Alice and Bob Marble Fight - Part 2

Alice and Bob are at either end of a (really long) track. Alice has 30 marbles and Bob has 20 marbles. They send all of their marbles towards each other in quick succession. Whenever 2 marbles collide they will just bounce back and start traveling in the opposite direction.

How many times do the marbles collide?


Assume that the marbles are of the same mass, and that the collisions are perfectly elastic. All marbles move at the same speed.

Image credit: Wikipedia Leonardo Poletto


The answer is 600.

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2 solutions

Ivan Koswara
Mar 1, 2015

Instead of colliding, let's assume that the marbles simply pass through each other, as the result is identical except for different labels on the marbles which we don't care.

But now there are trivially 30 20 30 \cdot 20 meetings: each of Alice's marble will meet each of Bob's marble exactly once. The number of meetings must be equal to the number of collisions, as a meeting is essentially a collision, thus there are 30 20 = 600 30 \cdot 20 = \boxed{600} collisions.

Exactly how I did it :) Elegant problem, elegant solution!

Dylan Pentland - 6 years, 3 months ago

I thought I would be the first one to post a solution with the pass-through approach and conservation of linear momentum!

Aalap Shah - 6 years, 3 months ago

Nice solution... Loved it !

Vaibhav Prasad - 6 years, 3 months ago

Great solution! The "pass-through" approach is the one that I'm thinking of.

Another followup question would be if their marbles are coloured, and then to ask how many coloured marbles did they get.

Calvin Lin Staff - 6 years, 3 months ago

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According to conservation of momentum: after all collisions have finished there must still be 30 marbles travelling away from Alice and 20 travelling away from Bob. Since the marbles don't change order, Bob gets his 20 back and then 10 of Alice's.

Matt McNabb - 6 years, 3 months ago
Ramesh Goenka
Feb 28, 2015

Can you explain why? Is there a simple solution / approach of thinking about this problem?

Calvin Lin Staff - 6 years, 3 months ago

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Indeed a simple approach ... !! the first term 'n' corresponds to no. of times the first ball released by both the persons will collide.. !! the second term 'n(m-n)' corresponds to the extra balls i.e (m-n balls ) that will collide against each other so that they travel towards the person having less no. of balls ( in this case its n) which is nothing but an extension to part 1 of this problem. And the last term depicts the no. of times the balls changes direction apart from extra balls. Another approach can be a recursive one when the problem of 30-20 balls can be reduced to 29-19, 28-18....... 11-1balls problem.

Ramesh Goenka - 6 years, 3 months ago

The equation you give simplifies to m n mn which agrees with Ivan Koswara's solution.

Siddhartha Srivastava - 6 years, 3 months ago

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