Weightless person in a cart on a rail In a problem I'm trying to solve I'm having a conflict between my understanding and the proposed solution of the problem.
In the problem, a cart of mass $m$ is put at rest in the start of a friction-less ("covered with ice") rail. A person of mass $m'$ is sitting inside it.
The question is (translated):

Inside the cart there are spring scales on which the person is sitting. His limbs are in the air and do not touch the cart or the scales.
What is the height of point $C$ above ground level if the person is weightless?


Now, the first thing that comes to my mind is that if the person is weightless, then only the cart itself is activating force on the rail:
$F_N=mg$
However, in the proposed solution to the problem it is:
$F_N=0$
Which means both the cart and the person are "just floating".
From here, I can of course solve the problem myself and find H, but it depends on what $F_N$ is. Some given information is omitted here in my question.
It is important to mention that the question is from an official national highschool test, and that the solution is NOT official.
Who's right and who's wrong?
I hope it's clear to understand. Thank you!
 A: I believe that the correct answer is indeed $9.5m$ because the person and the cart accelerate and decelerate together, so it is easier if you make them into a system. If you follow along you will see that the masses really don't matter.
I don't know if that answered your question, but I'll run through the problem anyways.
The key to solving this question is clearly centipetal force because the diagram makes it very clear that point $C$ lies along a circular piece of track.
Then you must ask yourself, what will make the person lose contact with the scale, and the answer to that is when there is 0 $F_n$. The centripetal force for any point on the ramp is the force of gavity minus the normal force ($F_c=F_g-F_n$), and in this case there is no normal force, so $F_c=F_g$
So I'm sure you know this equation: 
$$F_c=\frac{mv^2}{r}$$
And because $F_c=F_g$:
$$F_g=\frac{mv^2}{r}$$
$F_g=mg$:
$$mg=\frac{mv^2}{r}$$
The masses cancel out; then, simplify for $v^2$:
$$v^2=gr$$
Hang onto that answer, now to use conservation of energy to find the height:
$$∑E_i=∑E_f$$
The initial energy is purely gravitation, and at point $C$ it becomes kinetic and gravitational:
$$E_g=E_k+E_g$$
I'm sure you know the energy formulas $E_g=mgh$ and $E_k=\frac{1}{2}mv^2$ so:
$$mgh_i=\frac{1}{2}mv^2+mgh_f$$
$h_i$ is the initial height and $h_f$ is the final height (The one we're solving for).
Again all of the masses cancel out:
$$gh_i=\frac{1}{2}v^2+gh_f$$
Solve for $h_f$: 
$$2gh_i=v^2+2gh_f$$
$$2gh_i-v^2=2gh_f$$
$$h_f=\frac{2gh_i-v^2}{2g}$$
Now remember what $v^2$ was equivalent to?
$$h_f=\frac{2gh_i-(gr)}{2g}$$
The "$g$'s" now cancel out, so it actually doesn't matter what planet you are on!
$$h_f=\frac{2h_i-r}{2}$$
Now just plug in your known variables and solve!
$$h_f=\frac{2(12)-(5)}{2}$$
$$h_f=9.5$$
I hope this helped!
A: All you need is 
1.) the formula for the velocity of a falling body  $$v_C = \sqrt{2g\Delta h}  $$
2.) the formula for the equilibrium in orbital velocity is $$v_C = \sqrt{\frac{GM}{r}}  \rightarrow \sqrt{gr}  $$
Just set $x = h_C$ and solve the system

Inside the cart there are spring scales on which the person is
  sitting. His limbs are in the air and do not touch the cart or the
  scales. What is the height of point C above ground level if the person
  is weightless?

If it was not clear : during the fall the force on the person is $F_N=mg$, if at point C the person is weightless the centrifugal force (2) equals the force of gravity ($F_c(mv^2/r)=-mg$), and net force is $F_N=0$. That happens only when the tangential velocity (1) equals the square root of $gr$ :  $$\sqrt{2g\Delta h} = \sqrt{gr}  $$
I can't give you more details, solve the system and then check the forces at $h_C$

I still don't understand why they both are weightless. Why both of
  them disconnect from the cart?

Because the centrifugal force pulls them upwards, and when (2) > (1) ($v^2>gr$, he and the cart will both fly off the track.
A: Me2 has given an excellent description of the solution, but, since you seem to know the principles, I'll expose the solution in a more concise form. 
Since the unknown is the height at point C you set $x = h_C$ and you know 
a) that the velocity of a falling body is $$v_C = \sqrt{2g\Delta h}  \rightarrow \sqrt{20 * (12-x)}  \rightarrow x = 12 - \frac{v^2}{20}$$
b) that the (equilibrium) orbital velocity is $$v = \sqrt{\frac{GM}{r}}  \rightarrow \sqrt{gr} \rightarrow v^2=50 $$
and the solution is there h = 12 - 2.5.
