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Poof and Foop

This BoingBoing article raises the following question: (see the link for a nice illustration)

If a can of compressed air is punctured and the escaping air blows to the right, the can will move to the left in a rocket-like fashion.
Now consider a vacuum can that is punctured. The air blows to the left as it enters the can. After the vacuum is filled the can will

a. be moving to the left
b. be moving to the right
c. not be moving

What a great question! To clarify the problem, let’s say we don’t just puncture the right wall, but remove it altogether. So at t=0 we have a vessel with a left wall but without a right wall, and with vacuum inside. What happens next is pretty straightforward: gas molecules are hitting the left wall from the outside of the vessel, thus exerting a force to the right. No molecules are hitting the vessel from the right however, so we have a net force to the right. Hence at t=0 the vessel accelerates to the right.

The acceleration continues as the gas fills the vessel, until the first molecules that entered the vessel begin hitting the left wall — this time from the inside. The process of filling the vessel takes a finite amount of time, and therefore at the end of it the vessel is moving to the right with velocity V.

But that’s not the end of the story. Suppose the vessel is now completely filled with gas from the outside, so we have the same particle density with the same velocity distribution hitting the left wall from the inside as from the outside, and the vessel as a whole is moving to the right. To understand what happens next, let’s look at matters from the vessel’s point of view.

So we’re sitting on the vessel, completely still, and watching particles hitting our wall from the left and from the right. Whenever a particle hits the wall, we see it completely reverse its velocity and momentum. The vessel absorbs the difference in momentum and accelerates. But from our point of view, the particles to our left are slower than the ones to our right! That’s because, in the original frame of reference, the vessel is moving to the right — i.e. toward the particles on the right, and away from the particles on the left. As a result, the particles on the left are depositing less momentum with each hit on the wall — in other words, they are exerting a weaker force than those particles on the right.

We’ve shown that F<F’, so we have a net force to the left — slowing the vessel down. The slowing down will continue until V=0, at which time F=F’ and the process is complete. So the answer is that at the end of the process, the vessel isn’t moving, but is positioned to the right of its original position.

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