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(D) Let $v_1$ and $v_2$ be the velocities of the bodies of mass $m$ and $2m$ respectively after the collision.By the law of conservation of linear mom

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$\frac{1-2e}{1+e}$

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(D) Let $v_1$ and $v_2$ be the velocities of the bodies of mass $m$ and $2m$ respectively after the collision.By the law of conservation of linear momentum: $mv + (2m)(0) = mv_1 + 2mv_2$,which simplifies to $v = v_1 + 2v_2$ (Equation $1$).The coefficient of restitution $e$ is defined as $e = \frac{v_2 - v_1}{u_1 - u_2}$.Given $u_1 = v$ and $u_2 = 0$,we have $e = \frac{v_2 - v_1}{v}$,which implies $v_1 = v_2 - ev$ (Equation $2$).Substitute Equation $2$ into Equation $1$: $v = (v_2 - ev) + 2v_2$.$v(1+e) = 3v_2$,so $v_2 = \frac{v(1+e)}{3}$.Now,substitute $v_2$ back into Equation $2$: $v_1 = \frac{v(1+e)}{3} - ev = \frac{v + ev - 3ev}{3} = \frac{v(1-2e)}{3}$.The ratio of the velocities $v_1/v_2$ is $\frac{v(1-2e)/3}{v(1+e)/3} = \frac{1-2e}{1+e}$.

$A$ body of mass $m$ moving with a velocity $v$ collides head-on with another body of mass $2m$ at rest. If the coefficient of restitution between the two bodies is $e$,then the ratio of the velocities of the two bodies after collision is:

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