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 weight `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 1

Angular momentum imparted by the bullet `L = mv xx r = (10 xx 10^(-3)) xx 500 xx (1)/(2) = 2.5`
Also, `I = (ML^(2))/(3) = (12 xx 1.0^(2))/(3) = 4kg m^(2)`
As `L = I omega :. Omega = (L)/(I) = (2.5)/(4) = 0.625 rad//sec`.