A) \[{{m}_{2}}/({{m}_{1}}+{{m}_{2}})\]
B) \[{{m}_{1}}/({{m}_{1}}+{{m}_{2}})\]
C) \[{{m}_{1}}/({{m}_{1}}-{{m}_{2}})\]
D) \[{{m}_{2}}/({{m}_{1}}-{{m}_{2}})\]
Correct Answer: A
Solution :
For conservation of momentum, we have, \[{{m}_{1}}{{v}_{1}}=({{m}_{1}}+{{m}_{2}})v\]or\[v={{m}_{1}}({{m}_{1}}+{{m}_{2}}){{v}_{1}}\] Now the loss of energy\[=\frac{1}{2}{{m}_{1}}{{v}_{1}}^{2}-\frac{1}{2}({{m}_{1}}+{{m}_{2}}){{v}^{2}}\] \[\therefore \]Fraction of energy lost \[=\frac{\frac{1}{2}{{m}_{1}}v_{1}^{2}-\frac{1}{2}({{m}_{1}}+{{m}_{2}}){{v}^{2}}}{\frac{1}{2}{{m}_{1}}{{v}_{1}}}\] \[=1-[{{m}_{1}}+{{m}_{2}})/{{m}_{1}}]\times ({{v}^{2}}/{{v}_{2}}^{2})\] \[=1-[{{m}_{1}}+{{m}_{2}})/{{m}_{1}}]\times [({{m}_{1}}{{v}_{1}}^{2}/{{[{{m}_{1}}+{{m}_{2}})}^{2}}]\times (1/v_{1}^{2})\]\[=1-[{{m}_{1}}/[({{m}_{1}}+{{m}_{2}})]={{m}_{2}}/({{m}_{1}}+{{m}_{2}})\]
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