Problem in applying Newton's Second law I do have some problem in applying Newton's Second to systems with variable mass.
This is what I understand:

Mathematically, the law states that:
$$F=\frac{dp}{dt}=m\frac{dv}{dt} + v\frac{dm}{dt}$$
where $F$ is the net external force "on the system" and $\frac{dp}{dt}$ is the
rate of change of momentum "of the system".

If the mass of a system is changing, say is decreasing and the "removed" mass do possess a momentum. Then, in order to change the momentum of the "removed" mass, the system must have applied a force on the "removed" mass and in reaction the removed mass would also apply an equal and opposite force on the system,say $F$.
I am confused whether this force $F$, on the system, would be considered as an external force or an internal force (Since, as soon as the "removed" mass gets detached from system, it can apply an external force on the "main" system and is not the part of the "main" system).
If you say that even after being detached the "removed" mass remains a part of the system,then I would disagree because if this was the case then the mass of the system will not be considered to change at all.
If I am all correct then  see this post.
 A: The force $\mathbf F$ which you refer to in the equation
$$\mathbf F=m\frac{\mathrm d \mathbf v}{\mathrm dt}+\frac{\mathrm dm}{\mathrm dt}\mathbf v_{\rm relative}\tag{1}$$
is the net external force on the moving body (excluding the removed masses) and $\mathbf v_{\rm relative}$ is the relative velocity of the body with respect to the removed mass. In the case when the removed mass becomes stationary (in the ground frame) after being removed, the equation $(1)$ simplifies to
$$\mathbf F=m\frac{\mathrm d \mathbf v}{\mathrm dt}+\frac{\mathrm dm}{\mathrm dt}\mathbf v\tag{2}$$
where $\mathbf v$ is the velocity of the moving body. This net external force doesn't include the interaction force between the removed mass and the body. The interaction force between the removed mass and the body is taken care of by the term $\displaystyle \frac{\mathrm d m}{\mathrm dt}\mathbf v$.
This can be easily seen by the fundamental equation of Newton's second law which states that
$$\mathbf F_{\rm net, external}=\frac{\mathrm d\mathbf p}{\mathrm dt}\tag{3}$$
You might also want to look at this Wikipedia page on variable mass systems for more insight. Do note that the equation given in the Wikipedia article
$$\mathbf F_{\rm net,external}+\frac{\mathrm d m}{\mathrm dt}\mathbf v_{\rm relative}=m\frac{\mathrm d\mathbf v}{\mathrm d t}\tag{4}$$
is completely equivalent to the equation $(1)$ mentioned above in the answer (in equation $(4)$, $\mathbf v_{\rm relative}$ is the relative velocity of removed mass with respect to the moving body). Proving the equivalence is left as an exercise to the reader.
A: When we say "internal force", we mean forces that each part of our whole system exert on each other. If the ejected mass was not part of our system in the first place, then why would the mass of the system change at all? The mass that was initially part of the system was removed. So, it's obvious the total mass of the system changed. The equation $F=\,{d(mv)\over dt}$ is applied to our system whose mass is changing over time. I repeat, the jettisoned mass was part of the system and that's exactly why the mass of our system is changing. Now, when we consider the mutual internal force between our changing system and the removed mass, we freeze our clock to the time before the mass completely leaves the system. While the mass is pushed backwards, the mass simultaneously thrusts the system forward in the time it is been ejected until it is removed completely. Once completely ejected, it doesn't contribute to the thrust on the system anymore. External forces in general refer to forces other than this mutual interactive force between the individual masses of our system. It could be gravity, air drag or friction.
