JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A rod of mass \(M\) and length \(L\) is lying on a horizontal frictionless surface. A particle of mass ' \(m\) ' travelling along the surface hits at one end of the rod with velocity \('u'\) in a direction perpendicular to the rod. The collision is completely elastic. After collision, particle comes to rest. The ratio of masses \(\left(\frac{m}{M}\right)\) is \(\frac{1}{x}\). The value of ' \(x\) ' will be ..... .
- A \(5\)
- B \(4\)
- C \(14\)
- D \(23\)
Answer & Solution
Correct Answer
(B) \(4\)
Step-by-step Solution
Detailed explanation
Just after collision From momentum conservation, \({P}_{{i}}^{\circ}={p}_{{F}}\) \(m u=M v \ldots \ldots(i)\) From angular momentum conservation about \(O,\) \(mu. \,l\frac{L}{2}=\frac{M L^{2}}{12} \omega\) \(\Rightarrow \omega=\frac{6 {mu}}{{ML}} \quad \cdots . . (ii)\) From…
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