Question

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

Answer 1

When the solid sphere collides with the wall; it rebounds with velocity 'v' towards left but it continues to rotate in the clockwise direction.

So, the angular momentum

= mvR - $\left(\frac{2}{5}\right)m{R}^2\times \frac{v}{R}$

After rebounding, when pure rolling starts lrt the velocity be 'v' and the corresponding angular velocity is

$\left(\frac{v^{\prime }}{R}\right)$

So, $mvR-\left(\frac{2}{5}\right)m{R}^2=\left(2/5\right)m{R}^2\frac{v^{\prime }}{R}+m{v}^{\prime }\left(R\right)$

$v^{\prime }=\frac{3v}{7}$

So, the sphere will move with velocity $\frac{3v}{7}$