There's a spaceperson with a very bouncy ball and a rigid box in the form of a cube with one face missing. One day she throws the ball into the box and notices the ball bounces off each face exactly once before exiting through the missing face.
(The ball travels in a perfectly straight line, being unaffected by air resistance, spin or any other forces other than the reactions with the box. Also the ball bounces symmetrically such that the incoming angle is identical to the outgoing angle and again is unaffected by spin. Also, the box cannot be moved while the ball is in motion.)
How many different combinations are there of the order in which the ball can bounce off all five faces?
On returning to Earth our spaceperson notices that new combinations are possible.
(All conditions are the same except the ball is now affected by gravity.)
How many different combinations are there of the order in which the ball can bounce off all five faces now?
(In reply to
re(2): weighing in on the gravity solution (full solution?) by Jer)
I'm afraid I'm going to have to disagree.
The 2d analogue is a good idealet's continue with that.
I noticed that in your ascii drawing, you had a very nice parabola
after the bounce, but none before the bounce. Of course, I assume that
you only ommitted the parabola because it's very hard to draw two
parabolas in an ascii drawing. But the parabola before the bounce is
very important. Here's what your drawing would look like with both
parabolas.
______
XXXXXXXXXX 
X_  X __
 X X 
  X 
X   X 
X   X 
X  X 
X VX \
XXXXXXXXXX
As you can see (if my ascii is good enough) the parabola goes right
through the side of the box. The parabolas before and after the
bounce on the bottom must be equal sizes, because the force of gravity
is the same on a falling and rising ball. Since they are equal
sizes, one parabola cannot cross the other when they are going the same
direction (right/left) . In your drawing, that's just what they
did.

Posted by Tristan
on 20051121 19:05:33 