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Originally Posted by OngBonga
I kinda followed along with that trignonmetry. I dunno if I was imagining correctly, but I pictured a circle with two right triangles inside, sharing a hypotenuse. Triangle A has 16x the area that triangle B has, and with each collision, triangle B increases in size and triangle A decreases. After pi collisions, the areas of the trianlges are the same.
As we scale up, and the mass of the ball (area of the triangle relative to the other) increases hundred-fold, the number of collisions until we have equality, is pi increased ten-fold.
Fuck knows why, but it's cool.
The area of the triangles is a red herring. The side-lengths (which vary while the hypotenuse length does not) is all that matters to motivate the construction of a circle.
This is enough to invoke pi into this discussion without any ambiguity as to why it arises.
Draw a line whose length is the total KE of the 2 masses. Since one of the masses starts at rest, this is simply a line of length (1/2)m1_i*v1_i^2.
Call the ends of the line A and B, so that AB = KE_tot.
Each of those points is always an endpoint for exactly 1 side of any triangle for which the line AB is a hypotenuse.
For reference sake, call the 3rd vertex of the triangle C.
The locus of all points C which fit this construction creates a circle with the hypotenuse AB as a diameter.
I.e. a circle whose center is at the midpoint of AB and whose radius is AB/2 describes all the combinations of KE in m1 and m2 which share the same total KE.
In our problem, call the KE of mass 1 the length of side CA, and the KE of mass 2 the length of side CB.
In the initial condition, our "triangle" has CA = AB, and CB = 0, so it's not really a triangle, as such.
This tells us that all of the KE is contained in just one mass, mass 1, and that mass 2 is at rest with no KE.
After the first collision, CA is shorter, and CB is longer, but still under the constraint that CA^2 + CB^2 is the same is was before the collision.
I.e. the total KE is conserved, which is to say, the hypotenuse formed after the collision is the same length AB.
This is true again after the 2nd collision and all collisions.
Using this picture, the "allowed" relations between the energies of the 2 masses sweep out a semi-circle.
(It's a semi-circle because we're only talking about cons'n of E and we need to invoke cons'n of momentum to avoid squaring out the negative numbers.
When we invoke cons'n of momentum, it becomes a full circle, with negative velocities indicating motion in the opposite direction.)
All of this leads to a picture of a circle with a diameter drawn across and the lengths CA and CB which satisfy cons'n of energy all put point C on that circle.
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It's worth pointing out in all of this that we're ignoring the collisions between the smaller mass and the wall (infinite mass, if you like). This is justified, since the effect of the wall being perfectly rigid (infinite mass) is to never accelerate due to collisions with anything, and all of the KE in the small ball prior to the collision is present in the small ball after the collision. The ONLY outcome of the collisions with the small ball and the wall is to reverse the direction of the small ball, without changing its mass or speed.
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