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(D) The kinetic energy $(KE)$ of a body is given by $KE = \frac{1}{2} mv^2 = \frac{p^2}{2m}$,where $p = mv$ is the momentum.Since both bodies have the

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$M_2 V_2 > M_1 V_1$

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(D) The kinetic energy $(KE)$ of a body is given by $KE = \frac{1}{2} mv^2 = \frac{p^2}{2m}$,where $p = mv$ is the momentum.Since both bodies have the same kinetic energy,$KE_1 = KE_2$.Therefore,$\frac{p_1^2}{2M_1} = \frac{p_2^2}{2M_2}$.This implies $\frac{p_1^2}{p_2^2} = \frac{M_1}{M_2}$,or $\frac{p_1}{p_2} = \sqrt{\frac{M_1}{M_2}}$.Since the body with mass $M_2$ is heavier,$M_2 > M_1$,which means $\frac{M_1}{M_2} Consequently,$\frac{p_1}{p_2} Since $p = mv$,we have $M_1 V_1  M_1 V_1$.

If a lighter body of mass $M_1$ and velocity $V_1$ and a heavy body of mass $M_2$ and velocity $V_2$ have the same kinetic energy,then:

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