A) \[\frac{1-e}{1+e}\]
B) \[\frac{e-1}{e+1}\]
C) \[\frac{1+e}{1-e}\]
D) \[\frac{2+e}{e-1}\]
Correct Answer: A
Solution :
[a] As \[{{\operatorname{u}}_{2}} =0\,\,and\,{{m}_{1}}={{\operatorname{m}}_{2}}\]therefore from \[{{\operatorname{m}}_{1}}{{u}_{1}} +{{m}_{2}}{{u}_{2}} ={{m}_{1}}{{v}_{1}} + {{m}_{2}}{{v}_{2}}\] We get\[{{\operatorname{u}}_{1}}={{v}_{1}}+{{v}_{2}}\] Also, \[\operatorname{e}=\frac{{{v}_{2}}-{{v}_{1}}}{{{u}_{1}}}=\frac{{{v}_{2}}-{{v}_{1}}}{{{v}_{2}}+{{v}_{1}}}=\frac{1-{{v}_{1}}/{{v}_{2}}}{1+{{v}_{1}}/{{v}_{2}}}\] which gives\[\frac{{{v}_{1}}}{{{v}_{2}}}=\frac{1-\operatorname{e}}{1+e}\]
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