### 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 *ML*^{2}/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 m^{2} s^{-1} …(i)

Moment of inertia of the door:

I = ML^{2} / 3

= (1/3) × 12 × 1^{2} = 4 kgm^{2}

But α = Iω

∴ ω = α / I

= 2.5 / 4

= 0.625 rad s^{-1}