How Do Elastic Collisions on a Frictionless Ring Affect Object Velocities?

[luma]

[PLAIN]http://img219.imageshack.us/img219/6725/diagramt.png

So I was doing this usual physics type question thing which goes along the lines kind of like this:

2 balls are constrained to a frictionless ring and collide elastically. What is their after velocities? Easy question using,

$$\frac{1}{2}m_{1}u_{1}^{2}+\frac{1}{2}m_{2}u_{2}^{2}=\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}$$

$$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$$

I got the (correct) answers of -1 and 4 for the after speeds.

The thing that confuses me, is that since the collision is elastic, if we repeat it for the next collision that happens on the ring then we will get the same answers of -1 and 4... What does this mean? It's saying that the balls will just pass through each other?

 

[Subductionzon]

You might want to check your original "correct" answer. Just doing the math in my head your momentum is not the same before and after the collision.

 

[luma]

You might want to check your original "correct" answer. Just doing the math in my head your momentum is not the same before and after the collision.

Yes it is. The book even says it's correct.

Energy before = 50J
Energy after = 50J

Momentum before = 20 kg m/s
Momentum after = 6*4 + 4*(-1) = 24 - 4 = 20 kg m/s

What I'm asking is that since the momentum and energy are conserved, if you do the calculations for a subsequent collision you will get the same velocities. But that sounds absurd and nonsensical!

Like if A is traveling 4 m/s CCW (counter clockwise), and B travels 1 m/s CW, then after they loop round and meet each other again... they are still 4 m/s CCW and 1 m/s CW! Why? That must be nonsense as it implies the balls passed through each other

 

[tiny-tim]

hi luma!

I got the (correct) answers of -1 and 4 for the after speeds.

The thing that confuses me, is that since the collision is elastic, if we repeat it for the next collision that happens on the ring then we will get the same answers of -1 and 4... What does this mean? It's saying that the balls will just pass through each other?

No, you get two "mathematical" solutions in each case, (5,0) and (-1,4).

You reject the first solution as non-physical in the odd-numbered collisions (technically, it is physical, and would apply if the balls somehow missed each other), and the second solution in the even-numbered collisions …

the 6kg ball keeps stoppping. ​

 

[luma]

ahhh! yes, yes I did! very good! thanks :D Makes sense woo