If the momentum increases by 0.01%,what is th | vedclass

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(B) The kinetic energy $K$ is related to momentum $p$ by the formula $K = \frac{p^2}{2m}$.Since $m$ is constant,$K \propto p^2$.Taking the logarithmic

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$0.02$

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(B) The kinetic energy $K$ is related to momentum $p$ by the formula $K = \frac{p^2}{2m}$.Since $m$ is constant,$K \propto p^2$.Taking the logarithmic derivative,we get $\frac{\Delta K}{K} = 2 \frac{\Delta p}{p}$.Given the percentage change in momentum $\frac{\Delta p}{p} 	imes 100 = 0.01\%$.Therefore,the percentage change in kinetic energy is $\frac{\Delta K}{K} 	imes 100 = 2 	imes (0.01\%) = 0.02\%$.

If the momentum increases by $0.01\%$,what is the percentage increase in kinetic energy (in $\%$)?

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