Question

A bullet of mass $10g$ and speed $500m{s}^{-1}$ is fired into a door and gets embedded exactly at the centre of the door. The door is $1.0m$ wide and weighs $12kg$ . 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

Answer: 625

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