Let u and v be the initial and final velocities of ball 1 and u and v be the similar quantities for ball 2. Here, u = 0 and v = 0.
$\therefore\quad$ initial KE, $K_{i}=\frac{1}{2}mu^{2}_{1}+\frac{1}{2}mu^{2}_{2}=\frac{1}{2}mu^{2}_{1}$
and final KE, $K_{f} =\frac{1}{2}mv^{2}_{1}+\frac{1}{2}mv^{2}_{2}=\frac{1}{2}mv^{2}_{2}$
Loss of KE, $\Delta K=K_{i}-K_{f} =\frac{1}{2}mu^{2}_{1}-\frac{1}{2}mv^{2}_{2}$
According to question,
$\frac{1}{2}\left(\frac{1}{2}mu^{2}_{1}\right)=\frac{1}{2}mu^{2}_{1}-\frac{1}{2}mv^{2}_{2}$
($\because \,$ half of its KE is lost by impact)
or$\quad u^{2}_{1}=2v^{2}_{2}$ or $v_{2}=\frac{u_{1}}{\sqrt{2}}$
$\therefore\quad$ Coefficient of restitution,
$e=\left|\frac{v_{2}-v_{1}}{u_{1}-u_{2}}\right|=\frac{v_{2}}{u_{1}}=\frac{1}{\sqrt{2}}$