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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)

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Solution Add gravity... | Comment 8 of 26 |
Considering any pair of opposing walls of a cube, the ball should still bounce at regular intervals (since no energy is lost from the bounces) with or without gravity.  For this reason, I believe that gravity does not affect the answer of 16 reached by Charlie. 

That's my basic reasoning, but it is not as vigorous as I'd like it to be.  Here's my reasoning in greater detail:

First, there are a few assumptions I'm making.  I assume that the force of gravity is uniform, and that the ball bounces on no corners or edges.

Let's just say that the box is a closed unit cube, with all six faces.  One face serves as the starting and ending point of the ball.  The ball starts at any point on the face, and with any random velocity.  The velocity can be broken up into a vector with x, y, and z.  Each of the velocity vector's components corresponds to two faces.

If we were to let the ball bounce on all the sides, we would see that the interval between bounces on the faces corresponding to x never change.  Even with gravity, the ball will bounce back and forth at a set rate, even if the x component of the velocity is constantly changing.

Let's call the starting face the "Top" face, which is opposite the "Bottom" face.  Let's also say, without loss of generality, that the ball's journey takes exactly one second.  The B face is hit at exactly t=.5 seconds.  Sometime during this second, the South and North faces are each hit.  The interval between these bounces cannot be less than the interval before the first N/S bounce and after the second N/S bounce.  That means that the interval T is between 1/3 and 1 second.  Furthermore, one bounce is before the B bounce, and the other is after.  The same is true for the E/W faces.

There are no other relevant constraints.  The B face is always hit third.  The first two bounces have 8 possibilities (N or S?, E or W?, in what order?).  The last two bounces only have two possibilities ( in what order? ).  Therefore, there are 16 possibilities with or without gravity.

  Posted by Tristan on 2005-11-17 19:54:03
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