Question

A ball strikes a horizontal floor at an angle of $\theta =45^{\circ }$ . The coefficient of restitution between the ball and the floor is $0.5$ . The fraction of its kinetic energy lost in collision is

• A. $\frac{5}{8}$
• B. $\frac{3}{8}$
• C. $\frac{3}{4}$
• D. $\frac{1}{4}$

Answer 1

Answer: (B)

Let ball strikes at a speed $u$ the $K_1=\frac{1}{2}m{u}^2$
Due to collision tangential component of velocity remains unchanged at $u$ sin 45 ${}^{\circ }$ , but the normal component of velocity change to
$u$ sin45 ${}^{\circ }=\frac{1}{2}ucos45^{\circ }$
$\therefore$ Final velocity of ball after collision
$v={\left(\sqrt{{\left(usin45^{\circ }\right)}^2+{\left(\frac{1}{2}ucos45^{\circ }\right)}^2}\right)}^{}$
$=\sqrt{{\left(\frac{u}{\sqrt{2}}\right)}^2+{\left(\frac{u}{2\sqrt{2}}\right)}^2}=\sqrt{\frac{5}{3}}u.$
Hence final kinetic energy $K_2=\frac{1}{2}m{v}^2=\frac{5}{16}m{u}^2$ .
$\therefore$ Fractional loss in KE
$=\frac{K_1-K_2}{K_1}=\frac{\frac{1}{2}m{u}^2-\frac{5}{16}m{u}^2}{\frac{1}{2}m{u}^2}=\frac{3}{8}$