If the momentum of a body increases by 0.01%, | vedclass

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(B) The relationship between kinetic energy $E$ and momentum $P$ is given by $E = \frac{P^2}{2m}$.Since the mass $m$ of the body is constant,we have $

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$0.02$

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(B) The relationship between kinetic energy $E$ and momentum $P$ is given by $E = \frac{P^2}{2m}$.Since the mass $m$ of the body is constant,we have $E \propto P^2$.Taking the logarithmic derivative,we get $\frac{dE}{E} = 2 \frac{dP}{P}$.For small percentage changes,the percentage increase in kinetic energy is approximately twice the percentage increase in momentum.Given that the percentage increase in momentum is $0.01\%$,the percentage increase in kinetic energy is $2 	imes 0.01\% = 0.02\%$.

If the momentum of a body increases by $0.01\%$,its kinetic energy will increase by ........... $\%$

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