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JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2013
- Medium
- MCQ
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
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(0.625\, rad/sec\)
Step-by-step Solution
Detailed explanation
Angular momentum imparted by the bullet
\(L = mv \times r \)\( = \left( {10 \times {{10}^{ - 3}}} \right) \times 500 \times \frac{1}{2} = 2.5\)
\(I = \frac{{M{L^2}}}{3} = \frac{{12 \times {{1.0}^2}}}{3} = 4kg{m^2}\)
\(L = I\omega \)
\(\therefore \omega = \frac{L}{I} = \frac{{2.5}}{4} = 0.625\,rad/\sec \)
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