How did the ball know that a vertical circle follows a slope at the point $A$? A ball (mass $m$) slides from rest down a frictionless track which consists of a slope followed by a vertical circle.
Please see the picture below.
The original question asks for the normal force at $A$.
The answer is $(1 + \frac{2h}{r})mg$.
My question is here:
How did the ball know that a vertical circle follows a slope at the point $A$?
If in place of the circular track, a flat plane followed the slope, the answer were $mg$.
I think the ball has no eyes.

 A: 
I think the ball has no eyes.

This is my favorite sentence that I have ever seen on Physics SE, and you are absolutely right that the ball has no eyes. The ball does not know that a vertical circle is coming. The only thing the ball knows at any moment is what the track feels like at that point.
For the purpose of calculating the normal force, one needs to know the acceleration of the ball in the direction normal to the surface. That is determined by the curvature of the track at that point. Now, to make this problem simple, the author said that the rest of the track is all a circle with radius $r$. So we know at that point A the local radius of curvature is also $r$ because point A is on that circle. But it didn't have to be that way. The rest of the track could have been a more complicated shape, and it wouldn't matter. If the local radius of curvature at point A were still $r$, the answer would be the same.
To summarize: the author said the rest of the track is a circle to make it easy for you to determine the radius of curvature at the point A. But the ball itself doesn't care about the rest of the track at that moment.
A: The ball does not know that a vertical circle follows a slope. It is just moving along the path the question setter has restricted it to move on. It is just that the circular path is a path for which we know the standard results and are applying it. Therefore the question mentions that there is a circular path after the slope. In fact the question setter could have given a more complex path but then the question would lose its essence as it would depend on the student's knowledge of maths more than that of physics.
A: 
How did the ball know that a virtical circle follows a slope at the point $A$?

It does not.  Normal force at point A is due to a) ball weight and b) centrifugal force :
$$ N_A = F_w + F_c $$
Expressing weight and centrifugal force (which in absolute value is centripetal force) :
$$ N_A = mg + \frac {mv^2}{r} $$
Remember that mechanical energy must be conserved, so gravitational potential energy at height $h$ converts fully into ball kinetic energy at point A, :
$$ mgh = \frac {mv^2}{2} $$
From this we can extract ball speed at lowest altitude :
$$ v = \sqrt{~2gh~} $$
Substituting ball speed at point A, back into normal force expression, gives the answer :
$$ N_A = mg\left(1+\frac{2h}{r}\right) $$
A: Most probably your question is how does the ball know that it moves on a circle of radius $r$. It is a local observation (instantaneous), it does not have to be a full circle. Whats is important is the curvature of the trajectory at this point. Namely, let's say you parametrize the curve as
\begin{equation} 
\vec r = (x, f(x)),
\end{equation}
where $x$ is the horizontal coordinate and $f(x)$ the vertical one. Point $A$ corresponds to $(0,0)$ (it is just a choice). Now we compute the velocity:
\begin{equation} 
\vec v = \frac{d\vec r}{dt}=(\dot x, f'(x)\dot x)=\dot x(1, f'(x)).
\end{equation}
Again at $A$ we have obviously $\vec v _A = \dot x (1,0)$, as it should, for the velocity is horizontal. As for the acceleration we get
\begin{equation} 
\vec a = \frac{d^2\vec r}{dt^2}=(\ddot x, f'(x)\ddot x+f''(x)\dot x)
\end{equation}
Which at $A$ amounts to
\begin{equation} 
\vec a_A =(\ddot x, f''(0)\dot x).
\end{equation}
So in order to know the vertical acceleration you need to know the curvature of the curve at this point, in other words $f''(0)$. This one can always be chosen as $f''(0)=R^{-1}$ with $R$ being the radius of a circle that is tangential to the curve. In this case $R=r$
A: Interesting question.Here, the particle is in circular motion of variable radius of curvature even before it reaches A.
It is in fact called curvilinear motion. In general, in such a motion, direction of instantaneous velocity is tangential to the path, while the acceleration may have any direction. In the picture you provided, we can resolve it into two components at A:

*

*Tangential to velocity

*Along the velocity

Since Nₐ is acting along the direction perpendicular to velocity, we consider the component of acceleration tangential to velocity to be centripetal acceleration of constant radius as the curvilinear motion is now transforming into vertical circular motion.
Attached is a picture to help you understand.
A: Well the ball doesn't know anything and it doesn't need to know when to apply greater normal force or lesser.
Look at this picture :

When the ball follows the initial curve and lands on the horizontal path (let's say at some point B) it has only horizontal velocity and the normal force on it from the surface below it is equal to $mg$.
But what if we consider a point where the slope just starts becoming a part of circle (point A). The ball tends to move horizontally but the path doesn't want it to go through it and hence it applies a normal force on it of different magnitude (this time greater than $mg$ since the ball gets pressed on the curve path and thus more normal force) and in somewhat different direction than perfectly vertical and due to this greater force it goes upward following the circle.
So literally speaking the point where the normal force changed (and this change doesn't ask the ball for permission to change) was the point when the path just started to curve and this point in your figure is just after point A or the point A itself. Any point just before it will exert normal force equal to $mg$.
Hope it helps .
I am sorry for such a rough image.
