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  • 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 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: 

Option: 1

6.25 rad/sec


Option: 2

0.625 rad/sec


Option: 3

3.35 rad/sec


Option: 4

0.335 rad /sec


Answers (1)

best_answer

Angular \ momentum imparted by the bullet L=m v \times r=\left(10 \times 10^{-3}\right) \times 500 \times \frac{1}{2}=2.5\\ Also, I=\frac{M L^{2}}{3}=\frac{12 \times 1.0^{2}}{3}=4 kgm ^{2}$ \\ As $L=I \omega \therefore \Omega=\frac{L}{I}=\frac{2.5}{4}=0.625 rad / sec$

Posted by

Suraj Bhandari

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