Q131 / 180 · COMEDK 2026 (9 May Shift 2) · Physics
A bullet, fired into a door gets embedded exactly at it's centre, causing the door to rotate about it's vertical axis, practically without friction, with an angular velocity of $0.625\ rads^{-1}$. The door is 1.0 m wide and weighs 12 kg. If the mass of the bullet is 10g, find the speed with which it was fired. (Hint: The moment of inertia of the door about the vertical axis at one end is $\dfrac{ML^2}{3}$).
- A$342\ ms^{-1}$
- B$645\ ms^{-1}$
- Correct answerC$500\ ms^{-1}$
- D$124\ ms^{-1}$
Solution
By conservation of angular momentum about the vertical axis of the door:
$L_i = L_f$
$m v r = (I_{door} + I_{bullet}) \omega$
$m v \left(\dfrac{L}{2}\right) = \left(\dfrac{ML^2}{3} + m\left(\dfrac{L}{2}\right)^2\right) \omega$
Substituting the given values $M = 12\ kg$, $L = 1.0\ m$, $m = 0.01\ kg$, and $\omega = 0.625\ rads^{-1}$:
$0.01 \times v \times 0.5 = \left(\dfrac{12 \times 1^2}{3} + 0.01 \times 0.5^2\right) \times 0.625$
$0.005 v = (4 + 0.0025) \times 0.625$
Neglecting the small term $0.0025$ compared to $4$:
$0.005 v = 4 \times 0.625$
$0.005 v = 2.5$
$v = \dfrac{2.5}{0.005} = 500\ ms^{-1}$
$L_i = L_f$
$m v r = (I_{door} + I_{bullet}) \omega$
$m v \left(\dfrac{L}{2}\right) = \left(\dfrac{ML^2}{3} + m\left(\dfrac{L}{2}\right)^2\right) \omega$
Substituting the given values $M = 12\ kg$, $L = 1.0\ m$, $m = 0.01\ kg$, and $\omega = 0.625\ rads^{-1}$:
$0.01 \times v \times 0.5 = \left(\dfrac{12 \times 1^2}{3} + 0.01 \times 0.5^2\right) \times 0.625$
$0.005 v = (4 + 0.0025) \times 0.625$
Neglecting the small term $0.0025$ compared to $4$:
$0.005 v = 4 \times 0.625$
$0.005 v = 2.5$
$v = \dfrac{2.5}{0.005} = 500\ ms^{-1}$
Hence, the correct answer is option: 'C'.