Question

A bullet of momentum \( p \) is fired into a door and gets embedded exactly at the center of the door. The door is \( 1.0 \, \text{m} \) wide and weighs \( 12 \, \text{kg} \). It is hinged at one end and rotates about a vertical axis practically without friction. The angular speed of the door just after the bullet embeds into it is:

• A. \( \frac{p}{4} \)
• B. \( \frac{3p}{5} \)
• C. \( \frac{p}{8} \)
• D. \( \frac{3}{8p} \)

Hint:

In collision problems involving rotation about a hinge, conserve angular momentum about the hinge because impulsive hinge force has zero torque about the hinge.

Answer 1

The Correct Option is C

Step 1: Use conservation of angular momentum.
Since the hinge provides no torque about the hinge axis, angular momentum about the hinge is conserved.

Step 2: Find initial angular momentum of bullet.
The bullet hits the center of the door. Since the width of the door is \( 1.0 \, \text{m} \), distance of center from hinge is:
\[ r = \frac{1}{2} = 0.5 \, \text{m} \]

Step 3: Write initial angular momentum.
Momentum of bullet is \( p \), so angular momentum about hinge is:
\[ L_i = pr \]
\[ L_i = p \times 0.5 = \frac{p}{2} \]

Step 4: Moment of inertia of the door about hinge.
For a rectangular door rotating about one vertical edge:
\[ I = \frac{ML^2}{3} \]

Step 5: Substitute values of mass and width.
\[ M = 12 \, \text{kg}, \quad L = 1.0 \, \text{m} \]
\[ I = \frac{12 \times (1)^2}{3} \]
\[ I = 4 \, \text{kg m}^2 \]

Step 6: Apply conservation equation.
\[ L_i = I\omega \]
\[ \frac{p}{2} = 4\omega \]

Step 7: Calculate angular speed.
\[ \omega = \frac{p}{8} \]
Therefore, the angular speed is:
\[ \boxed{\frac{p}{8}} \]