Explanation

(b) Let ball strikes at a speed $u$ the $K_{1} = \frac{1}{2}mu^{2}$
Due to collision tangential component of velocity remains unchanged at $u$ sin 45${^\circ}$, but the normal component of velocity change to $u$ sin45${^\circ} = \frac{1}{2}u\cos{45{^\circ}}$
$\therefore$ Final velocity of ball after collision
$v = {\sqrt{\left( u\sin{45{^\circ}} \right)^{2} + \left( \frac{1}{2}u\cos{45{^\circ}} \right)^{2}}}^{\ }$
$= \sqrt{\left( \frac{u}{\sqrt{2}} \right)^{2} + \left( \frac{u}{2\sqrt{2}} \right)^{2}} = \sqrt{\frac{5}{3}}u.$
Hence final kinetic energy $K_{2} = \frac{1}{2}mv^{2} = \frac{5}{16}mu^{2}$.
$\therefore$ Fractional loss in KE
$= \frac{K_{1} - K_{2}}{K_{1}} = \frac{\frac{1}{2}mu^{2} - \frac{5}{16}mu^{2}}{\frac{1}{2}mu^{2}} = \frac{3}{8}$