Two rockets, A and B, are initially close together and on the same axis but facing in opposite directions Suppose we have two rockets, A and B, and they are initially close together and on the same axis but facing in opposite directions. Their masses are $m_A$ and $m_B$ respectively, excluding fuel. Rocket
A has fuel of total mass $M$ which is ejected from the rear at speed $u_0$. Rocket B has no
fuel. The exhaust from Rocket A is collected by Rocket B without loss. Both rockets
are initially at rest in space and are not affected significantly by gravity or friction. How would we derive an expression for the velocity of rocket B?
I'm trying to derive a "rocket equation" for rocket B, while in rocket A's rest frame by just considering the change in momentum, however this doesn't seem to lead to anywhere. Any thoughts?
 A: The law of conservation of momentum would apply here, as you would have guessed.
So the initial momentum of the eject will be equal to the final momentum of fuel and rocket combined. That mass will be $M+m_b$ and it will have a common velocity.
As Möbius points out, the mass of fuel depends on the time for which the rockets are kept in proximity. But we can find out the final velocity.
You can take the final velocity of the combined mass of fuel and the rocket as $v$.
So:
$$p_{fuel} = v(m_b + M)$$
$$Mu_0 = v(m_b + M)$$
$$\therefore v = \frac{Mu_0}{m_b+ M}$$
Let me know if anything is unclear or I have misunderstood your question.
A: Method pointed out by Sid:

By Law of Conservation of Energy:  The initial momentum of the eject
is equal to the final momentum of fuel and rocket combined. That mass
will be $M+m_b$ and it will have a common velocity.
You can take the final velocity of the combined mass of fuel and the
rocket as $v$. So: $$p_{fuel} = v(m_b + M)$$ $$Mu_0 = v(m_b + M)$$
$$\therefore v = \frac{Mu_0}{m_b+ M}$$

This method pointed out by Sid is a very useful method in cases when the whole mass is transferred to the body instantaneously.
However let us consider a tad bit more complicated case which is exactly why I am putting down this answer :
Suppose the ejected mass enters B with a rate of $\alpha $ $kgs^{-1}$
Now how do we find the velocity of the rocket at a time when some of the mass of fuel is transferred to the rocket while some is still left?
Firstly the velocity in this case will be variable and hence a function of time.Let us consider that we want to calculate the velocity at a time instant $'t'$.Let velocity at this instant be $v_b$.Now in this time total mass of fuel that has entered into B is $\alpha t$ .This mass initially had a velocity $u_0$ but after entering B it has common  velocity  $v_b$. .So final momentum of this mass of fuel and rocket  is : $(\alpha t + m_b)v_b$.
Final momentum of rocket + fuel = initial momentum of rocket + fuel(By Law of Conservation of Momentum) .... 1
So
Initial Momentum of rocket = 0 (since velocity os 0)
Initial Momentum of fuel = $\alpha t u_0$
Initial momentum of rocket + fuel system = 0 + $\alpha t u_0$ = $\alpha t u_0$
Final momentum of this mass of fuel and rocket  is : $(\alpha t + m_b)v_b$.
from 1 above :
$$ \alpha t u_0 = (\alpha t + m_b)v_b$$
$$ v_b = \frac{\alpha t u_0}{\alpha t + m_b}$$
Which is my answer
