An L shaped wedge of mass m having long horizontal base is placed on smooth horizontal surface as shown. A thin ring of mass m and radius R is set in motion on the rough horizontal part of the wedge with a linear velocity v and angular velocity $\omega$ as shown, such that $v=\omega R$. Ring collides with the smooth vertical part of wedge such that collision is perfectly elastic. After a sufficiently long time, the final velocity of wedge is $\frac{v}{K}$. Find the value of $K$.
(Assume wedge does not topple due to collision)
 

Observing the motion in COM frame. Then system will look like
After the collision, ring starts sliding and then starts pure rolling on wedge as shown. $\omega \text{'}R=2v$ (pure rolling)
 

Applying COAM of ring in centre of mass frame (after the collision) about point P on wedge

$$\begin{gathered} m{R}^2\omega +(-mR\frac{v}{2})=mRv\text{'}+m{R}^2\omega \text{'} \\ mR\frac{v}{2}=3mRv\text{'}\Rightarrow v\text{'}=\frac{v}{6} \end{gathered}$$

Therefore, the velocity of wedge is $\frac{v}{2}$ (in centre of mass frame) when pure rolling starts after the first collision. Then they again collides and ring again starts pure rolling on wedge and velocity of wedge is $\frac{v}{18}$ (in centre of mass frame) after $2^{nd}$ collision.
The process goes on and there will be infinite collisions and finally wedge comes at rest in centre of mass frame.
$\therefore$ Velocity of wedge = $\frac{v}{2}$ (in ground frame)