Q:

Newton's first law

I realised that Newton's first law has significance in that it defines the inertial reference frame which is the condition for Newton's second law to hold. This means, from second law we should be able to find an implicit reference to the first law (i.e. we should be able to reach the first law by manipulating expressions). For now, I am just going to state the first and second laws.

Newton's first law: $$\sum \vec{F} =0 \Leftrightarrow \frac{d \vec{v}}{dt}=0$$
Newton's second law: $$\vec{F}=\frac{d\vec{p}}{dt} $$

When the mass is constant (i.e. $dm/dt=0$), we often write $\vec{F}=m \vec{a}$. I understand this because if we plug in $\vec{F}=0$, then we can obtain $\vec{a} = d\vec{v}/dt=0$, which is what the first law states. (To emphasise again, I am aware that the first law is something that cannot be "derived" from the second law, but I am doing this to help you understand my question.)
Let's now say that the mass $m(t)$ is a function of time. Then,
$$\vec{F} = \frac{d(m(t)\vec{v}(t))}{dt} = \frac{dm(t)}{dt}\vec{v}(t) + m(t)\frac{d\vec{v}(t)}{dt} $$
This general case should not contradict with the first law. But, if I take $\vec{F}=0$, then
$$ m\vec{a} = -\frac{dm}{dt} \vec{v} $$
and therefore
$$ \vec{a} = -\frac{1}{m}\frac{dm}{dt} \vec{v} $$
But according to the first law the acceleration should be equal to 0, so it gives the condition (algebraically) either $dm/dt=0$ or $\vec{v}=0$. This means that either the mass should be constant or the velocity should be 0.

A:

Obviously the first law applies to an isolated system. For this the mass cannot change. Take the example of an evaporating mass, as in one of the answers. The vapour carries off mass but also momentum. So either you consider the total system to be the evaporating particle plus vapour, for which the sum of forces over all parts is zero, or you consider these to be interacting systems on which case there is a force F=ma according to the second law and for which the sum of forces is zero due to the third law. It amounts to the same. 

