Question

A solid sphere is rolling purely on a rough horizontal surface with speed of center $u=12$ m/s. It collides inelastically with a smooth vertical wall at a certain moment, the coefficient of restitution being $\frac{1}{2}$. How long (in sec) after the collision, the sphere will begin pure rolling?
[coefficient of friction between the sphere and the ground is $\frac{3}{35}$]

• A. 16
• B. 12
• C. 6
• D. 13

Answer 1

The correct option is C 6

Velocity of sphere after collision is $w_{0}6$.

Angular velocity just after collision is $w_1=12/R$.

For pure rolling, the final angular velocity is 6/R.

Taking torque about the center of cylinder, $\mu MgR=\frac{2}{5}M{R}^2\alpha$

So, $\alpha =5\mu g/2R$

$w_0=w_1-\alpha t\Rightarrow t=6$