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 ML²I³.)

Answer 1

Consider the door in the figure

[From perpendicular  axis theorem]

Mass of the bullet, m_B = 100gm

velocity of bullet, V = 500 m/s

Mass of door, M = 12kg

width of door, I = 1 m

Since there is no frictional torque on the doors, the angular momentum is conserved for door bullet system.

∴ L (initial) = L( final)

≈ 0.625 rad/sec.