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(B) Let the mass of the moving particle be $m$ and its initial velocity be $u$. The mass of the stationary particle is $m' = \frac{m}{n}$.Assuming a p

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$\frac{4 n}{(1+n)^2}$

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(B) Let the mass of the moving particle be $m$ and its initial velocity be $u$. The mass of the stationary particle is $m' = \frac{m}{n}$.Assuming a perfectly elastic head-on collision,we apply the conservation of linear momentum:$mu = mv_1 + \frac{m}{n}v_2 \Rightarrow u = v_1 + \frac{v_2}{n}$ (Equation $1$)Using the coefficient of restitution $e=1$ for an elastic collision:$v_2 - v_1 = u \Rightarrow v_1 = v_2 - u$ (Equation $2$)Substituting Equation $2$ into Equation $1$:$u = (v_2 - u) + \frac{v_2}{n} \Rightarrow 2u = v_2(1 + \frac{1}{n}) = v_2(\frac{n+1}{n})$$v_2 = \frac{2nu}{n+1}$The kinetic energy transferred to the stationary particle is $K' = \frac{1}{2} m' v_2^2 = \frac{1}{2} (\frac{m}{n}) (\frac{2nu}{n+1})^2 = \frac{1}{2} \frac{m}{n} \frac{4n^2 u^2}{(n+1)^2} = \frac{2mnu^2}{(n+1)^2}$.The initial kinetic energy is $K = \frac{1}{2} mu^2$.The fraction of kinetic energy transferred is $\frac{K'}{K} = \frac{\frac{2mnu^2}{(n+1)^2}}{\frac{1}{2} mu^2} = \frac{4n}{(n+1)^2}$.

$A$ moving particle collides with a stationary particle of mass $\frac{1}{n}$ times the mass of the moving particle. The fraction of its kinetic energy transferred to the stationary particle is:

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