A bullet of mass 10 g and speed 500 m s - 1 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. If ω is the angular speed (in rad s - 1 ) of the door just after the bullet embeds into it, then find the value of 10 ω . Ignore the mass of the bullet as compared to the

Step-by-step solution

Given, mass of bullet (m) = 10 g = 0.01 kg Speed of bullet (v) = 500 m s -1 Width of the door (l) = 1.0 m Mass of the door (M) = 12 kg As bullet gets embedded exactly at the centre of the door, therefore its distance from the hinged end of the door, ​ (r) = l ⁡ 2 = 1 2 m Angular momentum transferred by the bullet to the door, (L) = mv × r = 0.01 × 500 × 1 2 = 2.5 J-s Moment of inertia of the door about the vertical axis at one of its end, (I) = M l ⁡ 2 3 = 12 × (1) 2 3 = 4 kg-m 2 But angular momentum, (L) = Iω​ ∴ 2