It's a bad question. For one thing, answer (C) is utter nonsense. (Maybe that's a bit harsh. It might be just regular nonsense.) In order for something to convert gravitational potential energy into kinetic energy, it has to drop to a lower height under the influence of gravity. This does not happen during a collision. Collisions in physics are effectively instantaneous events; they occur at one point in space and time and then they're over and done with. There is no change in height by which GPE could be converted into KE during the collision. Whatever (kinetic) energy the balls run away with, they had to obtain it from the kinetic energy that the cart had coming into the collision.
Now, the kinetic energy of the cart at the point of the collision was converted from the gravitational potential energy that the cart had higher up the ramp. But that conversion was done by the cart alone; the balls had nothing to do with it.
The other reason I don't like this problem is that they don't tell you at which point on the ramp the cart has the speed of $5\text{ m/s}$. It's possible that the cart maintains a constant velocity as it goes down the incline, but that would require some mechanism to keep the cart from accelerating, and if some such mechanism is involved, it should be mentioned in the problem. If that is the case, the gravitational potential energy that the cart started out with would have been converted into some other form of energy, not kinetic. It might be heat, electricity, spring energy, etc. but there's no way to know unless they tell you what mechanism is keeping the cart from accelerating.
In a pinch, if you encountered this problem on the test and didn't have any opportunity to ask for clarification, I would just assume that $5\text{ m/s}$ is the speed at the end of the ramp, immediately prior to when the cart hits the balls. Why? The alternative is that the problem is unsolvable. If the speed of the cart coming into the collision is not $5\text{ m/s}$, you have no other information that would allow you to calculate what it is. (Self-check: do you understand why this is the case?)
Once you assume that the speed of the cart coming into the collision is $5\text{ m/s}$, you have a collision of 3 objects, each of which has a mass and initial and final velocities. All 3 masses, all 3 initial velocities, and two of the final velocities are known, so you should have enough information to solve for the third. If you don't find any solution, then the situation is impossible and the answer is (D); on the other hand, if you do find a solution for the final velocity of the cart, then that velocity will distinguish between choices (A) ($v_f = 0$), (B) ($v_f < 5\text{ m/s}$), and (C) ($v_f = 5\text{ m/s}$, if you ignore the stuff about energy being converted).
It's not the four-force that is conservative, but the Einstein definition of force,
$$ F= {dp\over dt}$$
This force for a particle in an electromagnetic or linearized gravitational field is conservative in the same way as in Newton's model: the force is
$$ F = qE$$
and the integral of a static E around a closed loop is zero, still in relativity. The reason is explained in this answer: a priori validity of $W=\int Fdx$ in relativity? . The integral of the force over the distance as Einstein defines it is still the work done in the relativistic system.
Best Answer
There are a few different ways of answering this one. For brevity, I'm going to be a bit hand-wavey. There is actually still some research going on with this.
Certain spacetimes will always have a conserved energy. These are the spacetimes that have what is called a global timelike (or, if you're wanting to be super careful and pedantic, perhaps null) Killing vector. Math-types will define this as a vector whose lowered form satisfies the Killing equation: $\nabla_{a}\xi_{b} + \nabla_{b} \xi_{a} = 0$. Physicists will just say that $\xi^{a}$ is a vector that generates time (or null) translations of the spacetime, and that Killing's equation just tells us that these translations are symmetries of the spacetime's geometry. If this is true, it is pretty easy to show that all geodesics will have a conserved quantity associated with the time component of their translation, which we can interpret as the gravitational potential energy of the observer (though there are some new relativistic effects--for instance, in the case of objects orbiting a star, you see a coupling between the mass of the star and the orbiting objects angular momentum that does not show up classically). The fact that you can define a conserved energy here is strongly associated with the fact that you can assign a conserved energy in any Hamiltonian system in which the time does not explicitly appear in the Hamiltonian--> time translation being a symmetry of the Hamiltonian means that there is a conserved energy associated with that symmetry. If time translation is a symmetry of the spacetime, you get a conserved energy in exactly the same way.
Secondly, you can have a surface in the spacetime (but not necessarily the whole spacetime) that has a conserved killing tangent vector. Then, the argument from above still follows, but that energy is a charge living on that surface. Since integrals over a surface can be converted to integrals over a bulk by Gauss's theorem, we can, in analogy with Gauss's Law, interpret these energies as the energy of the mass and energy inside the surface. If the surface is conformal spacelike infinity of an asymptotically flat spacetime, this is the ADM Energy. If it is conformal null infinity of an asymptotically flat spacetime, it is the Bondi energy. You can associate similar charges with Isolated Horizons, as well, as they have null Killing vectors associated with them, and this is the basis of the quasi-local energies worked out by York and Brown amongst others.
What you can't have is a tensor quantity that is globally defined that one can easily associate with 'energy density' of the gravitational field, or define one of these energies for a general spacetime. The reason for this is that one needs a time with which to associate a conserved quantity conjugate to time. But if there is no unique way of specifying time, and especially no way to specify time in such a way that it generates some sort of symmetry, then there is no way to move forward with this procedure. For this reason, a great many general spacetimes have quite pathological features. Only a very small proprotion of known exact solutions to Einstein's Equation are believed to have much to do with physics.