PhysicsISC Class 11 · Work, Energy and Power

Collisions: Elastic, Inelastic and Oblique, with Restitution

Updated 17 October 2026 · Fact-checked

In every collision the total linear momentum is conserved. In an elastic collision kinetic energy is also conserved and the coefficient of restitution e = 1. In a perfectly inelastic collision the bodies move together (e = 0). For a general collision, e = (speed of separation)/(speed of approach).

Momentum is conserved in both; in the elastic case the first ball stops and the second moves off, while in the inelastic case they stick and move slower.

Illustration drawn for QuestionBuddy.

Understand Collisions in One and Two Dimensions

Conservation laws. Linear momentum is conserved because the internal forces are equal and opposite. Kinetic energy is conserved only in elastic collisions; it decreases in inelastic ones.

Elastic head-on collision. With velocities u₁, u₂ before and v₁, v₂ after: v₁ = [(m₁ − m₂)u₁ + 2m₂u₂]/(m₁ + m₂) and v₂ = [(m₂ − m₁)u₂ + 2m₁u₁]/(m₁ + m₂). Equal masses exchange velocities. A light body hitting a heavy stationary one bounces back with nearly the same speed; a heavy body hitting a light one continues with nearly the same speed.

Coefficient of restitution. e = (v₂ − v₁)/(u₁ − u₂), with 0 ≤ e ≤ 1. For a ball dropped from height h that rebounds to h′, e = √(h′/h). After n bounces the height is e²ⁿh.

Perfectly inelastic collision. The bodies stick: v = (m₁u₁ + m₂u₂)/(m₁ + m₂), and the loss of kinetic energy is ½[m₁m₂/(m₁ + m₂)](u₁ − u₂)². The ballistic pendulum uses this: the bullet embeds in a block, then the block swings up by h = v²/2g.

Two dimensions. Momentum is conserved along each axis separately. For an elastic collision of equal masses with one at rest, the two bodies move off at right angles to each other.

Collisions in One and Two Dimensions: formulas and key facts

Momentum: $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$; Valid for every collision.
Restitution: $e = (v_{2} - v_{1}) / (u_{1} - u_{2})$; e = 1 elastic; e = 0 perfectly inelastic.
Perfectly inelastic: $v = (m_{1} u_{1} + m_{2} u_{2}) / (m_{1} + m_{2})$; Loss of K = ½ μ (u₁ − u₂)², μ = m₁m₂/(m₁ + m₂).
Bouncing ball: $e = \sqrt (h'/ h) ∣ h_{n} = e^{2 n} h$; Height after n rebounds.

Collisions in One and Two Dimensions in real life

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.

A bullet in a wooden block

Suppose a 20 g bullet moving at 300 m s⁻¹ strikes and embeds in a hanging 1.98 kg wooden block (g = 10 m s⁻²).

Find the speed just after impact, the height the block rises and the fraction of kinetic energy lost.

Try it first, then show the solutionSolution

Momentum: 0.02 × 300 = 2.00 × v → v = 3 m s⁻¹.
h = v²/2g = 9/20 = 0.45 m.
K before = ½ × 0.02 × 300² = 900 J; K after = ½ × 2 × 9 = 9 J; fraction lost = 891/900 = 99%.

Answer: 3 m s⁻¹; 0.45 m; 99% of the kinetic energy is lost.

Memory hook: Momentum is conserved in the collision; energy is conserved only afterwards, in the swing.

Tricks to remember Collisions in One and Two Dimensions

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

Equal masses, one at rest, elastic: swap
The moving body stops and the other takes its velocity.
Use it when: Billiard-ball style questions.
e of a bouncing ball
e = √(rebound height / drop height).
Use it when: Bounce problems without velocities.

Collisions in One and Two Dimensions: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Exchange of velocities · 1 mark · MCQBoard level

A body collides head-on and elastically with another of equal mass at rest. After the collision:
(a) both move with half the speed
(b) the first stops and the second moves with the original speed
(c) both stop
(d) the first reverses

Try it first, then show the solutionSolution

Putting m₁ = m₂ and u₂ = 0 in the elastic formulae gives v₁ = 0 and v₂ = u₁.

Answer: (b)

Example 2: Perfectly inelastic collision · 3 marks · Short answerBoard level

A 2 kg body moving at 6 m s⁻¹ collides with a 4 kg body at rest and they stick together. Find the common velocity and the loss of kinetic energy.

Try it first, then show the solutionSolution

2 × 6 = 6 × v → v = 2 m s⁻¹. (1 mark)
K before = ½ × 2 × 36 = 36 J; K after = ½ × 6 × 4 = 12 J. (1 mark)
Loss = 24 J. (1 mark)

Answer: v = 2 m s⁻¹; loss 24 J.

Example 3: Elastic head-on collision · 4 marks · Short answerHarder board level

A 3 kg body moving at 4 m s⁻¹ collides elastically and head-on with a 1 kg body at rest. Find the velocities after the collision and verify the conservation of momentum and energy.

Try it first, then show the solutionSolution

v₁ = (3 − 1)/(3 + 1) × 4 = 2 m s⁻¹. (1 mark)
v₂ = 2 × 3/(3 + 1) × 4 = 6 m s⁻¹. (1 mark)
Momentum: 12 before; 3 × 2 + 1 × 6 = 12 after. (1 mark)
Energy: ½ × 3 × 16 = 24 J before; ½ × 3 × 4 + ½ × 1 × 36 = 6 + 18 = 24 J after. (1 mark)

Answer: v₁ = 2 m s⁻¹, v₂ = 6 m s⁻¹; both conserved.

Example 4: Ballistic pendulum · 4 marks · Short answerHarder board level

A 20 g bullet moving horizontally at 300 m s⁻¹ embeds in a 1.98 kg block hanging from strings. Find the height through which the block rises, and the percentage of the bullet's kinetic energy that is lost. (g = 10 m s⁻²)

Try it first, then show the solutionSolution

0.02 × 300 = 2 v → v = 3 m s⁻¹. (1 mark)
h = v²/(2g) = 9/20 = 0.45 m. (1 mark)
K₁ = 900 J, K₂ = ½ × 2 × 9 = 9 J. (1 mark)
Loss = 99%. (1 mark)

Answer: h = 0.45 m; 99% lost.

Example 5: Oblique collision · 5 marks · Long answerHOTS / competency

A ball of mass m moving at 5 m s⁻¹ collides elastically and not head-on with an identical ball at rest. After the collision the first ball moves at 30° to its original direction. Find the speed and direction of each ball.

Try it first, then show the solutionSolution

For equal masses in an elastic collision with one at rest, the paths after the collision are at 90°. (1 mark)
So the second ball moves at 60° on the other side of the original line. (1 mark)
Momentum along the original line and perpendicular to it: speeds v_A = 5 cos30° = 4.33 m s⁻¹ and v_B = 5 cos60° = 2.5 m s⁻¹. (1 mark)
Check along x: 4.33 cos30° + 2.5 cos60° = 3.75 + 1.25 = 5. Check along y: 4.33 sin30° = 2.165 = 2.5 sin60°. (1 mark)
Check energy: 25 = 18.75 + 6.25 (in units of m/2). (1 mark)

Answer: First ball 4.33 m s⁻¹ at 30°; second ball 2.5 m s⁻¹ at 60° on the other side.

Where students lose marks in Collisions in One and Two Dimensions

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

Mistake: Conserving kinetic energy in an inelastic collision
Writing ½m₁u₁² + ½m₂u₂² = ½m₁v₁² + ½m₂v₂² for sticking bodies.
Do this: Only momentum is conserved unless the collision is stated to be elastic.
Usually costs: 2 marks.
Mistake: Sign errors in opposite-direction collisions
Adding speeds as positive when one body moves the other way.
Do this: Choose a positive direction and give the opposite velocity a negative sign.
Usually costs: 1 to 2 marks.
Mistake: Applying momentum conservation in the swing of the pendulum
Using momentum conservation for the rise after impact.
Do this: Momentum is conserved during the collision; energy is conserved during the swing.
Usually costs: 1 to 2 marks.

Exam tips

Draw before and after pictures with arrows and a positive direction.
Write momentum conservation first; add the energy or restitution condition second.
State clearly whether the collision is elastic or inelastic.
For 2-D problems resolve momentum along two perpendicular axes.

Practice questions on Collisions in One and Two Dimensions

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 · MCQBoard level
In a perfectly inelastic collision the relative speed of separation is:
(a) equal to the speed of approach
(b) half the speed of approach
(c) zero
(d) twice the speed of approach
Hint
The bodies move together.
Answer
(c) zero
Q2 · 2 marks · Short answerBoard level
A 5 kg body at 2 m s⁻¹ strikes a 3 kg body at rest and they stick together. Find their common velocity.
Hint
Conserve momentum.
Answer
v = (5 × 2)/8 = 1.25 m s⁻¹.
Q3 · 3 marks · Short answerHarder board level
A ball is dropped from 4 m on a floor and the coefficient of restitution is 0.5 (g = 10 m s⁻²). Find the height after the first and the second rebound.
Hint
hₙ = e²ⁿ h.
Answer
h₁ = 0.25 × 4 = 1 m. h₂ = 0.0625 × 4 = 0.25 m.
Q4 · 3 marks · Short answerHarder board level
A 2 kg body moving at 5 m s⁻¹ collides head-on with a 3 kg body moving opposite at 2 m s⁻¹ and they stick together. Find the common velocity and the loss of kinetic energy.
Hint
Take the first direction as positive.
Answer
v = (10 − 6)/5 = 0.8 m s⁻¹. K before = 25 + 6 = 31 J; K after = ½ × 5 × 0.64 = 1.6 J; loss 29.4 J.
Q5 · 4 marks · Short answerHarder board level
A 4 kg body at 6 m s⁻¹ catches up with a 2 kg body at 2 m s⁻¹ in the same direction. After the collision the 4 kg body moves at 4 m s⁻¹. Find the velocity of the 2 kg body, e, and the loss of kinetic energy.
Hint
Use momentum, then e = separation/approach.
Answer
28 = 16 + 2v → v = 6 m s⁻¹. e = (6 − 4)/(6 − 2) = 0.5. K before = 72 + 4 = 76 J; after = 32 + 36 = 68 J; loss 8 J.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a particle of mass m collides head-on and elastically with a stationary particle of mass 12m.
(i) Find the velocity of the first particle after the collision as a fraction of its initial velocity u. (1 mark)
(ii) Find the fraction of its kinetic energy it retains. (1 mark)
(iii) What fraction does it lose if the target also has mass m? (1 mark)
(iv) What happens to the retained fraction as the target mass becomes very large? (1 mark)
Hint
v₁ = (m₁ − m₂)u/(m₁ + m₂).
Answer
(i) v₁ = (1 − 12)/13 u = −11u/13 = −0.846u (it rebounds). (ii) (11/13)² = 121/169 = 0.716, so 71.6% is retained. (iii) With equal masses it stops, losing 100%. (iv) It tends to 100%: the light particle bounces back with almost its full speed.

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.Is momentum conserved in an inelastic collision?Show answer
Yes, always.
2.Define the coefficient of restitution.Show answer
Speed of separation divided by speed of approach.
3.What happens in an elastic collision of equal masses (one at rest)?Show answer
They exchange velocities.
4.Angle between paths after an oblique elastic collision of equal masses (one at rest)?Show answer
90°.

Collisions in One and Two Dimensions: frequently asked questions

Where does the lost kinetic energy go?

It becomes heat, sound and the energy of deformation of the colliding bodies. Total energy is conserved, only mechanical energy is lost.

Why is momentum conserved even in a perfectly inelastic collision?

The forces between the two bodies are internal and equal and opposite, so the total momentum of the pair does not change, whatever happens to the kinetic energy.

Can e be greater than 1?

No. That would mean the bodies separate faster than they approached and create kinetic energy from nothing in a normal collision.