A ball impinges directly on a similar ball at rest. The first ball...

Let $u_1$ and $v_1$ be the initial and final velocities of ball 1 and $u_2$ and $v_2$ be the similar quantities for ball 2. Here, $u_2=0$ and $v_1=0$. $\therefore \quad$ initial KE, $K_i=\frac{1}{2} m u_1^2+\frac{1}{2} m u_2^2=\frac{1}{2} m u_1^2$ and final KE, $K_f=\frac{1}{2} m v_1^2+\frac{1}{2} m v_2^2=\frac{1}{2} m v_2^2$ Loss of $\mathrm{KE}, \Delta K=K_i-K_f=\frac{1}{2} m u_1^2-\frac{1}{2} m v_2^2$ According to question, $\frac{1}{2}\left(\frac{1}{2} m u_1^2\right)=\frac{1}{2} m u_1^2-\frac{1}{2} m v_2^2$ ( $\because$ half of its $\mathrm{KE}$ is lost by impact) or $u_1^2=2 v_2^2$ or $v_2=\frac{u_1}{\sqrt{2}}$ $\therefore \quad$ Coefficient of restitution, $e=\left|\frac{v_2-v_1}{u_1-u_2}\right|=\frac{v_2}{u_1}=\frac{1}{\sqrt{2}}$