Question

A bullet of mass $10\,g$ and speed $500\,ms^{- 1}$ 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. If $\omega $ is the angular speed (in $rad\,s^{- 1}$ ) of the door just after the bullet embeds into it, then find the value of $10^{3}\,\omega $ . Ignore the mass of the bullet as compared to the door.

Answer 1

Given, mass of bullet $( m )=10 g =0.01\, kg$
Speed of bullet $(v)=500 \, m s ^{-1}$
Width of the door$( I )=1.0 \, m$
Mass of the door$( M )=12 \, kg$
As bullet gets embedded exactly at the centre of the door, therefore its distance from the hinged end of the door,
$( r )=\frac{l}{2}=\frac{1}{2} m$
Angular momentum transferred by the bullet to the door,
$(L)=m v \times r$
$=0.01 \times 500 \times \frac{1}{2}$
$=2.5\, J - S$
Moment of inertia of the door about the vertical axis at one of its end,
$( I )=\frac{ M l^{2}}{3}=\frac{12 \times(1)^{2}}{3}=4\, kg - m ^{2}$
But angular momentum, $(L)=I \, \omega$
But angular momentum, $( L )= I\, \omega$
$\therefore 2.5=4 \times\, \omega$
or $\omega=\frac{2.5}{4}$
$\omega=0.625\, rad \, s ^{-1}$