Q 11-05-128JEE MainJEE Main 2022 (27 Jun, Shift 1)Medium
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)
Answer: (C) $55.6\%$
In a head-on elastic collision of $m_1$ with a stationary $m_2$, the fraction of kinetic energy transferred is
$$f = \frac{4m_1m_2}{(m_1+m_2)^2}$$
With $m_2 = 5m_1$:
$$f = \frac{4(1)(5)}{(6)^2} = \frac{20}{36} = 0.556 = 55.6\%$$
Solution by Sreeraj P, M.Sc Physics