A uniform bar of length ‘L’ and mass ‘M’ lies on a smooth horizontal table a point mass ‘m’ moving on same horizontal plane with speed ‘v’ strikes the bar as shown and sticks to the bar. Immediately after collision angular velocity of the system is $\frac{nv}{7L}$. Then n =

Conservation of linear momentum,

$mv=M{v}_1+m{v}_2$...........(1)

With respect to center of mass of rod, angular momentum before collision is

$L_i=mv\frac{l}{2}$

Just after collision, the speed of center of mass alone is $v_1$. Its angular speed about center of mass of rod is $\omega$. The speed of point mass is $v_2$.

$\begin{array}{l}L{}_f=\left(\frac{M{l}^2}{12}\right)\omega +m{v}_2\frac{l}{2}\end{array}$

Using conservation of angular momentum,

$mv\frac{l}{2}=\hspace{0.17em}\left(\frac{M{I}^2}{12}\right)\omega +m{v}_2\frac{I}{2}$...............(2)

Also, Coefficient of restitution is zero. 

Thus, $v_2=v_1+\omega \frac{l}{2}$........................(3)

After solving, we get

$\omega =\frac{6v}{7l}$