A.2 · IB Physics
Momentum and collisions
Fire two blocks at each other along a frictionless track. Total momentum is conserved in every collision, but kinetic energy is conserved only when the collision is elastic. Change the masses, speeds and the coefficient of restitution.
Predict first
Start from the default settings (m₁ = 2 kg at 4 m s⁻¹ meets m₂ = 3 kg at −2 m s⁻¹). Change Collision type from Elastic to Perfectly inelastic. What is the common velocity after impact?
The blocks stick, so momentum conservation gives v = (m₁u₁ + m₂u₂) ÷ (m₁ + m₂) = (2×4 + 3×(−2)) ÷ 5 = 0.4 m s⁻¹. The sign is positive, so the pair moves right.
Check it in the model: Choose Perfectly inelastic in the Collision type menu and read v₁′ and v₂′ in the readouts (both 0.4 m s⁻¹).
block 1 block 2 total momentum
v₁′ = [(m₁ − e m₂)u₁ + m₂(1 + e)u₂] ÷ (m₁ + m₂)
v₂′ = [(m₂ − e m₁)u₂ + m₁(1 + e)u₁] ÷ (m₁ + m₂)
e = (v₂′ − v₁′) ÷ (u₁ − u₂)
Try this
- Use the Heavy target preset (m₁ = 1 kg hits a stationary 6 kg block, elastic). Predict the direction of the light block afterwards, then check v₁′.
- Keep m₁ = 2 kg, m₂ = 3 kg, u₁ = 4, u₂ = −2 and compare Elastic with Perfectly inelastic. Which quantity differs after impact, and which is unchanged?
- Choose Custom restitution and slide e from 1 to 0. Watch KE lost rise from 0 % while the momentum readout stays the same.
Exam tip: write down that momentum is conserved (no external force) before using any other equation, and quote kinetic energy conservation only if the question says the collision is elastic.