A) \[\frac{1-e}{2}\]
B) \[\frac{1+e}{2}\]
C) \[\frac{1-e}{1+e}\]
D) \[\frac{1+e}{1-e}\]
Correct Answer: D
Solution :
Given, \[{{m}_{1}}={{m}_{2}}=m,{{u}_{1}}=u\] and \[{{u}_{2}}=0\] \[\therefore \] \[{{v}_{1}}={{u}_{1}}=\frac{({{m}_{1}}-e{{m}_{2}})}{({{m}_{1}}+{{m}_{2}})}+{{u}_{2}}\frac{(1+e){{m}_{2}}}{({{m}_{1}}+{{m}_{2}})}\] \[=\frac{u(1-e)}{2}\] \[\Rightarrow \] \[\frac{{{v}_{1}}}{u}=\left( \frac{1-e}{2} \right)\]You need to login to perform this action.
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