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 ML²/3.)

Answer 1

Given, the mass of the bullet is 10 g and its initial speed is 500 m/s. The width of the door is 1.0 m and it’s weight is 12 kg.

The equation for the angular momentum imparted by the bullet on the door is,

L=mvr

Here, m is the mass of the bullet, v is its initial speed and r is the radial distance of the point of impact from the axis of rotation.

Substituting the given values in the above equation, we get:

L=( 10 g )( 500 m/s )( 1.0 m 2 ) =( 10 g× 1 kg 1000 g )( 500 m/s )( 1.0 m 2 ) =2.5  kg-m 2 /s

Since there is no non-conservative force acting on the system, the angular momentum remains conserved.

The expression for the moment of inertia of the door about the vertical axis at one end is,

I= M L 2 3

Here, M is the mass of the door and L is the width of the door.

The expression for the conservation of angular momentum is,

L=I( ω− ω 0 )

Here, I is the moment of inertia of the door about the vertical axis at one end, ω is the angular speed of the door just after the bullet embeds into it and ω 0 is the angular speed of the door just before the bullet embeds into it.

Substituting the value of I in the above equation, we get:

L=( M L 2 3 )( ω− ω 0 )

Initially, the angular velocity of the door is zero as it is at rest. Therefore,

ω 0 =0

Substituting the given values in the above equation, we get:

2.5=( 12× 1.0 2 3 )( ω−0 ) 2.5=( 12× 1.0 2 3 )ω ω=0.625  rad-s −1

Thus, the angular speed of the door just after the bullet embeds into it is 0.625  rad-s −1 .