A) \[1/\sqrt{3}\]
B) \[1/\sqrt{2}\]
C) \[\sqrt{2}\]
D) \[\sqrt{3}\]
Correct Answer: B
Solution :
Since, we know that in the case, when \[{{m}_{1}}={{m}_{2}}\] and \[{{v}_{1}}=0\] then, \[v_{1}^{}=\left( \frac{1+e}{2} \right){{v}_{2}}\] and \[v_{2}^{}=\left( \frac{1-e}{2} \right){{v}_{2}}\] Given that, \[{{K}_{F}}=\frac{3}{4}{{K}_{i}}\] \[\Rightarrow \] \[\frac{1}{2}mv_{1}^{2}=\frac{1}{2}mv_{2}^{2}=\frac{3}{4}\,\,\left( \frac{1}{2}m{{v}^{2}} \right)\] Substituting the values, we get \[{{\left( \frac{1+e}{2} \right)}^{2}}+{{\left( \frac{1-e}{2} \right)}^{2}}=\frac{3}{4}\] \[\Rightarrow \] \[{{(1+e)}^{2}}+{{(1-e)}^{2}}=3\] \[\Rightarrow \] \[2+2{{e}^{2}}=3\,\,\,\Rightarrow \,\,\,{{e}^{2}}=\frac{1}{2}\,\,\,\,\Rightarrow \,\,\,\,e=\frac{1}{\sqrt{2}}\]
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