A bullet of momentum $p$ is fired into a door and gets embedded exactly at the center of the door. The door is 1.0 m wide and weighs 12 kg . It is hinged at one end and rotates about a vertical axis practically without friction. The angular speed of the door just after the bullet embeds into it is (Assume the mass of bullet is negligible compared to the mass of door)

Let $M = 12$ kg be the mass of the door and $L = 1.0$ m be its width. The moment of inertia of the door about the vertical axis passing through the hinge is $I_{door} = \dfrac{1}{3} M L^2 = \dfrac{1}{3} \times 12 \times (1.0)^2 = 4$ kg m$^2$.

Let $m$ be the mass of the bullet and $v$ be its velocity such that $p = mv$. The bullet strikes the center of the door at distance $r = \dfrac{L}{2} = 0.5$ m from the hinge. The angular momentum of the bullet about the hinge just before impact is $L_{bullet} = p \times r = p \times 0.5 = \dfrac{p}{2}$.

Since there is no external torque about the hinge, the angular momentum is conserved. Let $\omega$ be the angular velocity of the door-bullet system after impact. The moment of inertia of the system is $I = I_{door} + I_{bullet} = \dfrac{1}{3} M L^2 + m r^2 = 4 + m(0.5)^2 = 4 + 0.25m$.

Assuming the mass of the bullet $m$ is negligible compared to the mass of the door ($m \ll M$), the moment of inertia is dominated by the door, $I \approx 4$ kg m$^2$.

Equating angular momentum: $L_{bullet} = I \omega \Rightarrow \dfrac{p}{2} = 4 \omega \Rightarrow \omega = \dfrac{p}{8}$.

Answer: $\dfrac{p}{8}$