A bullet of mass 10 g and speed 500 m s -1 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. If a; is the angular speed (in rad s -1 ) of the door just after the bullet embeds into it, then find the value of 
. Ignore the mass of the bullet as compared to the door.
Text Solution
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6.25
Given, mass of bullet (m) = 10 g = 0.01 kg
Speed of bullet (v) = 500 m s -1
Width of the door (1) = 1.0m
Mass of the door (M)= 12 kg
As bullet gets embedded exactly at the centre of the door, therefore its distance from the hinged end of the door,
Angular momentum transferred by the bullet to the door,
(L) = mv x r
= 0.01 x 500 x 
= 2.5 J-s
Moment of inertia of the door about the vertical axis at one of its end,
But angular momentum, 
or 
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