Question

A ball of radius $R=20cm$ has mass $m=0.75kg$ and moment of inertia (about its diameter) $I=0.0125kg$ $m^2$. The ball rolls without sliding over a rough horizontal floor with velocity $v_0=10{ms}^{-1}$ towards a smooth vertical wall. If coefficient of restitution between the wall and the ball is $e=0.7$, calculate velocity $v$ (in m/s) of ball long after the collision. ($g=10{ms}^{-2}$).

Answer 1

Since the ball is rolling without sliding, therefore, its angular velocity (${\omega }_0$), just before collision with wall is$\frac{v_0}{R}=\frac{10}{0.2}=50rad$ $s^{-1}$
Translational velocity of centre of ball, just after collision.
$e{v}_0=0.7\times 10=7$ ${ms}^{-1}$ (Rightwards)
Since the wall is smooth, therefore no tangential force is applied by the wall. Hence, angular velocity $w_0$ of ball remains unchanged during collision. Now surface of ball slides on floor to the right as shown in figure.
Let coefficient of friction between ball and the floor be $m$. Considering free-body diagram of ball while sliding:
For vertical forces, $N=mg$
For horizontal forces, $\mu N=ma$
$a=\mu g=10\mu$
Taking moment (about $O$) of force acting on the ball
$\mu NR=I\alpha$ or $\alpha =120\mu$
Long after the collision, there will be no sliding or it will be pure rolling. Let sliding stop after a time $t$ after collision, then final translational velocity, $v=(7-at)$ or $v=7-10\mu t$ and final angular velocity $\omega =\left({\omega }_0\right)+\alpha t$
$\omega =(120\mu t-50)rad$ $s^{-1}$ (clockwise)
But at that instant, $v=R\omega$
$97-10\mu t)=0.2(120\mu t-50)$
$\mu t=0.5$
$v=2$ ${ms}^{-1}$