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 ..... .

Pick an option to check your answer.

Verified Solution

Answer & Solution

Correct Answer

(B) \(4\)

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 \(e=\frac{\text { R.V.S }}{\text { R.V.A }}\)

\(1=\frac{V+\frac{\omega L}{2}}{{u}}\)

\({v}+\frac{\omega {L}}{2}={u}\)

\({v}+\frac{3 {mu}}{{M}}={u}\)

\(\frac{{mu}}{{M}}+\frac{3 {mu}}{{M}}={u}\)

\(\frac{4 {mu}}{{M}}={u}\)

\(\frac{{m}}{{M}}=\frac{1}{4}\)

\({X}=4\)

Same subject

From JEE Mains