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Box bounce (Posted on 2005-11-17) Difficulty: 4 of 5
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?

No Solution Yet Submitted by Sir Percivale    
Rating: 4.1429 (7 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(3): weighing in on the gravity solution (full solution?) | Comment 19 of 26 |
(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 2-d analogue is a good idea--let'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.

X_ --  X --__
- X- X -
- - X -
X - - X -
X - - X -
X - -X -
X VX \

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 2005-11-21 19:05:33
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