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: me moment of inertia of the door about the vertical axis at one end is ML²/ 3.)

Answer 1

Answer:

Explanation:

* mass of the door = M = 12 kg

* mass of the bullet = m = 10 g = 0.001 kg

* initial velocity of the bullet = u = 500 m/s.

* width of the door = l = 1.0 m.

* Moment of inertial =$I = \dfrac{ML^2}{3}\ =\ \dfrac{12\times 1^2}{3}\ =\ 4\ kg.m^2$

Bullet is embedded at the center of the door, therefore, radius of the rotation = R = 0.5 m

Let 'w' be the angular speed of the door after the collision of the bullet,

After the collision,

angular momentum of the door = $muR = 0.001\times 500\times 0.5\ =\ 0.25 kgm/s.$

$\Rightarrow I w\ =\ 0.25\\\Rightarrow w\ =\ \dfrac{0.25}{I}\\\Rightarrow w\ =\ \dfrac{0.25}{4}\\\Rightarrow w\ =\ 0.0625\ rad/s.$

Hence, the angular speed of the door after the collision of the bullet is 0.0625 rad/s.