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

If the velocity after the collision of the lighter particle is taken as $v_2$ and that of the heavier particle is taken as $v_1$

Applying conservation of momentum,

$$\begin{gathered} mu=m{v}_1+\frac{m}{n}{v}_2 \\ \Rightarrow u=v_1+\frac{v_2}{n} \end{gathered}$$

From coefficient of restitution,

$$\begin{gathered} v_2-v_1=u \\ \Rightarrow v_1=v_2-u \end{gathered}$$

Putting the value of $v_1$ in the above equation,

$v_2=\frac{2nu}{n+1}$

Therefore, the fraction of kinetic energy transferred to the lighter object will be,

$\frac{{\displaystyle \frac{1}{2}\frac{m}{n}{v_2}^2}}{{\displaystyle \frac{1}{2}m{u}^2}}=\frac{4n}{{\displaystyle {\left(1+n\right)}^2}}$