Will a can filled with vacuum move when we let in air? 
This picture is from L.C.Epstein's book Thinking Physics. The upper can is filled with compressed air, and, when an opening is made on the right, the air comes out and the can shoots left. The question is what happens to the lower can, filled with vacuum, when we similarly make an opening. Does it move left - right - not at all?
Epstein says that the lower can doesn't move at all, "except for a
momentarily slight oscillation about the center of mass". I'm not sure I understand this. The explanation is that the air incoming into the bottle provides force on the left inner wall to compensate for the lack of force on the opening, and this balances the force on the left outer wall from the outer air. Which seems convincing, but opens a path to more questions:


*

*Shouldn't the can still start moving from the moment we make the opening and until the air pressure inside the can is equalized with the outside air?

*If that in fact happens, why would it stop and return ("a momentary slight oscillation about the center of mass") and not simply continue moving right with the constant velocity it's acquired?

 A: The easiest way to answer this problem
is by thinking about conservation of momentum.
Consider the still closed can with vacuum inside, and the air outside.

The can is at rest, hence it has momentum $\vec{p}_\text{can} = \vec{0}$.
The air is at rest too, hence it has momentum $\vec{p}_\text{air} = \vec{0}$.
Therefore total momentum is 
$$\vec{p}_{\text{total}} = \vec{p}_{\text{can}} + \vec{p}_{\text{air}} = \vec{0}. \tag{1}$$

Now we open the hole in the can.

No matter what complicated things will happen now,
no external forces are involved.
Only internal forces are involved (air molecules colliding with each other,
and air molecules colliding with the can).
Therefore we know, the total momentum will remain zero as before in (1):
$$\vec{p}_{\text{total}} = \vec{p}_{\text{can}} + \vec{p}_{\text{air}} = \vec{0} \tag{2}$$
We know that air will shoot into the can from right to left, that means
$\vec{p}_\text{air}$ will point to the left.
From (2) we conclude, the can must get a momentum in the opposite direction,
i.e. to the right:
$$\vec{p}_\text{can} = - \vec{p}_\text{air}$$

We can also further predict what will happen when the
air settles down inside the can.
So let's assume after a while the air has come to rest.
This obviously means we have $\vec{p}_\text{air} = \vec{0}$ again.
With a similar reasoning like above (conservation of total momentum)
we can conclude: $\vec{p}_\text{can} = \vec{0}$,
which means the can will come to rest, too.
But because the can was moving to the right for a while,
its final position will be displaced to the right by a certain distance.

A: If the pressure difference between inside and outside is the same (but opposite) then the force on the can in both situations is equal and opposite. If the geometry is the same then the pressure difference evolves in the same way but with opposite sign. The motion of the can, opposite, is only different due to difference in the amount of gas resistance.
So I don't fully agree with Epstein on this. 
A: 
  
*
  
*Shouldn't the can still start moving from the moment we make the opening and until the air pressure inside the can is equalized with the outside air?
  

Indeed it will, but that takes very little time.  


  
*If that in fact happens, why would it stop and return ("a momentary slight oscillation about the center of mass") and not simply continue moving right with the constant velocity it's acquired?
  

In the case where the air leaves the can, after the air is ejected, it keeps the momentum.  So the can balances that permanent momentum change by moving in the other direction.
In the case where the air enters the can, the air inside is quickly brought to a halt.  There is no quantity of air moving to the left that the can will balance by moving to the right.  Instead, the left wall of the can couples these two masses together so that they each stop the other.
