Perfectly elastic collisions
Key idea: A perfectly elastic collision conserves momentum and kinetic energy. In one dimension, the relative speed of separation equals the relative speed of approach.
Offline pages
Saved on this device only. Connect at least once every 30 days to confirm Premium. Sign-out or a change of account removes saved copies.
Loading guide controls…
Guided lessons · coming soon
Guided lesson
Guided steps
- Use signed velocities to find how quickly the gap changes.
- Find the approach speed.
- Follow how momentum and kinetic energy give the second equation.
- Follow both equations and the kinetic-energy check.
- Find momentum, relative speed and both final velocities.
- Find both velocities for carts travelling the same way.
- Test whether the proposed collision is elastic.
- Review signed differences and both conservation conditions.
On this page
You stopped at:
ContinueWhen two bodies collide and bounce apart, conservation of momentum gives one equation, but there are two unknown final velocities. For a perfectly elastic collision, one where no kinetic energy is lost, there is a second, simpler condition: the bodies separate at the same relative speed as they approached. Together, the two equations give both final velocities.
Approach and separation speeds
Take right as positive, and call the body on the left body 1 and the body on the right body 2. Before the collision they approach each other, and afterwards they separate:
- relative speed of approach = u₁-u₂
- relative speed of separation = v₂-v₁
Use the signed velocities. If both bodies move right at 5 m s⁻¹ and 2 m s⁻¹, they approach at 3 m s⁻¹, not 7 m s⁻¹.
For a perfectly elastic collision between two bodies,
The relative speed of approach equals the relative speed of separation.
Check your understanding 1 Closing speed
A cart on the left moves at + 4.00 m s⁻¹ and a cart on the right moves at -2.00 m s⁻¹. What is their relative speed of approach?
Show answer
u₁-u₂ = 4.00-(-2.00) = 6.00 m s⁻¹: they are moving towards each other, so their speeds add.
Why the relative speeds are equal
A perfectly elastic collision conserves both momentum and kinetic energy. Rearranging each equation to put body 1 on the left:
Dividing the second equation by the first gives u₁ + v₁ = v₂ + u₂, which rearranges to u₁-u₂ = v₂-v₁. So the relative-speed condition is just conservation of kinetic energy in a simpler form. During the contact, some kinetic energy is stored briefly as elastic strain energy; “perfectly elastic” means it is all returned by the time the bodies have separated.
Worked example 1
Both final velocities
Problem
A 2.00 kg trolley moving right at 5.00 m s⁻¹ collides with a stationary 3.00 kg trolley. The collision is perfectly elastic. Find the velocities of both trolleys afterwards.
Show full solution
Conservation of momentum
Method
Take right as positive and write the momentum equation.Reason
Momentum is conserved in every collision with no external force.Working
Answer
2.00 × 5.00 + 3.00 × 0 = 2.00v₁ + 3.00v₂, so 10.0 = 2.00v₁ + 3.00v₂
Relative speeds
Method
Use the elastic condition.Reason
It supplies the second equation.Working
Answer
5.00-0 = v₂-v₁, so v₂ = v₁ + 5.00
Solve
Method
Substitute into the momentum equation.Reason
Two equations give the two unknowns.Working
Answer
10.0 = 2.00v₁ + 3.00(v₁ + 5.00), so 5.00v₁ = -5.00: v₁ = -1.00 m s⁻¹ and v₂ = +4.00 m s⁻¹.
Check the kinetic energy
Working
Answer
Before: (1/2)(2.00)(5.00)² = 25.0 J. After: (1/2)(2.00)(1.00)² + (1/2)(3.00)(4.00)² = 1.0 + 24.0 = 25.0 J. The first trolley rebounds at 1.00 m s⁻¹ and the second moves off at 4.00 m s⁻¹.
Quick check
The trolleys separate at 4.00-(-1.00) = 5.00 m s⁻¹, the same as their approach speed.
Guided practice 2
A head-on elastic collision
Problem
A 1.00 kg cart moving at + 4.00 m s⁻¹ collides head-on with a 3.00 kg cart moving at -2.00 m s⁻¹. The collision is perfectly elastic. Find both final velocities.
Write the momentum and relative-speed equations
Show the steps Optional working
Show the steps Optional working
Hints
Hint 1
momentum
Total momentum before = 1.00(4.00) + 3.00(-2.00) = -2.00 kg m s⁻¹.
Hint 2
relative speeds
v₂-v₁ = u₁-u₂ = 4.00-(-2.00).
Show solution step by step
The two equations
Method
Write momentum conservation and the relative-speed condition.Reason
Two unknowns need two equations.Working
Answer
-2.00 = v₁ + 3.00v₂ and v₂ = v₁ + 6.00
Solve
Working
Answer
-2.00 = v₁ + 3.00v₁ + 18.0, so v₁ = -5.00 m s⁻¹ and v₂ = +1.00 m s⁻¹.
Quick check
Kinetic energy before: 8.00 + 6.00 = 14.0 J; after: 12.5 + 1.50 = 14.0 J. None is lost, as the collision is perfectly elastic.
Check your understanding 2 Equal masses
Two carts of equal mass move right along a track. The rear cart moves at + 5.00 m s⁻¹ and the front cart at + 2.00 m s⁻¹. They collide perfectly elastically. Find their final velocities.
Show answer
The masses cancel in the momentum equation: v₁ + v₂ = 7.00. The relative-speed condition gives v₂-v₁ = 5.00-2.00 = 3.00. So v₁ = +2.00 m s⁻¹ and v₂ = +5.00 m s⁻¹: equal masses in a perfectly elastic collision swap velocities.
Check your understanding 3 Is this collision elastic?
For the same two carts, a learner proposes final velocities of + 3.00 and + 4.00 m s⁻¹. These conserve momentum and the carts separate. Is the collision perfectly elastic?
Show answer
No. The carts separate at only 4.00-3.00 = 1.00 m s⁻¹, less than their approach speed of 3.00 m s⁻¹. The kinetic energy also falls: in units of (1/2)m, from 5.00² + 2.00² = 29 to 3.00² + 4.00² = 25. Conserving momentum does not make a collision elastic.
Common mistakes
- Adding speeds to find the relative speed of approach when both bodies move the same way. Use the difference of the signed velocities.
- Using only conservation of momentum when there are two unknown final velocities.
- Thinking a collision is elastic because momentum is conserved. It is elastic only if kinetic energy is conserved as well.
- Mixing up the order: approach is u₁-u₂, separation is v₂-v₁.
Spotted an error? Report a correction
Syllabus and review details
- GCE A-Level H1 Physics 2027 · 2027
8867 Content Overview, PDF pages 8–9; Subject Content, PDF pages 10–18
Last reviewed: