Answer:
We know, \[K.E.=\frac{1}{2}m{{v}^{2}}\] \[\therefore E=\frac{1}{2}m{{x}^{2}}\] ????.(1) \[2E=\frac{1}{2}m{{(x+2)}^{2}}\] ????.(2) Dividing (1) and (2), we get \[\frac{2E}{E}=\frac{\frac{1}{2}m{{(x+2)}^{2}}}{\frac{1}{2}m{{x}^{2}}}\] \[\Rightarrow 2=\frac{{{(x+2)}^{2}}}{{{x}^{2}}}\Rightarrow 2{{x}^{2}}={{x}^{2}}+4x+4\] \[\Rightarrow {{x}^{2}}-4x-4=0\Rightarrow {{(x-2)}^{2}}-8=0\] \[\Rightarrow {{(x-2)}^{2}}=8\Rightarrow (x-2)=\sqrt{8}\] \[\Rightarrow x=\sqrt{8}+2=2.828+2=4.828\,m/s\] Case-1 (before) Case-2 (after) \[{{v}_{1}}=x\,\,m/s\] \[{{v}_{2}}=(x+2)m/s\] \[K.E{{.}_{1}}=E\] \[K.E{{.}_{2}}=2E\]
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