A ball of mass 'm' moving with a speed 'u' undergoes a head-on elastic collision with a ball of mass 'nm' initially at rest. The fraction of initial energy transferred to the heavier ball is

${\mathrm{K}}_{\mathrm{i}}=\frac{1}{2}{\mathrm{mu}}^2$

$$\begin{gathered} \mathrm{By} \mathrm{conservation} \mathrm{of} \mathrm{momentum}, \mathrm{mu}={\mathrm{mv}}_1+\mathrm{nm} {\mathrm{v}}_2 \\ \mathrm{u}={\mathrm{v}}_1+{\mathrm{nv}}_2 ...\left(\mathrm{i}\right) \\ \mathrm{Elastic} \mathrm{Collision}, \mathrm{e}=1 ; \mathrm{So} , {\mathrm{v}}_2-{\mathrm{v}}_1={\mathrm{u}}_1-{\mathrm{u}}_2 \\ \therefore {\mathrm{v}}_2-{\mathrm{v}}_1=\mathrm{u} ...\left(\mathrm{ii}\right) \\ \left(\mathrm{i}\right)+\left(\mathrm{ii}\right)\Rightarrow 2\mathrm{u}=\left(\mathrm{n}+1\right){\mathrm{v}}_2 \\ \Rightarrow {\mathrm{v}}_2=\frac{2\mathrm{u}}{\mathrm{n}+1} \\ \mathrm{Kinetic} \mathrm{energy} \mathrm{transferred} {\mathrm{K}}_{\mathrm{t}}=\frac{1}{2}\left(\mathrm{nm}\right){{\mathrm{v}}^2}_2, \\ {\mathrm{K}}_{\mathrm{t}}=\frac{2{\mathrm{nmu}}^2}{{\left(\mathrm{n}+1\right)}^2}, \frac{{\mathrm{K}}_{\mathrm{t}}}{{\mathrm{K}}_{\mathrm{i}}}=\frac{4\mathrm{n}}{{\left(\mathrm{n}+1\right)}^2} \end{gathered}$$