A bullet, fired into a door gets embedded exactly at it's centre, causing the door to rotate about it's vertical axis, practically without friction, with an angular velocity of 0.625 rads⁻¹ . The door is 1.0 m wide and weighs 12 kg. If the mass of the bullet is 10g, find the speed with which it was fired. (Hint: The moment of inertia of the door about the vertical axis at one end is ML^2 3 ).

Step-by-step solution

By conservation of angular momentum about the vertical axis of the door: L_i = L_f m v r = (I_ door + I_ bullet ) m v ( L 2 ) = ( ML^2 3 + m ( L 2 )^2 ) Substituting the given values M = 12 kg , L = 1.0 m , m = 0.01 kg , and = 0.625 rads⁻¹ : 0.01 v 0.5 = ( 12 1^2 3 + 0.01 0.5^2 ) 0.625 0.005 v = (4 + 0.0025) 0.625 Neglecting the small term 0.0025 compared to 4 : 0.005 v = 4 0.625 0.005 v = 2.5 v = 2.5 0.005 = 500 ms⁻¹