# 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.

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 $M L^{2} / 3$.)

Solution:

Mass of the bullet, $m=10 \mathrm{~g}=10 \times 10^{-3} \mathrm{~kg}$

Velocity of the bullet, $v=500 \mathrm{~m} / \mathrm{s}$

Thickness of the door, $L=1 \mathrm{~m}$

Radius of the door, $r=\frac{1}{2} \mathrm{~m}$

Mass of the door, $M=12 \mathrm{~kg}$

Angular momentum imparted by the bullet on the door:

$\alpha=m v r$

$=\left(10 \times 10^{-3}\right) \times(500) \times \frac{1}{2}=2.5 \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-1}$ $\ldots(i)$

Moment of inertia of the door:

$I=\frac{1}{3} M L^{2}$

$=\frac{1}{3} \times 12 \times(1)^{2}=4 \mathrm{~kg} \mathrm{~m}^{2}$

But $\alpha=I \omega$

$\therefore \omega=\frac{\alpha}{I}$

$=\frac{2.5}{4}=0.625 \mathrm{rad} \mathrm{s}^{-1}$