Question

A rubber ball is dropped from a height of $5\,m$ on a plane, where the acceleration due to gravity is not shown. On bouncing, it rises to $1.8\,m$ . The ball loses its velocity on bouncing by a factor of $\frac{\alpha }{5}$ . Write value of $\alpha $ .

Answer 1

According to principle of conservation of energy
Loss in potential energy = Gain in kinetic energy
$mgh=\frac{1}{2}mv^{2} \, \Rightarrow \, \, \, v=\sqrt{2 g h}$
If $h_{1} \, $ and $ \, h_{2}$ are initial and final heights, then
$v_{1}=\sqrt{2 \, g h_{1}}$ and $v_{2}=\sqrt{2 \, g h_{2}}$
Loss in velocity
$\Delta v=v_{1}-v_{2}=\sqrt{2 g h_{1}}-\sqrt{2 g h_{2}}$
$\therefore \, \, \, $ Fractional loss in velocity
$=\frac{\Delta v}{v_{1}}=\frac{\sqrt{2 g h_{1}} - \sqrt{2 g h_{2}}}{\sqrt{2 g h_{1}}}=1- \, \sqrt{\frac{h_{2}}{h_{1}}}$
$=1-\sqrt{\frac{1.8}{5}}=1-\sqrt{0.36}=\frac{2}{5}$