What percentage of kinetic energy of a moving particle is transferred to a stationary particle when it strikes the stationary particle of 5 times its mass? (Assume the collision to be head-on elastic collision)

Let the velocity of striking particle be ${\mathrm{u}}_0$

Then, ${\mathrm{mu}}_0={\mathrm{mv}}_1+5{\mathrm{mv}}_2{\mathrm{u}}_0={\mathrm{v}}_1+5{\mathrm{v}}_2 .....\left(\mathrm{i}\right)$

As, collision is elastic, So, $\mathrm{e}=1\Rightarrow \frac{{\mathrm{v}}_2-{\mathrm{v}}_1}{{\mathrm{u}}_0}=1$

$$\begin{gathered} \Rightarrow {\mathrm{v}}_2-{\mathrm{v}}_1={\mathrm{u}}_0 .....\left(\mathrm{ii}\right) \\ \mathrm{Adding} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right) , \mathrm{we} \mathrm{get} \\ 2{\mathrm{u}}_0=6{\mathrm{v}}_2\Rightarrow {\mathrm{v}}_2={\mathrm{u}}_0/3 \\ \mathrm{So}, \%∆\mathrm{K}.{\mathrm{E}}_2=\frac{{\displaystyle \frac{1}{2}}\left(5\mathrm{m}\right){\left({\displaystyle \frac{{\mathrm{u}}_0}{3}}\right)}^2-0}{{\displaystyle \frac{1}{2}}\mathrm{m} {\mathrm{u}}_0^2}\times 100=\frac{500}{9}=55.6\% \end{gathered}$$