A body of mass m moving with velocity v collides head on elastically with another body of mass 2m at rest. The ratio of kinetic energy of the colliding body before and after collision is

Kinetic energy of colliding body before the collision $=\frac{1}{2}{\mathrm{mv}}^2$

After the collision, its velocity changes.

$\mathrm{v}\text{'}=\frac{({\mathrm{m}}_1-{\mathrm{m}}_2)}{{\mathrm{m}}_1+{\mathrm{m}}_2}\mathrm{v}=\frac{\mathrm{m}}{3\mathrm{m}}\mathrm{v}=\frac{\mathrm{v}}{3}$

Kinetic energy after collision,

$\frac{1}{2}{\mathrm{mv}}^2=\frac{1}{2}\frac{{\mathrm{mv}}^2}{9}$

The ratio of KE = $\frac{{\displaystyle \frac{1}{2}{\mathrm{mv}}^2}}{{\displaystyle \frac{1}{2}\frac{{\mathrm{mv}}^2}{9}}}=9:1$

The correct answer is 9:1.