If the kinetic energy of a body changes by 20 | vedclass

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(D) The relationship between kinetic energy $K$ and momentum $p$ is given by $K = \frac{p^2}{2m}$.Let the initial kinetic energy be $K_i$ and the fina

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None of the above

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(D) The relationship between kinetic energy $K$ and momentum $p$ is given by $K = \frac{p^2}{2m}$.Let the initial kinetic energy be $K_i$ and the final kinetic energy be $K_f$. Given that the kinetic energy changes by $20 \%$,we have $K_f = 1.20 K_i$.Since $K \propto p^2$,we can write $\frac{K_f}{K_i} = \left( \frac{p_f}{p_i} ight)^2$.Substituting the values,$1.20 = \left( \frac{p_f}{p_i} ight)^2$.Taking the square root on both sides,$\frac{p_f}{p_i} = \sqrt{1.20} \approx 1.0954$.The percentage change in momentum is given by $\frac{p_f - p_i}{p_i} 	imes 100 = (1.0954 - 1) 	imes 100 = 9.54 \%$.Since $9.54 \%$ is not among the given options,the correct answer is $D$.

If the kinetic energy of a body changes by $20 \%$,then its momentum would change by:

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