If there is no net force, but the mass can change, can momentum remain unchanged? 
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?
 A: 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.
A: 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.
A: 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.
A: 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 $~\vec F=-\vec \nabla U~\ne \vec 0$
A: 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.
A: 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.
A: 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.
A: 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.
A: 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.
A: You are right, there is no scenario where speed of object changes when external force is zero (Rocket is not a correct example as explained in comments by @GiorgioP).
But first, we have to keep in mind that that when defining the momentum of a system, we assume that the total mass of system is constant in a time intreval. (Refer "An Introduction to Mechanics" by Kleppner and Kolenkow, pg 133-134, sec 3.5).
Hence, for a system, we have to consider all the pieces of it even if it disintegrates, to account for  the total constant mass. In fact, this is how rocket equation is derived.
Since in the given question, the body on which net force is zero is taken as the system , it's total mass cannot change. If for example it explodes into two, the velocity of the system is still unchanged. Its total mass cannot magically disappear. So C is also correct.
A: 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.
