Number of collisions; elastic collisions

[Gloyn]

Homework Statement

Hi!

I have found an interesting statement. It says, that if we have a system of two masses and a wall (all collisions will be elastic ones) with one mass (lets label it as 1) trapped between the other mass (2) and the wall and if there is no friction, then if ratio of masses m2/m1=1, then number of collisions between the masses will be 3, when m2/m1=100 then there will be 31 bounces, if m2/m1=100.000.000 then there will be 31415 collisions etc, the number of collisions will approximate more and more numbers in Pi. How do I explain it?

Homework Equations

Conservation of momentum, Newton's Law of Restitution

The Attempt at a Solution

I have calculated several velocities of masses to see if there is some simple rule. What I got is that the velocity of 2nd mass is:

V_1=v\cdot \frac{\alpha-1}{\alpha+1}

V_2=v\cdot \frac{\alpha^2+4\alpha-1}{(\alpha+1)^2}

V_3=v\cdot \frac{\alpha ^3 + 13\alpha^2+15\alpha-5}{(\alpha+1)^3}

So denominator seems to be fairly regular, but numerator is rather wild.

 

[voko]

You have forgotten to specify the initial velocities of the masses. With some conditions, such as the outer mass having a velocity greater than that of the inner mass, and directed outside, there will not be any collisions.

 

[Gloyn]

We push the outer mass towards the inner, but velocity is not given, I just assumed it to be v. The inner mass is at rest before the first collision occurs.

 

[voko]

Are collisions with the wall counted as collisions "between the masses"? I think they should be, otherwise you only get two collisions with equal masses.

 

[Gloyn]

Yes, of course, sorry for not being specific. I meant the total count of collisions. All are elastic, no friction between any objects whatsoever. And by alpha I denoted that outer mass is alpha times bigger than inner mass.

 

[voko]

Here is the link to the full treatment of the problem by its author: http://ics.org.ru/doc?pdf=440