The number of collisions in this setup, often referred to as the "pi-collision problem" or "digit dynamics," tends to be around 3140. This intriguing result is related to the mathematical constant π (pi).

m1 = m2 = 1, collisionCounter = 3 link

 

m1 = m2 = 100, collisionCounter = 31 link

m1 = m2 = 10000, collisionCounter = 314 link

"pi-collision problem"

"digit dynamics," tends to be around 314. This intriguing result is related to the mathematical constant π (pi).

The setup involves two blocks of significantly different masses (here, 1 kg and 10000 kg) sliding on a frictionless surface. One block starts moving towards the other, and they collide elastically. The number of collisions before they come to rest or one block moves away is proportional to the digits of π.

For two blocks with masses in the ratio 1:100ⁿ, the number of collisions will be approximately equal to the first n+1 digits of π. In this case, with n = 3 (since 10000 = 100²), the number of collisions is close to the first 3 digits of π, which is 314.

This remarkable connection arises from the way the collisions and velocities behave, involving iterative calculations that mimic the series expansion of π.

How to get to 1000000 kg? and collisionCounter = 3141 ?

Since this is a computationally intensive during the collisions between left wall and 2 balls, perhaps slowing down the velocities of ball2 in x direction to -0.0001 might allow to compute the correct number of collisions before it becomes unstable? Good luck! and let me know if it works! Enjoy!

This remarkable connection arises from the way the collisions and velocities behave, involving iterative calculations that mimic the series expansion of π.