17 and 19 Balls, CollidingDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/