A spherical ball of radius R and mass m collides with a plank of mass M kept on a smooth horizontal surface. Just before impact, the centre of the ball has a velocity $v_0$ vertically downward and angular velocity ${\omega }_0$ as shown in the figure. After the impact, the normal velocity is reversed with same magnitude and the ball stops rotating after the impact. The coefficient of friction between the ball and the plank is $\mu$ . 
Assume that the plank is large enough. The distance on the plank between first two impacts of the ball is $\frac{4v_{0}R{\omega }_0}{\beta g}(1+\frac{m}{M}).$  Find the value of  $\beta$

The forces during impact are as shown in figure
 
Let the horizontal velocities of the plank and the ball be v2 and v1 in opposite directions as shown in
Figure
 
From conservation of linear momentum
 $m{v}_1=M{v}_2\mathrm{.......}\left(i\right)$
Linear impulse of the ball in vertical direction = Charge in linear momentum in vertical direction
$J=2m{v}_0\mathrm{..........}\left(ii\right)$ 
 $[\therefore J=ndt=\Delta p]$
Linear impulse on the ball in horizontal direction =change in linear momentum in horizontal
Direction
 $J_f=m{v}_1\mathrm{.........}\left(iii\right)$
Angular impulse on the ball about centre of mass = change in angular momentum about centre of
Mass
 $J_{f}R=I{\omega }_0=\frac{2}{5}m{R}^2{\omega }_o\mathrm{.....}\left(iv\right)$
Solving Eqs (i), (ii), (iii) and (iv) get
 $v_1=\frac{2}{5}R{\omega }_{0}and{v}_2=\frac{m}{M}\left(\frac{2}{5}R{\omega }_o\right)$
Now, actual path of the ball is projectile, whose time of flight will be
$T=\frac{2v_y}{g}=\frac{2v_o}{g}$ 
Relative velocity of ball with respect to plank in horizontal direction is
$v_r=v_1+v_2=\frac{2}{5}(1+\frac{m}{M})R{\omega }_o$ 
So, the desired distance is
 $s=v_{r}T=\frac{4}{5}.\frac{v_{o}R{\omega }_o}{g}(1+\frac{m}{M})$