Question

A solid sphere rolling on a rough horizontal surface with a linear speed $v$ collides elastically with a fixed, smooth, vertical wall. The speed of the sphere after it has started pure rolling in the backward direction is

• A. $\frac{v}{7}$
• B. $\frac{2v}{7}$
• C. $\frac{3v}{7}$
• D. Zero

Answer 1

Answer: (C)

Before collision

$mvR-\left(\frac{2}{5}m{R}^2\right)\times \omega =m{v}^{\prime }R+\left(\frac{2}{5}m{R}^2\right)\times {\omega }^{\prime }$
$\Rightarrow mvR-\frac{2}{5}m{R}^2\times \frac{v}{R}=m{v}^{\prime }R+$
$\left(\frac{2}{5}m{R}^2\right)\times \frac{v^{\prime }}{R}$
on solving $\frac{3v}{5}=\frac{7v^{\prime }}{5}\Rightarrow v^{\prime }=\frac{3v}{7}$