= Solution
During the short <collision>, the pivot's <impulse> has zero <torque> about the pivot, so <angular momentum conservation> about that point gives
$$
mvL=I\Omega,\qquad I=\frac{ML^2}{3}.
$$
The bullet stops rather than becoming embedded: immediately after impact only the rod contributes to the <kinetic energy>. Because this is an <elastic collision>,
$$
\frac12mv^2=\frac12I\Omega^2.
$$
Eliminating $\Omega$ yields $mL^2=I$: the projectile mass equals the <effective impact mass of a pivoted body>, $I/L^2$. Hence \b[the bullet has one third of the rod's mass], and
$$
\boxed{m=\frac M3,\qquad \Omega=\frac vL.}
$$
Ordinary <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.
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