Fraction of original velocity in series of elastic collisions I need to find the answer to this question:

Three particles A, B, and C, with masses $m$, $2m$ and $3m \, \mathrm k \mathrm g$ respectively, lie at rest in that order in a straight line on a smooth horizontal table. The particle A is then projected directly towards B with speed $u \, \mathrm m \mathrm s^{-1}$. Assuming the collisions are perfectly elastic, what fraction of $u$ is the speed of the particle C immediately after the second impact?

I tried to solve it like this:
$$P = mv$$
Momentum before collision = momentum after collision, therefore
$$um = mV_a + 2mV_b$$ where $V_a$ and $V_b$ are the velocities of the particles after impact.
$$\implies u = V_a + 2V_b$$
I got to there and then tried rearranging to make $V_b$ the subject and using that in the next collision but I couldn't get anything out of it in the end. How can I solve this? 

The collisions are perfectly elastic. So you should write energy balance equation.
$$ K.E._{initial}= K.E._{final}  $$
So for the $1^{st}$ collision
$$\frac{1}{2}mu^2=\frac{1}{2}mV_a^2+ \frac{1}{2}(2m)V_b^2  $$
Now we get 2 unknowns $ V_a \hspace{0.2cm} and \hspace{0.2cm} V_b $ and 2 equations  namely momentum conservation and energy conservation .
Solve . Repeat for next collision.
