 Solution

During the short collision, the pivot's impulse has zero torque about the pivot, so angular momentum conservation about that point givesThe bullet stops rather than becoming embedded: immediately after impact only the rod contributes to the kinetic energy. Because this is an elastic collision,Eliminating $\Omega$ yields $m{L}^2=I$: the projectile mass equals the effective impact mass of a pivoted body, $I/L^2$. Hence the bullet has one third of the rod's mass, andOrdinary momentum conservation for the bullet and rod together would be incorrect: the fixed pivot supplies an external impulse. The short-impact approximation neglects the change in gravitational potential energy during impact.