Why is pi involved with this puzzle?

LifeIn

Well-known member
I ran across this math/physics puzzle on one of my favorite YouTube channels, 3Blue1Brown. The real surprise is why the answer involves pi even though it is not about circles. The puzzle starts like this:

Imagine a 1 kg block resting on a level frictionless surface a short distance away from an immovable wall on one side and another block on the other side. The other block is sliding toward the first block with a certain velocity. When the moving block collides with the stationary block, the collision is totally elastic, conserving both total momentum and total energy. The moving block will stop dead and the first block will begin moving toward the wall at the same velocity. That block continues until it bounces off the wall and returns to hit the other block. The other block begins moving away from everything forever and there are no more collisions. Now, the only relevant parameter we want to consider is the total number of collisions. In this case it was 3 collisions, right? (Stop and review to make sure.) Now we change the setup. Instead of having both masses be the same, suppose the initial moving block is 100 kg. Now when the 100 kg block collides with the 1 kg block, it looses only a small amount of speed. The small block speeds towards the wall, bounces off and returns to hit the 100 kg block again, slowing it down some more, but still only slightly. The 1 kg bounces off the wall a second time and returns to hit the big block once more. This continues, with both blocks get very close to the wall, with a lot of collisions in a short span of time, and then the big block finally gets turned around and is heading away from everything and there are no more collisions. It may be hard to work out the sequence of collisions and the velocities at each collision, but if you do, you will find there are exactly 31 collisions. I did promise pi, right? Do you see it yet? The in the first experiment there were 3 collisions. In the second one there are 31 collision. Any guess as to what the next experiment will be? Consider the moving block to be 10000 kg. It turns out it will take 314 collisions with a 1 kg block bouncing of a wall to turn that block around. I'll give you one more. If the moving block is 1000000 kg, it will take 3141 collisions with the 1 kg block bouncing off the wall to before they go their separate ways. That's right. When the bigger block is 100^n times as big as the little block there will be a number of collisions equal to the first (n+1) digits of pi. Is that freaky or what? There are no circles in the problem. Here is a hint as to why pi might be involved: At each collision, the total energy is conserved. Energy is (1/2)m v^2. With two masses and two velocities, the total energy is proportional to

m1 v1^2 + m2 v2^2 = constant

That should look a little like the equation of an ellipse. For a complete explainer by 3Blue1Brown
And if you want the details of the solution, they are
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I ran across this math/physics puzzle on one of my favorite YouTube channels, 3Blue1Brown. The real surprise is why the answer involves pi even though it is not about circles. The puzzle starts like this:

Imagine a 1 kg block resting on a level frictionless surface a short distance away from an immovable wall on one side and another block on the other side. The other block is sliding toward the first block with a certain velocity. When the moving block collides with the stationary block, the collision is totally elastic, conserving both total momentum and total energy. The moving block will stop dead and the first block will begin moving toward the wall at the same velocity. That block continues until it bounces off the wall and returns to hit the other block. The other block begins moving away from everything forever and there are no more collisions. Now, the only relevant parameter we want to consider is the total number of collisions. In this case it was 3 collisions, right? (Stop and review to make sure.) Now we change the setup. Instead of having both masses be the same, suppose the initial moving block is 100 kg. Now when the 100 kg block collides with the 1 kg block, it looses only a small amount of speed. The small block speeds towards the wall, bounces off and returns to hit the 100 kg block again, slowing it down some more, but still only slightly. The 1 kg bounces off the wall a second time and returns to hit the big block once more. This continues, with both blocks get very close to the wall, with a lot of collisions in a short span of time, and then the big block finally gets turned around and is heading away from everything and there are no more collisions. It may be hard to work out the sequence of collisions and the velocities at each collision, but if you do, you will find there are exactly 31 collisions. I did promise pi, right? Do you see it yet? The in the first experiment there were 3 collisions. In the second one there are 31 collision. Any guess as to what the next experiment will be? Consider the moving block to be 10000 kg. It turns out it will take 314 collisions with a 1 kg block bouncing of a wall to turn that block around. I'll give you one more. If the moving block is 1000000 kg, it will take 3141 collisions with the 1 kg block bouncing off the wall to before they go their separate ways. That's right. When the bigger block is 100^n times as big as the little block there will be a number of collisions equal to the first (n+1) digits of pi. Is that freaky or what? There are no circles in the problem. Here is a hint as to why pi might be involved: At each collision, the total energy is conserved. Energy is (1/2)m v^2. With two masses and two velocities, the total energy is proportional to

m1 v1^2 + m2 v2^2 = constant

....

At first glance I would have applied a solution using simple harmonic motion to analyze the oscillatory nature of the collisions. I also expect an initial damped transitory term characterized by a peak from the initial impact followed by a damped steady state wave represented by each collision until the motion is finished. Conservation of energy is expected even when frictionless surfaces are considered.

After this statement above, I'm going to watch the video.

Checking....

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