What percentage of kinetic energy of a moving particle is transferred to a stationary particle when it strikes the stationary particle of 5 times its mass?
(Assume the collision to be head-on elastic collision)

Solution

Physics sir · 4 steps

Find the struck particle's speed first, then compare its kinetic energy with what the striker started with.

1. Set up. Let the moving particle have mass m and speed u. It hits a particle of mass 5m at rest, head-on and elastically.
2. Speed after collision. For a 1D elastic collision with the target at rest, the target moves off with v₂ = 2m₁/(m₁ + m₂)u = 2m/6mu = u/3.

   K₂/K₁ = (1/2(5m)(u/3)²)/(1/2mu²) = 5/9 = 4m₁m₂/(m₁+m₂)²

3. Check. The striker rebounds with v₁ = (m - 5m)/6mu = -2u/3, so it keeps 4/9 of the energy. Then 4/9 + 5/9 = 1, as an elastic collision requires.
4. So. The fraction transferred is 5/9 ≈ 55.6%, which is option (c).

The idea to keep. In a head-on elastic collision with a target at rest, the fraction of kinetic energy transferred is 4m₁m₂/(m₁+m₂)². It is 100% only when the two masses are equal.

Answer from the NTA final answer key. Solution by PrepWiser.