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 ML2/3.) Question 7.24 System Of Particles And Rotational Motion

Answer 1

Given,
Mass of bullet (m) = 10g = 0.01 Kg
Speed of bullet (v) = 500m/s
Width of wire(L) = 1m
Mass of door ( M) = 12 Kg
A/c to question,
Gets embedded exactly at the centre of door , so , distance from the hinged end of the door,
(r) = L/2 = 1/2 = 0.5 m
Angular momentum transferred by the bullet to the door = mvr= 0.01 × 500 × 0.5 = 2.5 js______(1)

Moment of inertia of the door about the vertical axis at one end of its end ,
I = ML²/3 = 12 × 1/3 = 4kgm²
Now,
angular momentum = Iw
2.5 = 4×w { from equation (1)
w = 0.625 rad/s

Answer 2

Answer:0.625rad/sec

Explanation:by conservation of angular momentum