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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 ML2/3.)
Asked by Franziska Last Modified
Answer 
Tasneem
Mass of the bullet, m = 10 g = 10 × 10–3 kg
Velocity of the bullet, v = 500 m/s
Thickness of the door, L = 1 m
Radius of the door, r = 
Mass of the door, M = 12 kg
Angular momentum imparted by the bullet on the door:
α = mvr
Moment of inertia of the door:
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