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).
A simple counterexample:
Imagine two particles with opposite direction and equal speed. The center of mass does not move, yet the kinetic energy of the system is non-zero.
Now let both particles come to rest (by friction, hitting a wall, whatever). The kinetic energy is now zero, and total momentum has been conserved, while energy is not.
The crucial point is that kinetic energy depends on the square of velocity, $E_\mathrm{kin} = \frac{1}{2}mv^2$, and so is always positive - it cannot "cancel out" as momentum does, so momentum $\vec p = m \vec v$ can perfectly be conserved while the kinetic energy changes if the terms with "positive" and "negative" sign decrease or increase in a fashion that keeps the total momentum constant.
Best Answer
Momentum, energy, angular momentum, and charge are conserved locally, globally, and universally. One must remember that conservation locally (within a defined system) does not mean constancy. Constancy occurs only when the system is closed/isolated from the rest of the universe.
Conservation means that these quantities cannot spontaneously change. Let's consider momentum: the momentum of a system at a later time must equal the momentum at an earlier time plus the sum of the impulses applied to a system. The impluses in this sum could be adding or removing momentum from the system, but never creating nor destroying momentum: $$\vec{p}_{later}=\vec{p}_{before}+\Sigma\vec{J}_{during}.$$
For an isolated collision, without outside influence, $\vec{J}_{during}=0$, and $\vec{p}_{later}=\vec{p}_{before}$.
For the energy: $E_{later}=E_{before}+W+Q+\mathrm{radiation}$
For angular momentum: $\vec{L}_{later}=\vec{L}_{before}+\Sigma\vec{\Gamma}_{outside}$ ($\Gamma$ is torque on system)
For charge: $ Q_{later}=Q_{before}+\int I\;\mathrm{d}t$
In the case of kinetic energy, it is not universally conserved. It can appear and disappear as energy is transformed to different manifestations:heat internal energy, gravitational, electromagnetic, nuclear, all of which are energy. The total energy is conserved in a system (not necessarily constant), with the transfer agent being work/radiation/heat. The elastic collision is defined to be one in which the kinetic energy of the system remains constant.
Note that if you define a single object as the system of interest, neither the momentum nor the kinetic energy will remain constant during a collision with another object or while it falls in a gravitational field, but the momentum will be conserved (the object is subjected to impluses) and the energy of the object is conserved (outside forces do work).
Bottom line: Define a system, look for transfers of momentum (impulse), energy (work, etc), angular momentum (torque), and charge (current) into or out of the system. Then see if any of those conserved properties are also constant for your situation.
EDIT - Response to OP specific questions:
Yes. One may also call it partially elastic. If the coefficient of restitution is zero (0), the collision is completely inelastic.
Yes. Momentum is conserved in all collisions and explosions. And sometimes it might even be constant for short periods of time.