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.)
Solution
Given,
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 = m / 2
Mass of the door, M = 12 kg
Angular momentum imparted by the bullet on the door,
α = mvr
= (10 × 10-3 ) × (500) ×
= 2.5 kg m2 s-1 ...(i)
Moment of inertia of the door is given by,
This is the required angular speed of the door just after the bullet embeds into it.
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 = m / 2
Mass of the door, M = 12 kg
Angular momentum imparted by the bullet on the door,
α = mvr
= (10 × 10-3 ) × (500) ×
= 2.5 kg m2 s-1 ...(i)
Moment of inertia of the door is given by,
This is the required angular speed of the door just after the bullet embeds into it.
