According to newton laws if something is in uniform motion and equivalent force is equal to zero then its mass cannot change.

Our book says an object stays in state of rest or motion unless compelled by external force. Now if mass is changing and velocity is not changing there will be non-zero equivalent force, but if equivalent force is equal to zero then mass cannot change. I deduced this, so I am asking for confirmation

I am just a school kid. I just entered physics so please explain kindly and simply.

  • 3
    $\begingroup$ That... sounds like it might be right to me, but I don't quite understand why someone would say that or what they are trying to point out. Is there more context for why this was said? $\endgroup$
    – JMac
    Feb 22, 2019 at 15:52
  • $\begingroup$ our book says an object stays in state of rest or motion unless compelled by external force.now if mass is changing and velocity is not changing there will be non zero equivalent force but if equivalent force is equal to zero then mass cannot change I deduced this so asking for confirmation $\endgroup$
    – user221263
    Feb 22, 2019 at 16:36
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Feb 25, 2019 at 18:32

2 Answers 2


Based on the comments it seems like the OP will think this is invalid, but I wanted to put what I think for the review of others.

Let's say we have a "point mass"$^*$ that is at rest. The OP also seems to have issues talking about rockets, so let's just say this is a general object. This general object is ejecting mass from either side (along both the positive and negative x-axis directions, for example) at equal and constant rates.$^{**}$

The choice of our system is subjective, but I am going to choose the system as the object and the mass that has not been ejected. Or we could say the system is the mass that is only confined within the volume of the object at that instant in time. In any case, the system is obviously losing mass as mass is being ejected. We can also see that, due to symmetry, the object will not have a net force acting on it, and will remain at rest. Therefore, we have fond something is in uniform motion and equivalent force is equal to zero but its mass is changing. Due to this one counterexample, I would say the original statement is false.

As a note to the other answers, I believe the equation $\mathbf F=\mathbf{\dot p}$ is being applied incorrectly$^{***}$, but you don't need to even think about that equation with this example, at least for our purposes here.

$^*$I use quotes because, like in most classical physics problems, we usually treat extended objects as point masses when the spatial distribution of mass does not influence the problem we are trying to solve. The object we are looking at here can't truly be a point mass, because I don't see how true point masses can lose mass and form other true point masses. But the OP expressed concerns about using only point masses, so I just wanted to say that for the purpose of this question we can treat the objects as point masses, even though in reality they will have some sort of spatial distribution.

$^{**}$ This could be anything really. Maybe you could think of two twins back-to-back tied together each throwing tennis balls away from each other at a rapid rate. Choose your favorite example.

$^{***}$The correct application is found here. In this case we have $$F+u_+\frac{\text d m}{\text d t}+u_-\frac{\text d m}{\text d t}=m\frac{\text d v}{\text d t}$$ where $u_+$ and $u_-$ is the velocity of the ejected mass in the positive and negative x-axis directions respectively relative to the main object. In our case, $u_+=-u_-$, and $F=0$, and so we correctly end up with $$ m\frac{\text d v}{\text d t}=0$$ which is what was claimed. Therefore, the system has no net force acting on it and no change in velocity, and yet the mass is changing.

  • $\begingroup$ Indeed, perhaps a simpler example which can be understood even more intuitively is where the propellant of a rocketmoving along the x axis is ejected along the positive and negative y axes, with equal velocities with respect to the rocket. $\endgroup$
    – user191954
    Feb 24, 2019 at 13:48
  • $\begingroup$ @Chair yes I would have preferred that example, but the OP expressed concerns in the comments (somewhere) of only looking at objects at rest? But I guess they are the same situation, just viewed from a different inertial frame moving relative to the object. $\endgroup$ Feb 24, 2019 at 13:51
  • $\begingroup$ Yeah, this has to be a book issue, unless "equivalent force" means something less intuitive than "net force"... Which as a typed it I realized may not mean net force. If by "equivalent force" they were trying to suggest "real + apparent forces"; because the equivalent force from expelling mass isn't always considered a force. $\endgroup$
    – JMac
    Feb 25, 2019 at 14:06
  • $\begingroup$ @JMac yeah, who knows :) $\endgroup$ Feb 25, 2019 at 14:29

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.

  • $\begingroup$ you assumed a lot of things constant force,wrong velocity put in the chain rule formula .centre of mass concept is not used it is not so simple and I never said there was external force on system I meant equivalent force on the rest part is zero you are assuming wrong system $\endgroup$
    – user221263
    Feb 24, 2019 at 16:34
  • 2
    $\begingroup$ @SandeepGulati Your comment is wrong in all respects. I do not assume constant force. The $\vec{v}$ in the chain rule formula is the velocity of the object... what else should it be? The derivation is entirely consistent with conservation of momentum in the original center of mass frame. I am reluctant to comment any more until you define what "equivalent force" means. This is not a term in general use. Clearly define your system. Clearly define the forces and their agent. I can say no more until you and I are speaking the same language. $\endgroup$
    – garyp
    Feb 24, 2019 at 17:58
  • $\begingroup$ @garyp I have applied this to a counter example in my answer $\endgroup$ Feb 24, 2019 at 23:23
  • $\begingroup$ equivalent force you can say it as net force $\endgroup$
    – user221263
    Feb 25, 2019 at 3:50
  • $\begingroup$ @SandeepGulati Do have any questions about my answer? I really just want to try to help you understand what is going on here $\endgroup$ Feb 25, 2019 at 5:09

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