Two bodies of equal masses moving with momentum of first double that of the second collide completely inelastically. If the total final kinetic energy is 9 joule, the initial kinetic energy of the body of greater momentum is $\_\_\_\_\_\_$ joule

Let m represent each body's mass. Let u be the second body's velocity and 2u be the first body's velocity. Using the principle of momentum conservation, we have

$\mathrm{m}\left(2\mathrm{u}\right)+\mathrm{mu}=2\mathrm{mV}$

where V is the combined mass's post-collision speed. Based on the foregoing, we can write

$\mathrm{V}=\frac{3}{2}\mathrm{u}$

$\mathrm{final} \mathrm{kinetic} \mathrm{energy} =\frac{1}{2}\times 2\mathrm{m}\times {\mathrm{V}}^2=9\mathrm{J}$

${\mathrm{mV}}^2=\mathrm{m}{\left[\frac{3}{2}\mathrm{u}\right]}^2=9\mathrm{J}$

$\left(\frac{9}{4}\right){\mathrm{mu}}^2=9\mathrm{J}\Rightarrow {\mathrm{mu}}^2=4\mathrm{J}$

The body with more initial kinetic energy is given by

$\frac{1}{2}\times \mathrm{m}{\left(2\mathrm{u}\right)}^2=2{\mathrm{mu}}^2=2\times 4=8\mathrm{J}$

Hence the correct answer is 8J.