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, \[{{K}_{F}}=\frac{3}{4}{{K}_{1}}\] \[\Rightarrow \] \[\frac{1}{2}mv_{1}^{'2}=\frac{1}{2}mv_{2}^{'2}=\frac{3}{4}\left( \frac{1}{2}m{{v}^{2}} \right)\] 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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