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  • 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 a velocity 'u' in a direction perpendicular to the rod. The collision is comple

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 a 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 _______.
 

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best_answer

By linear momentum conservation ,

\begin{aligned} m u+0 &=m(0)+M\left(v_{2}\right) \\ v_{2} &=\frac{m u}{M} \rightarrow (1) \end{aligned}

By angular momentum conservation, about the centre of the rod

\begin{aligned} m u\left(\frac{L}{2}\right)+0 &=0+\left(\frac{M L}{12}\right) \omega \\ \omega &=\frac{6 m u}{M L} \rightarrow(2) \end{aligned}

For a perfectly elastic collision

K E_{i}=K E_{f}

\begin{aligned} &\frac{1}{2} m u^{2}+0=0+\frac{1}{2} I \omega^{2}+\frac{1}{2} M V_{2} \\ &\frac{1}{2} m u^{2}=\frac{1}{2} \times \frac{M L^{2}}{12} \times \frac{36 m^{2} u^{2}}{M^{2} L^{2}}+\frac{1}{2} M V_{2}^{2} \\ &\frac{1}{2} m u^{2}=\frac{3}{2} \frac{m^{2}}{M} u^{2}+\frac{1}{2} M V_{2}^{2} \end{aligned}

\begin{gathered} \frac{1}{2} m u^{2}\left[1-\frac{3 m}{M}\right]=\frac{1}{2} M v_{2}^{2} \\ \frac{1}{2} m u^{2} \frac{(M-3 m)}{M}=\frac{1}{2} M v_{2}^{2} \\ v_{2}^{2}=\left(\frac{M-3 m}{M^{2}}\right)mu^{2} \\ v_{2}=\sqrt{\frac{(M-3 m) m}{M^{2}} }u\rightarrow (3) \end{gathered}

From equations (1) and (3)

\frac{m}{M}=\frac{\sqrt{(M-3 m) m}}{M}

\begin{array}{r} m^{2}=M m-3 m^{2} \\ 4 m^{2}=M m \\ 4 m=M \\ \frac{m}{M}=\frac{1}{4} \end{array}

x= 4

Posted by

vishal kumar

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