When kinetic energy of a body becomes 36 time | vedclass

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(A) The relationship between kinetic energy $K$ and momentum $P$ is given by $K = \frac{P^2}{2m}$,where $m$ is the mass of the body.From this,we can w

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$500$

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(A) The relationship between kinetic energy $K$ and momentum $P$ is given by $K = \frac{P^2}{2m}$,where $m$ is the mass of the body.From this,we can write $P = \sqrt{2mK}$.Let the initial kinetic energy be $K_i$ and the final kinetic energy be $K_f = 36 K_i$.The initial momentum is $P_i = \sqrt{2mK_i}$ and the final momentum is $P_f = \sqrt{2mK_f} = \sqrt{2m(36K_i)} = 6\sqrt{2mK_i} = 6P_i$.The percentage increase in momentum is given by $\frac{P_f - P_i}{P_i} 	imes 100 \%$.Substituting the values,we get $\frac{6P_i - P_i}{P_i} 	imes 100 \% = \frac{5P_i}{P_i} 	imes 100 \% = 500 \%$.

When kinetic energy of a body becomes $36$ times of its original value,the percentage increase in the momentum of the body will be : (in $\%$)

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