Q 11-05-212JEE MainMedium
A ball of mass $m$ moving with speed $u$ collides head-on with a stationary ball of mass $2m$. The coefficient of restitution is $\dfrac12$. The fraction of the initial kinetic energy lost in the collision is
Answer: (C) $\dfrac12$
Concept. Use momentum conservation and the restitution equation $v_2 - v_1 = eu$.
Step 1. Equations.
$$v_1 + 2v_2 = u,\qquad v_2 - v_1 = \frac u2$$
Step 2. Solve. Adding twice the second to the first: $3v_2 = \dfrac{3u}{2} \Rightarrow v_2 = \dfrac u2$, and then $v_1 = 0$. The first ball stops.
Step 3. Energy.
$$K_f = \tfrac12(2m)\frac{u^2}{4} = \frac{mu^2}{4} = \frac12K_i$$
Half the kinetic energy is lost.
Answer. $\dfrac12$ (Option C).
Solution by Sreeraj P, M.Sc Physics