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(A) The relationship between momentum $P$ and kinetic energy $E$ is given by $P = \sqrt{2mE}$.Since $m$ is constant,$P \propto \sqrt{E}$.Taking the de

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$0.05$

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(A) The relationship between momentum $P$ and kinetic energy $E$ is given by $P = \sqrt{2mE}$.Since $m$ is constant,$P \propto \sqrt{E}$.Taking the derivative or using the approximation for small changes: $\frac{\Delta P}{P} \approx \frac{1}{2} \frac{\Delta E}{E}$.Given the percentage increase in kinetic energy is $\frac{\Delta E}{E} 	imes 100 = 0.1\%$.The percentage increase in momentum is $\frac{\Delta P}{P} 	imes 100 = \frac{1}{2} 	imes (0.1\%) = 0.05\%$.

If the kinetic energy of a body increases by $0.1\%$,the percent increase of its momentum will be .............. $\%$

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