The coefficient of restitution between a ball and the floor is ½. If the ball hits the floor at an angle  ${\text{θ = 45}}^{\text{0}}$, find the fraction of its kinetic energy lost in collision ( in J). 

Let ball strikes at a speed $u,\text{\hspace{0.17em}the}{K}_1=\frac{1}{2}m{u}^2.$ Due to collision, tangential component of velocity remains unchanged at $u,\sin {45}^0,$ but the normal component of velocity change to $u,\sin {45}^0=\frac{1}{2}u\cos {45}^0$

$\therefore$Final velocity of ball after collision $\upsilon =\sqrt{\left(u\sin {45}^0\right)+{\left(\frac{1}{2}u\cos {45}^0\right)}^2}=\sqrt{{\left(\frac{u}{\sqrt{2}}\right)}^2+{\left(\frac{u}{2\sqrt{2}}\right)}^2}=\sqrt{\frac{5}{8}}u$

Hence, final kinetic energy $K_2=\frac{1}{2}m{\upsilon }^2=\frac{5}{16}m{u}^2$

$\therefore$Fractional loss in KE 

$=\frac{K_1-K_2}{K_1}$

$\frac{\frac{1}{2}m{u}^2-\frac{5}{16}m{u}^2}{\frac{1}{2}m{u}^2}=\frac{3}{8}$