A) \[{{v}_{1}}={{v}_{2}}\]
B) \[{{v}_{1}}={{m}_{2}}\sqrt{\frac{2G}{d\left( {{m}_{1}}+{{m}_{2}} \right)}}\] \[{{v}_{2}}={{m}_{1}}\sqrt{\frac{2G}{d\left( {{m}_{1}}+{{m}_{2}} \right)}}\]
C) \[{{v}_{1}}={{m}_{1}}\sqrt{\frac{2G}{d\left( {{m}_{1}}+{{m}_{2}} \right)}}\]\[{{v}_{2}}={{m}_{2}}\sqrt{\frac{2G}{d\left( {{m}_{1}}+{{m}_{2}} \right)}}\]
D) \[{{v}_{2}}={{m}_{2}}\sqrt{\frac{2G}{{{m}_{1}}}}\] \[{{v}_{2}}={{m}_{2}}\sqrt{\frac{2G}{{{m}_{2}}}}\]
Correct Answer: B
Solution :
We choose reference point, infinity, where total energy of the system is zero. So, initial energy of the system = 0 Final energy \[=\frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{2}^{2}-\frac{G{{m}_{1}}{{m}_{2}}}{d}\] From conservation of energy, Initial energy = Final energy \[\therefore \]\[0=\frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{2}^{2}-\frac{G{{m}_{1}}{{m}_{2}}}{d}\] or\[\frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{1}}v_{2}^{2}=\frac{G{{m}_{1}}{{m}_{2}}}{d}\] ?.(1) By conservation of linear momentum \[{{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}=0\]or\[\frac{{{v}_{1}}}{{{v}_{2}}}=-\frac{{{m}_{2}}}{{{m}_{1}}}\Rightarrow {{v}_{2}}=-\frac{{{m}_{1}}}{{{m}_{2}}}{{v}_{1}}\] Putting value of \[{{v}_{2}}\]in equation (1), we get \[{{m}_{1}}v_{1}^{2}+{{m}_{2}}{{\left( -\frac{{{m}_{1}}{{v}_{1}}}{{{m}_{2}}} \right)}^{2}}=\frac{2G{{m}_{1}}{{m}_{2}}}{d}\] \[\frac{{{m}_{1}}{{m}_{2}}v_{1}^{2}+m_{1}^{2}v_{1}^{2}}{{{m}_{2}}}=\frac{2G{{m}_{1}}{{m}_{2}}}{d}\] \[{{v}_{1}}=\sqrt{=\frac{2Gm_{2}^{2}}{d({{m}_{1}}+{{m}_{2}})}}={{m}_{2}}\sqrt{=\frac{2G}{d({{m}_{1}}+{{m}_{2}})}}\] Similarly\[={{v}_{2}}=-{{m}_{1}}\sqrt{\frac{2G}{d({{m}_{1}}+{{m}_{2}})}}\]You need to login to perform this action.
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