Conservation of Momentum — Rear-End Collision (Worked Example)

Given

• m₁ = 1200 kg, u₁ = 15 m/s

• m₂ = 1500 kg, u₂ = 0

• cars stick together after impact

Find

Common velocity v after collision.

Principle

In a collision with no external impulse, total momentum is conserved: momentum before equals momentum after.

Formula

Solution

1. Momentum before: $p = 1200 \times 15 + 1500 \times 0 = 18{,}000\ \mathrm{kg\cdot m/s}$

2. Combined mass: $M = 1200 + 1500 = 2700\ \mathrm{kg}$

3. Conserve: $2700v = 18{,}000$

4. Solve: $v = 18{,}000/2700 = 6.67\ \mathrm{m/s}$

Unit check

kg·m/s divided by kg gives m/s — a velocity, as required.

What this means physically

The wreckage crawls away at 6.67 m/s (24 km/h) — less than half the striking car's speed, because the stationary car's mass had to be dragged along. The 'missing' kinetic energy became crumpled metal and heat.