COMEDK · Physics · 7. Center of Mass Momentum and Collision
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 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 is (Assume the mass of bullet is negligible compared to the mass of door)
- A \(\dfrac{3 p}{5}\)
- B \(\dfrac{p}{8}\)
- C \(\dfrac{3}{8 p}\)
- D \(\dfrac{p}{4}\)
Answer & Solution
Correct Answer
(B) \(\dfrac{p}{8}\)
Step-by-step Solution
Detailed explanation
Let \(M = 12\) kg be the mass of the door and \(L = 1.0\) m be its width. The moment of inertia of the door about the vertical axis passing through the hinge is \(I_{door} = \dfrac{1}{3} M L^2 = \dfrac{1}{3} \times 12 \times (1.0)^2 = 4\) kg m\(^2\). Let \(m\) be the mass of the…
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