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 is a bit more complicated. You should be able to land without the use of energy but you have to pass your momentum to something - and this passing on of momentum will be the same in both cases. Let us consider a "spherically-symmetric-chicken-in-a-vacuum" model to see this explicitly.
Momentum conservation and propelled mass
The spacecraft of mass $M$ is hovering just a bit above the surface of the planet at zero vertical velocity (it is hovering above the surface to be able to propel fuel downwards). Suddenly, it propels a single pulse of fuel with velocity $v_{\rm f}$, mass $m$ and total momentum $p_{\rm f} = m v_{\rm f}$ downwards. Due to momentum conservation, the spacecraft will start to move upwards with a momentum with equal magnitude. If the pulse is just enough for the spacecraft to escape the planet with velocity $v_{\rm e}$ it must hold that $$p_{\rm f} = m v_{\rm f} = M v_{\rm e}\,,$$ or that the velocity of propulsion of the fuel must be $$v_{\rm f} = \frac{M}{m} v_{\rm e}\,.$$
But your assumptions are that $M \gg m$ which in combination with the previous formula leads to $v_{\rm f} \gg v_{\rm e}$. Even though the model is vastly simplified, the basic conclusions are valid. In more complicated models you will continuously loose mass so that the propulsion will have varying effects but the "before vs. after" equation will look the same on average because momentum conservation always rules.
The very same relation would be valid for a descent. Say you have fallen from rest at infinity up to the surface of the planet and due to energy conservation now have a velocity equal to $v_e$ - but downwards. Momentum conservation! You have to throw the momentum away not to get smashed into pieces, so you propel all your momentum into a single fuel pulse downwards. The momentum of this pulse must be again equal to $p_{\rm f}$ so all the previous relations hold.
Energy balance
It is true you have to propel the same amount of mass at the same speed in both cases, but it is also interesting to see what happens to energy. In the takeoff you have before the pulse
$E_0=0$
and after the process a lot more kinetic energy due to the motion $$E_{1} = \frac{p_{\rm f}^2}{2m} + \frac{p_{\rm f}^2}{2M}$$ I.e., the smaller the $m$, the more energy you use in the takeoff. However, in the descent you have before the pulse $$E_0 = \frac{p_{\rm f}^2}{2M}$$ and after the pulse $$E_{1} = \frac{p_{\rm f}^2}{2m}$$ so in principle, it should be possible to save a lot of energy just by an elastic "passing-on" of the momentum. This is actually something we know very well from practice where we have the atmosphere to give the momentum to via a parachute or such.
From this "fuel-pulse" mode you can see that if you are able to pass your momentum to a mass of about your magnitude, in principle you should be able to land without burning a drop of fuel (I.e. without exerting any stored chemical/electric/nuclear energy).
The fact is that even if you just want to use mass and energy stored in your spacecraft, you will not be accelerating it all from zero but from the escape velocity $v_{\rm e}$ to $v_{\rm f}$. You can thus use your fuel to push and propel some dead carriage because your engines will actually propel at their original speed plus $v_{\rm e}$ in the planet rest frame.
CONCLUSION: You have to propel roughly the same mass at roughly the same speed both during take-off and landing, but you do not have to burn so much fuel during landing.
Just a note on the assumption of no mass change for a reaction engine. For a Moon to Earth escape we have $v_{\rm e}\approx 11 000\,\rm m/s$ which is way above the possibilities of a classical bipropellant rocket with maximum $v_f \approx 5000 \, \rm m/s$. Even if you used some of the state of the art ion thrusters, you would not reach more than a few times more than $v_e$. But that is in conflict with the conclusion $v_{\rm f} \gg v_{\rm e}$ from the momentum conservation balance and the assumption $M \gg m$. So the assumption of a small propelled mass $m$ in comparison with the mass of the spacecraft $M$ is not realistic.
Best Answer
These are 'superelastic' collisions - where energy is released.
Yes the smaller mass would gain velocity due to conservation of momentum, but only provided the energy released in the separation pushed the smaller mass 'forwards' in the same direction as the motion
If, for example, the skateboarder jumps forwards off the skate board he (the heavier body) can go forwards at a higher velocity and either slow down, stop, or force the skateboard to go backwards in the opposite direction depending on how hard he pushes back on the skateboard as he jumps.
As pointed out by Sofia in comment, momentum is conserved, but in these cases the total kinetic energy of the bodies increases.
I suppose strictly speaking these events are not collisions - more dissociations - but the are definitely superelastic