Perfectly elastic collisions and kinetic-energy change
Key idea: H2 Physics lessons on force–time impulse, one-dimensional momentum conservation, elastic interactions and kinetic-energy change.
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The core idea
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Big question: What distinguishes a perfectly elastic collision from other collisions?
Momentum is conserved in any isolated collision, but a perfectly elastic one also has relative speed of approach equal to relative speed of separation and unchanged total kinetic energy. Use momentum and the relative-speed condition as two independent equations; otherwise kinetic energy commonly becomes deformation, internal energy or sound.
Add the defining condition for a perfectly elastic collision
For two bodies in a one-dimensional perfectly elastic collision, relative speed of approach equals relative speed of separation. With a consistent direction convention, this can be written as u₁ − u₂ = v₂ − v₁ for an arrangement in which body 1 approaches body 2.
This condition is used with momentum conservation to find two unknown final velocities. A coefficient of restitution is not required in this course.
Check your understanding: Bodies approach at 7.0 m s⁻¹ relative speed. What must their separation speed be after a perfectly elastic collision?
7.0 m s⁻¹.
Keep momentum and kinetic-energy tests distinct
A closed system conserves momentum in every interaction when external impulse is negligible. Total kinetic energy is unchanged only for a perfectly elastic collision; otherwise some kinetic energy normally becomes deformation, internal energy or sound.
Use signed velocities for momentum and relative velocity, but use squared speeds for kinetic energy. Compare totals over every body in the chosen system.
Check your understanding: Can momentum be conserved while kinetic energy decreases?
Yes. Momentum can remain constant while kinetic energy transfers to other stores.
Key ideas to keep
- A coefficient of restitution is not required in this course.
- Calculate total kinetic energy using every object in the chosen system.
- A negative velocity still contributes positive kinetic energy because velocity is squared.
See the reasoning
Worked example
Solve an elastic collision with two independent equations
Question: A 2.0 kg trolley at +5.0 m s⁻¹ collides perfectly elastically with a stationary 3.0 kg trolley. Find both final velocities.
Step 1: Conserve momentum
Why: External impulse is negligible during the collision.
Working: 2(5) = 2v₁ + 3v₂.
Step 2: Use the relative-speed condition
Why: Perfect elasticity supplies the second independent equation.
Working: 5 − 0 = v₂ − v₁, so v₂ = v₁ + 5.
Step 3: Solve and check
Why: Both equations must hold simultaneously.
Working: 10 = 2v₁ + 3(v₁ + 5), giving v₁ = −1.0 m s⁻¹ and v₂ = +4.0 m s⁻¹.
Answer: The 2.0 kg trolley rebounds at 1.0 m s⁻¹ and the 3.0 kg trolley moves forward at 4.0 m s⁻¹.
Check: Approach and separation speeds are both 5.0 m s⁻¹, and total kinetic energy is 25 J before and after.
Another worked model
Question
A 1.0 kg trolley at +8.0 m s⁻¹ collides perfectly elastically with a stationary 3.0 kg trolley. Use momentum and the relative-speed condition to find both final velocities.
Check the worked solution
Momentum gives 8 = v₁ + 3v₂. Relative speed gives 8 = v₂ − v₁. Hence v₁ = v₂ − 8, so 8 = 4v₂ − 8 and v₂ = +4.0 m s⁻¹; v₁ = −4.0 m s⁻¹. Initial and final kinetic energies are both 32 J.
Use a hint if needed
Practise with support
Try this
A 2.0 kg trolley at +4.0 m s⁻¹ sticks to an identical stationary trolley. Find the common velocity and kinetic-energy change.
Hint: Conserve momentum first; sticking means the interaction is perfectly inelastic.
Check your answer
2.0(4.0) = 4.0v, so v = +2.0 m s⁻¹. Initial kinetic energy = 16 J and final = ½(4.0)(2.0²) = 8.0 J, a decrease of 8.0 J.
Now work without the hint
Practise independently
Your turn
Explain how to test whether a one-dimensional two-body collision is perfectly elastic without using coefficient of restitution.
Check your answer
First verify total signed momentum is unchanged for the closed system. Then compare relative speeds: speed of approach before must equal speed of separation after. Equivalently, total kinetic energy must also be unchanged; any kinetic-energy decrease means the collision is not perfectly elastic.
Avoid these traps
Common mistakes
Common mistake
Every collision conserves kinetic energy because total energy is conserved.
What is wrong with this reasoning?
Show better thinking
Total energy is conserved, but kinetic energy can transfer to internal energy, sound and deformation. Total kinetic energy is unchanged only for a perfectly elastic collision.
Common mistake
A coefficient of restitution is needed to solve a perfectly elastic collision.
What is wrong with this reasoning?
Show better thinking
It is outside this syllabus requirement. Use momentum conservation and relative speed of approach equal to relative speed of separation directly.
Write for the examiner
Exam guidance
Solve momentum first, then perform a separate kinetic-energy check and state the conclusion in words.
Exam-style practice [8 marks]
A 0.50 kg trolley at +6.0 m s⁻¹ collides perfectly elastically with a stationary 1.0 kg trolley. Find both final velocities and verify the kinetic-energy condition.
Plan before you answer
- Write signed momentum conservation.
- Use relative speed of approach = separation.
- Check both kinetic-energy totals.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
Momentum gives 3.0 = 0.50v₁ + 1.0v₂. Perfect elasticity gives 6.0 = v₂ − v₁. Solving gives v₂ = +4.0 m s⁻¹ and v₁ = −2.0 m s⁻¹. Initially, K = ½(0.50)(6.0²) = 9.0 J. Finally, K = ½(0.50)(2.0²) + ½(1.0)(4.0²) = 1.0 + 8.0 = 9.0 J, confirming the elastic energy condition.
Come back in three days
Check what stayed with you
Recall question 1
State the relative-speed condition for a perfectly elastic collision.
Check the answer
Relative speed of approach equals relative speed of separation.
Recall question 2
What condition is excluded from this course?
Check the answer
Use of the coefficient of restitution.
Recall question 3
Which quantity is normally conserved in every closed-system collision?
Check the answer
Total momentum.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. The coefficient of restitution is not required. Use momentum conservation, the perfectly elastic relative-speed condition and a direct comparison of total kinetic energy.
- GCE A-Level H2 PhysicsTopic 6(d) / Topic 6(e) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027