Elastic & Inelastic Collisions

Key idea: Use momentum conservation and impulse to solve 1D collision problems, and distinguish elastic, inelastic, and completely inelastic collisions (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Use impulse and momentum conservation in one-dimensional elastic and inelastic collisions.

1. Definitions (Must Know)

A. Collision (impulsive interaction)

A collision is a short interaction where two bodies exert large forces on each other for a short time, so their momenta change significantly.

B. Elastic collision

An elastic collision is one where:

  • total momentum is conserved, and
  • total kinetic energy is conserved.

C. Inelastic collision

An inelastic collision is one where:

  • total momentum is conserved, but
  • total kinetic energy is not conserved (some is transferred to other energy stores, e.g. heating, sound, deformation).

D. Completely (perfectly) inelastic collision

A completely inelastic collision is an inelastic collision where the objects stick together after impact, so they have the same final velocity.

E. Impulse

Impulse is the change in momentum:

vector J = Δ vector p

See: Momentum & Impulse.

2. Key Ideas (What Earns Marks)

  • Treat 1D collisions with a sign convention (e.g. right is +).
  • Momentum conservation is a system law: ∑ p_before = ∑ p_after provided external impulse is negligible.
  • Kinetic energy is conserved only in elastic collisions.
  • Completely inelastic: “stick together” ⇒ v₁ = v₂ after collision.
  • Perfectly elastic (1D): speed of approach = speed of separation: |u₁-u₂| = |v₂-v₁|
  • Average collision force comes from impulse: F_avg = (Δ p)/(Δ t)

3. Detailed Explanations

A. When can you conserve momentum?

Momentum is conserved for a system when the resultant external force is zero, or when the external impulse is negligible during the short collision time.

This is why collision questions often say “smooth surface” or “no external forces”.

B. 1D collision setup (exam workflow)

  1. Choose a positive direction.
  2. Label initial velocities as u₁, u₂ and final velocities as v₁, v₂ (include signs).
  3. Apply momentum conservation: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
  4. Use the collision type to get a second equation (if needed).

C. Completely inelastic collisions (stick together)

If the objects stick, they move together with a common velocity v:

m₁u₁ + m₂u₂ = (m₁ + m₂)v

Kinetic energy decreases because energy is transferred to other stores during deformation.

D. Perfectly elastic collisions (1D shortcut)

For a perfectly elastic collision in 1D, you use:

  • momentum conservation, and
  • speed of approach = speed of separation: |u₁-u₂| = |v₂-v₁|

This is equivalent to the “relative speed of approach equals relative speed of separation” statement in the syllabus.

E. Impulse and force in collisions

Collision forces vary with time, so questions often ask for an average force:

F_avg = (Δ p)/(Δ t)

For the two colliding objects, Newton’s 3rd law implies the impulses are equal in magnitude and opposite in direction.

Newton’s third-law force pair during a collisionOver the same contact interval, force on body 1 by body 2 is a positive pulse while force on body 2 by body 1 is an equal negative pulse. Their signed areas are opposite impulses.tFF₁₂F₂₁ = −F₁₂same contact interval for both bodies
Scroll diagram horizontally to read all labels.
At every instant, F₁₂ = −F₂₁. The bodies therefore receive equal-magnitude, opposite-direction impulses during the same collision.

4. Common Mistakes

  • Forgetting momentum is a vector (dropping signs in 1D).
  • Conserving momentum for one object instead of the system.
  • Assuming “collision = elastic” (elastic must be stated or proven with KE/relative-speed rule).
  • Using N = mg or friction assumptions without checking the question (external forces can break momentum conservation).
  • Mixing up impulses: each object’s impulse is opposite, but total system impulse is zero (if isolated).

5. Exam Tips

  • Write your sign convention first: “take rightwards as positive.”
  • If the objects stick, replace v₁ and v₂ with one unknown v immediately.
  • For elastic collisions, the relative-speed rule is often faster than using kinetic energy directly.
  • After solving, do a quick sanity-check:
    • does a heavier object change speed less?
    • do directions (signs) make sense?

6. Worked Examples

Modelled example 1

Completely inelastic: common velocity

Core

Problem

A 2.0 kg cart moves at + 4.0 m s⁻¹ while a 3.0 kg cart moves at -2.0 m s⁻¹. They collide and stick together. Find their common velocity.
Study the worked solution
  1. Keep the stated velocity signs

    Method

    The initial momenta oppose each other.

    Reason

    Momentum is a vector, so the cart moving in the negative direction contributes negative momentum.

    Working

    pᵢₙᵢₜᵢₐₗ = 2.0(4.0) + 3.0(-2.0)
  2. Use the sticking condition

    Method

    Both carts share one final velocity v.

    Reason

    A completely inelastic collision joins the objects into a combined 5.0 kg mass.

    Working

    2.0(4.0) + 3.0(-2.0) = (2.0 + 3.0)v
  3. Solve and interpret

    Method

    v = +0.40 m s⁻¹.

    Reason

    The initial positive momentum slightly exceeds the negative momentum.

    Working

    8-6 = 5v ⇒ v = +0.40 m s⁻¹

Guided practice 2

Kinetic energy lost in a completely inelastic collision

About 6 min

Problem

A 5.00 kg object moving at 15.0 m s⁻¹ collides with a stationary 10.0 kg object. They stick together. How much kinetic energy is lost?

Try this before viewing the solution

Unit: J

Hints

Hint 1: find the shared speed first
The combined 15.0 kg mass has the same momentum as the system had before impact.
View solution step by step
  1. Find the common velocity

    Method

    v = 5.00 m s⁻¹.

    Reason

    Momentum is conserved for the two-object system during the collision.

    Working

    5.00(15.0) + 10.0(0) = 15.0v ⇒ v = 5.00 m s⁻¹
  2. Calculate the initial kinetic energy

    Method

    E_(k,i) = 562.5 J.

    Reason

    Only the 5.00 kg object is moving initially.

    Working

    E_(k,i) = (1/2)(5.00)(15.0)² = 562.5 J
  3. Calculate the final kinetic energy

    Method

    E_(k,f) = 187.5 J.

    Reason

    After sticking, the full 15.0 kg mass moves at the common speed.

    Working

    E_(k,f) = (1/2)(15.0)(5.00)² = 187.5 J
  4. Find the kinetic-energy loss

    Method

    375 J is transferred from kinetic energy.

    Reason

    An inelastic collision conserves total energy but not the system’s kinetic energy.

    Working

    E_(k,i)-E_(k,f) = 562.5-187.5 = 375 J

Common misconception 3

Impulse and average force (collision time given)

Find and correct the mistake

Learner claim

A 0.20 kg ball moving at + 6.0 m s⁻¹ rebounds from a wall at -4.0 m s⁻¹ after 0.010 s. A learner uses the speed difference 6.0-4.0 and obtains an average-force magnitude of 40 N. Diagnose the method and find the correct magnitude.

Try this before viewing the solution

Unit: N

View solution step by step
  1. Identify the sign error

    Method

    The velocity change is -4.0-6.0 = -10.0 m s⁻¹.

    Reason

    The final velocity is negative, not a positive speed to subtract from the initial speed.

    Working

    Δ v = v-u = -4.0-(+6.0) = -10.0 m s⁻¹
  2. Calculate the impulse

    Method

    Δ p = -2.0 N s.

    Reason

    Impulse equals the ball’s signed momentum change.

    Working

    Δ p = 0.20(-10.0) = -2.0 N s
  3. Calculate average-force magnitude

    Method

    |F_avg| = 200 N.

    Reason

    Divide the impulse magnitude by the contact time.

    Working

    |F_avg| = 2.0/0.010 = 200 N

Examiner practice 4

Concept check: impulses on the two vehicles

2 marks

Examination question

A lorry and a motorcycle collide head-on. Which statement is always true during the collision? 1. The motorcycle experiences a larger force. 2. The impulse magnitudes on the vehicles are equal. 3. Each vehicle’s momentum is conserved. 4. Total kinetic energy is conserved. Select the statement and justify it. [2 marks]

Try this before viewing the solution

Always-true statement

View solution step by step
  1. Select statement 2

    1 mark

    Method

    The impulse magnitudes on the lorry and motorcycle are equal.

    Reason

    Their interaction forces form a Newton’s-third-law pair.

    Working

    |F_(L → M)| = |F_(M → L)|
  2. Connect force to impulse

    1 mark

    Method

    The impulses are equal in magnitude and opposite in direction.

    Reason

    The equal and opposite forces act over the same contact time.

    Working

    J_(L → M) = -J_(M → L)

Challenge 5

Perfectly elastic collision (use the relative-speed rule)

Minimal support

Independent transfer

A 1.0 kg trolley moving at + 5.0 m s⁻¹ collides in one dimension with a stationary 2.0 kg trolley. The collision is perfectly elastic. Find both final velocities.

Try this before viewing the solution

Unit: m s^-1
Unit: m s^-1

Hints

Hint 1: identify the two constraints
Use one equation for system momentum and one for the perfectly elastic relative-speed condition.
View solution step by step
  1. Write momentum conservation

    Method

    5 = v₁ + 2v₂.

    Reason

    External impulse is negligible for the two-trolley system.

    Working

    1.0(5.0) + 2.0(0) = 1.0v₁ + 2.0v₂
  2. Use the elastic relative-speed condition

    Method

    5 = v₂-v₁.

    Reason

    For this one-dimensional perfectly elastic collision, speed of separation equals the initial speed of approach.

    Working

    u₁-u₂ = v₂-v₁ ⇒ 5 = v₂-v₁
  3. Solve simultaneously

    Method

    v₁ = -1.67 m s⁻¹ and v₂ = +3.33 m s⁻¹.

    Reason

    Substituting v₂ = v₁ + 5 into momentum conservation determines both signed velocities.

    Working

    5 = v₁ + 2(v₁ + 5) ⇒ v₁ = -5/3, v₂ = 10/3 m s⁻¹

7. Mind Stretchers

Mind stretcher 1: When momentum conservation can failExtension

A ball hits the ground and rebounds. You model the ball alone as the system.

Is momentum conserved for this “system”? Explain.

Show Answer

No. The ball experiences a large external impulse from the ground during the collision, so momentum is not conserved for the ball alone.

If you treat “ball + Earth” as the system, momentum can be conserved (external impulse negligible).

Mind stretcher 2: Elastic vs inelastic: what changes?Extension

Two objects collide in 1D. Total momentum is conserved.

What additional condition must hold for the collision to be elastic?

Show Answer

Total kinetic energy must be conserved (no net transfer of kinetic energy to other energy stores).

Mind stretcher 3: Optional (Enrichment)Extension

A. Coefficient of restitution (not required)

Some courses define:

e = (speed of separation)/(speed of approach)

In this syllabus, you only need the special case for perfectly elastic collisions (relative speed approach = relative speed separation). You do not need to use e.

B. 2D collisions (beyond core scope)

For 2D collisions, you conserve momentum separately in perpendicular directions (e.g. x and y). These problems usually need extra information beyond momentum alone.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027