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If you drop a ball down a upward curving parabolic ramp, what is the expected movement of the ball? Assume the ball is set into motion with no external force. Would it go all the way from point A to point B or halfway to point B and come to rest at the middle of the curve?

enter image description here

I am also trying to see some evidence of such motion.. however most of the examples I see are more related to an object falling in a parabola or other examples on a downward projectile.

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I would add that you assume no external forces beside gravity. Because with no external forces at all (and a non-zero initial velocity), the ball will follow the curve up to B and just continue on upward from there. If you add gravity, you'll either get oscillation or a 'prison break' from the ball, depending on the magnitude of the initial velocity. – Wouter Jan 30 '13 at 17:34
You are asking a very specific question without giving many details as to the setup. – Nic Jan 30 '13 at 17:56
@Wouter : Why shoud the ball even start to slide down, if there is now gravity? If there are really no forces, the ball will in my opinion never start to move at all. – Noldig Jan 30 '13 at 21:59
@Noldig That's where the part "(and a non-zero initial velocity)" of my comment comes in. Of course, if the initial velocity isn't in the right direction, the ball might not even meet the curve at all. So I should have specified not only non-zero but also in the direction parallel to the derivative of the curve in $A$ and into the curve. – Wouter Jan 30 '13 at 22:24
I understand that as the ball goes down the ramp, it acquires some energy and it carries itself to point B. Assuming (a) no resistance from the ramp (b) no initial velocity - the ball isnt being let go from a certain height. It is just being held near the point A and let go. without any added manipulations and based just due to gravity, would it still go all the way to B? – user20375 Jan 30 '13 at 23:13

Assuming point A is some non zero point and point B is x=0, it depends on if you have friction or not. Without friction, the ball will oscillate back and forth about B, reaching the same height as A on the other side, before going back to point A.

With a bit of friction, it will oscillate in a similar manner, but losing height with each oscillation until it comes to rest at point B.

With a lot of friction, it will go directly to point B and stop.

GRAPH assumes orange line intersects point A, red line intersects point B.

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The OP has provided an image which differs slightly from the assumption you used. – dmckee Jan 30 '13 at 17:28

From $A$ to the bottom of the curve, there is very little difference between your case and that of a falling object. If there is no friction and the initial velocity/momentum of the ball is zero, the ball will gain momentum until it reaches the bottom of the curve. The only difference with the case of a ball falling vertically is the path it follows. Energetically, there is no difference. (see also the wikipage for a Conservative force) Due to this specific path, there is an important difference with the case of a falling ball, because it allows the ball to smoothly change the direction of its motion and start climbing. During its ascent it loses momentum until it reaches $B$ exactly.

You can easily see that $B$ is reached exactly by looking at the energy of the ball. Initially it has no kinetic energy, $$T=mv^2/2 = 0,$$ and a certain amount of gravitational potential energy equal to $$U_G = mgh$$ if $h$ is the height of point $A$ above some horizontal reference plane. Note that we can choose this plane wherever we want, since potential energy is only defined up to an arbitrary additive constant. So let's choose our plane of reference tangent to the bottom of the curve. This way, the ball (treated as a mass-point, i.e. assuming it has no dimensions) has a potential energy $U_G=0$ at the bottom of the curve.
Since there is no dissipation of energy through friction and the total energy should be preserved, the initial potential energy $U_G$ must have been transformed completely into kinetic energy (indeed, the ball has a finite velocity and in fact its maximum velocity at the bottom of the curve). This kinetic energy is subsequently transformed completely into potential energy, until all of the kinetic energy is used up and the potential energy is maximal again, which happens when the ball exactly reaches $B$.

More symbolically: at point $A$ the total mechanical energy is $$T+U_G = U_G = mgh$$ and at the bottom of the curve it is $$T'+U'_G = T' = mv^2/2.$$ Since energy must be preserved, it follows that $$T' = U_G$$ or, equivalently, $$v' = \sqrt{2gh}.$$ Now the ball starts climbing again, thereby increasing its potential energy at the expense of its kinetic energy. It reaches a height $h''$ where it comes to a stop for an infinitely short period of time before coming back down again. Now, the height the ball reaches is determined by its kinetic energy at the start of its ascend due to - again - preservation of energy, which demands that $$U''_G = T'$$ or, equivalently, $$v' = \sqrt{2gh''}.$$ We now have two equations for $v'$ and clearly these can only be compatible if $h'' = h$, meaning that the ball ascends to exactly the same height that it started its descent from in the beginning.

Now all of this can be repeated for the motion of the ball once it has reached $B$. This will lead to the conclusion that the ball goes all the way back to $A$, and back to $B$, and back to $A$, ... oscillating between $A$ and $B$ forever. Of course in any realistic setting, there is at least some friction, causing the ball to dissipate some of its energy (into heat). So we cannot use the above equations for energy preservation$^1$ and the ball will arrive at the bottom of the curve with less kinetic energy than its initial potential energy. This results in a lower maximum height on the $B$ side of the curve and in the ball eventually coming to a halt at the bottom of the curve.

$^1$ Note that there is still preservation of energy in the system even if there is friction. Part of the mechanical energy $T + U$ is merely being transformed into heat $Q$, so $E_{tot} = T + U + Q$ is still preserved.

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