17 and 19 Balls, Colliding

Date: 03/25/2003 at 16:26:10
From: Melissa
Subject: Math puzzle

There are 17 equally spaced spaced balls on the left side of a 
horizontal line, each moving to the right at constant speed S. There 
are also 19 equally spaced balls on the right side of the same line, 
each moving to the left at constant speed S. The balls are all the 
same size and mass and the spacing between the balls on the left is 
the same as the spacing between the balls on the right. When two balls 
collide they will reverse directions, but each will maintain the same 
speed S.
 
How many collisions will occur in the described system?

It looks as if there can be only 17 collisions, since the number of 
balls on the two sides is not equal ( i.e, 17 vs. 19).

Date: 03/26/2003 at 06:32:51
From: Doctor Jacques
Subject: Re: Math puzzle

Hi Melissa,

Look at what happens in a single collision. We start with two balls A 
and B moving against each other:

   A -->  <-- B

and we end up with the balls fleeing in opposite directions:

   A <--  --> B

If the balls had been travelling on parallel tracks, we would have no 
collisions at all, but the picture would be almost the same:

Before crossing:

   A -->  <-- B

After crossing:

   B <--  --> A

Note that the only difference is the name of the balls. We may imagine 
that each ball has a sticker on it with its name, and that we exchange 
stickers at each collision (this does not change the physics of the 
collisions).

Now, the problem is the same as if we have two parallel tracks: one 
with the balls going to the left, and one with the balls going to the 
right:

  A1 -> A2 ->.... A17 ->
                      <- B1.... <- B19

In this case, the end result is pretty obvious: all the A's will 
continue to the right, and all the B's will continue to the left.

For each collision in the original problem, one A ball crosses one B 
ball in the parallel track version. Now, note that, in the parallel 
track version:

* A given A ball cannot cross the same B ball more than once, since
  all balls continue in the same direction.

* We started with all A's on the left, and ended with all A's on the
  right.

This means that each A ball has crossed each B ball exactly once, and 
the number of crossings is the same as the number of collisions in 
the original problem.

Does this help?  Write back if you'd like to talk about this 
some more, or if you have any other questions.

- Doctor Jacques, The Math Forum
  http://mathforum.org/dr.math/