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 \[\Rightarrow \]\[\frac{3\sqrt{2}}{2}L\omega -2{{v}_{0}}+\frac{L\omega }{2\sqrt{2}}=0\] |
\[\Rightarrow \]\[L\omega \left( \frac{3\sqrt{2}}{2}+\frac{1}{2\sqrt{2}} \right)=2{{v}_{0}}\]\[\Rightarrow \]\[L\omega \left( \frac{6+1}{2\sqrt{2}} \right)=2{{v}_{0}}\] |
\[\Rightarrow \]\[\frac{7L\omega }{2\sqrt{2}}=2{{v}_{0}}\]\[\Rightarrow \]\[\omega =\frac{4\sqrt{2}{{v}_{0}}}{7L}\] |
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