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 \[=\frac{\frac{1}{2}{{m}_{1}}v_{1}^{2}-\frac{1}{2}({{m}_{1}}+{{m}_{2}}){{v}^{2}}}{\frac{1}{2}{{m}_{1}}{{v}_{1}}}\] \[=1-[({{m}_{1}}+{{m}_{2}})/{{m}_{1}}]\times ({{v}^{2}}/{{v}_{1}}^{2})\] \[=1-[({{m}_{1}}+{{m}_{2}})/{{m}_{1}}]\times {{m}_{1}}{{v}_{1}}^{2}/[{{({{m}_{1}}+{{m}_{2}})}^{2}}]\times (1/{{v}_{1}}^{2})\]\[=1-[{{m}_{1}}/[({{m}_{1}}+{{m}_{2}})]={{m}_{2}}/({{m}_{1}}+{{m}_{2}})\]
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