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(B) Let the mass of both spheres be $m$. The initial velocity of the first sphere is $u_1 = 3u$ and the second sphere is $u_2 = 0$.Let $v_1$ and $v_2$

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

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(B) Let the mass of both spheres be $m$. The initial velocity of the first sphere is $u_1 = 3u$ and the second sphere is $u_2 = 0$.Let $v_1$ and $v_2$ be the velocities of the first and second spheres after the collision,respectively.By the law of conservation of linear momentum: $m(3u) + m(0) = mv_1 + mv_2$,which simplifies to $v_1 + v_2 = 3u$ (Equation $1$).By the definition of the coefficient of restitution $e$: $e = \frac{v_2 - v_1}{u_1 - u_2} = \frac{v_2 - v_1}{3u - 0}$,which gives $v_2 - v_1 = 3ue$ (Equation $2$).Adding Equation $1$ and Equation $2$: $2v_2 = 3u(1 + e) \implies v_2 = \frac{3u(1 + e)}{2}$.Subtracting Equation $2$ from Equation $1$: $2v_1 = 3u(1 - e) \implies v_1 = \frac{3u(1 - e)}{2}$.The ratio of the velocity of the second sphere to the first sphere is $\frac{v_2}{v_1} = \frac{3u(1 + e) / 2}{3u(1 - e) / 2} = \frac{1 + e}{1 - e}$.

$A$ sphere of mass $m$,moving with velocity $3u$,collides head-on with another identical sphere at rest. If $e$ is the coefficient of restitution,what will be the ratio of the velocity of the second sphere to that of the first sphere after the collision?

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