Question

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)

• A. 50.0%
• B. 66.6%
• C. 55.6%
• D. 33.3%

Answer 1

Answer: (C)

55.6%

Answer 2

For a head on elastic collision

$v_2=\frac{mv_1}{m+5m}+\frac{mv_1}{m+5m}$

=$\frac{2v_1}{6}$ or $\frac{v_1}{3}$

Initial kinetic energy of first mass

=$\frac{1}{2}mv_1^2$

Final kinetic energy of second mass

=$\frac{1}{2}×5m\bigg(\frac{v_1}{3}\bigg)^2$

=$\frac{5}{9}\bigg(\frac{1}{2mv_1^2}\bigg)$

$⇒$ kinetic energy transferred = $55\%$ of initial kinetic energy of first colliding mass