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 $K_{1}=\frac{1}{2}mu^{2}$
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}ucos 45 ^\circ $
$\therefore $ Final velocity of ball after collision
$v=\left(\sqrt{\left(u sin 45 ^\circ \right)^{2} + \left(\frac{1}{2} u cos ⁡ 45 ^\circ \right)^{2}}\right)^{ \, }$
$=\sqrt{\left(\frac{u}{\sqrt{2}}\right)^{2} + \left(\frac{u}{2 \sqrt{2}}\right)^{2}}=\sqrt{\frac{5}{3}}u.$
Hence final kinetic energy $K_{2}=\frac{1}{2}mv^{2}=\frac{5}{16}mu^{2}$ .
$\therefore $ Fractional loss in KE
$=\frac{K_{1} - K_{2}}{K_{1}}=\frac{\frac{1}{2} m u^{2} - \frac{5}{16} m u^{2}}{\frac{1}{2} m u^{2}}=\frac{3}{8}$