question_answer

A rigid massless rod of length L connects two particles each of mass m. The rod lies over a horizontal frictionless table. Another particle of mass m and velocity \[{{v}_{0}}\] struck one of two particles attached with rod.

Angular velocity of rod about its centre of mass just after collision will be (take, collision elastic)

A) \[\frac{\sqrt{2}{{v}_{0}}}{7L}\]

B) \[\frac{2\sqrt{2}{{v}_{0}}}{7L}\]

C) \[\frac{3\sqrt{2}{{v}_{0}}}{7L}\]

D) \[\frac{4\sqrt{2}{{v}_{0}}}{7L}\]

Correct Answer: D

Solution :

Linear momentum is conserved,
\[m{{v}_{0}}=2m{{v}_{1}}-m{{v}_{2}}\]
\[\Rightarrow \]   \[2{{v}_{1}}-{{v}_{2}}={{v}_{0}}\]
\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\upsilon }_{2}}-2{{\upsilon }_{1}}-{{\upsilon }_{0}}\]	..?(i)

Angular momentum is also conserved, so \[m{{v}_{0}}\left( \frac{L}{2}\sin 45{}^\circ  \right)=2m{{\left( \frac{L}{2} \right)}^{2}}\omega -m{{v}_{2}}\left( \frac{L}{2}\sin 45{}^\circ  \right)\]
\[\Rightarrow \]   \[{{v}_{0}}=\sqrt{2}L\omega -{{v}_{2}}\]	?(ii)
As, collision is elastic.
Relative velocity of separation = Relative velocity of approach
\[\Rightarrow \]\[{{v}_{2}}+{{v}_{1}}+\frac{L}{2}\omega \times \frac{1}{\sqrt{2}}={{v}_{0}}\]		?(iii)
Putting the value of \[{{v}_{2}}\] from Eq. (i) in Eq. (iii), we get \[2{{v}_{1}}-{{v}_{0}}+{{v}_{1}}+\frac{L\omega }{2\sqrt{2}}={{v}_{0}}\]
\[3{{v}_{1}}-2{{v}_{0}}+\frac{L\omega }{2\sqrt{2}}=0\]

Value of \[{{v}_{1}}\]from Eqs. (i) and (ii), we get \[\Rightarrow \]\[\frac{3\sqrt{2}}{2}L\omega -2{{v}_{0}}+\frac{L\omega }{2\sqrt{2}}=0\]

\[\Rightarrow \]\[L\omega \left( \frac{3\sqrt{2}}{2}+\frac{1}{2\sqrt{2}} \right)=2{{v}_{0}}\]\[\Rightarrow \]\[L\omega \left( \frac{6+1}{2\sqrt{2}} \right)=2{{v}_{0}}\]

\[\Rightarrow \]\[\frac{7L\omega }{2\sqrt{2}}=2{{v}_{0}}\]\[\Rightarrow \]\[\omega =\frac{4\sqrt{2}{{v}_{0}}}{7L}\]