These two related questions (this one and this one) has me thinking again about variable mass systems. Here's a way of looking at it that seems (at least, it seems to me) to be fairly clear, but complicated. I apologize to the original poster. This is not a simple explanation, but the comments have gotten complicated. Still, I do believe that your original question is too fuzzy. The system and the situation not clearly set out. That has caused people to guess what you mean, and the discussion went off the rails.
I might try to apply the result at the bottom of this answer to your question, but I can't right now. If I have enough time and energy later I might edit this post. Or someone might add comments that make a clear connection to your question and my answer.
Imagine an object of mass $m$. We are going to study it in the frame of reference where it is originally at rest. We are going to allow it to lose mass, so its mass will change with time: $m(t)$. What happens?
Let's break off a very small piece of the object. The piece that breaks off has a mass $\Delta m$ where $\Delta m \ll m$. If the piece simply breaks off, but doesn't move, nothing has really changed. Imagine a microscopic crack between the piece and the object: the system is exactly the same. In order to lose mass, the piece has to move away. We'll say that it moves away with velocity $\vec{u}$.
The piece carries away momentum $\vec{u}\Delta m$, and as a consequence the object's momentum changes by $-\vec{u}\Delta m$. It's velocity changes. It accelerates. It has experienced a force we'll call $\vec{F}$. By Newton's third law, the piece experiences a force $-\vec{F}$. As with all interactions, the process of breaking off does not happen instantaneously. (In fact, in nature nothing happens instantaneously. Please don't ask if that's true of wavefunction collapse: I don't know. :-). ) At any rate, the process of losing the mass takes some time $\Delta t$, so the object experience an impulse of $\vec{F}\Delta t$.
What is the change of velocity of the object after the piece breaks off? The velocity can be found by considering the impulse on the object. The impulse is equal to the change in momentum $$\vec{F}\Delta t = m(t) \Delta \vec{v}$$
Did I write the mass correctly? Should I use the mass before the piece fell off or after? It doesn't matter because we imagine that $\Delta m$ is so small, the mass of the remaining part hasn't changed significantly. Perhaps it's more satisfying if we note that the change to the mass of the object is a second order effect, and we are going to ignore such things in a few moments.
The total momentum of the system remains zero $$ p_\mathrm{system} = 0 = \vec{u}\Delta m + \Delta\vec{v}m(t)$$
There are a few ways we could proceed from here. I choose to allow a complication: an external net force $\vec{F}_\mathrm{ext}$. This force imparts an impulse to the entire system during the same time interval in which the piece was breaking off and causes the final momentum of the system to be non-zero $$I_\mathrm{system} = \vec{F}_\mathrm{ext}\Delta t =\Delta \vec{p}_\mathrm{system}$$ Including this result in what we had earlier
$$ \vec{F}_\mathrm{ext}\Delta t = \Delta \vec{p}_\mathrm{system} = \vec{u}\Delta m + \Delta\vec{v}m(t)$$
Divide by $\Delta t$ and let $\Delta t \rightarrow 0$ so that all small quantities become infinitesimally small (here's where the second order effects are dropped.)
$$\vec{F}_\mathrm{ext} = \frac{\mathrm{d}\vec{p}_\mathrm{system}}{\mathrm{d}t}= \vec{u}\frac{\mathrm{d}m}{\mathrm{d}t} + \frac{\mathrm{d}\vec{v}}{\mathrm{d}t}m(t)$$
This is not what you get when you apply the chain rule to Newton's second law. The chain rule gives
$$\vec{F}_\mathrm{ext} = \frac{\mathrm{d}\vec{p}_\mathrm{system}}{\mathrm{d}t}= \vec{v}\frac{\mathrm{d}m}{\mathrm{d}t} + \frac{\mathrm{d}\vec{v}}{\mathrm{d}t}m(t)$$
The math of the chain rule is correct, but the physical assumptions that might allow it to be used are not correct. Using the chain rule here is not valid.
