question_answer

If the momentum of a body is increased by 50%, then what would be the percentage increase in its kinetic energy?

Answer:

\[{{P}_{2}}=1.5\,\,P\] \[K{{E}_{2}}=?\] \[KE=\frac{{{P}^{2}}}{2m}\Rightarrow KE\propto {{P}^{2}}\] \[K{{E}_{2}}=\frac{1.5P\times 1.5\,P}{P\times P}\times K{{E}_{1}}=2.25\,K{{E}_{1}}\] %Change in \[KE=\left( \frac{2.25\,K{{E}_{1}}}{K{{E}_{1}}}-1 \right)\times 100\]                 \[=1.25\times 100=125\]