Question

A solid sphere rolls on a rough horizontal surface with a linear speed $7\text{m/s}$ and collides elastically with a fixed smooth vertical wall. Then, find the speed of the sphere when it has started pure rolling in the backward direction in $\text{m/s}$.

Answer 1

(I) Just before collision:
Let the linear velocity be $v$ and angular velocity be $\omega$ (clockwise).
& $v=\omega R$ [pure rolling]

(II) After elastic collision:
Linear velocity will be reversed and angular velocity will remain same since there is no external torque (wall is frictionless).

(III) When it starts pure rolling in backward direction, let the linear velocity be $v^{\prime }$ and angular velocity be ${\omega }^{\prime }$ (anti-clockwise).
Then, $v^{\prime }={\omega }^{\prime }R$

Angular momentum is conserved about the point of contact [since friction passes through the point]

Angular momentum just after collision (II)
$L=mvR–I\omega$
$=mvR–\frac{2}{5}m{R}^2\times \frac{v}{R}=\frac{3}{5}mvR$

When pure rolling starts (III), angular momentum is,
$L^{\prime }=m{v}^{\prime }R+I{\omega }^{\prime }$
$=m{v}^{\prime }R+\frac{2}{5}m{R}^2\times \frac{v^{\prime }}{R}=\frac{7}{5}m{v}^{\prime }R$

So, we have $L=L^{\prime }$
$\Rightarrow v^{\prime }=\frac{3v}{7}$
Linear speed $v=7\text{m/s}$ (Given)
So, $v^{\prime }=\frac{3\times 7}{7}=3\text{m/s}$