Part 2: Superball
A small ball is thrown toward an incline. This (amazing) ball bounces perfectly.
It is thrown to the right horizontally from the point (0,5) and follows the path of the parabolic equation y=5  (x^2)/18.
The incline follows the equation y=x/2.
Find the equation of the path the ball takes after its first bounce off of the incline.
(If you wish, find some of the subsequent bounces.)
(Assume that gravity pulls straight down on the ball. Also assume the ball is not spinning so that it is perfectly reflected.)
(In reply to
re: potential path (spoiler) by Mindrod)
I agree that the point of impact is (6,3) and that the slope of the ball's trajectory is at that point 2/3. The angle between the ball's path and the plane is then arctan 2/3 + arctan 1/2, similarly to the reasoning in Bouncy Bouncy part 1. The reflected path also has this angle, but its overall angle relative to the ground must have another arctan 1/2 added to it, so it is arctan 2/3 + 2 arctan 1/2, which is about 86.82016988013577 degrees, rather than straight up.
The question now is to figure what parabola with that slope the ball's trajectory is on. Its slope at (6,3) is 18 (the tangent of the above angle). So if the parabola is y = ax^2 + bx + c, and the slope is y' = 2 a x + b, we get
12 a + b = 18
and
36 a + 6 b + c = 3
which are not enough to solve for a, b and c.
The solution must involve the fact that the total of potential energy plus kinetic energy must remain constant. When the direction changes at (6,3), the overall velocity must be the same, but less component is in the x direction. The component in the x direction remains constant for all of one parabola and is the only contributor to the kinetic energy at the crest of each parabola.
The potential energy is proportional to the height above the ground, and the kinetic energy is proportional to the square of the velocity.
Edited on February 11, 2006, 12:49 am

Posted by Charlie
on 20060211 00:48:46 