Lesson 1 · 16 min
Momentum and collisions
When forces are brief and ugly, integrate them. The integral is impulse; the leftover is momentum.
Why crash structures exist
Newton’s second law is also ΣF = dp/dt. Integrate both sides and the impulse J = ∫F dt equals the change in momentum Δp. A long crash pulse (crumple zone) is the same Δp with a smaller peak F. Airbags are this sentence, in nylon.
Isolated means external impulse is negligible during the event. Internal collision forces cancel by the third law.
Restitution
The coefficient of restitution e is the ratio of relative speed after to relative speed before, along the line of impact. e = 1 is elastic (KE conserved as well as p). e = 0 is perfectly inelastic (they stick). Most engineering collisions sit in between: rail buffers, sporting impacts, packaging drops.
Together with momentum conservation this solves for both final speeds.
Worked example
Coupling cars
- Given: m₁ = 20 t at 1.2 m/s
- Given: m₂ = 30 t at rest
- Given: e = 0 (they couple)
Find: Common speed after
- Stick: (m₁ + m₂)v = m₁ u₁
- v = (20 × 1.2) / 50 = 24/50
0.48 m/s, with a large kinetic-energy loss into the coupler and noise.
Check
In an isolated collision, which quantity is always conserved?
Bench
Collision
Σp conserved
- p in
- 5.00 kg·m/s
- p out
- 5.00 kg·m/s
- KE in
- 17.5 J
- KE out
- 17.5 J
v₁ = -2.00 m/s, v₂ = 3.00 m/s. KE retained 100%.