eli5: why do colliding blocks (on a frictionless surface with no resistance of any kind) compute pi?

256 views

I was watching a video; the amount of collisions made between two colliding blocks on a frictionless surface with walls on either side that have infinite mass always equals the digits of pi. So say one block is 100kg (block A) and the other is 1kg (block B) and we are assuming perfectly elastic collisions, the total amount of collisions before block Ahits the opposite wall would be 31. We keep increasing block A’s mass and the numbers go up: ex. 314 at 10 000kg, 3141 at 1 000 000kg.

After that video I tried to understand *why* this happens but I am no mathematics expert. If we’re being honest here “elastic collisions” was a stretch for me haha I had to reach back to my high school physics memories.

In: 323

4 Answers

Anonymous 0 Comments

presumably you watched this video https://www.youtube.com/watch?v=HEfHFsfGXjs

There is a follow up video that covers why this is the case https://www.youtube.com/watch?v=jsYwFizhncE

But the TLDW of it is, if you write out the physics equations you end up with 2 equations, one of them is a circle, and 1 of them is a line, the way these 2 interact is by bouncing the line along a specific distance down the circle. So in the limit, this is basically asking for half the circumference of the circle, which is 2PiR, R is 1, and we are asking for half, so its just pi

You are viewing 1 out of 4 answers, click here to view all answers.