Q:

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? 

A:

Notice. in perfect elastic collision the coefficient of restitution is taken $e=1$ 
Now, we have 
1.) Collision of A & B: let $V_A$ & $V_B$ be the velocities of A & B just after collision then using law of conservation of linear momentum, we get
$$mu+0=mV_A+2mV_B\tag 1$$
Now, using newton's equation of collision we get $$\frac{V_B-V_A}{u-0}=e=1\tag 2$$  On solving (1) & (2), we get $$V_A=\frac{-u}{3}, \ V_B=\frac{2u}{3}$$
2.) Collision of B & C: let $V_B'$ & $V_C$ be the velocities of B & C just after collision then using law of conservation of linear momentum, we get 
$$(2m)\frac{2u}{3}+0=2mV_B'+3mV_C\tag 3$$
Now, using newton's equation of collision we get $$\frac{V_C-V_B'}{\frac{2u}{3}-0}=e=1\tag 4$$  On solving (3) & (4), we get $$V_B'=\frac{-2u}{15}, \ V_C=\frac{8u}{15}$$
Hence, we get $$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{Velocity of C after second collision:}\ V_C=\frac{8u}{15}\ ms^{-1}}}$$

