Momentum and coefficient of restitution
Combine momentum conservation with Newton’s experimental law for direct or oblique impacts of smooth spheres and fixed surfaces.
Exam scope and assessment notes
Mapped to Cambridge International AS & A Level Further Mathematics 9231. These original explanations and self-checks are independent study support, not Cambridge questions, marks or grades. The syllabus assumes prior knowledge of the relevant 9709 Mathematics content; check each component's prerequisites with your teacher. Full-paper assessment and human subject-specialist review remain outstanding.
Before you begin
- Cambridge expects learners to know the whole Cambridge International A Level Mathematics (9709) content. Paper 1 further pure mathematics draws on 9709 Papers 1 and 3. Confirm your exam year and component route with your school.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What makes a Further Mathematics solution convincing?
What you will learn
- Define coefficient of restitution and identify elastic or inelastic limits
- Use momentum conservation and restitution in direct impacts
- Resolve oblique impacts along the line of centres
Restitution compares relative speeds along the impact line
For a one-dimensional collision where A approaches B, e=(speed of separation)/(speed of approach)=(vB−vA)/(uA−uB), using a consistent positive direction and the pre-collision ordering uA>uB.
Newton’s experimental law has 0≤e≤1 in the standard model. e=1 is perfectly elastic; e=0 means the bodies have no relative separation speed immediately after impact (perfectly inelastic in the direct-impact model).
Use momentum and restitution as a pair
For an isolated system, total momentum along the line of impact is conserved. Pair this equation with the restitution relation to find two unknown final velocities.
For smooth spheres in an oblique collision, the impulse acts along the line of centres: normal components change, while tangential components are unchanged. For a fixed smooth surface, only the normal component changes; apply restitution to that component.
Pause and explain
Which habit best checks your work on momentum and coefficient of restitution?
Worked example
Masses 2 kg and 1 kg collide directly. Before impact their velocities are 3 and 0 m s⁻¹; e=0.5. Find their velocities after impact.
Show the worked solution
- Momentum gives 2vA+vB=6.
- Restitution gives vB−vA=0.5(3−0)=1.5.
- Solve the equations to obtain vA=1.5 and vB=3.
Answer The 2 kg mass moves at 1.5 m s⁻¹ and the 1 kg mass at 3 m s⁻¹ in the original direction.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A 1 kg sphere moving at 6 m s⁻¹ hits a stationary 2 kg sphere directly. If e=0.5, find both final velocities in the original direction.
Give me a hint
Use momentum and e=(v₂−v₁)/6.
Compare my reasoning
- Momentum: v₁+2v₂=6.
- Restitution: v₂−v₁=3.
- Solve to get v₂=3 and v₁=0.
The 1 kg sphere stops; the 2 kg sphere moves at 3 m s⁻¹.
Look for these in your work
- Momentum conserved
- Restitution applied with correct order
- Both velocities stated
A sphere strikes a fixed smooth wall at 10 m s⁻¹ along the normal, with e=0.7. Find its velocity component after impact along that normal, taking toward the wall as positive before impact.
Give me a hint
The wall is fixed, so separation speed along the normal is e times approach speed in the opposite direction.
Compare my reasoning
- Approach speed is 10 m s⁻¹ toward the wall.
- The rebound speed is 0.7(10)=7 m s⁻¹ away from the wall.
- Away is negative under the stated sign convention.
Normal velocity after impact is −7 m s⁻¹.
Look for these in your work
- Approach speed identified
- Restitution magnitude applied
- Sign matches rebound direction
Two smooth spheres collide obliquely. Which components of velocity can change during impact, and why?
Give me a hint
Resolve each velocity into components parallel and perpendicular to the line of centres.
Compare my reasoning
- The impulse acts along the line of centres for smooth spheres.
- Therefore the normal components can change during impact.
- There is no tangential impulse, so tangential components remain unchanged.
Normal components may change; tangential components are unchanged because the impulse is along the line of centres.
Look for these in your work
- Smooth-contact assumption used
- Impulse direction stated
- Component effects distinguished
Common mistakes
- Reversing approach and separation speeds
- Applying restitution to tangential components in a smooth oblique impact
- Assuming kinetic energy is conserved when e<1
What can you explain now?
What does e=0 mean for a direct collision immediately after impact?
Compare with the explanation
The relative separation speed is zero
The bodies have no relative speed along the line of impact.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Why are momentum conservation and the coefficient of restitution both needed to find post-impact velocities?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.