Consider a uniform rod of mass M = 4m and length l pivoted about its centre. A mass m moving with velocity v making angle $\mathrm{\theta }=\frac{\mathrm{\pi }}{4}$to the rod’s long axis collides with one end of the rod and sticks to it. The angular speed of the rod-mass system just after the collision is

Take into account the pivoting rod and point mass m system.

At point O, the rod pivots around its axis.

The only cause of the system's rotational momentum prior to impact is point mass m.

${\mathrm{L}}_{\mathrm{initial}}=\mathrm{mvr}=\mathrm{mv}\frac{l}{2}\mathrm{sin\theta }$

After a collision, the bulk sticks to the rod.

The combined system's moment of inertia around O equals

${\mathrm{I}}_{\mathrm{final}}={\mathrm{I}}_{\mathrm{rod}}+{\mathrm{I}}_{\mathrm{mass}}=\frac{\mathrm{M}{l}^2}{12}+\mathrm{m}{\left(\frac{l}{2}\right)}^2=\frac{4\mathrm{m}{l}^2}{12}+\frac{\mathrm{m}{l}^2}{4}=\frac{7\mathrm{m}{l}^2}{12}$

The angular momentum is conserved since the system is not subject to any external torque.

${\mathrm{L}}_{\mathrm{initial}}={\mathrm{L}}_{\mathrm{final}}$

$\Rightarrow \mathrm{mv}\frac{l}{2}\mathrm{sin\theta }=\frac{7\mathrm{m}{l}^2}{12}\mathrm{\omega }$

$\Rightarrow \mathrm{\omega }=\frac{3\sqrt{2}\mathrm{v}}{7l}$

Hence the correct is $\frac{3\sqrt{2}v}{7l}.$