A rod AC Of length $l$ and mass m is kept on a horizontal smooth plane. It is free to rotate and move. A particle of same mass m moving on the plane with velocity $v$ strikes rod at point B making angle ${37}^0$  with the rod. The collision is elastic. After collision:

The ball has V’ component of its velocity perpendicular to the length of the rod immediately after the collision. 

u is the velocity of CM of the rod and $\omega$ is angular velocity of the rod just after collision.  

The ball strikes the rod with speed $v\cos {53}^0$ in perpendicular direction and its component along the length of the rod after the collision is unchanged.

Using for the point of collision.

Velocity of separation = Velocity of approach

$\frac{3V}{5}=(\frac{\omega l}{4}+u)+V\text{'} ....................\left(1\right)$

Conserving linear momentum (of rod + particle) in the direction  $\perp$ to the rod,
$$\begin{gathered} mV\frac{3}{5}=mu-mV\text{'} ....................\left(2\right) \\ 0=0+[u\frac{l}{4}-\frac{m{l}^2}{12}\omega ]\Rightarrow u=\frac{\omega l}{3} \\ \Rightarrow u=\frac{24V}{55},W=\frac{72V}{55l} ....................\left(3\right) \end{gathered}$$

Time taken to rotate by $\pi$ angle,  $t=\frac{\pi }{\omega }$
 

In the same time, distance travelled   $=u_2 t=\frac{\pi l}{3}$
Using angular impulse-angular momentum equation,
$$\begin{gathered} {\int }_{ }^{ }Ndt\frac{l}{4}=\frac{m{l}^2}{4}.\frac{72V}{55l}\omega \\ \{\because {\int }_{ }^{ }Ndt\frac{l}{4}=\frac{24mV}{55}\} \end{gathered}$$

  [using impulse-momentum equation on the rod  ${\int }_{ }^{ }Ndt=mu=\frac{24mV}{55}$ ]