Collisions: impulse, momentum and kinetic energy
Key idea: Use a declared system and sign convention to connect force–time area to momentum change, then distinguish momentum conservation from kinetic-energy conservation.
Before you start: Momentum & ImpulseElastic & Inelastic Collisions
By the end, you can
- Find impulse as signed area under a force–time graph and use it as momentum change.
- State and apply conservation of momentum to a closed system in one-dimensional elastic and inelastic interactions.
- Use relative speed of approach equal to relative speed of separation for a perfectly elastic two-body collision.
- Explain why momentum remains conserved in a closed system while kinetic energy usually changes.
Starting-point self-check
1. Check your starting point
Attempt all three groups without notes and mark the first graph area, system boundary or collision condition you could not justify. Use the recorded topic diagnostic above when you want scoring and a personalised repair plan.
Perfectly elastic collisions and kinetic-energy change 6(d)–(e)
Question 1
A 1.0 kg trolley at +6.0 m s⁻¹ collides elastically with a 2.0 kg trolley at rest. Afterwards their velocities are −2.0 m s⁻¹ and +4.0 m s⁻¹. Check the relative-speed condition and total kinetic energy.
Check the model response
Relative speed of approach = 6.0 − 0 = 6.0 m s⁻¹; separation = 4.0 − (−2.0) = 6.0 m s⁻¹. Initial kinetic energy = ½(1.0)(6.0²) = 18 J; final = ½(1.0)(2.0²) + ½(2.0)(4.0²) = 18 J.
repair
2. Repair the six common breaks
Use only the repair matching an error, then repeat the corresponding graph or system model.
Perfectly elastic collisions and kinetic-energy change 6(d)–(e)
Check this idea
Misconception: Every collision conserves kinetic energy because total energy is conserved.
Repair: 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.
Check this idea
Misconception: A coefficient of restitution is needed to solve a perfectly elastic collision.
Repair: It is outside this syllabus requirement. Use momentum conservation and relative speed of approach equal to relative speed of separation directly.
worked example
3. Follow three worked models
Follow how each solution declares graph sign, system boundary and collision type before calculating.
Perfectly elastic collisions and kinetic-energy change 6(d)–(e)
Model 1
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 model response
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.
guided practice
4. Guided practice
Use each hint only to choose the signed graph area, closed system or elastic condition.
Perfectly elastic collisions and kinetic-energy change 6(d)–(e)
Question 1
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 the model response
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.
independent practice
5. Independent practice
Solve without repair notes and show the sign convention, system condition and energy comparison.
Perfectly elastic collisions and kinetic-energy change 6(d)–(e)
Question 1
Explain how to test whether a one-dimensional two-body collision is perfectly elastic without using coefficient of restitution.
Check the model response
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.
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Perfectly elastic collisions and kinetic-energy change 6(d)–(e)
Question 1
A 0.50 kg trolley at +10 m s⁻¹ collides perfectly elastically with a stationary 1.50 kg trolley. Afterwards their velocities are −5.0 m s⁻¹ and +5.0 m s⁻¹. Verify momentum, relative speed and kinetic energy.
Check the model response
Initial momentum = 5.0 kg m s⁻¹; final = 0.50(−5.0) + 1.50(5.0) = 5.0 kg m s⁻¹. Approach speed = 10 m s⁻¹ and separation speed = 5.0 − (−5.0) = 10 m s⁻¹. Initial and final kinetic energies are both 25 J.
Re-test practice
7. Delayed re-test practice
Return after at least three days and solve these fresh contexts without reopening earlier responses. The recorded plan enforces the delay and uses a separate re-test family for selected-response skill-group evidence.
Perfectly elastic collisions and kinetic-energy change 6(d)–(e)
Question 1
A 2.0 kg trolley at +7.0 m s⁻¹ collides perfectly elastically with a 1.0 kg trolley at −2.0 m s⁻¹. Use momentum and relative speed to find both final velocities, then check kinetic energy.
Check the model response
Momentum gives 12 = 2v₁ + v₂. Approach speed is 7.0 − (−2.0) = 9.0 m s⁻¹, so v₂ − v₁ = 9.0. Solving gives v₁ = +1.0 m s⁻¹ and v₂ = +10 m s⁻¹. Initial and final kinetic energies are both 51 J.