A ball of mass $m$ is attached to a cord of length L, pivoted at point O, as shown in figure. The ball is released from rest at point A, swings down and makes an inelastic collision with a block of mass $2m$ kept on a rough horizontal floor. The coefficient of restitution of collision is $e=2/3$ and coefficient of friction between block and surface is $\mu .$ After collision, the ball comes momentarily to rest at C when cord makes an angle of $\theta$ with the vertical and block moves a distance of $3L/2$ on rough horizontal floor before stopping. The values of $\mu$ and $\theta$ are, respectively,

 

Velocity of the ball just before collision is $v_0=\sqrt{2gL}$

Applying momentum conservation along horizontal direction (because momentum is conserved  in collision along the line of impact), we get, $m{v}_0=-m{v}_1+2m{v}_2$

Applying coefficient of restitution equation, we get, $e=\frac{2}{3}=\frac{v_1+v_2}{v_0}$

Solving the above two equations, we get, $v_1=\frac{v_0}{9}and{v}_2=\frac{5v_0}{9}$

As the block moves a distance of $3L/2$ before coming to rest, so from work-energy theorem, 

$0-\frac{1}{2}\left(2m\right)v_2^2=-\mu \times 2mg\frac{3L}{2}\Rightarrow v_2^2=3\mu gL$

$\frac{25}{81}\times 2gL=3\mu gL\Rightarrow \mu =\frac{50}{243}$

For ball, KE is converted to gravitational potential energy after collision, so, $0-\frac{m{v}_1^2}{2}=-mg\times L(1-\cos \theta )$

$\cos \theta =\frac{80}{81}$