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(C) Let the initial kinetic energy be $K_1$ and the final kinetic energy be $K_2$.Given that the kinetic energy decreases by $19\%$,so $K_2 = K_1 - 0.

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(C) Let the initial kinetic energy be $K_1$ and the final kinetic energy be $K_2$.Given that the kinetic energy decreases by $19\%$,so $K_2 = K_1 - 0.19 K_1 = 0.81 K_1$.We know the relation between kinetic energy $(K)$ and momentum $(p)$ is $K = \frac{p^2}{2m}$.Therefore,$\frac{K_2}{K_1} = \left( \frac{p_2}{p_1} ight)^2$.Substituting the values: $0.81 = \left( \frac{p_2}{p_1} ight)^2$.Taking the square root on both sides: $\frac{p_2}{p_1} = \sqrt{0.81} = 0.9$.This means $p_2 = 0.9 p_1$.The percentage decrease in momentum is given by $\frac{p_1 - p_2}{p_1} 	imes 100\%$.$= \frac{p_1 - 0.9 p_1}{p_1} 	imes 100\% = 0.1 	imes 100\% = 10\%$.

The kinetic energy of an object decreases by $19\%$. What is the percentage decrease in its momentum?

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