Answer:
Given, \[{{v}^{2}}=2{{v}_{1}}\] Percentage change in K.E. = ? \[K.E{{.}_{1}}=?K.E{{.}_{2}}=?\] Change in \[K.E.=?\] We know, \[K.E.=\frac{1}{2}m{{v}^{2}}\] \[K.E{{.}_{1}}=\frac{1}{2}mv_{1}^{2}K.E{{.}_{2}}=\frac{1}{2}mv_{2}^{2}=\frac{1}{2}m{{(2{{v}_{1}})}^{2}}\] \[=4\left( \frac{1}{2}mv_{1}^{2} \right)\] \[\therefore \]Change in \[K.E.=K.E{{.}_{2}}-K.E{{.}_{1}}=4\,\,K.E{{.}_{1}}-\] \[K.E{{.}_{1}}=3\,\,K.E{{.}_{1}}\] Percentage change in K.E. \[=\frac{change\,\,in\,\,K.E.}{initial\,\,K.E.}\times 100\] Therefore, percentage change in K.E is 300%.
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