Ants on a pole

Classical Mechanics Level 1

33 ants are placed on a pole of length $\SI{1.0}{\meter}.$ They each have negligible length, and they each crawl with a velocity of $\SI[per-mode=symbol]{1}{\centi\meter\per\second}.$ If two ants meet head on, they both turn around and immediately continue crawling.

If an ant reaches either end of the pole, the ant will drop off the pole. What is the longest possible time (in seconds) until all the ants drop off the pole?

The answer is 100.

2 solutions

Laszlo Mihaly
Aug 4, 2017

If we have one ant, the longest possible time is 100 seconds. That is needed if the ant starts at one end of the pole, and crawls to the other end. This remains the answer for any number of ants.

To see that, let us look at what happens when two ants meet: ant #1 traveling to the right and ant #2 traveling to the left. After the meeting, ant #2 will be traveling to the right and ant #1 will go to the left. Notice, however, that it does not matter how did we number or name the ants, there is still one ant going to the left and another one going to the right. In effect the situation is the same as if the ants would just crawl over each other.

Beautiful problem! :-) There is a great analogy between this and the basic derivation for the pressure of an ideal gas in the kinetic theory of gases. The ants here like the idealized gas molecules are executing perfectly elastic collisions! I always thought of this interaction as if the molecules were going right through each other (There is no distinction between them). So the problem boils down to a single ant starting from one end and going right through all other ants on its way till it falls off the other end. :-)

Ujjwal Rane - 3 years, 10 months ago

Can we see it as elastic collisions between balls of equal masses? In that case, the velocities exchange just after the collision.

Rohit Gupta - 3 years, 10 months ago

But in this case, if you apply the law of momentum conservation, then nothing will be changed due to they have negligible mass & second is that they have same speed with opposite direction

Shaishav Asmaniwala - 3 years, 10 months ago

In elastic collisions with same masses, the velocities of the colliding objects get exchanged. This is valid if both of them are very light or very heavy as well.

Rohit Gupta - 3 years, 10 months ago

What an Idea !

Swapan Das - 3 years, 9 months ago

Lucas Viana Reis
Aug 14, 2017

For any number of ants on the pole, if one of the ants sit on an end, it will always take 100s before the last ant falls.

Let's look at the ant on the end of the pole. Now, it's possible to think of collisions where the speed of the two objects is equal and conserved simply as if they are passing through each other, as there is no distinction between one before and the other after after the collision. So, no matter how many ants are there in the rest of the pole, this ant will have to cross the entire pole before it falls, and starting in any other position is worse. So, the time it takes to cross the pole is (1m) / (1cm/s) = (100cm) / (1cm/s) = 100s

What if all the ants on the left half are moving towards left and all the ants in the right half are moving towards the right? Then all the ants will get off the pipe in a maximum of 50 sec, isn't it?

Rohit Gupta - 3 years, 9 months ago

That's why the question asks for the longest possible time.

Dan Hewitt - 3 years, 9 months ago

No matter with how many ants its done, and where we place the ants. The only condition is that one of the ants is placed in the edge towards inside. Then it will always take 100 seconds.

Agustin B - 3 years, 9 months ago

The question didn't mention that the ants were placed at the same time, so if you place an ant the border, facing the first ant, everytime the first ant is about to complete crossing the pole, the correct answer should be 3300s.

R L - 3 years, 9 months ago

Ants on a pole

The answer is 100.

2 solutions

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