Conservation of momentum in collisions

In a closed system (no outside forces such as friction acting) the total momentum before an event equals the total momentum after it.

total momentum before = total momentum after
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Here u is a velocity before and v a velocity after. Each object's momentum changes in a collision; only the total is fixed.

Momentum is conserved in every collision, sticky or bouncy. Kinetic energy is a different story (lesson 5).

When the objects stick, they share one velocity afterwards, and the total mass is the sum.

1. A 1.5 kg trolley at 0.8 m/s hits a stationary 0.5 kg trolley and sticks
2. before: 1.5 × 0.8 + 0.5 × 0 = 1.2 kg m/s
3. after: (1.5 + 0.5) × v = 1.2
4. v = 1.2 ÷ 2.0 = 0.6 m/s

Coming towards each other? Give one velocity a minus sign first. A 60 kg skater at 3 m/s and a 40 kg skater at 2 m/s the other way: total = 180 − 80 = 100 kg m/s, so v = 100 ÷ 100 = 1 m/s in the first skater's direction.

If they bounce, each has its own velocity after.

Write the equation with what you know and solve for the one unknown.

1. A 0.17 kg cue ball at 2 m/s hits a stationary 0.16 kg ball and slows to 0.4 m/s
2. before: 0.17 × 2 = 0.34 kg m/s
3. after: 0.17 × 0.4 + 0.16 × v = 0.068 + 0.16v
4. 0.34 = 0.068 + 0.16v, so v = 0.272 ÷ 0.16 = 1.7 m/s

Dividing by only one mass when the objects stick, or forgetting the momentum the first ball still has, are the usual slips.

Choose Stick together. Set Mass 1 to 3 kg and Velocity 1 before to 2 m/s, Mass 2 to 1 kg and Velocity 2 before to 0. Work out the velocity after (total momentum ÷ total mass), type it as your prediction, then press Start. Compare Total momentum before and Total momentum after. Then switch to Elastic and run it again: is the total still the same?