A-Level Impulse and the conservation of momentum and energy: worked solution

Question

A 0.50 kg trolley moving at 2.0 m/s strikes a stationary 0.20 kg pendulum bob hung on a light 1.5 m string. The collision is elastic and the bob swings up. Find the bob speed just after the collision.

Let v₁ be the trolley's velocity and v₂ the bob's velocity immediately after the collision, both measured as positive in the trolley's original direction of travel. Step 1: apply conservation of momentum. Total momentum before the collision equals total momentum after, since no external horizontal force acts during the impact. Before: (0.50 × 2.0) + (0.20 × 0) = 1.0 kg m/s (the bob is stationary, so it contributes nothing) After: 0.50v₁ + 0.20v₂ So 1.0 = 0.50v₁ + 0.20v₂ ... equation (1) Step 2: use the fact that the collision is elastic. For an elastic collision the relative speed of separation equals the relative speed of approach. (This is equivalent to conserving kinetic energy but avoids the squared terms, so it is much quicker.) Speed of approach = 2.0 − 0 = 2.0 m/s Speed of separation = v₂ − v₁ So v₂ − v₁ = 2.0 Rearranged: v₁ = v₂ − 2.0 ... equation (2) Step 3: substitute equation (2) into equation (1) to eliminate v₁. 1.0 = 0.50(v₂ − 2.0) + 0.20v₂ Expand the bracket: 1.0 = 0.50v₂ − 1.0 + 0.20v₂ Collect the v₂ terms: 1.0 = 0.70v₂ − 1.0 Step 4: solve for v₂. Add 1.0 to both sides: 2.0 = 0.70v₂ Divide both sides by 0.70: v₂ = 2.0/0.70 = 2.857... So the bob leaves the collision at 2.857, that is about 2.9 m/s, in the direction the trolley was moving. (For reference, equation (2) then gives v₁ = 2.857 − 2.0 = 0.857 m/s, so the trolley continues forwards more slowly, which is consistent.)