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All external forces on a body cancel out. Which statement must be correct?

A) The body does not move.

B) The momentum of the body remains unchanged.

C) The speed of the body remains unchanged.

D) The total energy (kinetic and potential) of the body remains unchanged.

I understand that A is incorrect because the forces may still cause rotation. B is correct [according to the Mark scheme (MS) as well] since the momentum is the product of mass and velocity since net force is zero, there is no change in velocity hence no change in momentum.

C I am confused with. How can it be wrong? Under what condition can speed change?

The MS quotes:

The key to the answer is in the word ‘must’. At first sight, both B and C seem possible, but C is not correct if the mass of the body is changing, whereas B is always correct under any circumstances.

I understand that if mass changes the speed of a body can change. Also, that, if mass changes the kinetic energy and the potential energy may change.

But wouldn't the "mass can change" argument also applies to momentum? Since p = mv, if mass changes momentum can also change. How is B the correct answer?

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    $\begingroup$ Because here body should be seen as an unchanging piece of something. In my opinion this kind of test are unclear and causes this possible misunderstanding. In the sense that the student who knows more is more prone to check what the sheet sees as wrong or viceversa. $\endgroup$
    – Alchimista
    Jan 17, 2021 at 15:43
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    $\begingroup$ Accordingly to this page on meta physics.meta.stackexchange.com/questions/714/… about homework and homework-like questions, linked in the Help Center, "This site it's a place to get specific conceptual physics questions answered." I have a hard time seeing how this question fails to meet this requirement, even if the origin is homework. I think that people flagging questions should occasionally refresh their ideas about the existing rules. $\endgroup$ Jan 18, 2021 at 7:33
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    $\begingroup$ This is just a terribly written question... yes, if an object breaks into pieces, then the speeds of the pieces can change, even if there was no net external force. (For example, things can explode.) But the question talks about "the" speed of "a" body, which seems to imply the body is a single object, not multiple pieces. Really, the correct answer is to not learn physics from whatever source produced this question. $\endgroup$
    – knzhou
    Jul 30, 2022 at 7:27
  • $\begingroup$ @knzhou And not to mention the constant interaction between the pieces and billions and billions of photons. $\endgroup$ Jul 31, 2022 at 16:37
  • $\begingroup$ This isn't technically related to your question, but it's an important conceptual misunderstanding that warrants mentioning, imho: A is not wrong just because of rotation. It's just plain false even without rotation. If the forces are in equilibrium, the body can't change how it moves (or more technically, it can't change its momentum), but if it's already moving it will continue moving with the same momentum. $\endgroup$ Jul 31, 2022 at 18:06

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The answer book is simply wrong. Both B and C are true. Within the context of Newtonian mechanics, I maintain that a “body” is rigid, and its mass is constant.

The answer book claims the speed can change without external forces if the mass changes; but the mass only decreases if you allow the “body” to split into multiple pieces, and redefine the “body” as a subset of the pieces. If you do that, neither B or C is true in general. On the other hand, for the mass to increase, you must add an additional piece to the “body”. To do this without exerting a force, the extra piece must have the same velocity. In this scenario, C would be true, but not B. So even in this very dubious interpretation of what a “body” can be, the answer book is wrong.

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  • $\begingroup$ Strictly speaking, you are right. But plenty of textbooks speak about the dynamics of variable mass bodies. In such a context, it is usual not to include the change of momentum due to the gained/expelled mass in the external forces. $\endgroup$ Aug 6, 2022 at 5:50
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Translation

$$\frac{d}{dt}(m\,\vec v)=\frac{d}{dt}(\vec p)=\vec 0\quad\Rightarrow$$

linear momentum $~\vec p~=$ constant

and from the equations of motion

$$m\,\frac{d\vec v}{dt}=\vec 0\quad\Rightarrow$$

the velocity vector $~\vec v~$ is constant so the speed $~v~$ is constant

rotation

$$\frac{d}{dt}(I\,\vec\omega)=\frac{d}{dt}(\vec L)=\vec 0\quad\Rightarrow$$ the angular momentum $~\vec L~$ is constant

answer D is wrong because the force is $~\vec F=-\vec \nabla U~\ne \vec 0$

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So, in the question we should assume that the mass is allowed to change but all external forces cancel.

How can the mass change with no external forces?

Here is an example. The body spontaneously chooses to emit a gamma ray.

Then answer A is wrong because the body does move.

Answer B is wrong because the body's momentum has changed. (Unless you want to consider the body and its gamma ray together as one body. Good luck with that.)

Answer C is wrong because the speed changes.

Answer D is wrong because the total energy of the body changes.

The MS should state that the mass does not change.

Which answers are wrong when the mass does not change?

Answer A is wrong because it's entirely arbitrary which constant velocity we assign to the body. We can say it moves no matter what.

Answer C is wrong. The body could be a bolas -- three weights on strings that spin around a center, and at the center there is a mechanism which can reel in the strings or let them out. With no external forces the body can choose on its own to change its angular speed.

Answer D is wrong. The body could be very hot, and inside a cavity inside a bunch of dry ice. Radiation from its surroundings is constant in all directions and all external forces are balanced, but that radiation does not balance the radiation the body gives off. The body loses energy.

Does that make B wrong too? As it cools its individual molecules lose momentum, but does that mean the body as a whole loses momentum? Imagine it's a hot steel wire that spins fast. As it cools it contracts, so it will automatically spin faster and that doesn't affect momentum. I don't think there's any benefit from that kind of trick.

So does the momentum of all the molecules in the object, affect the momentum of the object?

My first thought is no. The linear momentum of all those molecules has to cancel out apart from the linear momentum of the object as a whole. If they slow down they'll still cancel out, it will just be less of it cancelling.

Can they lose angular momentum when they radiate? I'm not sure, but does it matter? If their angular momentum cancels out before they get colder, it will cancel out afterward.

What if their angular momentum did not cancel before? Say the object is a container with transparent walls, filled with a gas where a whole lot of the gas molecules have the same rotation angle. Not at equilibrium. They contribute a lot to the angular rotation of the whole object. The object is surrounded by something that uniformly bathes it in radiation, and that radiation is randomly absorbed by gas molecules that then vibrate differently in random directions. Does that change the angular momentum of the object? Or does the container start to rotate to make up for the loss by the gas?

Or maybe there's simply no way for radiation to affect a molecule's angular momentum? But of course it can. A polar molecule with a charge difference on either end. An electromagnetic wave that pulls opposite charges in opposite directions, of course that can affect rotation.

My conclusion is that if I got this question on an undergraduate physics test it would encourage me to stay away from physics.

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Thinking of it, you are absolutely right...

How can a body lose mass:

In most cases we encounter, mass is lost from a body as smaller masses. Rocket propulsion, Evaporation, sand leaking from a trolley, etc are all examples of a body losing mass.
where body = trolley, rocket, etc.

What's interesting is that neither the momentum, nor mass nor velocity of these bodies is conserved. Conservation of momentum holds for a system.

So its the momentum of the system = (trolley+lost sand) , (rocket+fuel ejected), etc. that stays conserved.

A body can also lose mass as energy. Say a body converts a small amount of its mass to energy. This extra energy will then be radiated as em-waves, which also carry momentum. The momentum of the body will still not be conserved as it is the total momentum that stays conserved.

So...

their argument - that momentum of a body will always be conserved in the absence of force (even if mass changes), is vague. If mass is changing, momentum is also changing unless we talk about the whole system.

Still the correct answer will never be velocity, since the velocity is never conserved, even for a system as whole. In the freight cart problems, the sand leaks out with zero velocity. Instead if the sand was thrown backwards, like a rocket engine, speed of the trolley would increase.

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A mining cart is rolling on the tracks. Suddenly you drop a rock into the cart. No net force was ever exerted horizontally, yet the cart will slow down since it has to pull the newly added mass up to speed - the current kinetic energy is now spread over more mass, so the speed drops a bit.

This is an example of zero net force, $F_{net}=0$.

  • A) and C) are obviously not correct, since there is motion with a changing speed.

  • B) can be analysed via Newton's 2nd law, $$F_{net}=\frac{dp}{dt}.$$ A time derivative is what we understand as a change. Since we know that $F_{net} =0$ then the momentum change is also zero, meaning no change taking place. So B) is correct.

  • D) can be analysed via the kinetic energy formula and the momentum definition, $$K=\frac12 mv^2\quad \text{and} \quad p=mv. $$ Since we now know that the momentum $p$ doesn't change, then e.g. doubling $m$ must be accompanied by $v$ being halved. That will change the kinetic energy, though, due to the squaring of $v$. So D) is also not correct.

So, I have here given an example of a scenario where only B) is correct. Your question asks for which option that necessarily always must be correct - with just one example I have now shown that A), C) and D) clearly are not always true. We are down to just a question whether $B)$ always is true. Since there must be one correct answer, then you can now pick B). Or you could argue for why Newton's 2nd law explains that B) always is true.


Even though A), B) and C) are not generally true since we above have a counter example, then you mifyr still be able to find examples where some of them are true.

Your freight-truck example with gradually leaking mass is indeed one such example where the speed stays constant despite the decreasing mass. Note that there is a categorical difference between my example of rock-into-cart and your example of sand-leaking-from-truck; they are not equivalent examples.

  • When dropping a rock into a moving cart, I am not adding kinetic energy (the potential energy from the drop will be absorbed into the surface underneath). Instead, the rock will have to be dragged up to speed by the cart, and due to energy conservation (we only have the energy that we've got - no more or less) then the current amount of kinetic energy will thus be spread over this now larger total mass, reducing the speed slightly.

  • When leaking sand from the moving truck, then notice that the sand grains still carry their kinetic energies with them after leaving. In my rock-into-cart example, no energy was added or removed so energy conservation could help explain some speed changes. But in your example here, kinetic energy is leaving the system, and that loss in total energy might perfectly balance out the loss in total mass.

Same goes for your example of cutting-a-train-in-half. So, your examples show scenarios of constant speed during mass loss. My example above show a case of speed change during mass loss. Thus, you can't as a general rule say that the speed stays constant during mass loss.

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I have edited my answer to correct a mistake in the original version when discussing point C. However, I also took the occasion to add something to the discussion of the dynamics of mass varying systems.

Answer A is wrong non only because there could be a rotation but, for a much more basic reason, because the body can move with a constant velocity.

Answer D is false. A simple counterexample is the case of a body falling in a fluid at the terminal velocity: external forces do equilibrate, but the (gravitational) potential energy is decreasing.

Answers B and C could be correct in the hypothesis of a constant mass body, which would not be incompatible with the question's wording. However, in such a case, there would be no unique correct answer.

According to the MS indication, the person who wrote the question was thinking about the case of a varying mass body.

However, applying the correct formulae for varying mass dynamics shows that both the answers B and C may be wrong without additional information.

Clearly, those who wrote the question and answers were thinking that every varying mass problem is described by the equation $$ \frac{{\mathrm d}{\bf p} }{{\mathrm d} t} = {\bf F_{ext}}. \tag{1} $$ Unfortunately, it is (or should be) well-known that such an equation is not unrestrictedly valid in Classical Mechanics. There are cases where it cannot be used. Probably the simplest is the case of a body losing (or acquiring) mass isotropically. In that case, the correct equation of motion is $$ m(t)\frac{{\mathrm d}{\bf v} }{{\mathrm d} t} = {\bf F_{ext}} \tag{2} $$ and $m {\bf v}$ varies even if the external force is zero. The most general equation of motion for a varying mass body is the following (see, for instance, A.Sommerfeld, Mechanics. Lectures on Theoretical Physics, Vol.I, New York, 1952, p.28): $$ m(t)\frac{{\mathrm d}{\bf v}}{{\mathrm d} t}= {\bf F_{ext}}+{\bf u}\frac{{\mathrm d}m}{{\mathrm d} t}, \tag{3} $$ where ${\bf u}$ is the relative velocity of the escaping (or incoming) mass with respect to the center of mass of the body moving with velocity ${\bf v}$. Equations $(1)$ and $(2)$ are special cases of $(3)$, respectively in the case of ${\bf u}=0$ or ${\bf u}=-{\bf v}$.

Now it is easy to see why both answers B and C are not of general validity.

In the case of answer C, if we are in the case of a mass loss (or increase) with relative velocity ${\bf u}\neq 0$, in the absence of external forces, there will be an acceleration $$ \frac{{\mathrm d}{\bf v}}{{\mathrm d} t}= {\bf u} \frac{{\mathrm d}\log m}{{\mathrm d} t}. $$ and a change of speed proportional to $$ {\bf v} \cdot \frac{{\mathrm d}{\bf v}}{{\mathrm d} t}= {\bf v} \cdot {\bf u} \frac{{\mathrm d}\log m}{{\mathrm d} t}. $$ Therefore, the validity of answer C is not general.

In the case of answer B, if we are in the case of a mass loss (or increase) with relative velocity ${\bf u}= 0$ (isotropic mass variation), in the absence of external forces, the velocity remains constant, but momentum ($m(t){\bf v}$) does not.

In conclusion, either the question has two correct answers (in the case of constant mass) or no correct answer without additional information.

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  • $\begingroup$ The latter point is important. The question is, if you have an “object” whose mass is changing, how do you notice forces on it? Assuming that you use the first two Newton’s laws, it is in a state of force balance if it remains at constant velocity, period. So for its mass to change its momentum must change. If it is an asteroid flying through a field of dust and it slows down from accumulated dust, we would surely want to say that there is a drag force, especially as there must be energy loss in those sticky collisions—but the momentum does not change. $\endgroup$
    – CR Drost
    Jan 17, 2021 at 19:34
  • $\begingroup$ @CRDrost, this could be another question more than a comment. $\endgroup$ Jan 17, 2021 at 20:48
  • $\begingroup$ The momentum equation always holds, if one describes the system correctly. Mass cannot appear or disappear - rather we are talking about a composite body, parts of which are moving in different directions. The total momentum is conserved, but the momenta of the parts change, because they act on each other with forces. This is usually discussed in details in the context of the rocket equation. $\endgroup$
    – Roger V.
    Jan 29, 2021 at 17:28
  • $\begingroup$ @Vadim, I am discussing the momentum of a body losing or acquiring mass, not the total momentum. The point is the relationship between the change of momentum and forces. Newton's equation with the derivative of momentum does not always give that. If you feel this is wrong, please show me how to deal with the case of an isotropic mass expulsion that is not the same as the rocket. $\endgroup$ Jan 29, 2021 at 18:00
  • $\begingroup$ @GiorgioP the example of uniform circular motion isn't a valid example for answer $c$ because there is always an external centrifugal force working on it and the question includes no external force. $\endgroup$ Jul 30, 2022 at 14:20
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In Newtonian mechanics and relativistic mechanics both B and C are true, and A and D are false. If we define, net force as the derivative of momentum for a body we have:

$$\frac{\text{d}\boldsymbol{p}}{\text{d}t} = \boldsymbol{F}_{net} = \boldsymbol{0}$$

then the linear momentum can be defined as:

$$\boldsymbol{p} = \frac{m\boldsymbol{v}}{\sqrt{1-\epsilon\cdot v^2/c^2}}$$

where $\epsilon = 0$ for Newtonian mechanics and $\epsilon = 1$ for special relativistic mechanics. Computing the derivative, we have:

$$\frac{\text{d}\boldsymbol{p}}{\text{d}t} = \frac{\dot{m}\boldsymbol{v}}{\sqrt{1-\epsilon\cdot v^2/c^2}} + \frac{m\boldsymbol{\dot{v}}}{\sqrt{1-\epsilon\cdot v^2/c^2}} + \frac{m\boldsymbol{v}}{[1-\epsilon\cdot v^2/c^2]^{3/2}}\frac{\boldsymbol{\dot{v}}\cdot\boldsymbol{v}}{c^2}$$

so the change in mass can be compensated by a change in velocity. For the unidimensional case:

$$0 = \frac{\dot{m}v}{\sqrt{1-\epsilon\cdot v^2/c^2}} + \frac{m\dot{v}}{[1-\epsilon\cdot v^2/c^2]^{3/2}}$$

Then, we have:

$$\frac{\dot{v}}{v}\frac{1}{\sqrt{1-\epsilon\cdot v^2/c^2}}=-\frac{\dot{m}}{m} \quad \Rightarrow \quad \frac{\sqrt{1-\epsilon\cdot v^2/c^2}+1}{\sqrt{1-\epsilon\cdot v^2/c^2}-1} =\frac{C_0^2}{m^2(t)}$$

so that any variation in mass can be compensated by a variation in velocity.

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The question is wrong and it has some mistakes. Let me explain why?

Option A isn't correct because the body can rotate or it can move with a constant velocity when all the external forces are taken to zero.

Option D is not correct and it has so many counter examples like when a body is falling with terminal velocity in a fluid its potential energy is decreasing.

Now the contradiction starts between option B and option C.

B and C both are right for normal conditions according to Newton's laws of motion.

But if it comes to the point of varying mass then both B and C are incorrect. As Momentum is the product of mass and velocity, change in an object's mass moving with some velocity absolutely causes change in its momentum. So, option B is wrong. Option C is also wrong as an object without any external force working on it gains some mass its velocity will change because the momentum needs to be conserved (the additional mass should be moving with the same velocity of the object or it will exert some force on the object).

Now, it looks like the question is wrong.

For my opinion, it should refer net momentum. According to conservation of momentum, net momentum of bodies is always conserved if all external forces are zero. And then, the correct option is B.

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If the net force is zero, $dp/dt=0$, even for the relativistic case.

But what if the object was losing mass over time?

You should calculate the total momentum of all particles of the body, say, if it is a rocket, you should take into account the momentum of the used propellant as well. Then the total momentum does not change.

why is option C not accepted?

Let us consider a rigid body consisting of two massive material points A(t) and B(t) connected by a rigid massless rod in plane Q. Let us assume that initially the points are at rest and force F(t) acts on point A and force -F(t) acts on point B, and the direction of F(t) is always orthogonal to the rod connecting A(t) and B(t), and F(t) is always in plane Q. Then the speeds of points A(t) and B(t) will change with time and those two speeds (not velocities!) will always be equal. Then one can either assume that the speed of the body is the same as that of the material points and it changes with time, or one can assume that there is no reasonable definition of the speed of a rotating rigid body, but in this case C is also incorrect.

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You are basing your statement on math rather than physics.Let's say a rocket loses some mass of its fuel every second.Then the rocket is accelerated because momentum must be conserved.

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    $\begingroup$ Which momentum is conserved? Would you call the compound system (rocket+gas) a body? $\endgroup$ Jan 17, 2021 at 15:41
  • $\begingroup$ No just the rocket's momentum because it loses mass. $\endgroup$
    – user285949
    Jan 17, 2021 at 16:32
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    $\begingroup$ Then the external force on the body is not zero. The total system is made by rocket and expanding gas. The latter is responsible for an external force on the rocket. $\endgroup$ Jan 17, 2021 at 16:55
  • $\begingroup$ @GiorgioP i dont take into account that force exerted by the gas to the rocket.I only take into account the acceleration of the rocket due to the reduction of its mass. $\endgroup$
    – user285949
    Jan 17, 2021 at 18:04
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    $\begingroup$ There is no force due to "reduction of the mass only". The conservation of momentum you are using to analyze the situation directly comes from the vanishing of the internal forces of the gas+rocket system. $\endgroup$ Jan 17, 2021 at 18:42

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