Question

A bullet of mass 10 g and speed 500 m/s is fired into a door and gets embedded exactly at the centre 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. Find the angular speed of the door just after the bullet embeds into it.
(Hint: The moment of inertia of the door about the vertical axis at one end is $M{L}^2/3$.)

Answer 1

Mass of the bullet, $m=10g=10\times {10}^{-3}kg$
Velocity of the bullet, v = 500 m/s
Thickness of the door, L = 1 m
Distance of the centre of door from the hinge axis, $r=\frac{1}{2}m$
Mass of the door, M = 12 kg
Angular momentum imparted by the bullet on the door:
$\alpha =mvr=(10\times {10}^{-3})\times \left(500\right)\times \frac{1}{2}=2.5kg{m}^2{s}^{-1}\dots \left(i\right)$
Moment of inertia of the door:
$I=\frac{1}{3}M{L}^2=\frac{1}{3}\times 12\times (1{)}^2=4kg{m}^2$
But $\alpha =I\omega \therefore \omega =\frac{\alpha }{I}=\frac{2.5}{4}=0.625rad{s}^{-1}$