PhysicsCBSE Class 11 · Work, Energy and Power

Elastic and Inelastic Collisions in One Dimension Class 11

Updated 17 October 2026 · Fact-checked

In every collision with no external impulse, momentum is conserved. In an elastic collision kinetic energy is also conserved, and the final velocities are v₁ = [(m₁ − m₂)u₁ + 2m₂u₂]/(m₁ + m₂) and v₂ = [(m₂ − m₁)u₂ + 2m₁u₁]/(m₁ + m₂). In a perfectly inelastic collision the bodies move together and the maximum kinetic energy is lost.

Momentum is conserved in both collisions, but only the elastic one keeps kinetic energy; sticking together loses it.

Illustration drawn for QuestionBuddy.

Understand Elastic and Inelastic Collisions in One Dimension

A collision is a short interaction in which the internal impulsive forces are far larger than any external force, so the total momentum is conserved in all types.

Elastic collision. Kinetic energy is also conserved. For masses $m_{1}, m_{2}$ with initial velocities $u_{1}, u_{2}$ along a line:
$ $v_{1} = \frac{( m _{1} - m _{2} ) u _{1} + 2 m _{2} u _{2}}{m _{1} + m _{2}}, v_{2} = \frac{( m _{2} - m _{1} ) u _{2} + 2 m _{1} u _{1}}{m _{1} + m _{2}} .$ $
The relative velocity of approach equals the relative velocity of separation, $u_{1} - u_{2} = v_{2} - v_{1}$ . Special cases (with $u_{2} = 0$ ): equal masses exchange velocities; a light body hitting a heavy one at rest rebounds with nearly its own speed; a heavy body hitting a light one at rest barely slows and drives the light one at nearly twice its speed.

Fraction of kinetic energy transferred from $m_{1}$ (speed $u_{1}$ ) to $m_{2}$ at rest in an elastic collision is $\frac{4 m _{1} m _{2}}{( m _{1} + m _{2} ) ^{2}}$ , which is greatest (100%) for equal masses. This is why light moderators (hydrogen) slow fast neutrons better than heavy ones.

Perfectly inelastic collision. The bodies stick: $v = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}$ . The kinetic energy lost is $\frac{1}{2} \frac{m _{1} m _{2}}{m _{1} + m _{2}} (u_{1} - u_{2})^{2}$ , the largest possible loss consistent with momentum conservation.

Partly inelastic. The bodies separate but kinetic energy is lost. A ball dropped from $h_{1}$ that bounces to $h_{2}$ has speeds $2 g h_{1}$ and $2 g h_{2}$ ; the fraction of kinetic energy lost is $1 - h_{2} / h_{1}$ .

Ballistic pendulum. A bullet of mass $m$ embeds in a block of mass $M$ hanging as a pendulum which rises a height $h$ . Collision: $m v = (m + M) V$ (momentum). After the collision energy is conserved: $V = 2 g h$ . Energy is lost in the collision, not conserved during it.

Elastic and Inelastic Collisions in One Dimension: formulas and key facts

Momentum conservation: $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$; All collisions.
Elastic: v₁: $v_{1} = \frac{( m _{1} - m _{2} ) u _{1} + 2 m _{2} u _{2}}{m _{1} + m _{2}}$; Kinetic energy conserved.
Elastic: v₂: $v_{2} = \frac{( m _{2} - m _{1} ) u _{2} + 2 m _{1} u _{1}}{m _{1} + m _{2}}$; Kinetic energy conserved.
Perfectly inelastic: $v = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}$; Bodies move together.
Energy fraction transferred: $\frac{Δ K _{2}}{K _{1}} = \frac{4 m _{1} m _{2}}{( m _{1} + m _{2} ) ^{2}}$; Target at rest, elastic.
Kinetic energy lost (inelastic): $Δ K = \frac{1}{2} \frac{m _{1} m _{2}}{m _{1} + m _{2}} (u_{1} - u_{2})^{2}$; Perfectly inelastic.

How to solve Elastic and Inelastic Collisions in One Dimension questions

Write momentum conservation first; add the energy condition only if the collision is elastic.

1Choose a positive direction and give each initial velocity a sign.
2Write conservation of momentum for the system.
3If elastic, add conservation of kinetic energy (or the relative-velocity rule) and solve for the two final velocities.
4If perfectly inelastic, use a single common velocity.
5Compute kinetic energy before and after to find the loss.
6After the collision, use energy methods or kinematics for what happens next (pendulum, sliding).

Elastic and Inelastic Collisions in One Dimension 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.

Railway wagons coupling

Suppose a 20-tonne wagon moving at 3 m/s couples with an identical wagon at rest on level track.

Find the speed after coupling and the fraction of kinetic energy lost.

Try it first, then show the solutionSolution

v = (20 × 3)/(40) = 1.5 m/s.
K before = ½ × 20000 × 9 = 90000 J; after = ½ × 40000 × 2.25 = 45000 J.
Fraction lost = 45000/90000 = 1/2.

Answer: 1.5 m/s; half the energy is lost.

Memory hook: Equal masses sticking lose half the kinetic energy.

A bouncing ball

Suppose a ball dropped from 2.0 m rebounds to 1.25 m (g = 10 m/s²).

Find the fraction of its kinetic energy lost at the bounce.

Try it first, then show the solutionSolution

Fraction lost = 1 − 1.25/2.0 = 0.375.

Answer: 37.5%.

Memory hook: Heights give the energy fraction directly.

Tricks to remember Elastic and Inelastic Collisions in One Dimension

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

Equal masses swap
Elastic, equal masses: velocities exchange.
Use it when: Quick answers.
Momentum always, energy sometimes
Momentum conserved in every collision; energy only in elastic.
Use it when: Which law to use.
Approach = separation
In elastic collisions u₁ − u₂ = v₂ − v₁.
Use it when: A faster route than energy.
Stuck means inelastic
Perfectly inelastic: one final velocity.
Use it when: Spotting the type.

Elastic and Inelastic Collisions in One Dimension: solved examples

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

Example 1: Perfectly inelastic collision · 2 marks · Very short answerBoard level

A 3 kg block moving at 4 m/s collides with a 1 kg block at rest, and they stick together. Find their common velocity and the kinetic energy lost.

Try it first, then show the solutionSolution

v = (3 × 4)/(3 + 1) = 3 m/s. (1)
K before = ½ × 3 × 16 = 24 J; after = ½ × 4 × 9 = 18 J, so loss = 6 J. (Check: ½ × (3/4) × 16 = 6 J.) (1)

Answer: 3 m/s; 6 J.

Example 2: Elastic collision of unequal masses · 3 marks · Short answerHarder board level

A 2 kg body moving at 6 m/s overtakes a 3 kg body moving at 1 m/s in the same direction, and the collision is elastic. Find both velocities after the collision and verify kinetic energy conservation.

Try it first, then show the solutionSolution

v₁ = [(2 − 3) × 6 + 2 × 3 × 1]/5 = 0. (1)
v₂ = [(3 − 2) × 1 + 2 × 2 × 6]/5 = 25/5 = 5 m/s. (1)
K before = ½ × 2 × 36 + ½ × 3 × 1 = 37.5 J; after = 0 + ½ × 3 × 25 = 37.5 J. (1)

Answer: 0 and 5 m/s; kinetic energy conserved.

Example 3: A bouncing ball · 3 marks · Short answerHarder board level

A ball dropped from 1.8 m rebounds to a height of 0.8 m (g = 10 m/s²). Find its speeds just before and just after the bounce and the fraction of kinetic energy lost.

Try it first, then show the solutionSolution

v before = √(2 × 10 × 1.8) = 6 m/s. (1)
v after = √(2 × 10 × 0.8) = 4 m/s. (1)
Fraction lost = 1 − 16/36 = 5/9 ≈ 56%. (1)

Answer: 6 m/s and 4 m/s; 5/9 lost.

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

A 20 g bullet is fired horizontally into a 1.98 kg wooden block hung by strings. The block with the bullet rises 0.2 m (g = 10 m/s²). Find the bullet's speed and the kinetic energy lost in the collision.

Try it first, then show the solutionSolution

After impact V = √(2gh) = √(2 × 10 × 0.2) = 2 m/s. (1)
Momentum: 0.02 v = 2 × 2, so v = 200 m/s. (1)
K before = ½ × 0.02 × 200² = 400 J; just after = ½ × 2 × 4 = 4 J. (1)
Loss = 396 J (99%), turned into heat and deformation. (1)

Answer: 200 m/s; 396 J.

Example 5: Energy transfer and moderation · 4 marks · Long answerHOTS / competency

In an elastic head-on collision a moving particle of mass m₁ strikes a stationary particle of mass m₂. (a) Show that the fraction of kinetic energy transferred is 4m₁m₂/(m₁ + m₂)². (b) Find it for m₂ = 3m₁, m₂ = 2m₁ and m₂ = 12m₁. (c) Which target slows the particle most effectively?

Try it first, then show the solutionSolution

(a) v₂ = 2m₁u/(m₁ + m₂), so ½m₂v₂²/(½m₁u²) = m₂ × 4m₁²/(m₁ + m₂)² ÷ m₁ = 4m₁m₂/(m₁ + m₂)². (1)
(b) m₂ = 3m₁: 12/16 = 75%. m₂ = 2m₁: 8/9 ≈ 88.9%. (1)
m₂ = 12m₁: 48/169 ≈ 28.4%. (1)
(c) The fraction is 100% for equal masses and falls for very different masses, so a target of comparable mass slows the particle most. (1)

Answer: (a) derivation; (b) 75%, 88.9%, 28.4%; (c) a target with mass close to the projectile's.

Where students lose marks in Elastic and Inelastic Collisions in One Dimension

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

Mistake: Using the elastic formulae for a sticking collision.
Not reading the type of collision.
Do this: If bodies stick, use a single final velocity.
Usually costs: The answer mark.
Mistake: Forgetting the sign of a velocity in a head-on collision.
No sign convention.
Do this: Opposite directions carry opposite signs.
Usually costs: Method mark.
Mistake: Saying kinetic energy is conserved in every collision.
Confusing total energy with kinetic energy.
Do this: Only elastic collisions conserve kinetic energy.
Usually costs: Concept mark.
Mistake: Using energy conservation during the bullet-block collision.
Applying the pendulum rise to the impact.
Do this: Use momentum for the impact and energy for the swing.
Usually costs: 2 marks.

Exam tips

Name the type of collision in the first line.
Use the relative-velocity rule to simplify elastic problems.
Verify with a kinetic energy check; it earns the last mark.
In ballistic pendulum questions separate the two stages.
Give the direction of motion with the sign of the answer.

Practice questions on Elastic and Inelastic Collisions in One Dimension

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
A body of mass m moving at velocity u collides elastically head-on with an identical body at rest. After the collision
(a) both move at u/2 (b) the first stops and the second moves at u (c) the first rebounds at u (d) both stop
Hint
Equal masses exchange velocities.
Answer
(b).
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Momentum is conserved in an inelastic collision although kinetic energy is not.
Reason (R): The impulsive forces between the colliding bodies are internal, so the net external impulse is negligible.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Internal forces cancel in pairs.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
A 0.1 kg ball hits a wall normally at 10 m/s and rebounds elastically at the same speed. Find the impulse on the ball and the change in its kinetic energy.
Hint
Reverse the velocity.
Answer
Impulse = 0.1 × 20 = 2 N s; ΔK = 0.
Q4 · 3 marks · Short answerHarder board level
A 5 kg body moving at 4 m/s collides head-on with a 3 kg body moving at 2 m/s in the opposite direction, and they stick. Find the common velocity and the kinetic energy lost.
Hint
Opposite directions have opposite signs.
Answer
v = (20 − 6)/8 = 1.75 m/s in the direction of the 5 kg body; K before = 40 + 6 = 46 J, after = ½ × 8 × 1.75² = 12.25 J, loss = 33.75 J.
Q5 · 4 marks · Long answerHOTS / competency
A 2 kg block moving at 5 m/s collides elastically head-on with a stationary 3 kg block. The surface has μ = 0.2 for both blocks (g = 10 m/s²). Find the velocities just after the collision and the distance between the blocks when both have stopped.
Hint
Elastic formulae first, then v² = 2μgd.
Answer
v₁ = (2 − 3) × 5/5 = −1 m/s; v₂ = 2 × 2 × 5/5 = 4 m/s. Deceleration = μg = 2 m/s². The 3 kg block stops after 16/4 = 4 m; the 2 kg block (moving back) after 1/4 = 0.25 m. Separation = 4.25 m.

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.Conserved in all collisions?Show answer
Linear momentum.
2.Conserved only in elastic?Show answer
Kinetic energy.
3.Result for equal masses (elastic, target at rest)?Show answer
Velocities are exchanged.
4.Energy lost in perfectly inelastic?Show answer
½ × reduced mass × (relative speed)².
5.Ballistic pendulum stages?Show answer
Momentum at impact, energy during the swing.
6.Elastic: relative speed of approach?Show answer
Equals the relative speed of separation.

Elastic and Inelastic Collisions in One Dimension: frequently asked questions

Why is momentum conserved even when energy is lost?

The internal forces are equal and opposite, so the total momentum is unchanged; the energy goes into heat, sound and deformation.

Is a collision between a ball and the floor elastic?

Usually not; the ball rebounds to a lower height, showing kinetic energy is lost.

Can the kinetic energy lost exceed the initial kinetic energy?

No; at most all the kinetic energy in the centre-of-mass frame can be lost.

When do two bodies exchange velocities?

In an elastic collision when the masses are equal.

Why is hydrogen a better moderator than lead?

A hydrogen nucleus has nearly the neutron's mass, so a neutron can transfer most of its energy in one collision.