Complex Collisions
Suppose you have a frictionless surface with a similarly frictionless wall at one end. The other end goes off to infinity. You start with a block of 1,000,000,000,000 kg. mass, and a second block, nearer to the wall, of 1 kg. If the first block is launched at the second, assuming all collisions are perfectly elastic, how many collisions will take place?
Note: (Just comment on any ambiguity, reporting will not be necessary, just tell me what is wrong) Inspiration from 3Blue1Brown
The answer is 3141592653589.
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1 solution
As you told, if the masses of the striking bodies be equal , then there will be π number of collisions. Does it not sound absurd? It should be ⌊ 1 0 k × π ⌋ number of collisions.
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Umm, yes to avoid the half collision which does not make sense. But it will be floor of ⌊ 1 0 6 × π ⌋
1 0 6 × π = 3 1 4 1 5 9 2 . 6 5 3 5 9 ≈ 3 1 4 1 5 9 3 , so the answer given is wrong? Also why is π here? I did not understand.
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I think the problem changed to 12 digits instead of 6. Do you know @Alak Bhattacharya ? Anyways, my proof will not even be a fraction of what 3blue1brown explains. So better you watch this video: here
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I saw the video now. It states that it should be ≈ 1 0 0 d − 1 × π , d = Mass of big block .
A proof will be too lengthy. In short, if the mass of the large block is 1 0 k of the other one, there will be ⌊ 1 0 1 2 × π ⌋ number of collisions