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)

Step-by-step solution

After collision, Applying momentum conservation, m u = m v 1 + 5 m v 2 ⇒ u = v 1 + 5 v 2       . . . 1 Coefficient of restitution, e = 1 = v 2 - v 1 u ∴   u = v 2 - v 1       . . . 2 Adding 2 u = 6 v 2 ⇒ v 2 = u 3 and u = u 3 - v 1 ∴   v 1 = u 3 - u = - 2 u 3 Percentage change in kinetic energy = 1 2 mv 1 2 - 1 2 mu 2 1 2 mu 2 × 100 = v 1 2 - u 2 u 2 × 100 = 4 u 2 9 - u 2 u 2 × 100 = - 5 9 × 100 = 55 . 6 %