A bullet of mass 10 g and speed 500 m/s is fired into a door and gets… | JEE Main 2013 (9 Apr, Shift 1)

Q80JEE Main 2013 · 9 APR · SHIFT 1Medium

Angular Momentum & its Conservation

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. The angular speed of the door just after the bullet embeds into it will be:

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Answer: (b)

Angular momentum about the hinge is conserved: $mv\frac{d}{2} = I\omega$ .
$mv\frac{d}{2} = 0.01(500)(0.5) = 2.5\ \text{kg m}^2\text{/s}$
$I = \frac{1}{3}Md^2 + m\left(\frac{d}{2}\right)^2 = \frac{12}{3} + 0.0025 \approx 4\ \text{kg m}^2$
$\omega = \frac{2.5}{4} = 0.625\ \text{rad/s}$
So the answer is (b) 0.625 rad/sec.

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