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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?

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    $\begingroup$ If rocket B collects all of the fuel from rocket A, then its change in momentum is equal to the momentum of that fuel. That should tell you everything you need to know about how rocket B moves. $\endgroup$
    – Eric Smith
    Apr 21, 2021 at 22:15
  • $\begingroup$ Yes, but do i consider a change in momentum $dm$$u_0$ or do I model it as one big chunk of fuel $M$ ? $\endgroup$
    – jambajuice
    Apr 21, 2021 at 22:16
  • $\begingroup$ That depends if you want to know just the final velocity of B, or if you want to know $v_B(t)$ $\endgroup$
    – JEB
    Apr 22, 2021 at 1:44

2 Answers 2

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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

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  • $\begingroup$ $\alpha t = M$ after the entire mass has been transferred, so it comes out again to $\frac{M u_0}{M + m_b}$. However you have correctly pointed out that my answer is an oversimplification, for this I thank you. $\endgroup$
    – Sid
    Apr 22, 2021 at 6:24
  • $\begingroup$ Yeah, I think this method is correct, and much more simple. I was thinking about how a small mass $dm$ would impart momentum onto the bigger rocket. So we would end up with an expression $dM_a u_0+ M_bV_b = (M_b + dM_a)(V_b +dV_b)$, and then rearrange to find the equation for the velocity, which I think still leads to me the answer - albeit a bit more complicated. $\endgroup$
    – jambajuice
    Apr 22, 2021 at 7:41
  • $\begingroup$ Glad to here it helped you $\endgroup$
    – Möbius
    Apr 22, 2021 at 9:41
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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.

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