Collisions: Complete A Level H2 Physics Cheatsheet
A Level Physics (H2), Chapter 6 · 9478 Topic 6 · Read time: ~10 minutes
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Key Formulas
Impulse
Impulse is the area under a force-time graph. It equals the change in momentum it causes.
For the same change in momentum, a longer contact time means a smaller average force. That is why airbags, crumple zones and bending your knees on landing all reduce injury.
Trick: For an irregular force-time graph, count squares or split it into simple shapes, the same way you find displacement from a v-t graph.
Conservation of momentum
Principle: the total momentum of a system remains constant provided no resultant external force acts on it.
Momentum is a vector. Choose a positive direction and give every velocity a sign before substituting.
Momentum is conserved in every collision or explosion of a closed system, whether elastic or not.
Trap: Kinetic energy uses speed squared, so signs don't matter there, but they matter a lot for momentum. A trolley bouncing back has negative momentum.
Elastic and inelastic collisions
Perfectly elastic: momentum and total kinetic energy are both conserved. For two bodies in one dimension, the relative speed of approach equals the relative speed of separation.
Inelastic: momentum is conserved but some kinetic energy is converted to other stores (thermal, sound, deformation). Perfectly inelastic: the bodies stick together.
Even though momentum is always conserved in a closed system, some change in kinetic energy usually takes place.
Trick: For an elastic collision, the relative speed equation is linear and much faster than solving with kinetic energy. Use it with the momentum equation as a pair of simultaneous equations.
Worked Example
Trolley P (2.0 kg) moves at 6.0 m s⁻¹ to the right and hits stationary trolley Q (4.0 kg). Afterwards, P moves at 2.0 m s⁻¹ to the left.
(a) Calculate the velocity of Q after the collision. [2]
Right positive: (2.0)(6.0) + 0 = (2.0)(−2.0) + 4.0 v_Q
12.0 = −4.0 + 4.0 v_Q → v_Q = 4.0 m s⁻¹ to the right
(b) Show that the collision is perfectly elastic using relative speeds. [2]
Relative speed of approach = 6.0 − 0 = 6.0 m s⁻¹
Relative speed of separation = 4.0 − (−2.0) = 6.0 m s⁻¹
They are equal, so the collision is perfectly elastic.
(c) The collision lasts 0.050 s. Calculate the average force on Q. [2]
Δp of Q = 4.0 × 4.0 − 0 = 16 kg m s⁻¹
F = Δp / Δt = 16 ÷ 0.050 = 320 N to the right
Why this question is a good test of the topic: P bouncing back looks like energy was lost, but the relative speed check shows it wasn't. Part (c) links impulse to force, a common Paper 2 extension.
Frequently Asked Questions
What is impulse?
The area under a force-time graph, equal to the change in momentum of the body.
When is momentum conserved?
In any interaction where no resultant external force acts on the system, for both elastic and inelastic collisions.
How do I know if a collision is perfectly elastic?
Total kinetic energy is unchanged, or equivalently, the relative speed of approach equals the relative speed of separation.
Do I need the coefficient of restitution?
No. The 9478 syllabus states that knowledge of the coefficient of restitution is not required.
Why do airbags reduce injury?
They increase the time over which the person's momentum changes, so the average force on the person is smaller.
Momentum and collisions not clicking?
Small group A Level H2 Physics classes at TGC Academy's Bishan, Bukit Timah and Potong Pasir centres, helmed by Andrew Seah, MOE Award-Winning Teacher and Marshall Cavendish textbook author, with a team of tutors.