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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.

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    $\begingroup$ I agree with your reasoning, and I think that a similar conclusion can be reached non-mathematically by arguing that Newtons laws give conservation of momentum, so if an object's Mass were to magically change the object would have to be at zero velocity in order for momentum to be conserved. A body that can magically change its mass without releasing that mass into other bodies (water leaving a bucket) does not exist in real life, so this isn't an issue in practice. By the way your question lacks a "?" anywhere. What is the question? $\endgroup$
    – Dast
    Commented Dec 16, 2019 at 10:40
  • $\begingroup$ You have applied Newton's second law incorrectly for variable mass systems: en.wikipedia.org/w/… $\endgroup$ Commented Dec 16, 2019 at 12:57
  • $\begingroup$ @AaronStevens I read the link. But why should we apply it to the entire system? If we say $\vec{F}_{net}=\frac{dm}{dt} \vec{v} + m \frac{d \vec{v}}{dt}$ then I intuitively feel like 1) this formula is for rocket and 2) it is ok to consider the rocket only. Also, I wonder, considering the entire system, whether we should apply (for example) gravity to the mass of the system, i.e. $\vec{F}_{ext}=m \vec{g}$. $\endgroup$ Commented Dec 24, 2019 at 23:18
  • $\begingroup$ If you read the link, then you should understand why applying the product rule to $F=\dot p$ is incorrect for variable mass systems. $\endgroup$ Commented Dec 25, 2019 at 2:28
  • $\begingroup$ I don't understand even after reading that so I'm asking $\endgroup$ Commented Dec 25, 2019 at 8:19

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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.

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  • $\begingroup$ Thank you. This answer is the best and clarifies things for me. Does that mean the second law only applies to an isolated system? In the comment under my question, I replied to Aaron Stevens, and I would also like you to have a look at that question. $\endgroup$ Commented Dec 24, 2019 at 23:21
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Imagine a body moving with some velocity $\vec{v}$ which slowly evaporates. There are no external forces. It's mass decreases with time, it's velocity remains constant, it's momentum decrease as well. The equation $\vec{F}=\frac{d\vec{p}}{dt} $ obviously doesn't hold.

The Newton's second law is simply not applicable here. It's applicable to a closed system of bodies - in this case you can calculate external forces and total momentum of these bodies - but it's not applicable to cases when some parts get into or out from the described system.

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The problem is that Newton's Second Law of motion says that net force is directly proportional to the rate of change of momentum ($\frac {dp}{dt}$) and $F=ma$ is the special case where mass doesn't change. For me I believe that conversation laws are more fundamental and hence under the absence of a force ($F=0$) the momentum must be conserved and hence velocity must increase or decrease as mass decreases or increases.

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I think the point the OP is raising is that if it were possible for the mass of an object to be truly variable, and not just the result of absorbed or ejected mass (eg as rocket exhaust or evaporation), then the conservation of momentum would be possible only if the object had zero velocity. However, that could not hold true generally, since the velocity depends upon the frame of reference. Given that, mass must be conserved.

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