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 weight `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 `ML^(2)//3)`

Explanation

To solve the problem, we will use the principle of conservation of angular momentum. The steps are as follows:

### Step 1: Identify the initial and final states
Initially, the bullet is moving with a speed of 500 m/s, and the door is at rest. After the bullet embeds itself in the door, the system (door + bullet) will rotate together.

### Step 2: Calculate the initial angular momentum
The initial angular momentum (L_initial) is due only to the bullet since the door is at rest. The angular momentum can be calculated using the formula:
\[ L = m \cdot v \cdot r \]
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