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Suppose I'm considering both the rocket, its unburnt fuel and the exhaust all of them combined as my system. Also suppose that the rocket was initially at rest. Then according to momentum conservation, shouldn't I get, $$(m_f)(v_f) - (m_e)(v_e)=0$$ Where, $m_f$=final mass of rocket+unburnt fuel and $m_e$=mass of exhaust=$(m_f - M)$ with M being the total mass.

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  • $\begingroup$ Why do you think you can't derive the rocket equation from that? Have you seen another derivation that you can follow? $\endgroup$ – jacob1729 Dec 31 '18 at 14:15
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I assume that $v_{\rm f}$ is the final velocity of the rocket relative to an observer (on the ground) and that the equation is set up at that final time.

The problem is your velocity $v_{\rm e}$ as you have not defined it.

Let the velocity of the rocket at a time $t$ be $v(t)$.

If $v_{\rm e}$ is the velocity of the exhaust gases relative to the rocket then the velocity of the exhaust gases relative to the observer $v(t) -v_{\rm e}$ vary with time.

If $v_{\rm e}$ is the final velocity of the exhaust gases relative to the observer the problem is that the exhaust gases that have been emitted during the time of acceleration of the rocket do not all travel at the same speed relative to the observer, their speed varies with time.

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