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. | Learn NCERT solution | Education portal Class 11 Physics | Study online Unit-7 System Of Particles And Rotational Motion



Q.24:- 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:-



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) × (1/2)  =  2.5 kg m2 s-1    …(i)
Moment of inertia of the door:
I = ML2 / 3
= (1/3) × 12 × 12 = 4 kgm2
But α = Iω
∴ ω = α / I
= 2.5 / 4
= 0.625 rad s-1