This question already has an answer here:
Let $m$ be the mass of a rocket in free space (including fuel) at the time $t$. Now suppose that the rocket ejects a mass $\Delta m$ during the interval $\Delta t$ with velocity $v_e$ relative to the rocket.
I was told that the correct ansatz is as follows:
$$ mv = (m - \Delta m) \cdot (v + \Delta v) + \Delta m \cdot (v + \Delta v - v_e). $$
If you work this out, it reduces to:
$$ m \Delta v - \Delta m v_e = 0. $$
However some books start with the equation
$$ mv = (m - \Delta m) \cdot (v + \Delta v) + \Delta m \cdot (v - v_e) $$
which reduces to
$$ m \Delta v - \Delta m v_e - \Delta m \Delta v = 0. $$
Then you need to "neglect" the higher order term $\Delta m \Delta v$ to get the correct result.
Is there any good conceptual reason why the first ansatz is the more correct one?
However if you use $\tilde m = m(t+ \Delta t)$ in the derivation, you get the correct result without the need of neglecting higher order terms in the line of the second derivation:
$$ (\tilde m + \Delta m) \cdot v = \tilde m (v + \Delta v) + \Delta m (v - v_e) $$
so why is the following equation the conceptually wrong ansatz:
$$ (\tilde m + \Delta m) \cdot v = \tilde m (v + \Delta v) + \Delta m (v + \Delta v - v_e)? $$