kWh is energy
kilowatt × hour = energy; 1 kWh = 3.6 MJ (one 'unit' of electricity).
Use it when: Unit conversions.

Work, Energy and Power: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Work done by the centripetal force · 1 mark · MCQExam level

The work done by the centripetal force on a body in uniform circular motion over one revolution is:
(a) $2 π r F$
(b) $π r F$
(c) zero
(d) $\frac{m v ^{2}}{2}$

Try it first, then show the solutionSolution

The force is along the radius and the displacement at every instant is along the tangent: θ = 90°, so W = 0 (the speed does not change).

Answer: (c)

Example 2: Friction round a closed path · 1 mark · Assertion-reasonExam level

Assertion: The work done by friction on a body taken round a closed path is not zero.
Reason: Friction is a non-conservative force.
(a) Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Try it first, then show the solutionSolution

Friction always opposes the motion, so it does negative work on every part of the path; this is what makes it non-conservative.

Answer: (a)

Example 3: Work from an F-x graph · 2 marks · Short answerHarder exam level

A force acting on a body rises uniformly from 0 to 10 N as the body moves from x = 0 to x = 4 m, and then stays at 10 N up to x = 6 m. Find the total work done. [2]

Try it first, then show the solutionSolution

Triangle from 0 to 4 m: ½ × 4 × 10 = 20 J
Rectangle from 4 to 6 m: 2 × 10 = 20 J; total = 40 J

Answer: 40 J

Example 4: Bullet through a plank · 3 marks · Short answerHarder exam level

A 20 g bullet moving at 400 m s⁻¹ stops after penetrating 10 cm into a thick block. If it is fired with the same speed into a 6 cm plank of the same wood, with what speed does it emerge? Assume the same constant resistance. [3]

Try it first, then show the solutionSolution

Resistance: F × 0.10 = ½ × 0.02 × 400² = 1600 J, so F = 16 000 N.
Work against resistance in 6 cm = 16 000 × 0.06 = 960 J; KE left = 1600 − 960 = 640 J.
$v = \frac{2 \times 640}{0.02} = 253 m s^{- 1}$

Answer: About 253 m s⁻¹

Example 5: Block hitting a spring · 3 marks · Short answerHarder exam level

A 2 kg block slides at 4 m s⁻¹ on a smooth floor and hits a horizontal spring of constant 800 N m⁻¹. Find the maximum compression, and the speed of the block when the compression is 0.1 m. [3]

Try it first, then show the solutionSolution

All KE becomes spring PE: ½ × 2 × 16 = ½ × 800 × x², so x² = 0.04 and x = 0.2 m.
At x = 0.1 m: spring PE = ½ × 800 × 0.01 = 4 J, so KE = 16 − 4 = 12 J.
½ × 2 × v² = 12, v = 3.46 m s⁻¹

Answer: 0.2 m; 3.46 m s⁻¹

Example 6: Elastic collision with a target at rest · 5 marks · Long answerHarder exam level

(a) A body of mass $m_{1}$ moving at $u_{1}$ collides head-on and elastically with a body of mass $m_{2}$ at rest. Derive their velocities after the collision. [3]
(b) Show that if the masses are equal, they exchange velocities. [1]
(c) What fraction of its kinetic energy does a neutron (1 u) lose in a head-on elastic collision with a carbon nucleus (12 u) at rest? [1]

Try it first, then show the solutionSolution

(a) Momentum: $m_{1} u_{1} = m_{1} v_{1} + m_{2} v_{2}$ , so $m_{1} (u_{1} - v_{1}) = m_{2} v_{2}$ … (1)
KE: $m_{1} (u_{1}^{2} - v_{1}^{2}) = m_{2} v_{2}^{2}$ … (2). Dividing (2) by (1): $u_{1} + v_{1} = v_{2}$ .
Substituting in (1): $v_{1} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} u_{1}$ , $v_{2} = \frac{2 m _{1}}{m _{1} + m _{2}} u_{1}$ .
(b) If $m_{1} = m_{2}$ : $v_{1} = 0$ and $v_{2} = u_{1}$ : the velocities are exchanged.
(c) Fraction transferred = $\frac{4 m _{1} m _{2}}{( m _{1} + m _{2} ) ^{2}} = \frac{48}{169} = 0.28$

Answer: (c) About 28% of its KE

Example 7: Ballistic pendulum · 3 marks · Short answerHOTS / competency

A 10 g bullet is fired into a 990 g wooden block hanging from a long string. The block, with the bullet embedded, swings up through a vertical height of 0.2 m. Find the speed of the bullet and the fraction of its KE lost in the impact. (g = 9.8 m s⁻²) [3]

Try it first, then show the solutionSolution

After impact, energy is conserved in the swing: $V = 2 g h = 2 \times 9.8 \times 0.2 = 1.98 m s^{- 1}$ .
During impact, only momentum is conserved: 0.01u = 1.0 × 1.98, so u = 198 m s⁻¹.
Fraction of KE lost = $\frac{M}{m + M} = \frac{0.99}{1.00} = 99%$

Answer: About 198 m s⁻¹; 99% of its KE is lost

Example 8: Power of a pump · 3 marks · Short answerHOTS / competency

A pump raises 600 kg of water per minute through a height of 25 m and delivers it at 10 m s⁻¹. Find the output power. If the pump is 80% efficient, what is its input power? (g = 9.8 m s⁻²) [3]

Try it first, then show the solutionSolution

Energy per minute = mgh + ½mv² = 600 × 9.8 × 25 + ½ × 600 × 100 = 147 000 + 30 000 = 177 000 J
Output power = 177 000/60 = 2950 W
Input = 2950/0.8 = 3690 W

Answer: 2.95 kW output; about 3.7 kW input

Case-study practice (Section D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 1 + 2 + 1 marks.

Trolleys on an air track

Suppose, on a frictionless air track, a 0.4 kg trolley moving at 0.5 m s⁻¹ collides head-on with a 0.6 kg trolley at rest. In one run the trolleys carry Velcro pads and stick together; in another they carry springy bumpers and the collision is elastic.

(i) · 1 mark
Find the common velocity when they stick together.
Answer
Answer: 0.2 m s⁻¹
0.4 × 0.5 = 1.0 × v, v = 0.2 m s⁻¹.
(ii) · 1 mark
How much kinetic energy is lost in that collision?
Answer
Answer: 0.03 J
Before: ½ × 0.4 × 0.25 = 0.05 J; after: ½ × 1.0 × 0.04 = 0.02 J; loss 0.03 J.
(iii) · 2 marks
Find both velocities after the elastic collision.
Answer
Answer: −0.1 m s⁻¹ and +0.4 m s⁻¹
$v_{1} = \frac{0.4 - 0.6}{1.0} \times 0.5 = - 0.1$ m s⁻¹ (rebounds); $v_{2} = \frac{0.8}{1.0} \times 0.5 = 0.4$ m s⁻¹. KE after = 0.002 + 0.048 = 0.05 J ✓.
(iv) · 1 mark
Which quantity is conserved in both runs?
Answer
Answer: Total linear momentum (0.2 kg m s⁻¹)
No external horizontal force acts in either run.

A spring-loaded toy launcher

Suppose a toy launcher has a spring of constant 400 N m⁻¹. It is compressed by 5 cm and launches a 10 g ball vertically upward. (g = 9.8 m s⁻², ignore air resistance)

(i) · 1 mark
Find the energy stored in the spring.
Answer
Answer: 0.5 J
½ × 400 × 0.05² = 0.5 J.
(ii) · 1 mark
Find the launch speed of the ball.
Answer
Answer: 10 m s⁻¹
½ × 0.01 × v² = 0.5, v = 10 m s⁻¹.
(iii) · 1 mark
How high does the ball rise?
Answer
Answer: About 5.1 m
h = 0.5/(0.01 × 9.8) = 5.1 m.
(iv) · 2 marks
How much extra work is needed to compress the spring from 5 cm to 10 cm?
Answer
Answer: 1.5 J
½ × 400 × (0.10² − 0.05²) = 200 × 0.0075 = 1.5 J.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

A force $F = (6 x + 2)$ N acts on a body along the x-axis. Find the work done as the body moves from x = 0 to x = 2 m. (2 marks)

Step	Marks
$W = \int_{0}^{2} (6 x + 2) d x = [3 x^{2} + 2 x]_{0}^{2}$	1
W = 12 + 4 = 16 J	1

A 1000 kg car climbs a slope rising 1 m for every 20 m along the road at a steady 10 m s⁻¹ against a resistance of 500 N. Find the power of the engine. (g = 9.8 m s⁻²) (3 marks)

Step	Marks
Component of weight along the slope = mg sin θ = 1000 × 9.8/20 = 490 N	1
Total force needed = 500 + 490 = 990 N	1
P = Fv = 990 × 10 = 9900 W ≈ 9.9 kW	1

State and prove the work-energy theorem for a variable force. A 2 kg body speeds up from 3 m s⁻¹ to 7 m s⁻¹; find the net work done on it. (5 marks)

Step	Marks
Statement: net work done on a body = change in its kinetic energy	1
$F = m \frac{d v}{d t} = m v \frac{d v}{d x}$	1
$W = \int F d x = \int_{u}^{v} m v d v$	1
$W = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}$	1
W = ½ × 2 × (49 − 9) = 40 J	1

Where students lose marks in Work, Energy and Power

The slips examiners see most often, and what to write instead.

Mistake: Using ½kx² for the work between two extensions
Students compute ½k(x₂ − x₁)² instead of the difference of the stored energies.
Wrong working
From 5 cm to 10 cm with k = 800 N m⁻¹: ½ × 800 × 0.05² = 1 J.
Right working
W = ½k(x₂² − x₁²) = 400 × (0.01 − 0.0025) = 3 J.
Do this: Write U at each extension, then subtract.
Usually costs: 1-2 marks.
Mistake: Conserving kinetic energy in an inelastic collision
In ballistic-pendulum and sticking problems, students equate KE before and after the impact.
Wrong working
½mu² = ½(m + M)V² for the bullet entering the block.
Right working
Impact: momentum only, mu = (m + M)V. Swing afterwards: energy, ½(m + M)V² = (m + M)gh.
Do this: Split the problem into the collision (momentum) and the motion after it (energy).
Usually costs: 2-3 marks.
Mistake: Forgetting the work done by friction or gravity in the work-energy theorem
Students use only the applied force, not the net force, so the answer is too large.
Wrong working
Pushing a box with 50 N over 4 m against 20 N friction: ΔK = 200 J.
Right working
ΔK = (50 − 20) × 4 = 120 J.
Do this: List every force and its work (with sign) before adding.
Usually costs: 1 mark.
Mistake: Treating kWh as a unit of power
The 'kilowatt' in the name misleads students.
Wrong working
A heater uses 2 kWh of power.
Right working
A 2 kW heater used for 1 hour uses 2 kWh of energy = 7.2 × 10⁶ J.
Do this: kW is power; kWh is energy.
Usually costs: ½-1 mark.
Mistake: Dropping signs of velocities in collisions
When the lighter body rebounds, students give a positive answer and the momentum check fails.
Wrong working
0.4 kg at 0.5 m s⁻¹ hits 0.6 kg at rest: $v_{1}$ = 0.1 m s⁻¹ forward.
Right working
$v_{1} = - 0.1$ m s⁻¹: the lighter trolley rebounds.
Do this: Keep the sign from the formula and check momentum at the end.
Usually costs: 1 mark.

Practice questions on Work, Energy and Power

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
1 kWh is equal to:
(a) 3.6 × 10³ J
(b) 3.6 × 10⁵ J
(c) 3.6 × 10⁶ J
(d) 1000 J
Hint
1000 W × 3600 s.
Answer
(c)
Q2 · 1 mark · MCQExam level
In a perfectly elastic collision:
(a) only momentum is conserved
(b) only KE is conserved
(c) both momentum and KE are conserved
(d) neither is conserved
Hint
Definition of elastic.
Answer
(c)
Q3 · 1 mark · Very short answerExam level
What is a conservative force? Give one example.
Hint
Path independence.
Answer
A force whose work between two points does not depend on the path (zero round a closed path), e.g. gravity or a spring force.
Q4 · 1 mark · Very short answerExam level
Give one example where a force does negative work.
Hint
Opposes motion.
Answer
Friction on a sliding block (or gravity on a ball rising upward).
Q5 · 2 marks · Short answerExam level
Show that $K = \frac{p ^{2}}{2 m}$ . By what factor does K change if the momentum is doubled?
Hint
Substitute v = p/m.
Answer
K = ½mv² = ½m(p/m)² = p²/2m. Doubling p makes K four times as large.
Q6 · 2 marks · Short answerHarder exam level
Bodies of mass 1 kg and 4 kg have equal kinetic energies. Find the ratio of their momenta.
Hint
$p = 2 m K$ .
Answer
$p_{1} : p_{2} = 1 : 4 = 1 : 2$ .
Q7 · 2 marks · Short answerHarder exam level
A force $F = (2 \hat{i} + 3 \hat{j})$ N moves a body from the point (1, 2) m to (4, 6) m. Find the work done.
Hint
s = (3, 4).
Answer
W = F · s = 2(3) + 3(4) = 18 J.
Q8 · 2 marks · Short answerHarder exam level
Find the work done in stretching a spring of constant 500 N m⁻¹ from an extension of 2 cm to 6 cm.
Hint
Difference of ½kx².
Answer
½ × 500 × (0.06² − 0.02²) = 250 × 0.0032 = 0.8 J.
Q9 · 2 marks · Short answerHarder exam level
A 1200 kg van moves at a steady 15 m s⁻¹ up a slope of 1 in 30 (measured along the road) against a resistance of 600 N. Find the power needed. (g = 9.8 m s⁻²)
Hint
Add mg sin θ to the resistance.
Answer
mg sin θ = 1200 × 9.8/30 = 392 N; F = 600 + 392 = 992 N; P = 992 × 15 = 14 880 W ≈ 14.9 kW.
Q10 · 3 marks · Short answerHarder exam level
A 0.2 kg pendulum bob on a 1 m string is released from rest with the string at 60° to the vertical. Find its speed and the tension in the string at the lowest point. (g = 9.8 m s⁻²)
Hint
h = l(1 − cos 60°).
Answer
h = 0.5 m; v = √(2 × 9.8 × 0.5) = 3.13 m s⁻¹; T = mg + mv²/l = 1.96 + 0.2 × 9.8 = 3.92 N.
Q11 · 3 marks · Short answerHarder exam level
A 2 kg ball moving at 5 m s⁻¹ along the x-axis collides with a 3 kg ball at rest. After the collision the 2 kg ball moves at 1 m s⁻¹ along the y-axis. Find the velocity of the 3 kg ball and state whether the collision is elastic.
Hint
Conserve momentum along x and y separately.
Answer
x: 10 = 3vₓ, vₓ = 3.33 m s⁻¹; y: 0 = 2(1) + 3v_y, v_y = −0.67 m s⁻¹. Speed 3.40 m s⁻¹ at 11.3° below the x-axis. KE: 25 J before, 1 + 17.3 = 18.3 J after, so it is inelastic.
Q12 · 5 marks · Long answerHarder exam level
(a) A body of mass $m_{1}$ moving at u collides with a body of mass $m_{2}$ at rest, and they stick together. Show that the fraction of kinetic energy lost is $\frac{m _{2}}{m _{1} + m _{2}}$ . [2]
(b) Define a conservative force and show that the work done by gravity in taking a body round a closed path is zero. [2]
(c) Express 1 kWh in joule. [1]
Hint
v = m₁u/(m₁ + m₂).
Answer
(a) $K_{f} = \frac{1}{2} (m_{1} + m_{2}) (\frac{m _{1} u}{m _{1} + m _{2}})^{2} = \frac{m _{1}}{m _{1} + m _{2}} K_{i}$ , so the fraction lost is $1 - \frac{m _{1}}{m _{1} + m _{2}} = \frac{m _{2}}{m _{1} + m _{2}}$ . (b) A conservative force does path-independent work. Going up a height h gravity does −mgh, coming down it does +mgh (horizontal parts do none), so the total is zero. (c) 1 kWh = 1000 × 3600 = $3.6 \times 1 0^{6}$ J.
Q13 · 3 marks · Short answerHOTS / competency
A uniform chain of mass 2 kg and length 1 m lies on a smooth table with one-third of its length hanging over the edge. Find the work needed to pull the hanging part back onto the table. (g = 9.8 m s⁻²)
Hint
Lift the centre of mass of the hanging part.
Answer
Hanging mass = 2/3 kg; its centre of mass is 1/6 m below the table top. W = (2/3) × 9.8 × (1/6) = 1.09 J.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.When is the work done by a force zero?Show answer
When there is no displacement or the force is perpendicular to the displacement.
2.State the work-energy theorem.Show answer
The net work done on a body equals the change in its kinetic energy.
3.PE stored in a spring?Show answer
½kx².
4.Name a non-conservative force.Show answer
Friction (or air resistance).
5.1 horsepower in watt?Show answer
746 W.
6.What happens when equal masses collide head-on elastically, one at rest?Show answer
They exchange velocities.
7.What is conserved in an inelastic collision?Show answer
Linear momentum (and total energy), but not kinetic energy.
8.Instantaneous power in vector form?Show answer
P = F · v.

Best order to study it

Learn W = Fs cos θ with positive, negative and zero work.
Learn work by a variable force (integral and F-x area) and the spring's ½kx².
Prove the work-energy theorem and use it for stopping problems.
Learn conservative and non-conservative forces, potential energy and conservation of mechanical energy (falling body, oscillating spring).
Learn power, P = Fv, and units (W, hp, kWh); practise vehicles and pumps.
Derive one-dimensional elastic-collision velocities, learn the special cases, then practise inelastic and oblique collisions.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you give examples of positive, negative and zero work?
Can you find work from F(x) by integration and from an F-x graph?
Can you prove the work-energy theorem?
Can you state two properties of a conservative force?
Can you show that mechanical energy is conserved for a falling body and a spring?
Can you find the power of a pump or a car on a slope?
Can you derive the velocities after a one-dimensional elastic collision?
Can you solve an oblique collision by resolving momentum?

Work, Energy and Power: frequently asked questions

Do I need to derive E = mc²?

No. The syllabus says E = mc² with no derivation. You should know what it means and use it, for example that 1 g of mass is equivalent to 9 × 10¹³ J.

Are two-dimensional collisions in the ISC Class 11 syllabus?

Yes. The syllabus includes elastic and inelastic collisions in one and two dimensions, with oblique collisions named explicitly. Conserve momentum along x and y separately.

Which one-dimensional collision results must I learn?

The velocity equations for unequal masses with the second body at rest, and the special cases: equal masses exchange velocities, a heavy body hitting a light one, and a light body hitting a heavy one.

Is mechanical energy conserved when friction acts?

No. Friction converts mechanical energy into heat. Total energy is still conserved, so the loss of mechanical energy equals the work done against friction.

Why is mgh valid only near the Earth's surface?

It assumes g is constant over the height h. For heights comparable with the Earth's radius, use the gravitational potential energy −GMm/r, studied in Gravitation.

Can kinetic energy be negative?

No. Since K = ½mv² and both m and v² are non-negative, kinetic energy is never negative. Potential energy can be negative, depending on where its zero is chosen.

Based on: CISCE ISC 2028 Physics syllabus (Classes XI and XII); CISCE Regulations for the ISC Examination Year 2028 (Class XI internal examination)