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

By law of conservation of momentum P_i = P_f

$\Rightarrow {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}mV\hspace{0.17em}+\hspace{0.17em}O\hspace{0.17em}=\hspace{0.17em}mV}}_{\text{1}}\text{\hspace{0.17em}+\hspace{0.17em}}\frac{\text{m}}{\text{n}}{\text{V}}_{\text{2}}$

$\Rightarrow {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}V\hspace{0.17em}=\hspace{0.17em}V}}_{\text{1}}\text{\hspace{0.17em}+\hspace{0.17em}}\frac{{\text{V}}_{\text{2}}}{\text{n}}$

As nothing is cleared about type of collision, consider collision as elastic.

So, $\text{e\hspace{0.17em}=\hspace{0.17em}1\hspace{0.17em}\hspace{0.17em}}\Rightarrow \frac{{\text{V}}_{\text{2}}{\text{-V}}_{\text{1}}}{\text{V}}\text{=1\hspace{0.17em}}\Rightarrow {\text{\hspace{0.17em}V}}_{\text{2}}{\text{-V}}_{\text{1}}\text{\hspace{0.17em}=\hspace{0.17em}V}$

So, ${\text{V}}_{\text{2}}\text{-}\left(\text{V-}\frac{{\text{V}}_{\text{2}}}{\text{n}}\right)\text{\hspace{0.17em}=V}$

${\text{V}}_{\text{2}}\left(\text{1+}\frac{\text{1}}{\text{n}}\right)\text{\hspace{0.17em}=\hspace{0.17em}2V}$

${\text{V}}_{\text{2}}\text{\hspace{0.17em}=\hspace{0.17em}}\frac{\text{2Vn}}{\text{n+1}}$

So, $\text{ΔK\hspace{0.17em}=\hspace{0.17em}}\frac{\text{1}}{\text{2}}\frac{\text{m}}{\text{n}}\text{×\hspace{0.17em}}\frac{{\text{4V}}^{\text{2}}{\text{n}}^{\text{2}}}{{\left(\text{n+1}\right)}^{\text{2}}}$

So, $\frac{\text{ΔK}}{{\text{K}}_{\text{f}}}\text{\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}}\frac{\frac{\text{1}}{\text{2}}\frac{\text{m}}{\text{n}}\left(\frac{{\text{4V}}^{\text{2}}{\text{n}}^{\text{2}}}{{\left(\text{n+1}\right)}^{\text{2}}}\right)}{\frac{\text{1}}{\text{2}}{\text{mV}}^{\text{2}}}\text{=}\frac{\text{4n}}{{\left(\text{1+n}\right)}^{\text{2}}}$