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 firction. 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.]

Answer 1

Here, 

L = 1 m,

m = 12 kg,

m = 10 g = 0.1 kg

and v = 500 m/s

According to the law of conservation of angular momentum, angular momentum of the bullet-door system should remain conserved. Before impact,only bullet is moving. Therefore, initial angular momentum of the system.

vector L_i = mg x L/2.......(i)

[Bullet is aimed at the centre of the door]

After the bullet gets embedded, suppose that the system acquires angular velocity equal to ω. Let I be the moment of inertia of the bullet-door system about an axis at one end of the door. Then the final angular momentum vector L_f = Iω.

Now, I = M.l. of the bullet about an axis through one end of door + M.I. of the door about the same axis,