simple really .... if the airs getting pulled out then the can will want to pull in that air so
        __
air <-()__) the can will need the air back in it so it will suck the same way as soon as the - preshure is released (sorry cant spell im spanish)
anyways cann would be moving left (sorry if this is wrong i got linked from a difrent site)

the can will not move because of the probability of the cans weight divided by the pressure of the air comming in the can.

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i think its got to do wth the pressure acting on the can. when its not punctured, that time pressure is the same all thruout its soides. when punctured and as long as the air is getting in the can will move to rite with a acceleration, n there after wneh the air is completely in it will move with a constant velocity. thats coz 1ce the air is in pressure will be again same throughtout, as earlier n no unbalanced force acting. i will describe the motion to rite side as the pressure on left side is more than on right side coz of the tiny hole on the rite side ( pressure meaning air pressure)
 if found ne thing wrong wth my reasoning plz inform me. i wd learn from it.. sagar_gambhir@rediffmail.com

I assume that there is friction--a lot of friction! Therefore, by the time the can fills with air, it is no longer moving.

Consider the can and the air around it as the system.

The crux is that since their is no ground friction , there is no HORIZONTAL external force on this system.Hence the position of the COM of the system remains on the same vertical line at all time.

Now, as the air starts to enter the hole the COM OF THE AIR shifts to the left and hence the can (being a rigid body) should shift to the right. BUT, when the can is completely filled, the COM of the air molecules would have returned to thir starting position, and hence the COM of the can must also have returned to its initial position.

All said, i believe the can would move to the right initially but at some point turn back and return to its original position AND stay there.

The physics of this solution seems fine to me , but there could exist a flaw. Would someone please do this experiment and write what actually happens?

Edited by author 01-20-2007 05:10 AM
The amount of motion depends on the mass of the can, air pressure, and size of the hole. (assuming no friction)

How far would the can move?

I think the air's momentum equals the can's momentum. I think the can moves right while filling and then stops from the air colliding with the left wall.

Don't forget that the can does not suck in air. The external air pressure forces air into the can.

The air is accelerated by constant pressure from the right. Think of a plunger with some mass (a frictionless disk that starts in contact with the right side of the can) getting pushed in by the incoming air. It mass gets accellerated by the incoming air (air pressure means force on the disk -- Force= PressurexArea) until it hits the left wall. The same thing happens to the air when the disk is not present.

Meanwhile the imbalance of left and right forces on the can (due to the area of the hole) causes the can to accellerate right. Note that the air forced into the can is through a hole whose size determines how fast the can moves right. It's not hard to imagine that the energy imparted to the air is the same as the energy imparted to the can since both are caused by the same hole. Then they meet and stop.

BTW Bob, my sentence about heat amd momentum was totally off.

Bob,

There is no problem with the air continuing to move leftwards - and there is no need for some kind of 'pulse' to do so either, as diffusion will result in some kind of a bulk motion of the air as a whole. Imagine throwing a pebble into the sea - eventually the pebble will transfer all of its momentum to the water, get tired, and flop to the bottom - and the momentum which it once had will be shared amongst many water molecules in some complex fashion. Same thing here, so I don't think there is anything to worry about here. Also, there is nothing wrong with air molecules going off to infinity - that's exactly what it would do if there were nothing else in the way. If you are an astronaut doing a space walk in the middle of nowhere, and happen to be holding a cricket ball which you throw in some direction, it will just keep going off to infinity, while you will go spinning off in the opposite direction (rather more slowly, I should imagine, unless you have been on a diet) to 'negative' infinity. There is nothing mysterious about that.

And of course the air outside must move - at the very least it must replace the air that gets sucked into the can. I think the motion would be very complex - and I agree that there may not be much net momentum of the air whichever way the can ends up going, but there most certainly will be motion, and lots of it. There is plenty of potential energy to distribute after all, in various forms, including kinetic. Note that even heat will eventually be converted into kinetic energy through dissipation.

Your momentum argument seems incorrect - when the air enters the can, it will do so with much greater velocity than the the can's rightward motion, so I don't think you can naively reach the conclusion that the air's momentum is less than the can's (though admittedly I do think that it probably is).

meBigGuy - you argue that the air molecules entering the can first will be 'pushed' to greater velocities by air molecules entering after them - this is true, but this increase in momentum will be exactly balanced by the loss in momentum of the molecule that hit it, so the total momentum imparted remains the same.

Bob - I doubt the can would reach as far as Fiji. The polar bears and penguins can be well separated if you like, but would you prefer that they starved to death rather than let nature take its course. Even cute and furry animals get eaten by predators - it is a fact of life. Of course we can make a conscious choice way one way or the other if we so wish as there are alternatives, but you wouldn't deny me the pleasure of eating the odd penguin now and then, would you? I am sure penguin korma would taste quite nice. And what about starfish - do they not have rights too? Or are they not cute enough?

Tyler - The can will certainly move - don't know for sure which way (probably rightwards, based upon earlier arguments), and be reassured that there is no problem with the air moving leftwards.

I was wrong, it will not move, please disregard all previous posts, its been fun.

Edited by author 01-18-2007 04:24 AM
Hello?

Energy can convert to whatever form it likes, but momentum is momentum. If net rightward motion develops in any system, net leftward motion must develop outside of that system. This can only be caused by some force of interaction (both ways), or by some net flow of matter, across the system boundary. Apologies if I'm misinterpreting you, but if you can't understand and agree with that then we're a bit stuck, no?

Edited by author 01-18-2007 01:11 AM
Tyler

I am not sure how you can assume the air molecules rebound elastically off the left side of the can. I don't know exactly what they do, just don't understand the basis for your assumption.

So, you are saying that the plunger analogy is not valid because the nature of the entering air column can't be compared to a plunger.

Think about shooting a pressure stream of air at a can. It shoots left. That is what is happening here.
 
Your basic assumption is that the nature of entering air causes energy loss due to turbulence and reflections. Therefore the air's energy gained by the leftward pressure is not applied to the left side of the can. And, the difference in area between the left and right means the external rightward pressure is no longer balanced, so the can moves right after filling.

Bob's assumption that something moving right means something else must be moving left need not hold if energy is converted to heat.

But, I think the collision is much less elastic than you think. One data point is the post that described flouresent tubes shooting left after their right ends were broken off.

So, how can you prove your assumption of an elastic collision imparting little or no leftward momentum?

Edited by author 01-17-2007 04:52 PM
tyler -

Near-elastic collisions mean minimum kinetic energy loss, but maximum momentum transfer. Which means the faster-than-average, denser-than-average stream going into the can packs more punch than anything else going on. I see no reason why they shouldn't balance - I think you're underestimating it.

all -

I think I've got an argument for it not being in motion at all. See what you think.

I've been thinking about a system consisting of the can, its contents, and all the air to the right of it (think of the space that would be mapped out by an infinite translation of the can to the right). If the can is going to keep move right, something outside of this system has got to keep moving with equal momentum to the left. (Air to the right of the can can't keep moving left without leaving this system, unless a strange leftward pulse shuffles rightward to infinity!) Bear in mind that any air moving left that enters or hits the can will give up its leftward momentum to the can, not carry it away as is required. The air is (or at least may well be) a lot lighter than the can, so the movement of air to the left must be either much faster or much greater in volume than the movement of the can.

I can't see why any air outside of this system should move a great deal at all, especially not to the left. The volume of air required to fill the space where the can was when it moves is tiny in comparison to the volume of the can, unless the can is very light indeed - so that doesn't do the job.*

Is anyone following me here?

If you can agree that for the can to end up with motion to the right, there must be a leftward flow of air around the can (not to the right of it) that is much greater than the space left by the can as it moves, then would you agree that, unless there's some identifiable physical cause for such a movement, we can infer that the can does NOT continue in motion once full?

I'm thinking so. I'm thinking that there is no physical reason, i.e. no force or pressure gradient, that would cause air to start flowing left, and to continue to flow left without entering this system, as a result of the puncture appearing. The boundaries of this system are potenitally at a considerable distance from the hole.

So I don't think it can sustain any motion at all. If it does, it'll be on much, much tinier scale to the 'poof' case. It may end up drifting very slowly left after all (post 394).

Away from Greenwich.
Towards Fiji.


*Consider the ratio of the mass of the can to the mass of the air in it when full. The momentum of air moving back to fill the space as the can moves forward is less than the momentum of the can moving forward by the same ratio. So if the can moves right, this effect alone cannot make up the missing leftward momentum - we need some other reason for the flow of air.

Yes, Bob that force to the left you are talking about is the slowly increasing force of pressure (P=F/A). It is less than the force of pressure outside. When the initial jet of air bounces of the back wall this is the beginning of the pressure rise and it is very small. This initial jet does not stop and bounce the back wall off of it, giving it all of its energy, it rebounds elastically, with very little momentum loss. This tiny momentum loss the can feels as the tiny initial rise in internal pressure. This rebounding air is now colliding with all the incoming air, the situation becomes very chaotic very quickly. Because of all the random collisions the incoming jet sees, its force is essentially acting simultaneously at all times on all of the inside surface areas of the can. This is the internal pressure of the can. It is always less than the force felt on the outside until the pressures are the same. During this time, that inbalance of forces is what accelerates the can, it will also accelerate slightly the random air molecules that are hitting the back of the can during that time. This is how the can and the air inside will both be travelling together, the air is chaotic in all directions but the bulk motion is with the can. Now when the pressures equalize, I cannot see any unbalanced forces that are stopping motion other than air friction on the outside of the can.

Sab, I think it is incorrect to say that the forces acting on the inside left side of that can are the same is the same as the force that was operating on the right side of the can before the puncture. The air entering the can increases in velocity as it enters and travels (it has a pressure pushing it to the left) and then it actually builds momentary pressure against the left side of the can when it hits (due to inertia gained). I think this force will cause the rightward motion of the can to stop, much as in my space analogy. Also, consider my earlier plunger analogy.

If, for some reason that I don't understand yet, the momentum increase in the entering air column and the momentum increase of the can are not equal (excluding friction and turbulance), then there will be a resultant motion after filling. I think the increases are the same because the pressure and area in both cases are equal. But, there could be a subtlety of air entering a vacumme through an orfice that I don't understand.

Unfortunately, the polar bear dies due to Artic temperature increases.

Edited by author 01-16-2007 06:48 PM
tyler - Must schmust. Air fills in space and provides a force to the left. There's more of it per square inch through the hole than the molecules hitting the surface, and it's going in faster. If you've got a reason to say it's still not big enough to balance the forces to the right, then give it. If you haven't, then there ain't no must about it.

Still, I think you're right - I think it does move right. Even if the book says it doesn't (as Paul seem to imply before he vanished). But I don't see any good reason yet.

But I'll give you the pressure thing - gas pressure is completely and utterly caused by momentum, I can't deny that - so long as you promise not to use thermodynamic pressure in calculations for wildly anisotropic streams into a vacuum.

Sab - you ain't causing no trouble.

Don't forget, though, that 'pressure' (which is completely and utterly caused by momentum) is the sum of the rate of change of momentum per unit area of gas molecules hitting the surface PLUS those rebounding from the surface. The molecules come in from all angles at the can whether there's a hole or not; but when there's a hole, they don't rebound as much from the right. That's why I thought the can moves right (post 341)

Question is: how much more gas goes through the hole than would have hit the surface if the hole hadn't been made, and how fast does it go through? Paul (who vanished) (post 384) convinced me that both are greater. I done da equations like (post 390). But I don't see why a=b and c=d.

As you say, the chaos inside and around doesn't matter if we look at momentum flowing in and out of the system.

Penguins? At the North Pole? With the polar bears? Sir, you have a cruel streak.
I do hope they are not evacuated penguins, as a polar bears could puncture one in a split-second. Before you could say J.

Double quadruple half whatever it's still proportional. An increase in pressure is an increase in the aggregate momentum of all the molecules and vice versa. We can monitor the change in average momentum of the molecules by measuring the pressure or temperature, which are related by the ideal gas law as you say.
In summary,
Air moves left into can : Can must move right : Air is filling in space left behind moving can (this counts as continued momentum of air to left) : can must continue right until friction wins.

Sorry, I didn't mean to cause any trouble. Forget about the disappearing mass of the puncture. Regarding timescales, for any reasonable size of puncture, I would guess that the can would fill with air before you can say 'Jack Robinson' or even just 'J', because of the immense pressure of the Earth's atmosphere. The fact that the can has not already been crushed means that it must be made of some seriously strong (but according to you, very light!) stuff. [I wrote this message before I read your latest one, s forgive any repetitions/apparent plagiarism].

Note also that the air entering the can will impart the same (leftwards) momentum on the left side of the can as it would have done on the right side had the puncture not been there. The only effect of the puncture is to generate a small delay in momentum transfer as the air has to now travel the distance from the puncture to the left side of the can before hitting it. As the can fills with air, the average delay in momentum transfer will tend to zero. It is during this short period of time that the can will have a net rightwards force acting on it that will cause it to accelerate in that direction. I don't see any way that the force due to the air entering the puncture could be any greater than it would have been had the puncture not been there at all, which means to say that the net rightwards force (neglecting friction) will decrease monotonically to zero - so in the frictionless case, I would expect the can to end up moving rightwards with constant velocity (and of course the surrounding air to have a net negative velocity). [The reason I thought a spherically symmetric system would be appropriate for the thought experiment was because it is easier to visualise the air going round and round with constant (angular) velocity].

Okay, now I have read your post, I would say that there is probably no way that v2 could be negative - the total leftward force on the can (ignoring friction) will always be less than or equal to the rightward force on the can, and friction being what it is, cannot only slow the can down, and certainly cannot make it change direction. (Even though the velocity distribution of the air molecules entering the puncture is different from the bulk, I don't see how this can change/increase the momentum imparted to the can). Even though the air in the can is in a state of chaos, momentum must be conserved, so for our purposes, the chaotic motion does not matter.

And yes, how can one have a frictionless ice rink without polar bears? Penguins too.

Edited by author 01-16-2007 05:01 PM
Ok Sab,

I'm not with you on the centre of gravity shift and dissolving with acid thing - it complicates things further: what's happened to the mass, where's it moved to and how? A puncture can take place without any shift of the centre of gravity, and I propose that we take it as such. Generally if something punctures, like a can, mass doesn't disappear. If you'd like an image, imagine a little pair of vacuum-sealed lift doors opening, very quickly, on a small patch on the side of the can. As they go in opposite directions, there's no shift of centre of gravity, no jolt or torque or anything. But wouldn't it be better just to imagine a hole appearing, without anything else changing, and keep it as simple as possible?

But that aside, I seem to totally agree with everything in your post, including the lack of a conclusion!

The can will certainly accelerate right at the very beginning (before the first molecules have crossed the can), reaching some velocity v1 to the right at time t1 when the first (fastest) molecules strike.

The force will rapidly decrease as the stream of molecules entering the can begins to pile into the wall and (later) into the air contained in the can. Whether or not this force will become negative, causing the can to decelerate until it reaches a standstill by the time the can is full, we don't seem to have a definitive answer to. So all we've got is that there are several forces acting during the filling of the can, and the rightward velocity approaches some value v2 as the can approaches being full, say at t2.

We know the can - and the air - will have stopped at the very end (due to air resistance and the self-damping of turbulence). Say at t3.

(Both t2 and t3 are subject to an arbitrary definition of when an exponential process can be said to 'end' - but we can define them precisely if we need to; and I can explain why we don't need to if you need me to.)

However, what we want to know is whether v2 is (a) negative, (b) positive or (c) zero.

What would be the 'force' applied by a stream of molecules, moving slightly faster than the mean speed in the gas and entering the hole at a rather higher flux (numbers per second per unit area) than they would normally strike the sides, hitting the inside back of the can? It's not described well by the concept of gas pressure, which is a parameter of semi-equilibrium states. Later, as the can starts to fill, what happens when the slowing stream piles into the gas that's already there? What happens is chaotic - turbulence and a big mess. Inside the can is chaos.

Here then (for tyler and anyone else interested in forces) are what I think are the main forces acting during filling:

F1: a constant force externally on the left, pushing right, which is simply the atmospheric pressure over the area of the hole on the opposite side.
F2: an unknown force internally pushing left as the air streams in, which varies with time in an unknown way but settles to a value equal and opposite to F1 when the can is full.
F3: an unknown reduction in the leftward force externally on the left - the pressure acting in the vicinity of the hole - which also contributes in the right hand direction, and decreases to zero when the can is full.
F4: air resistance, acting in the opposite direction to any motion, and much smaller than the above (during filling) unless the hole is very tiny. We can ignore this before t2 (unless everything else cancels perfectly to zero, in which case we can bring it back in.)

These forces act over time. Integrating each one over time will give the impulse of that force; when we know that we can see how the impulses of F2 and F3 (which act to the left) compare with the impulse of F1 (which acts to the right), and that'll tell us whether or not the can is moving at time t2.

It's all a bit difficult.

Alternatively, we can use the fact that impulse equals change in momentum, and look instead at the flow of momentum into and out of the system. It's not mysterious, it's exactly the same thing, and it makes a lot more sense. All we need is some much more elementary properties of the flow of air through the hole and externally in the vicinity of the hole during the filling of the can.

I just don't know how to do it, as I don't know what form a(t), b(t), c(t) and d(t) will take (ref post 390).

So... whaddya think?

Sab - are there any polar bears in your model? I like polar bears.

Ah - I see where you're coming from tyler. Yes, pressure is defined in terms of force per unit area, and force is defined as rate of transfer of momentum, so they are related.

The reason they are not proportional, though, is that if you increase the momentum of molecules, you also increase their speed, which means more of them hit per second. If dt is the mean time between collisions in which a mean momentum of dD is transferred, doubling momentum means we double dD and we half dt, and so, all else being equal, the pressure quadruples.

Hence pressure ends up being proportional to the mean kinetic energy, which in turn is proportional to the temperature, as per the old PV=nRT thing.

Bob, pressure and momentum are directly related:
To avoid confusion I will use D to represent momentum instead of the convetional P, which will represent pressure.
P=pressure
dD=change in momentum
F=force
dt=change in time
A=area

F=dD/dt Force equals the change in momentum over time
P=F/A Pressure equals a force acting over an area

So if F=PA and F=Dd/dt
Then PA=dD/dt
Thus change in momentum directly proportional to pressure.

Sab, I think you touched on an important point. Many have argued that the can moves and then needs to stop to satisfy a momentum balance. However, when the can is in motion, there is a equal diplacement of air flowing around and filling up the space behind it as it moves, thus still satisfying a momentum balance. This is why objects in motion tend to stay in motion (in a frictionless universe) even in fluids. So if you say the can moves, it does not suddenly stop once the air stops flowing into it. There is still air flow around and behind the can as it moves to satisfy the balance. Then only friction will stop it not some mysterious conservation of momentum force.

Hello Sab!

First quick read through your post and I haven't a clue what you're on about at all, but I sort of expected that. Greenwich? I'll have a proper look this evening.

mBG (& Sab) - Re friction / air resistance / etc. please see my response to Huge in post 365 re timescales. The problem as stated has only a can and a presumably infinite extent of air (as no pressure gradients, boundaries or topological oddities are specified), and asks only whether or not the can is moving once it's filled.

We're free to assume there is no gravity if we like (which makes sense considering there's nothing else in the system) - and indeed it's silly not to, as if we don't, the can will blatantly be moving down. So the ice rink is useful as an image, and inasmuch as friction is similar to air resistance anyway - but without losing sight of the problem as posed.

I'm getting a little lost.

When you start considering surface friction and turbulance the problem becomes very complex. Energy is lost (in the form of heat?) from both of these.

There is probably more turbulance inside the can than in the left side air due to the can moving, so this would probably cause the can to tend to move right after filling since the incoming air would lose momentum.

The surface friction would cause the can to move right less while filling, so would tend to cause the can move leftward after filling since the internal air (turbulant though it may be) would travel further and gain more momentum.

So, which is more significant? I think the problem could be adjusted to make either direction correct.

So, what if we only consider a frictionless surface and a small enough hole to avoid turbulance. Is it agreed, in that ideal system, that the can would initially move right and then stop? Or, am I missing something.

Okay Bob, I have thought a bit more about this, and this is my halfpenny's worth on the matter.

Think of the system (initially assumed to be stationary) as being the Earth surrounded by air arranged about it in a spherically symmetric manner, and with the can hovering above the North Pole (say) breaking the spherical symmetry. Suppose that the can is oriented such that its 'right' side (i.e the one to be punctured) is facing due south along the great circle passing through Greenwich. Let's just ignore the downward force on the can due to gravity as it unnecessarily complicates things. Perhaps it is best to think of the surface of the Earth as a frictionless ice rink for this purpose.

Then in the unpunctured case, the centre of mass of the system (assuming an appropriately symmetric can) will be in the centre of the can just above the North Pole. After the (circular, say) puncture has been created at time t=0 (I won't ask how, but let's assume the small circular piece of tin is quickly dissolved by some strong acid), then the centre of mass of the system at t=0 is actually slightly further from Greenwich than in the unpunctured case (i.e. to the left of the centre of the can if the puncture is on its right). Since there are no external forces on the system (ignoring the downward force of gravity), the centre of mass in the final state, whatever that happens to be, must be slightly to the left of the centre of the initial position of the can. Note also that the initial momentum is zero so the net final momentum of the can and the surrounding air must also be zero. Note also that there is quite a bit of potential energy initially, which must eventually be converted to kinetic and internal energy.

Presumably everyone is agreed that the net force on the newly punctured can from the air on its left side will cause it initially to accelerate rightwards (i.e. towards Greenwich). When the can is eventually filled with air, there will be no net force on it, except due to friction from the air outside, which will eventually cause it to slow down to a halt if it is moving. When it stops, the configuration of the air must be such that its net momentum is zero, and that it is distributed in such a way as to ensure that the centre of mass is where it should be. The motion of the air will in general be very complex.

So which way does the can end up going after its initial rightward acceleration? Well, once the air entering the puncture starts hitting the inside of the left wall of the can, the net force causing the can to accelerate rightwards will start to decrease, until eventually there is no net force (If the force on the left side of the can pushing it rightwards is more or less constant - which I think it would be - then I cannot imagine a scenario in which the total leftward force due solely to the air entering the can can be greater than the total rightward force, which might cause a net leftward acceleration). Actually that is not quite correct - there will eventually be a net leftward force due to external frictional forces opposing the rightward motion of the can, which will contribute to eventually slowing the can down to a halt. To summarise, the can will initially accelerate rightwards towards Greenwich, reach some limiting velocity and finally decelerate slowly to zero velocity due to friction.

Conservation of momentum in turn means that the air surrounding the can must have a net motion leftwards (i.e. away from Greenwich). I am no expert on fluid dynamics, but this can possibly be partly understood by the low pressure region created around the puncture as air enters it, sucking more air from right to left, but there will also be a low pressure region in the wake of the rightward moving can, which will also tend to such air towards it - whatever the case I can only imagine that the motion of the air will be very complex complete with vortices etc. The frictional force which reduces the momentum of the can will also serve to reduce the momentum of the surrounding air.

I am happy to be corrected, but that's what comes to mind.

Pressure isn't a great deal to do with average momentum in a gas. If you double the average momentum in a gas, you won't simply double the pressure - you're likely to roughly quadruple it. Pressure is also a quantity that only really has a definition for a gas in a state at least reasonably resembling equilibrium, which is not what we have.

Here the question asks whether or not the can is moving at the end - ie. whether or not there has been a net transfer of momentum to it. Which makes it essentially a momentum question. Furthermore, as the only other thing in the universe is air, it's essentially a question about whether or not the momentum transferred between the can and the air adds up to zero or not.

Just like a can punctured identically in both sides simultaneously has absolutely no reason to move at all, and will not, despite the big changes in energy, pressure, flow of air, etc., there is no reason why the end result of this process need be motion. Unless, that is, you have a valid momentum argument to show it.

In the case of the apple falling, when the string is cut, the forces are well out of balance in one direction so long as it doesn't hit anything. Here there are a number of forces acting in a number of different directions. If we looked at an apple that falls for a distance under gravity and is then also subject to some force opposing gravity (for example an apple landing in some sand) - there may well be a perfectly good reason why the result at the end of the time-frame in question should be a motionless apple.

However, I can't see what that reason is either.

If it really does take statistical mechanics to unravel this, then unless there's a proper statisctical mechanic on hand to give us an outline of what is involved, I'll have to just go away and try and work out how to do it. I've done a fair bit of statistical mechanics - just none in the last, er, xx years is all. And even if I could do that calculation, I may still be left thinking, "yeah, but WHY?" And I'd not know if I'd done something wrong. I'm sure there are folk lurking out there who know and can say why they know...

Bob, since you seem to insist we look at the air (though I believe just looking at the can will solve it). Lets look at ALL of the air's momentum for the time frame of this problem.
The momentum that the air gains rushing into the can slowly turns into the what the can 'sees' as internal pressure. Pressure can be defined as an average of all the little air molecules 'momentum'. The internal pressure(average momentum) for the time frame is always less than the pressure(average momentum) of the air acting on the outside.
So for the time frame involved, the pressure(average momentum)outside is greater than the pressure(average momentum)inside. In order for the momentums to be balanced at the end of time frame, something else (hey, maybe the can!) must be moving.

Bob, your apple on a string analogy is apt. As you say, cutting the string, at t=0 'now the forces are out of balance', which is very much the same as saying, 'now the momentums are out of balance'. In order to balance the momentum, the apple drops, right? Now, lets suppose some time later, we magically turn off gravity, which is analogous to the can being in a state of equal pressure in/out. No more unbalanced acceleration due to forces. Does the apple then stop having its momentum gained from the time span of the forces acting? No, even when gravity is gone, the apple will still be moving.

tyler's just made me realise that my a=b condition (for the impulses balancing and the can not moving in the end) can't be exactly true for every moment during the filling of the can. The fact that no molecules are leaving the can at t=0 would then imply that twice as many are entering as would normally impact on that part of the surface. This can't be true at t=0, as they haven't had time to speed up yet.

This doesn't make a great deal of difference to the overall rate of momentum flow into the can, though. The molecules would be accelerated to their streaming-in speed within a few mean free paths, which is a tiny tiny distance compared to the size of the can (mfp for air is 68nm, according to Wikipedia).

meBigGuy

Yes, then it would move left (assuming Paul's right to say the overall impulses of the can filling should balance, which I'm not entirely convinced about). This is relevant to the question as it's stated, because although there is no solid block preventing the can from moving right, there is some air resistance. This is why I initially said the can would jump right, but end up with a very tiny momentum to the left (post 103), due to this small external resistance to the motion of the jump.

Edited by author 01-15-2007 01:21 PM
I see your point, tyler, but at the moment immediately after the can has punctured, the velocity, momentum and position of every air molecule is identical to what they were at the moment immediately before the can was punctured, when the gas was in a state of equilibrium with the unpunctured can. So it is a valid initial condition for the state of the air. And if we're assuming that it's possible for the mass, postion and velocity of the can to be unaffected by whatever causes the hole to appear (so the question is purely about the physics of a can filling), then it's certainly a valid initial condition for the can too.

A given 'state' of a dynamic system generally means the positions and momenta of the bodies involved. We can introduce a force at a given time without affecting that state at all in that instant. Forces act over time, changing momenta at a given rate. At t=0 in the new system, the force will have had no effect.

An example might be: suspend an apple (or an orange) from the ceiling on a thread. We specify the state of the system exactly, by measuring where it is and observing that it is not moving (it's in equilibrium). Then at t=0 we cut the thread. Now the forces are out of balance, and they act on that same initial state (an apple at height h with mass m and velocity 0 at time 0), changing it over time.

I think that it is wrong to present the initial condition as that of the can starting without a hole and pressures in equilibrium. This idea has lead the educated physicists here to the conclusions that 'the can not moving before must mean can can't be moving after', Or 'It will move then must stop'.
Once the hole is put in the can you have a completely different system than without a hole. The initial condition needs to be the very moment the can is punctured, before air starts to move. Yes, the can has not moved yet, but the imbalance of forces need to factored into the momentum balance, ie: It is not starting from equilibrium, which means the system should have momentum in the end. Alot of this momentum ends up in the turbulent eddies being created both in and outside the can, in fact all of it eventually ends up there as friction slowly stops the can from moving. But in the instant that the pressure inside and outside the can equalize, you will have a can, full of turbulent air, still moving to the right, creating more turbulent air around it, until all of the initial potential energy of the system is 'lost'. There is no force that I can see that cancels the initial imbalance, or stops the can once it moves.
You cannot look at the can being balanced by pressure forces (without a hole) as the initial condition. Otherwise you are comparing a can without a hole in the beginning to a can with a hole in the end.
Apples and oranges.

My space argument is pretty rough, I'll agree. For example, with the can, it is actually the air molecules (air pressure) on the outside left that push the can right, not the vacuume pulling the can right. And, the air column isn't sucked into the can, it is actually pushed by the air pressure on the right.

A vacummue doesn't suck things, it just provides an empty space for things to get pushed into. The lower the outside pressure, the lower the velocities (momentum) attained by the pushing (the limit being no flow if there is a vacuume outside).

Back to my favorite twist on this (making foof more like poof)
If the can is blocked from moving right, the right side air pressure acts on the column of air for a longer time (since the air travels a longer distance), accelerating it to a higher velocity, and then the can moves left after filling due to the momentum of the air (which actually pressurizes the air in the can also, causing some poof).

Edited by author 01-14-2007 05:25 PM
Still going! Might be on my own, doesn't matter.

I've done a little scribbly calculation.

Let's say the number of molecules hitting the surface per second in the vicinity of the hole before the puncture is N. (We can define 'vicinity' however we wish, so as not to exclude any significant edge effects.)
At time t after the puncture, let's say that:
the number approaching that area per second is N + a(t), and
the number leaving that area per second is N - b(t).
(these might be entering/leaving through the hole, or merely hitting/rebounding from the surface in the vicinity of the hole.)

Let's also say the average perpendicular component of velocity of molecules hitting the surface before the puncture is v
that of molecules entering the vicinity of the hole is v + c(t) to the left, and
that of molecules leaving the vicinity of the hole is v - d(t) to the right.

If, throughout the time of the filling of the can,

a=b and c=d ,

then the can will not be moving at the end.

This equality seems very reasonable to me, on vague symmetry grounds - that's as much as I can say.

The result comes from looking at momentum flow arriving and leaving the surface of the can and through the hole (taking what's in the can as being part of the can).

I like meBigGuy's description of the essential difference between poof and foop, first paragraph. It's not exactly the situation here, as we've got air flowing outside the box that could carry momentum off, but it knocks the 'symmetry' right on the head. Brilliant.

Paul – ok as far as your perpetual motion argument…

First, here's what I'm happy to agree with:

Yes, when the hole is first opened, there'll be a short burst of force to the right until the incoming molecules made it across the inside of the can. Much of this burst will be counteracted by the air molecules then slamming into the back. However, I'm going to leave this as I don't believe it helps us get anywhere with answering the question of whether or not it's still moving when it's full.

I agree also with your remarks about acceleration. In the macroscopic picture, the air is accelerated into the can by the pressure difference. In the microscopic picture, it's more interesting - I hadn't really thought that the molecules themselves would be accelerated (after all, a vacuum does not accelerate anything, and they're be moving freely in), but I can accept that the ones entering will be faster than average because (a) slow ones are less likely to enter anyway (as I said in 103), and (b) if a molecule of medium speed was heading for the entry it is more likely to be struck from behind and increase in speed than to be struck in a way that will slow it down. So yes, it does seem that the molecules entering the hole will be faster than the average molecules striking the surface. And this acceleration will decrease to zero as the internal pressure approaches the external.

Then, yes, the question is, as you say, is it "0+0=0" or is it "x-x=0". Is there a net transfer of momentum (or, equivalently, is there a non-zero impulse) in the period between the time of the puncture and the time when the can becomes full.

But then you say if the air has any net momentum to the left, ie is still moving after the can has filled, then you'd get a pressure gradient that could give you a perpetual motion machine? Is that your argument?

In that case, imagine a (non-punctured) can with velocity v to the right through a medium of air. The can is slowed by air resistance to a velocity of zero, passing on its momentum to the air. The air now has a net momentum to the right. By your reasoning, would the air then move to the right hand side of the universe, creating a pressure gradient that accelerates the can to the left hand side of the universe and thereby creates perpetual motion? That sounds a bit silly to me. So perhaps I’m missing something of what you mean?

Moving back to our original filled can - having a can move to the right, with air gaining equal and opposite momentum to the left, does not create a perpetual motion machine. It's a perfectly reasonable state of affairs. After a while, the can will slow down due to air resistance (at least because collisions with molecules on its leading side will be greater than collisions on its trailing side), and the rightward momentum it has gained will be transferred back to the air. Once it's slowed to a stop, the net momentum of the air will again be zero, nothing moves to the edge of the universe and there's nothing supernatural to worry about. The can may still have been moving long after it was filled.

Indeed, if the result is a small air flow to the left, wouldn’t that increase the resistance to motion of the can to the right anyway, slowing it down all the more?

I'm not sure what led you to think I'd neglected conservation of momentum in my earlier posts. I was surprised because I thought the air hitting the inside during filling would pack the same punch as if it had hit the outside. It seems counter-intuitive that it shouldn't.

However, I can see the flaw in my argument (295) about the flow of air molecules in and out of the system. I had assumed that, during filling, the molecules entered and left the can with the same average perpendicular component of velocity as those that strike the surface, and I can now see that this will not be the case. Those that enter will do so with more momentum to the left, and those that leave against a pressure difference will do so with less momentum to the right. So I can see that there will certainly be less momentum to the right than my argument. I can also see that to determine whether or not this would yield zero would get more complicated than I want it to. Which, for me, throws the whole thing open again!

So I don’t know any more. Aha! – a result!

Your perpetual motion argument isn’t compelling – but it may be that, as you say, it’s not really possible to argue this properly without statistical mechanics. Which would be a shame.

The difference between poof and foof is illustrated by the weight in space example. When I pull a weight toward me in space, we collide and stop (foof). When I throw a weight, we continue to move apart (poof).

If , after pulling the weight, we do not collide, the weight and I will cross and continue to move apart. This would happen if the left side of the can were a 1 way permeable membrane that allowed the accelerated air to escape.

To me the interesting example is when the can is held in place and then released "just" as it fills. Now there is non zero momentum as the can moves to the left.

In space, this would be like the rock was anchored as I pulled the rope, and then we both continued to move after we collided.

Ah, the final puzzle of it. Why is foop not like poof? It's something to do with the geometry of the problem. The fact that the air that moves must necessarily interact with the box, and thus zero out the resultant velocity? It bugs me because if were to draw a graph relating the velocity of the box at time T to the pressure in the box, the graph would have a kink in it at the equal pressure point.

I found it difficult to quantify the "wind" effect - or the air "slamming" into the can - but what I found compelling about the piston reasoning is imagine a very thin piece of tissue paper covering the hole. When the hole is popped, the tissue paper is accelerated left - same force, different mass to the can. However it still picks up the same momentum, since it accelerates more, due to less mass ie momentum = F*m/m *dt. So the tissue paper has the same momentum as the can. Now replace the tissue paper with a couple of air molecules near the hole and you have the same reasoning.

Woo - that's more like it!

I'll have a proper look at that tomorrow.

Bob D:

Your momentum analysis is somewhat unnecessarily complicated, but mostly correct. The "surprise" you mention in post 295 is due to the fact that you forgot this system is isolated and momentum must be conserved.

The reason that the total impulse of the air hitting the inside of the container is the same as the total impulse outside (and this applies only when the system has equilibrated) is because air will move into the can so long as there is a difference in pressure between the inside air and the outside air. At equilibrium, the pressures are the same. Pressure is force/area so equal pressure implies equal force. Force is impulse/time so equal force implies equal impulse.

So lets look at it from an impulse/momentum point of view. Before the can is punctured, air molecules are striking it randomly from all directions with random speeds. On average (we won't worry about Brownian motion of the can here), the total impulse delivered to the can from all these molecules is zero. Consequently, the total momentum of the can does not change -- it started zero and it remains zero.

Now open the hole. For a short time, there are more molecules striking the left side of the can than the right, where the hole is. Again, they have random directions and velocities, but the fact that there are more of them hitting the left side means that the net impulse on that side is momentarily greater than on the right side. The can moves to the right.

Once the can is fully filled with air, the situation is physically the same as it was before the hole was opened so again the momentum of the can is constant. The only question is what happens during the filling process. Does the can end up with a non-zero momentum to the right? In fact, it should not. To show this rigorously is not particularly easy -- you'd have to roll out the machinery of statistical mechanics (or at least, I don't immediately see an easier way to do it). So let's see if we can at least get the essential idea.

First, we need to clear up "an egg's" confusion. Something IS compelling the air to squirt into the can -- namely, the pressure exerted by the rest of the air outside. They don't just drift in. Air molecules interact with each other too. When you open the hole, there is an unbalanced force on the air molecules right next to the hole -- they are receiving an impulse from collisions with the air molecules to their right, but no longer receiving an impulse from the can to their left. The result is that they are actually accelerated into the can. During the filling process, that acceleration decreases exponentially to zero as additional molecules entering the hole are beginning to experience an impulse from the air that entered earlier.

The net effect is that you have fewer molecules but moving with a higher average velocity when they strike the opposite wall of the can inside. So the question, then, is will this basically be a wash, with the increased velocity compensating for the decreased numbers to yield the same net force, or does it not?

To do a detailed calculation would be difficult, but we can argue from global conservation laws that it probably is not a wash. If the system of can + air starts with zero net average momentum, then so long as there are no external interactions, it must also end with zero net average momentum. There are two ways to accomplish this: either the can and the air independently have zero momentum (so 0+0=0), or they have equal but opposite momenta (so x - x =0). If the can ends with net momentum to the right, that requires the air to have net momentum to the left, so it moves over to the left side of the universe. This would create a pressure gradient accelerating the can further to the right and voila! Perpetual motion of the first kind -- a clear violation of conservation of energy (note: total energy, not necessarily total mechanical energy) and so the second possibility is excluded. Whatever happens at the microscopic level, it must in the end lead to a final state in which the can and the air independently have zero average momentum. So when the system reaches equilibrium again, the can cannot be moving.

Note that the answer Epstein gives in his book is seriously incomplete.

One might wonder, then, why this argument doesn't apply to "poof" -- where does it acquire its non-zero kinetic energy? I will leave this as an exercise for the reader. The key idea is that the poof and foop situations are in fact not mirror images, though they look like it superficially.

sorry, positive number

A. force on the outside of the left face of the can=atmospheric pressure x the area of the face (this is reasonably constant throughout)

B. initial air pressure on the other side of that face=0
C. initial force on that face=0

D. final force on the inside of the face=atmospheric pressure x the area of the face (this is as big as it can get, remember, all those particles slamming into the right side of the can are only going at the same average speed as the ones on the other side, and there are fewer of them at first, finally, there are the same number)

resultant initial force= A-C =A
resultant final force = A-D=0

the effects on the other face of the can are the same except that the other face has a hole in it so it has less area and therefore will experience a lesser effect, substitute "area of the face" with "area of the face-area of the hole" and as long as the hole is a positive integer the effect will be less.

hope this helps

Paul -

I'm interested to know what reason you have to believe that the total impulse of the air hitting the inside of the container is the same as the impulse of the air hitting the outside. It may seem intuitively reasonable at first, but looking in detail at the exchanges of momentum involved I found a different answer altogether. It's certainly not satisfactory to merely say "the air entering the hole crosses the interior of the can and slams into the left wall, pushing it back the other way" and then conclude that no final motion is imparted. I may well be wrong (I think I was when I first piled in here) but it'll take more than some authoritative 'weighing in' to convince me! So - some solid physics, please.

I've set out my broad reasoning in post 341, based purely on the flow of momentum of air molecules approaching and leaving the region of the can, and more rigorously in post 295.

The air "slamming" into the back of the can will always provide a force less than or equal to the force from the pressure on the outside of the can, nothing is compelling the air to slam in there, it is just drifting into the available space. I see no reason that the can should move to the left. If, as I said, the friction between the can and the surface on which it rests it sufficiently small, the can will move right due to an unbalanced force on the can and then the air "slamming" into the back of the can will simply act to balance those forces after a short time. I fail to see where a force greater than the force of atmospheric pressure on the left will come from as the maximum force that the air can provide is that due to atmospheric pressure when the can is finally full. Am I missing something?

I sent this to Mark about a week or so ago, but didn't post it here because . . . well, it just wouldn't be fair for a physicist to weigh in at the very beginning. But he told me enough time had passed that I could add it now.


OK, the can moves to the right briefly because the pressure on the left side is greater than the pressure on the right immediately after the puncture. But soon the air entering the hole crosses the interior of the can and slams into the left wall, pushing it back the other way. Once the can is full of air, the pressure is the same at all points inside and out so no further acceleration is possible. So a brief wiggle, and then nothing.

Now extend the experiment. Does the can wiggle both ways (i.e. first right and then back left) or does it go the the right and then stop? Would anything change if the left wall of the can were made out of stretchy rubber instead of something rigid? Suppose you did the hole entire experiment with the can sitting on a scale -- what would the weight reading be throughout the experiment? Suppose the can were placed in a container that was not much bigger than the can, so that there is a significant drop in air pressure as the volume of the can is opened up -- how does it move then?

To anyone who finds this sort of thing interesting, I would also recommend Epstein's other book, Relativity Visualized, as well as Paul Hewitt's cartoon puzzles each month in The Physics Teacher.

By the way, the people who have noted that mechanical energy (KE + PE) is not conserved in this problem are to be congratulated -- you are on to something. Total energy must, of course, always be conserved in any thermally isolated system. However, the entry of this gas into the can is not a quasistatic process. There will be considerable turbulence and that means the entropy will increase -- the process is not reversible. Entropy is energy that is not available for doing work, so it vanishes from the mechanical part of the system.

meBigGuy I find your arguement compelling - I can't see anything wrong with it. This means my ideas of pressure are not applicable bacause the system is not in a static state? The right side of the left wall must get a force in excess of air pressure for a while.

I agree with burn. The air rushing in from the right will cause a lower air pressure on that side, causing the higher pressure air on the left to push the can to the right as it moves into the lower pressure area.

of course the can will want to move in the direction of the puncture (to the right in the illustration). sucking the air out of the space to it's right will cause a lower air pressure on that side (assuming it happens fast enough), and will, therefore, have a suction effect. blowing air out will cause you to move *away* from the puncture direction, sucking air will cause you to move *towards* the puncture direction

Mark-- Do we have a real answer yet?
I'm having trouble sleeping over this mess and must have the puzzle resolved!

The momentum change of the incoming particles on the back of the can will only equal the momentum change of the particles hitting the left side of the can when the can is full of air, so briefly, there will be an imbalance. Forces, momentum, energy, they all yield the same answer.

Since the question is about momentum, the principals of conservation of momentum need to be observed. In the first instance, the gas escaping from the container adds to the overall momentum of the gas outside the container to the right, and the container matches that addition with it's increased momentum to the left.
On the other hand, the gas entering the container in the second instance trades momentum with the container as it enters, but once again returns it, as it makes contact with the back wall of the container. There is no change in momentum of the gas outside the container, so there can be no change in momentum to the container.

This is very easy, the can will move to the right, just as a vacuum cleaner hose attempts to move away from your hand. When the can is punctured by the mysterious puncturing entity that doesn't block the hole or impart any momentum on the can in either direction, there will be, for a very small time, an unbalanced force on the can, air pressure on the left and no air pressure on the right, as air enters the can it will just tend to restore that balance, so, briefly, the can will experience a force to the right, and if it is light enough and friction between the can and the surface are small enough, it will move to the right. It may be that the pressure in the first can was far greater than the difference between air pressure and a vacuum, in which case the can will stay still, or that the hole is too small and the expanding gas, coming through a small hole will cool the aperture sufficiently to block it with ice from the water vapour in the air, but in the event of those cases not being so, it will move as described. Have fun with this.

I suspect, however, that to construct a can that could support a total vacuum without collapsing and be punctured without a change in momentum, and be moved by the ensuing force might be a very tricky proposition indeed.

That's a handy site.

You should really drop the conservation of energy thing though, for exactly the reason you give - inelastic collisions ensure that mechanical energy (PE+KE) is not conserved, much of it is lost and we don't know how much. Imagine if the can was punctured simultaneously on both sides. The situation is perfectly symmetrical, so can't result in motion. All the PE of the situation is quickly lost to heat and turbulence (which will itself dissipate quickly into heat).

Sorry, that link didn't go to the equations I was looking at, look for thrust of a rocket in the mechanical links.

Bob I am looking at this site-http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
This is the momentum balance equation that developes the formula for the thrust of a rocket. You can see they have two velocity components from different reference frames. The vacuum can is basically a rocket in reverse, only it is taking on mass instead of ejecting it. I am having trouble with the sign conventions and how to apply it to the can. But I think that change in mass and velocity of air relative to can both being non zero, there has to be thrust, therefore motion. The only doubt I have is the can colliding with the incoming air is an inelastic collision, which consumes kinetic energy.
What do you think?

MeBigGuy -

Hey - wonderful! - that's exactly what I thought to start with (posts 103 & 205) and nobody had a clue what I was on about. I thought it was a great theory.

But I changed my mind (295 & 341).

Matware -

You missed the obvious thing. I'm a bit tired of obvious things now.

Taking a different view on it, conservation of energy.

Case 1: The can has potential energy stored as the difference in pressure between the inside and outside of the can. You puncture it, and the potential energy gets converted to kinetic energy, and it buggers off to the left. Easy.

Case 2: The can has potential energy stored as the difference in pressure between the inside and outside of the can. (wow, how can that be!). You puncture the can, and the energy has got to be converted into kinetic energy once again. Now the can will move in the opposite direction to the mass. The mass is flowing into the can (to the left) so the can will move to the right.

Now after the pressure equalises (in both cases) the can should act like a body in free motion, so the cans will keep on moving until the eventually hit the singer on stage, and she walks off in a huff.

I'd love to have it pointed out to me how I'm wrong, and I missed the obvious thing.

As for everybody who has used the word 'push' to describe why the can moves, I just want to slap you.

Tarral -

You got a response is about moving air because you asked where the balancing momentum is. All there is here is a can and some air. What were you expecting?

If you've already decided that the air won't be moving after the can is full, then the can can't be moving either, because of your decision. But if you were to work out whether or not anything else is moving rather than decide beforehand, it might turn out different.

There is an observer here, in an inertial reference frame. A problem asking 'what happens' needs the possibility of an observer in order to have an answer. We could attach the observer to the can, but then it'd be a silly question as the can wouldn't move relative to him no matter what you did - you'd have to ask 'could you feel it move', which is the same as 'is the observer in a non-inertial frame'. It all becomes very silly and pointless. You could attach your observer to an air molecule! If you want to keep it simple and get some sense out of a situation, put your observer in an inertial reference frame unless there's a very good reason not to.

As the question says, the observer is in his own mental physics lab. He could reinvent physics if he wants, which is what most people do! In which case it's a bit pointless arguing. But I do anyway. Cos I'm a bit daft.

Tyler -

I'm not sure I understand what you mean - in order to solve what? It moves relative to an inertial reference frame - isn't that all there is to it?

Huge -

You're right - it is ambiguous because there are two timescales involved here. The first is the timescale of the can filling, and the second is the timescale of the can slowing due to air resistance. We can't actually tell which is bigger from the problem. If the can is huge, and the hole is tiny, then any movement will quickly be damped, so by the time it's full it won't be moving at all. So there is an assumption here, that the can fills much faster than the time it would take for air resistance to slow it, and that 'after the can has filled' refers to a length of time after the puncture that is greater than the filling timescale and less than the slowing timescale. I think this is a reasonable assumption because (a) if you choose any other meaning to the word 'after', the problem is trivial, and (b) if it were a tiny hole it wouldn't go 'foop', and (c) the diagram shows it floating in the air (ie nothing else to slow it down) and with the hole being quite large.

Zwack -

Your universe is very nice. We want to fill the vacuum can, then see what's moving. In your universe, the air coming out of an air-filled can in a vacuum would hit the can on the nose and slow it to zero by the time it has 'filled' your 'universe'. So you're not topologically inverting the right problem.

Good stuff though.

Was going to respond to Matt (308) and Poopy (301) but I guess they're not here any more. Hello Matt and Poopy.

Edited by author 01-12-2007 05:46 AM
Assuming no static frictional forces, the can initially moves to the right, then stops when the accelerated air molecules hit the left side of the container. After filling, both are halted.

I think the following example illustrates this very well.

Imagine that the vacumme was created by pulling out a plunger. When you let go of the can, it sucks back onto the plunger (moving right). Imagine releasing the can and plunger at the same time. The plunger moves left, the can moves right, and all stop when the can is filled (they gain momentum in opposite directions, then cancel when they collide). The air acts very much like the plunger.

Now, static friction on the can might cause another effect. The force moving the can to the right might not be enough to overcome static friction in that direction, so the can does not move. When the air molecules hit the left side of the can, there may be enough force to overcome static friction in that direction (much more momentum in the air column since the can did not move). The result is that all the energy is now applied to leftward momentum, and the can will start moving to the left after it fills. This explains why one poster reported flouresent light tubes shooting to the left when the ends were broken off.

tyler, what are you smoking? This is an observer-less problem. No reference frames involved.

Edited by author 01-12-2007 02:51 AM
It says "after the can is filled". I got a response that talks about moving air. The air won't be moving anymore once the can is full. So the can is not moving either.

Yes, Bob you are right, in that case, I think velocities seen from your external(inertial) reference frame still need to be referenced to velocities relative to the can in order to solve.

I agree the word after is a bit misleading, but I think the intent is somehting like "assuming very low friction, is there anything else that would bring the can to a halt once it was filled, if it started moving"

As for momentum, as it's said below, the air is deprived of some right-wards momentum when it pushes the can - ie, on average, it gets some(just enough!) left-wards momentum to compensate. This pocket of air, without as many right-moving particles as it had before, will diffuse around the volume. Another way of looking at it is a wind to the left just like a rocket. If you put a rocket in a box, the exhaust would hit the wall, and transfer momentum to the box. So in a very large box, the air carries the momentum. In a smaller box, it is transferred to the box after a period of time.

Somebody pays me to edit questions like this. If it wasn't the author's intent for the answer to be no movement, I'd ask author to rewrite removing *after*. Actually, the word *after* makes the answer trivial, so it should have been deleted. But hey, this flurry of responses is entertaining!

BobD...

I disagree. The universe in my thought experiment has a specific volume but no specific edge. Think of it as the equivalent of a pacman or asteroids screen if you must. You approach one "edge" and appear on the opposite edge"

In this case the difference between the two scenarios is that the can has effectively moved so that it is around the edges in the second example, but in the centre in the first.

Think of it as a topological inversion. What was the inside of the can is now the outside and vice versa.

Z.

Edited by author 01-11-2007 02:08 PM
ok, i can't go through all 356 posts that precede me, so perhaps i'm restating someone else's argument. but maybe i'm simplifying it. or making it worse. who knows.

--the can will move to the right.

first, I think you can discount the semantics of the wording "which direction will the can be moving *after* it fills with air" although that seems at least inadvertently confusing. if the can has any momentum after it fills, it will quickly be stopped by friction (either from the surface it's standing on, or the air itself.) but ok, it'll keep moving for a second.

my reasoning for rightward movement is that while air is rushing into the hole, the molecules on the right side of the can will be moving faster than those on the left, which creates less air density, which creates a pressure drop, which (like an airplane wing) will cause the object to want to move towards the lower pressure. (more specifically, the higher pressure air on the left side will be pushing harder than the low pressure air on the right, causing movement)

this feels like the obvious answer, which makes me second guess myself. you kind of want the can to move to the left to make it interesting, but I still vote "right."

Which systems are you referring to? Surely no problem to assume there are no external forces on the system of can + all air molecules.

Here is where all conservation of momentum arguements are being misled, please note giant IF in the following definition:

Conservation of momentum is a fundamental law of physics which states that the momentum of a system is constant IF there are no external forces acting on the system. It is embodied in Newton's first law (the law of inertia).

The sum of the momentums in this problem have to add up to the inbalance of forces present when hole is first opened. Since there is no starting momentum, there has to be an ending momentum to balance equation. They do not add up to a constant:

momentum after = momentum before + momentum caused by forces acting.

Most arguements here seem to use:
momentum after = momentum before
This is wrong.

Alright for lack of simple terms to use I will get technical again. Let's think outside the box for a moment. Instead of your control volume being BOX shaped around the can or around the universe, the control volume should trace the outside wall of the box, turn inside at the hole and trace the inside wall and come back out, it's sort of c shaped, see? Now you have no air crossing boundries to worry about. Sum the forces acting on the control volume from the air outside and in. You should see a resultant force to the right, which will accelerate the box and the air entering it. When the pressures equalize, which is the question at hand, box full of air will stop accelerating, but not stop moving. Answer B.

Edited by author 01-11-2007 10:41 AM
Zwack,

If you take the universe to be double the size of the can, then you have a very different question indeed. You have one in which the inpouring of gas continually deprives the 'outside' of leftward-moving molecules. This doesn't happen in a vast universe, as the leftward-moving molecules will just keep coming, and (as I've explained below) it's actually the rightward-moving molecules in the vicinity of the hole that are depleted the most.

It's the force of air molecules leaving the outside left surface of the punctured can that makes it move. Nothing happens on the other side of your 'universe', so it won't move in the same way. The can has an inside and an outside, and both sets of surfaces are subject to changes in air behaviour as the system evolves - your universe does not.

Symmetry arguments are wonderful - but you need a very good symmetry to use them.

Physics_dude...

Point 1 is incorrect.

System: Can in a universe.

Consider the universe as being a bounded container. Now make it double the volume of the can to simplify matters and you have the complete system.

Z.

OK I had the right idea, but the wrong result...

Let us assume that the Universe is bounded, and is double the volume of the can.

Assuming that the effects due to friction are negligible, and that the "air hitting the far end of the can" is also negligible (I'll get to why in a second)...

In the first drawing the air is released out of the can into the universe. This causes a jet of air which pushes the can in the opposite direction. The can moves to the left.

In the second drawing the air is released out of the universe into the can. This causes a jet of air which pushes the universe in the opposite direction (Remember that all motion is relative to your frame of reference, I am using the vacuum as my reference point here). If the Universe moves to the right then the can moves to the left. As the volume of the universe outside the can and the volume of the can are equal in this thought experiment then this is just the inverse of the previous problem.

Why am I assuming that we can ignore the air hitting the end of the can? Because we can. The air moving into the can initially will be headed right to left. Some of that will hit the sides of the can, while some will hit the end. The particles that hit the end will impart some momentum to the can, but they will also bounce off and start moving to the right. The next set of particles will enter the can, and some will hit the sides, and some will hit the end, and some will hit the particles bouncing around in the can and... Very quickly we will have Brownian motion inside the can and the effects of the air hitting the end of the can can be ignored. If you're really concerned about it, remember that what little momentum is imparted by an air particle to the end of the can will push the can to the left. The same way that it is already moving as the Universe is being pushed to the right.

When the can is full, the movement will gradually decrease due to air resistance, so the correct answer is that the can moves to the left.

Z.

Edited by author 01-11-2007 07:05 AM
Taral (346) -

The opposite movement to conserve momentum is the air molecules bouncing off the left-hand side of the can opposite the hole. Air molecules are not bouncing off the right side where the hole is.

When the can is full, as many are coming out as going in, so that's effectively the same as air molecules bouncing off the hole, but while the can is filling, there's a definite imbalance of air molecules leaving the can to the left.

physics_dude (344)

Your final equation tells us that the can stops accelerating once the air stops flowing into it.
But while air was flowing, the acceleration was always to the right.
So once it's full, it's not accelerating any more, but it's moving to the right.

(or at least it would be if you'd used your right-is-positive convention for V in equation 5, but you didn't, hence your confusion at the end.)
The equation also uses V on both sides for two different quantities, which doesn't help.

But I agree with Huge, your point 7 is a little dodgy, as you can't say atmospheric pressure is acting on the inside of the can through the hole - the pressure inside the can is not atmospheric. Pressure is a useful concept in static or nearly static situations - beyond that the definition gets a bit awkward, which is why I've stuck to molecules and momentum. Immediately after the puncture, there are only half the number of molecules in the vicinity of the hole as there are elsewhere in the atmosphere, because only those moving left are present. The rightward-moving ones that would normally be present near the wall, having bounced off it (this bouncing is a crucial part of what pressure is), are not there at all.

Also you can't say 'mass is constant' in point 9, as the can is taking in air.

But I kind of agree with what you were trying to do.

It will move to the left (duh) Its elementary. Just think about it. a vacuum just like a compressor create a pressure differential, granted that the difference between the two pressures is great enough the can will move.

>The can cannot possibly keep moving after the pressures are equalized. Where is the opposite movement to conserve momentum?

It is indeed the edge of the air (container) that also moves. By saying it is under pressure you are implying the gas is contained. If you removed the container at the instant you opened the hole, then you would get a mass of air moving to the left(ie, rocket situation) to counter the can moving to the right. mmm crack, mmm.

Edited by author 01-11-2007 02:24 AM
The can cannot possibly keep moving after the pressures are equalized. Where is the opposite movement to conserve momentum?

(Those arguing that there must be a surroundings which could take up the difference in momentum are simply on crack. You cannot just "add" stuff to the system to make your conservation equations work. What if, for example, the air goes on forever?)

7) Simplify: The sum of the forces on the can is 0, since it is atmospheric pressure acting equally on all sides of the can, therefore:

This is just wrong. On the left side, the pressure acts on the surface. On the right side it acts on the (surface - area of hole).

Edited by author 01-10-2007 10:59 PM
1) System: The can

2) Identify Forces: Atmospheric pressure on all sides of the can

3) Identify Mass flows in and out: Mass flow in

4) Define Axis: Up is positive y, Right is positive x

5) Conservation of Linear momentum:
dP
-- = Sum(Forces) + Sum(Mass flow in * velocity in) - Sum(Mass flow out * velocity out)
dt

6) Simplify: There is no mass flow out, therefore:
dP
-- = Sum(Forces) + Sum(Mass flow in * velocity in)
dt

7) Simplify: The sum of the forces on the can is 0, since it is atmospheric pressure acting equally on all sides of the can, therefore:
dP
-- = Sum(Mass flow in * velocity in)
dt

8) Simplify: There is one mass flow in with one velocity.
dP
-- = m_flow_in * V
dt

9) Momentum, P, is equal to mass*velocity. Mass is constant, so it can be removed from the derivative on the left. Velocity can be assumed to be changing because air is not always entering at the same rate.
dV
-- * mass = m_flow_in * V
dt

10) Simplify the equation to solve for acceleration, which is dV/dt
dV
-- = m_flow_in * V / mass
dt

11) From this, acceleration looks to be positive, which means the can is moving to the right. However, recall that velocity is a vector, which includes a magnitude and direction. That velocity is to the left because air is moving from the right to the left into the can.

Therefore, we can conclude that the can will move to the left.

Edited by author 01-10-2007 10:36 PM
answer c: the can will not be moving. after reading the most recent page full of responses (15 or so), i'm not even worried about the english of "after the pressure has equalized." simply put, the air escaping from the pressurized can is acting on the environment outside of the can (pushing against air that is already there), causing the can to move to the left. in the case of the vacuumed can, the air rushing in is acting on the can itself, but is likely pushing little harder on the far end of the can then on any other surface as the pressure equalizes, so likely not enough to move the can to the left. also, during pressure equalization, there is slightly less pressure on the outside of the right side of the can than the outside of the left (as the external pressure gradient equalizes), again probably nowhere near enough to cause the can to move to the -right-. and these even act against each other. but ultimately the air rushing -into- the can is not acting on the can relative to its surroundings, so the can will not move.

How does the horizontal pressure vary with time ?

----------------------------------------------------------
High pressure |left wall| Vacuum High pressure
-----------------------------------------------------------

Here I use 5 for max pressure, X for wall, 0 for vacuum
as the air diffuses into the hole.

t0: 555555X0000000055555555
t1: 5555555X000001455555555
t2: 55555555X00134555555555
t3: 555555555X1455555555555
t3: 5555555555X555555555555

At the end point, X has no net force and is moving to the right.

Does the pressure on the right ever get above 5, and hence
provide some leftward decelleration?
I don't see how - but I guess I could be wrong.

Edited by author 01-10-2007 09:09 PM
Golly. I go away for a few hours and I come back to mayhem.

Ok - the only thing providing forces here is air, so let's look at the air. Air coming, and air going.

We make a hole in the right hand side of a can, containing a vacuum, surrounded by air.

Air is just molecules moving about all over the place. Some hit the can, and pressing on it, always inwards. Some enter the hole instead, but they will still hit either the inside of the can or another molecule of air that is already inside, pressing in exactly the same way as if it had hit the side.

The result is that air is always pressing on the can+contents equally from all directions, same number of molecules from all side, same forces, even with the hole. This is true before the puncture, it's true while it's filling, and it's true after it's filled. There might be a very brief jump to the right when the hole is first made, cos the first molecules entering from the right take a little time to hit anything inside the box, but they all hit something in the end. Air is coming from all sides. None of it can or will set it in motion in any direction.

BUT: if air molecules HIT the sides of the can, they also have to LEAVE. Air doesn't stick to cans. Why? Because molecules are jiggly critters. An air molecule that has hit the side very quickly jiggles its way off the side of the can and speeds off, pressing the can back as it does so, like a bird taking off presses back on a branch.

So all the air molecules LEAVING the can are pressing it inwards too. Before the box is punctured, this happens equally from all sides. *While the can is filling, it does NOT - the leaving molecules press less from the right side, because there are less of them (there's a hole with not a lot coming out of it).* Once the can is full, air molecules again leave the can pressing equally from all sides.

During the time that the can is filling, MORE molecules are LEAVING the can and kicking back against it from the LEFT side than from the right, because of the hole and the lack of molecules leaving through it. THIS is the force that propels the can off to the right, and it's the ONLY force acting.

Ok, there's air resistance, and this will slow it down eventually. But it's too late - when the can is filled, it will be moving to the right.

Anybody convinced by this?
I think it's bloody marvellous.

I linked.

The vaccumed can will move to the left. It is sucking air, not like it's sucking a curtain or a ribbon or something that will draw it the other direction.

Since the top and bottom forces cancel (I think we all agree), you can picture a tube, instead of a volume. Also, assume the hole covers the whole right-side wall.
So the equivalent diagram, once the hole has been opened is:

-----------------------------------------------------------
High pressure |left wall| Vacuum High pressure
-----------------------------------------------------------

Does the left-wall accellerate to the right ? Yes
What stops it ? Friction (sometime after the pressure has equalised)
So b.

Please explain in laymans terms what force stops it, conservation of momentum arguments mean nothing to joe six-pack. Even very educated people have made bad arguments here using sophisticated formulae, including myself. Remember the KISS principle, Keep It Simple, Stupid.
If it moves, what stops it?
If it doesn't move, what stops it?
Has anyone gotten the definitive answer?
I would be interested in %'s of this group who picked a,b,or c.
I think the answer is B.

After the vacuum is filled the can will

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

The answer to the question lies in the ENGLISH of the question, not the SCIENCE of it. It all comes down to the word after.

They didn't specify "frictionless" anything.
We live in the real world where there is friction.
There is a point where there can be absolutely no less molecules in a given area, this is absolute vacuum.
Sea level has 14.7 P (pounds) of pressure on every Si (square inch). We commonly refer to this as 0 psi, because it is our base-line of what is normal for us. Any more pressure above 14.7psi gets read on our gauges as whatever it really is compared to absolute vacuum minus 1 atmosphere of pressure. Absolute vacuum is a theoretical idea and has never been achieved by the best scientists in the world, and does not even exist in the whole of the known universe because no matter how sparse, there is always something within any measurable volume. Given this, there can be at most a 15.7 psi pressure difference between a theoretical 'absolute vacuum' and the highest atmospheric pressure ever recorded on the face of the earth. To contrast this point air is regularly compressed to multiple times atmospheric pressure for use in pneumatic machines, my air compressor right now has a little more than 8 atmospheres of pressure in it, my tires between two and three depending on how hot they are, and my turbocharger system one at a good boosting, and these are considered low pressure.

So, a "can of compressed air" can have many atmospheres of pressure yielding a huge pressure difference, however a vacuum can will only hold at most 1 atmosphere of pressure difference.
Rocket propulsion is a non-linear acceleration, meaning on a graph with acceleration as the X axis and time as the Y, the acceleration/time line would curve upward to the right, in an exponential fashion. IE: at the end of the first second, its not really going anywhere in comparison to at the end of the 10th second.

Given the above data, that a vacuum canister can only support up to 15.7 psi of pressure difference because we cant make the atmosphere weigh any more than it does, and the fact that a can of compressed air can hold a pressure difference of many times that of a vacuum can it is known that the potential energy of a can of compressed air is going to be much higher than its vacuum counterpart. Furthermore, vacuum, and air rushing in to fill that vacuum does not produce thrust in any significant amount.

So, assimilate all the above data and its easy to see that even IF there was some sort of thrust produced by a pressure difference of at most 15.7 psi, and the vacuum vessel was punctured some how (while leaving a gaping hole and not sealing itself on the puncturing device) AFTER the vacuum is filled the can will not be moving due to the real world environment that it is in.

People, read the question again. It says

"The air blows to the left as it enters the can. After the vacuum is filled the can will:"

*After* the vacuum is filled. Not _while_ the vacuum is being filled.

If you think about it you will quickly realize that the can will not be moving after the air inside and outside the can is of equal pressure. It will have stopped if it was moving, or even if it never moved while the air was filling the vacuum it surely will be stopped AFTER the vacuum is filled.

It's a trick question. No need for Mathematica.

You've all done very well!

One example is a rocket ship losing mass, the other is a rocket ship taking on mass. Taking on mass will not stop the rocket, just gradually slow it down. When either rocket is out of fuel (first one empty, other one full, so to speak), they will still have momentum built up, and continue moving, not STOP.

Can won't move after filling.
Original pull to right will be exactly offset from inertia of air now compressing (at fill point) to left of can.

the can won't move. The air around the can is not under high pressure, so when the vacuum container is open, air will move in to equalize pressure,moving from higher pressure to low pressure. This will happen more or less pretty slowly. If the air around the can was compressed air, under higher than normal pressure, then after opening the can, it might move to the left.

Take a different approach. Lets put the can with pressure up against a wall. Imagine how much push (force) against the can you would have to add in order to keep the can against the wall without any pressure venting. That should be somewhat equal to the force of the venting action.

Now put the can with vacuum up against a wall. It holds itself against the wall. Start pulling on the can until the force of you pulling separates the can from the wall. The force you used should be somewhat equal to the force that resulted from the vacuum.

Now, think about it...

Edited by author 01-10-2007 01:24 PM
The can will move to the right.

Reasoning: The atmospheric pressure will push down on the can equally on both sides. However, there will be a brief time after the can is opened that the pressure will have nothing to push up against on the right side, but will on the left side, thus pushing the can to the right. When the gas reaches the left side of the inside of the can, there will be equal pressure on both sides of the can thus no more net force is being applied to the can. Assuming no friction, the can will continue sailing to its right at a constant velocity.

Of course the assumption of no friction doesn't make any sense since we know the can is moving through an atmosphere. So, the can will eventually come to rest.

Edited by author 01-10-2007 11:21 AM
After sitting with pen and paper for a bit, and a little help from mathematica I am confident that the net effect will be that the can will move to the right and come to rest.

The initial pressure difference will accelerate the can towards the right, but air resistance will push back to the left, and it will push harder the faster the can is moving. In addition, the pressure difference will rapidly vanish and with it the initial acceleration. Regardless, there will be *some* initial acceleration to the right, and the can will be displaced some distance to the right before coming to rest again.

Without reading posts.... I believe the can will move to the right, where the hole is punctured.

Fluids moving have a lower pressure then those that remain still. The moving air to the right, by the hole creates a low pressure zone, while the left side of the can remains unchanged. This creates a net force to the right, pushing the container to the right.

before reading any of the above posts:
my basic understanding of the Bernoulli principal is that the degreased pressure in front of the open side of the can will move it to the right...

I think the vacuum (better, higher pressure outside) will accelerate a volume of the air into the can; this takes a force, pulling the air in(the air has mass and does move, for sure) Also, the area of the inlet aperture, is subracted from the equal static forses on the ends, so net air pressure pushes it to the right. As the forces must equal zero ( net) I think the can moves to right, the unbalanced force is used to accelerate the can. The how of that is complex, re forces inside the can does not matter at first ? I see it as if the can is climbing an imaginary but real piston, the air mass..

Inertia of air mass compressability and decell of resulting moving air mass inside may move can the other way at the end?

Can will jump?

Michael Norman is right, the correct answer is actually C even though it's not apparent at first.

If you're in space and throw a 100lbs weight to your right, you will be moving to the left. I think everyone agrees with that and that's the first experiment, as the bottle ejects the air to the right, this makes the bottle move left and continue moving left.

But the second experiment is very different, because the bottle is not throwing the air, it is just shifting it to the left (into its cavity.) If you're in space and you shift a 100lbs weight from your right to your left, your body will move to the left, but since you are not letting go of the weight, your body will stop moving to the right as soon as you stop the weight from moving to the left. There is no friction involved.

The bottle in the second experiment is taking the air from the right and shifting it to the left a bit and then stopping it inside itself. The bottle is NOT throwing the air, it is just shifting it around.

So the bottle does shift to the right when it moves the air from the right to the left, but just like the man in space moving the weight around, the bottle comes to a complete stop when it stops moving the weight. It's not because of friction.

"but this air also pushes on the inside of the can with an equal force"

But it takes time for the gas to fill the can and equilibrium to be re-established. During this time the pressure on the inside of the can is increasing but it is *not* equal to the pressure from the outside. So the can is pushed right.

When the pressure does eventually equalise the push stops, but the can will continue to move right for a short time due to inertia. This should be an obvious result - your car doesn't stop moving the instant you take your foot off the gas, after all.

The can will move to the right. When the can is punctured the area of top exposed to external pressure is reduced by the area of the puncture. The area of the bottom is not changed. The can was originally in equilibrium, so that after puncture, the can experiences a net positive force on the bottom, accelerating the can to the right.

The can will not move.

Reasoning:
in the top diagram, the can will expel gas continuously at a decreasing rate till the pressure from the can inside and outside is equal... this gas output will push on the air ourside, making a resultant force to the left, thereby pushing the can.

HOWEVER:- In the lower diagram, the air that attempts to fill the vacuum will cause the can to gain some force to the right, much as a jet engine does via sucking at the air, but this air also pushes on the inside of the can with an equal force. Since the addition of two equal opposing forces is null, the can will not move.

But what if the can holds a carbonated beverage my friends, would this alter the shift in mass? A carbonated drink, entering the can also, would this vacuum be altered by the state of the liquid. HMMM, Indeed. one too ponder this is.

Edited by author 01-10-2007 03:38 AM
The can will not move. While I may not properly explain the physics involved I recall reading of a similar problem in "Genius" a book about Richard Feynman. (I solved it intuitively and correctly at the time) Feynman's problem placed a garden hose in a pool of water. Turn on the water in the hose and the hose will move opposite the direction of the water flow. Conversely, what happens if the hose draws water in? The hose remains motionless because water is pulled in from all directions. Similarly, the can moves when air escapes because force is exerted in one direction. When the vacuum in the can is broken the can remains stationary because the air flows in from all directions around the can. Again my explanation may not be perfect from the standpoint of the physics involved, but I do believe the two problems and the solutions are analogous.

Dogan, it is my pleasure to tell you that I can rip your lengthy post ( /m306 ) to shreds with this short paragraph. You claimed: "It is obvious that after the gas fills the can the [centre of mass] of gas shifts to 'left'." No it isn't. Before the experiment, the gas was merely all around the can; afterwards, it is all around and inside, too, i.e. equally distributed throughout the system. If it shifts anywhere, then it shitfs towards the centre of the vacuum you had at the original position of the can. Now that I've clearly shown that this claim is nonsense, all the rest of your reasoning, which was resting on top of it, falls down like a house of cards. Boy, that was fun.

Funnily enough, I almost fully agree with the conclusion at which you arrived :-)

Oh, and if you want to argue other points that are plain bullshit...I will tell you this without trying to use language to make myself look more intelligent than I am nor to belittle you:

1. True entropy (as defined by total heat/energy loss) is bullshit.
2. The only time that exists is now, meaning there is no such thing as time travel.
3. Absolute zero is an arbitrary number. There is no point where all "matter" stops.
4. A black hole is not a hole.
5. The big bang was NOT an explosion nor the beginning of all things...rather it was an event. The universe had no beginning and will have no end.
6. There is only one universe. The notion of multi-verses is pure nonsense. The word Universe has uni which means one...it's just a word to describe ALL things.
7. The great FSM is your noodley master that knows all...hehe

Firstly, the can will accelerate to the right - simple pressue. A bunch of stuff pushing from the left, not so much from the right. Look at the snapshot and add up the forces. Nothing to do with momentum.
There is no "wind" effect - or more accurately, the pressure of the incoming air will be less than or equal to the pressure from outside (because it is "pressured in" by the outside), so the net effect will still be a force to the right, or zero.

Momentum/mass arguements do not apply becuase the air is under pressure - therefore it must be pusing against something somewhere else - trasnferring momentum whenever it wants. eg, consider the can and the air in a box - at some point, you could have the can moving one way and a (smaller mass of) air moving the other and the *box* also moving to conserve momentum.

Wow Michael, you must have a really small penis in order to spout such arrogant nonsense. The only thing you need to pad is your underwear.

It's a hypothectical question, Brainiac, with no truly defined variables except the existence of a can and air pressure. THAT'S it. Period. Fini. Done. The End.

We know the can will be encouraged to move because of the pressure differences occuring on either side of it. How far it will move would depend upon the friction between the can (affected by the can mass)and it's surroundings and how much force is exerted by the pressure changes. Those same variables will dtermine when it will stop.

Just in case you missed it: The lower pressure created by the moving air on the right side of the can WILL make the can move in that direction.

PERIOD.

Get a life. You aren't as smart as you think you are.

Now with padding:
-----------------------------------------
Object____|__Force__|_Duration_|_Impulse
-----------------------------------------
Can_______|__P x A__|____t_____|__P x A x t
Air_______|_-P x A__|____t_____|_-P x A x t
Can/Air-__|___NA____|____NA____|_P x A x t +
System____|_________|__________|_(-P x A x t) = 0

P = Ambient pressure
A = Area of hole
t = Duration of experiment, from puncture to equalized pressure

Some preliminaries:
1) Viscosity acts over time, and decreases as the velocities involved go to zero. The question is asking what is happening at the *instant* after the can has filled with air. Therefore, those saying that the can will be stopped due to friction can be dismissed right out of the gate for not having done their homework.

2) Center of mass arguments only work when (and here they work marvelously) you consider that the can and the inblown air are moving at the same velocity at the end of the problem. You see, if you were to ignite a rocket in an inertial reference frame in deep gravity-free space (realistically, there is no such thing) the momentum of the rocket/propellant system never changes. The rocket goes in one direction, and the propellant goes in another, but the combined momentum of the rocket/exhaust system never changes. They rocket and the exhaust have momentum vectors that (if the rocket starts from rest) are equal in magnitude but opposite in sign. In our case, the "exhaust" gases are inside the can, and so have the same magnitude but opposite sign, *and* are identical. The only quantity of anything that is the same as its negative is the zero quantity. Therefore, the can/air system has no net momentum. None. Zip. Zero. Nada. Standing still at the end of the experiment.

3) There are a lot of bad physicists chiming in. For example, people who think or say that this is a trick question, merely because they have misread or misunderstood it. People who think temperature matters to the impulse and momentum equation. People who don't know what a control volume is or don't know how to use one. Note, these may be great people in many other areas of life. These are probably charming, caring, otherwise intelligent people who just happen to be out of their depths when it comes to physics and fluid mechanics. My fluids professor, who shall remain nameless here, used to say that no layperson would even think to claim that they understand brain surgery, but won't hesitate for a second to chime in on a fluid dynamics problem. He's right. It's amazing the false confidence that non-specialists have in this realm. Physics in general as well, but especially physics where it involves fluids.

4) Using even the method outlined in (2) there may still be some doubt, simply because it lacks rigor. (You could arm-wave about the motion of the fluid changing the pressure distribution over the surface of the can.) The rigorous way to look at this, I repeat, is to use the principle of impulse and momentum applied to a control volume surrounding the can. When you do this, you see that , (A) the can experiences a net unbalanced force to the right for the duration of the experiment, due to the hole as per Dogan (306), that is simply the
Area of the hole times the external pressure. The direction of this force vector is to the right. Furthermore, you have to keep track of the momentum of the mass entering the system through the hole. Again, you do this through the principle of impulse and momentum. The gas entering the can experiences (B) a force equal to the area of the hole times the external pressure for the duration of the experiment. The direction of this force vector is to the left. This quantity, force times time, is called the impulse, and it has units of momentum.

Preliminaries dispensed.

In tabular form, taking + to the right:
-----------------------------------------
Object | Force | Duration | Impulse
-----------------------------------------
Can | P x A | t | P x A x t
Air | -P x A | t | -P x A x t
Can/Air- | NA | NA | P x A x t +
System | | | (-P x A x t) = 0

P = Ambient pressure
A = Area of hole
t = Duration of experiment, from puncture to equalized pressure

At the instant that the pressures equalize, the system is not moving, for exactly the reasons I have stated in tabular form above. Really. This is not my opinion. I don't offer this tentatively for your consideration or approval. This isn't even an experimental result only provable to within the errors of your equipment nor is it subject to neglecting viscosity, changes in temperature as the gas decompresses, entropy generation, or any other consideration outside of an external force acting on the system. This is *the* answer. The momentum of the can air system is identically (think mathematics, not twins) the same as it was at the start of the experiment.

I hope this helps. I actually hope that maybe this will inspire some of the posters here to take up the study of physics generally, and fluid dynamics particularly, at a major university. It's a fascinating study that will, nonetheless, grind away at your ego even as it makes you a better problem solver. At the very least, I hope some among those posting here or reading the commentary will take seriously the prospect that hubris is not just for American Presidents* and characters in Greek myths.

Sincerely,
Mike "Suitably humble in other areas" Norman
B.S.A.A.E., University of Washington

*Any of them, not just the current one.

You're all wrong. I recant my previous posts stating that:

a: The can WILL move to the right because of the lower pressure to its right.
b: The can will stop when there is enough friction to make it stop.
c: Most assumed equations are BS because of the assumed variables.

Instead, the Flying Spaghetti Monster will kick the can and breed pirates so that global warming will decrease.

The FSM says that it can not allow cans to exist with vacuums for the FSM is omnipresent and there is no such thing as nothing.

Hail my noodley master.

hehehe

The can will go left. Conservation of momentum. Particles rush left, hit the can, transfer momentum to the can.

Edited by author 01-09-2007 10:04 PM
You are correct mpk, the can Will shift to the right in the course of the puzzle. However, it will not continue to move, as the air particles in front of our moving can will serve to decelerate the can. Why? Because the particles in front of the moving can are imparting slightly more energy to the can than those over taking it. Additionally any particles moving more slowly than the can in its direction of travel are over taken and these collisions will deplete the can's kinetic energy. This effect will occur regardless of the density or temperature of the gas. Thus the can is opened, shifts(nominally) to one side and stops.

mpk,

When the can with the vacuum inside is punctured and air rushes there is no stipulation that the momentum be conserved, they start as two seperate systems and end up as two seperate systems since the can has a hole in it. Momentum is only conserved for a closed system such as the can filled with pressurized air.

Bob--

Your assumption that your kinetic energy from a 3 story drop is mostly converted to heat is incorrect. The earth does, indeed bounce off of you like a billiard ball. There's a little bit of a discrepancy between your mass and that of our planet, so it doesn't move very far.

Edited by author 01-09-2007 07:28 PM
Tom M (304):

Imagine drawing a dotted line around the can system (some distance from the can). Gas molecules are crossing that boundary and imparting forces form all directions to the can in an equal and opposite way (pressure). Since the molecules can not impart any momentum to the can they "bounce away" (again crossing the dotted boundary away from the can) with the same magnitude of momentum they started with. The system is in equalibrium since there is equal "input" to the system from all sides, and an equal "output" with molecules bouncing away in all directions.

Now the can is punctured, for some time, there will be molecules crossing the dotted boundary and not bouncing off and away (rather they go into the can and bounce around). If you sum up all the molecules approaching and deflecting from the hole, and non hole, sides of the can, the hole side will have more arrivals and less deflections while the can fills.

This momentum is exactly conserved by the can moving to the right.

Hello!
Physics PhD candidate from Turkey here. That's a nice question and I also would like to comment on that.
 
Let me write my answer first: When the can starts to fill up it will start to move to the right and when it fills up compeletely it will come to a halt.

I think it can be proved in two different ways. The first proof is straightforward for physicists. The second proof is for those who would like to know about how the machinery of this works.

First proof: "Center of Mass (COM) Argument"
Think of the whole "gas + can" system. It is what we call a "closed" system, that it is not subject to any "external" forces. In this situation the COM of the system should stay same before and after all kind of crazy things happens IN that system. It is obvious that after the gas fills the can the COM of gas shifts to "left". So the COM of can should shift to "right" because the COM of "gas + can system" doesn't change place. That means that the can started from rest moved to the right and when all is done it STOPS.
For a detailed explanation of how the dynamics (forces, acceleration, velocity, etc) works see the second proof.

Second Proof:
OK! Here it goes: The can is bombarded from all sides by air molecules. But because they hit the can from all sides it is in equilibrium. When you put a hole on the can from one the right side some molecules that were supposed to hit that place go in the can so there is an excess pressure on the left side. That means a net force on the can: That means acceleration: That means it is gaining velocity to the right. But when the filling molecules reach and start to strike the left wall (from inside) the excess pressure will be balanced (it is a fact from thermodynamics). So there will be no net force. That means no acceleration. That means it will KEEP its velocity (once it gained from the initial acceleration). Now it is moving with constant velocity to the right. OK but how is it going to stop eventually?
When it fills up compeletely it will be moving to the right so it will "feel a wind" from ahead and that wind is going to put a net force on it: It decelerates and comes to a halt.
(Actually all those things happens not by order. They happen all together which makes things "slightly" more complicated, but the general idea doesn't change because it SHOULD MOVE for them to happen)

Now! What about friction? Well it is in there already. Friction in fluids are proportional to the velocity of the object with respect to the fluid (gas here). I called the slowing down effect: the wind. That's actually nothing but friction. So friction is not a big deal. The big deal is: I have assumed the temperature of the system is constant. That's important to make things easy (Although it doesn't have a direct effect on the answer of the problem): IT MOVES TO RIGHT AND STOPS WHEN IT FILLS.

Arguments like "negligible", "small" depends on your expectations from the experiment. It will move 1 micrometer? nanometer? picometer?
Will it MOVE? YES! And then STOP! How much move? Depends on every single other parameter that you can think of about the system (masses, volume, shape etc... That's hell of a question... for me at least :-))

One last interesting remark: If the size of the can is very very very small. I mean if it can be considered as a micrometer size "dust". Than it will do crazy things. It will jiggle and wiggle in all directions. That is called Brownian Motion. (In 1905 Einstein remarked that this observed phenomenon in microscopes is the direct proof for the existence of atoms)

The can isn't moving at the end. From conservation of momentum, the only way it could be moving is if the gas had non-zero net momentum at the end. Since the gas is in equilibrium at the end of the problem, it can't have any momentum. The rocket comparison is a nice way of throwing people off, the difference is that because the gas is being emitted into a vacuum it does not achieve equilibrium and can have momentum off to the right. This allows the rocket to have non-zero velocity after emitting all its propellent. Cute problem.

Answer: C, the can will not move.

The principle needed to solve this problem is conservation of momentum. For the can filled with gas the can and the molecules of air constitute a closed system in which the momentum must be conserved. So as the can is punctured and the molecules rush to the right the can must move to the left so that the total momentum is conserved (Think about sitting motionless on a skateboard and throwing a heavy weight in one direction, the initial momentum of you and the weight, i.e. 0, will be conserved and you will move opposite the weight). This is the same principle used for rocket propulsion in space. Now for the empty can, the can and the air molecules do not compose a closed system and the momentum will not be conserved. So when the can is punctured and the air rushes in (which is not the can sucking or pulling the molecules in but simply the normal motion of the molecules causing them to find their way into the hole) the can will stay put.

IMHO there is no acceleration to the left, rather a braking effect that eventually reduces the acceleration to the right to zero.

Anyway... Most of this discussion is amusing but irrelevant. Read the question again... This seems a classic case of deliberately misleading, irrelevant data in a test question, and it seems that most posters have taken the bait.

No Poopy no, you don' wanna argue with Bob D.

Bob D,

There is a problem with your argument -- the elastic collision of particles at the "prehole" spot on the can cannot impart mv momentum, since you would be getting energy for free this way. Instead, the particles give up half of their kinetic energy, i.e., the energy imparted is mv^2/4. Actually, this argument is still not quite right, since the particles incident to the prehole are not coming in at 90 degrees all the time, so their momentum is mv, where v is the net component of horizontal motion, while the v in the energy calculation is the square root of the sum of the squares of the velocities. These two quantities differ by a constant factor.

Here's a different way to think about it. Suppose you completely and instantly removed the right side of the can. For a moment, there is a force on the left side of the can which is not countered by pressure on the right, causing it to accelerate to the right. After the air slams into the inside of the can (i.e., once the system has reached equilibrium) the pressure on the inside of the can counters this force exactly, so there is no acceleration. During reestablishment of equilibrium, which takes some amount of time, the inside of the can slams into particles of air, slowing it down until it stops. So, initially, there is an acceleration to the right, accompanied by movement to the right, followed by acceleration to the left, accompanied by a drop in velocity until the can no longer moves.

Now, when the hole in the can isn't as big as the entire side, the situation is the same, except that the initial acceleration is smaller and the decceleration occurs more slowly.

I know what you mean.
I think mine has moved off to the left, and my skull is floating off to the right.
(Or is it the other way round??)

Bob, at least we agree it moves. To the right.
My brain hurts.

Whoops. Sorry, knew I'd do something silly. I used 1 fl oz for the can's volume.

Make that 7.2 ft/s for the can, giving it 0.17 J of kinetic energy, in contrast to the 36 J of potential energy it starts with.

Anyway less than half a percent of the initial potential is converted to kinetic energy of the can.

(And, strictly, I've got a feeling that I should divide the speed of sound by the square root of three if we're in three dimensions, but I'll leave that.)

Thanks tyler, but in this case, with your figures, the can would be given a velocity of 0.6 ft per second (unless I've done something silly - I did it a bit quick).

I've assumed the mean velocity of gas molecules entering to be the speed of sound, 330m/s, and the density of the air to be 1.3kg/m3. (I can only work in metric, sorry.) This gives the momentum of the gas entering the can. If this is the final momentum, we divide it by the mass of the can to get the velocity of the can. The can ends up with 0.001 J of energy.

The potential energy before, pV, is 3J.

Like I said, the can don't get much of it.

The rest goes to heat and to the kinetic energy of the air moving leftwards (which will dissipate to heat pretty quickly too as this air's momentum is diluted throughout the gas)

Bob D, Behold the power of the conservation of energy balance.

Potential energy of system before=Kinetic energy after + heat loss

Volume displaced*Outside Pressure = 1/2 mass*velocity^2 + heat loss

assume standard soda can=12 fl oz
mass=0.03lb
atm pressure=14psi

solving for velocity

Final velocity of can = 11.8 ft/s - any velocity loss to heat(friction)

Edited by author 01-09-2007 05:17 PM
I think I've now conclusively left the school of "can moves to the left, slows as the can fills, and ends up with a small velocity to the right", because I think I was wrong to assume that the momentum imparted by the gas that enters the box is the same as it would have been if it had hit the front (ie if the hole hadn't been there). I now think the can will move decisively to the right.

Here's my method. It might not be everyone’s ideal web-browsing reading material, so sorry about that, but – well – you did ask! If you prefer, jump to the second-to-last paragraph, which sums it all up in two short sentences – that might be enough to convince you.

If the hole wasn't there, the picture is obviously a zero-motion situation. Let's call that picture N (for no hole), and the situation described as picture H. If that's our zero, then it follows that the momentum of the system described (picture H) is precisely the difference between the momentum imparted by the molecules that enter in picture N and those that would have entered in picture H had there been a hole.

(Strictly speaking, we should also adjust for any other external forces that might change when the puncture is made - viz. the effects of air resistance when the can moves, and for any other pressure / turbulence / bulk motion of air in the external vicinity of the hole as a result of the air stream going in, but I can (if you like) argue that these will be small in comparison to what does, in fact, I think, happen.)

In picture H, when a molecule enters the box, it carries with it an average momentum of mv into the can. (v is the average leftward component of velocity of the molecules passing through the hole) This momentum, ultimately, will be retained by the can+contents. We are giving the can+contents a total momentum of Mv, where M is the final total mass of air that enters the can, to the left.

However, in picture N, when a molecule hits the surface where the hole would be if there had been one, it also carries an identical momentum mv to the left, which it imparts to the external surface of the can. It then LEAVES the surface of the can, again with an average momentum of mv, but this time to the right. As it leaves the surface of the can, by Newton's 3rd, the can experiences an equal and opposite transfer of momentum to the left. The molecule therefore imparts a leftward momentum of 2mv to the can. Adding up all the molecules that would have entered had there been a hole, we find that a leftward momentum of 2Mv is imparted to the can by these molecules in picture N.

(I'm pretty sure that the average rightward component of momentum of molecules leaving this surface will be the same as those arriving. I tried constructing diagrams of situations where this wasn't the case, and they seemed to allow me to invent perpetual motion machines, which generally indicates that they're wrong.)

Now, picture N is the zero motion situation - this is what's required to keep the can stationary. In picture H, when there’s a hole, we don't have the required 2Mv to the left - we only have Mv to the left. Therefore a net momentum of Mv TO THE RIGHT must be transferred to the can by the air around it as a result of the puncture. This momentum is acquired by the can as it is filling, and will be retained as long as it can keep moving against the resistance of the air in which I imagine it to be floating.

This to me is counter-intuitive, but there it is. It wasn't what I was expecting. The momentum received by the can seems to be exactly equal to the momentum of the gas entering it, just as the original can full of air punctured in a vacuum receives a momentum equal to the gas leaving it. I’m surprised.

This leaves the question of where this momentum comes from. Is it from the air? If so, why would the air end up moving to the left? Here’s what happens:

In picture N, molecules that hit the wall of the can where the hole would have been will bounce off, carrying their momentum rightwards into the external gas. In picture H, these molecules go into the can, and become part of the can+contents system. In the no-motion scenario, they would have contributed to the rightwards momentum of air molecules outside; in the punctured can scenario the air outside is deprived of this rightwards momentum. The air outside will be moving left as a result of the puncture. We don't need to use words like 'sucked' to the left, but that's what happens.

It might be easier to see that there will be more molecules leaving the left hand external surface than the right hand one for the duration of the filling of the can. Air is pushed off to the left, into the distance; the full can moves off to the right.

I've had fun with this. I'll keep looking, see if anyone can puncture any holes in it. Meanwhile, thanks for the discussion so far!

When air pressure on the outside of can is balanced (left side vs. right), the can stays still.

When suddenly there is an imbalance (no pressing on the right side and same ol' pressure from the left), the can moves to right.

Read the question again: It is "After the vacuum is filled the can will..." IMHO, after the vacuum is filled the pressure balance will be in a static state, and therefore the can will not move.
While filling is another story. IMHO during filling there is some force pushing the can to the right, because the pressure outside the can near the opening will be less than the ambient pressure. However this would be rather weak and it may not be enough to overcome the friction between the can and its environment (e.g. the surface on which it is placed). The effect of the gas hitting the bottom of the can should be negligible. But in any case that was not the question ;)

Julien, your assumption that gas creates pressure in all directions is correct. Think of this, even if the gas 'jets' to the opposite side first, it cannot 'push' that side with any more force than the pressure that supplied the jet. Since this pressure is the same all around the can, there would be a opposing force to that jet. What happens is you essentially get a diminished jet hitting the bottom of the can, with a larger force still outside the can, pushing it to the right.

Thanks Bob (290). I think I see what you're saying.
The assumption in my explanation (which I think is wrong) is that when the gas enters the can it instantly creates pressure in all directions. But as some folks pointed out, this is not true until the entering particles bounced on the bottom of the bottle.
This means that for a while, there is actually more particles hitting the box towards the left, than towards the right.
The question is whether that's enough to stop the box, or will it simply slow it down a bit... :-(

If you look at the initial state from the point of view of potential, it seems that there is potential and that it has to go somewhere (energy conservation). I don't know how to quantify that potential though. But from that point of view, it seems that the box should still move at the end of the experiment (again ignoring friction).

Julien - see my comment at 278. Pressure is only one part of this problem - the bulk movement of gas itself imparts a force to the box (force = rate of change of momentum), and it's directed decidedly to the left. Is it bigger or smaller than the pressure imbalance that you describe, or is it the same?

I think that in the vacuum case (2), the box will move to the right as the gas fills it, then when the gas pressure inside is the same as the pressure outside the box continues to move to the right.

Gas applies pressure, which is a force (resulting from the gas particles bouncing off the surface at random). That force increases with the amount of gas: vacuum applies no pressure. Also that force increases with the surface that is exposed: a smaller surface receives less force.

If you consider each side of the box in the vacuum case, you have pressure on all outward facing sides, but no pressure on the inside. Also, the side with the hole receives a little less pressure the opposite side. If you look at all the forces/arrows, you will see that they are not balanced. The box accelerates to the right.

Because the gas particles moves at random, gas will start entering the box through the hole. Until the pressure inside the box reached equilibrium with the outside pressure, more particles will accidentally run through the hole inward than outward. The inside pressure increases.

While the inside gas pressure has not quite reached the outside pressure, it will apply forces on the inside of the box, but these forces are not as strong as their pair (on the other side). If you look at all the forces/arrows, you will see that they are still not balanced (because of the hole), although not quite as badly as when the process started. The box continues to accelerate to the right.

Once the pressure inside is equal to the pressure outside (which will happen after some time), the forces/arrows can be grouped in pairs of equal and opposite forces. They cancel each other out and the result force is zero. The box stops accelerating, and continues its movement if you ignore friction.

The same logic explains why the box filled with gas starts moving to the left (case 1).

Hold the evacuated can still (clamp it).
Draw a box around it.
Pierce it.
Air is moving across the box surface with speed
v and density p, so there is a momentum flow
across the box surface. This is opposed by
a reaction force at the clamp. The momentum
flow is the opposite of that obtained from the
pressurized can case, so the reaction force is
in the opposite direction. Repeat the experiment,
but remove the clamp. Now the can moves, the
opposite direction of the pressurized can case.

Edited by author 01-09-2007 03:28 PM
In a real situation the orifice will not be perfecly symmetrical and aligned to the center of the can, therefore there will be a torque. It will rotate and move in an unpredictable fashion. How about that.

Don't neglect friction please. If there's air there's friction.
I know in school teachers told you to neglect friction, but forget about that. The can is not in a vacuum.

eolon -

"A rocket taking off from a launch pad, a flat surface just a meter, say, from the exhaust, will have the same reaction as a rocket taking off with the launch pad surface 2 kilometers away." That can't be true. Try blowing at a wall, and you'll feel the air hit your face.

 "The inside surface opposite the hole plays no part in the system". That can't be true either - a moment after the puncture, it's bombarded by the molecules entering the can / balloon while the rest of the inside walls are not.

Edited by author 01-09-2007 03:14 PM
You're right, tyler - if I hit the ground and get squished, energy is involved in squishing and rearranging me in energy-consuming ways, so it doesn't all go to heat. But well over 99% will. Think of something that doesn't bounce or break apart, like a bag of tough ball bearings landing on a steel floor from three storeys up - there's not much rearranging will happen there that you couldn't do by merely putting it calmly on a desk. Everything else is lost as heat. Heat exchange is most often the largest component of an energy problem, especially when we have a gas free to leave and enter an open system, so neglecting heat loss here would lead you very far astray indeed.

Edited by author 01-09-2007 03:11 PM
Pretend the cans are spherical; it makes me more comfortable with the inevitable balloon analogy.

The pressure balloon has equal force arrows at every point on the inside surface; for simplicity, we will only draw 4, in 2 dimensions, at North, East, South, and West. if we make a hole in the balloon at the North, that arrow is gone and we are left with just the forces at E, S, and W. The balloon will head South.

Now, do the same thing with a a vacuum balloon. Now the force arrows are outside the balloon. make a hole at the North arrow. The forces left are S, E, W. Because they are outside the vacuum balloon, the balloon moves North, or towards the hole.

You can ignore the incoming air hitting the South wall of the vacuum balloon in the same way that we can ignore the effect of the air in the pressure example striking the "outside" air, or hitting a surface outside the hole, like the ground. A rocket taking off from a launch pad, a flat surface just a meter, say, from the exhaust, will have the same reaction as a rocket taking off with the launch pad surface 2 kilometers away.

The same is true with the vacuum balloon. Consider a 10 km wide vacuum balloon with a tiny pinhole versus a 1 cm vacuum balloon with a 1 mm hole in it. both will move in the direction of the hole. The inside surface opposite the hole plays no part in the system.

.::.

Bob, Conservation of energy does require some lost motion becomes heat, but in neglecting heat loss due to friction, that energy has to end up in some other form. In your example, three stories of potential energy turn into three stories of kinetic energy right before you hit the ground. A moment later the kinetic energy is gone, but energy itself cannot be lost. Obviously you don't bounce the earth of of you like a billiard ball(maybe a tiny bit but let's leave the scientific numbers out of this), but that energy is not 'gone'. Its the energy that ends up breaking your bones and squishing your guts, and yes, some turns into heat, but it all needs to be accounted for and cannot be destroyed.

I've thought about it in several different ways. Please disassemble them if you wish.

Consider a piston, where the can is the shaft. Drawing out the piston compresses the air and drives it out, propelling it to the left. Then, drawing the piston in drives in air, which causes the can to come to a stop. It must come to a stop by accelerating to the right, otherwise you would be able to infinitely accelerate the can by continuously drawing in and out air. This drawing in of the piston is analogous the foop case. Thus, B. This is what I believe right now.

Consider a can as a single point mass, whether or not it has particles or not. For the poof case, when particles stream out, it is as if the mass is throwing out particles to the right, which propels it to the left. For the foop case, it is as if it were amassing particles moving left, which propells it to the left. Thus, A.

Consider an anti-momentum particle. Thus, in the foop case is as the poof case, except the the can is filled with anti-momentum particles and the outside is empty. Steady release of these particles results in the opposite of the release of regular momentum particles, thus propelling it to the right. Thus, B.

Edited by author 01-09-2007 02:17 PM
I read the answer to this one in some book about about Richard Feynman or some such. It was stated slightly different using a rotating sprinkler submerged in water. It doesn't move since when the air leaves the box it crosses the system boundary (the system boundary is an imaginary box just outside the real box) it leaves it in a streamlined fashion with a net momentum to the right. However, when the air is entering the box, it can cross the system boundary from all directions and even the opposite end of the box. The net momentum entering the system is zero.

Edited by author 01-09-2007 02:11 PM
No, tyler - conservation of energy just means any lost motion becomes heat. If I jump out of a three storey building, I approach the ground with lots of energy. A moment later, it's gone. The Earth takes 100% of my momentum, but none of my kinetic energy. (well, I say none, about 10^-21 percent, perhaps)

YES mpk, you have decribed the same free body diagram that I came up with, resulting in same conclusions.
Bob D I respectfully disagree with your assertion that Conservation of energy does not apply here. It is actually the best argument against the "no motion" or "it stops" theories here. We are given three choices for answers, two of which are motion, one is not. You can safely assume the potential energy turns into kinetic energy for 2/3 answers. All that potential energy has to end up somewhere, (it is not lost to friction, which we should, as reasonable natural philosophers, neglect anyway) So to assume it stops the can, 1/2 the energy first moves it then 1/2 the energy must stop it.
That argument, quite simply, both sucks and blows. So the only conclusion can be kinetic motion and please reference mpk 276 comments free body diagrams to see forces in the right(ie: correct)direction.

mpk (276) I like your style. However we don't have a straightforward situation of a static pressure gradually increasing in the box, always equal in all directions - we have a situation of air molecules streaming in in one direction. Each one will come in carrying a definite amount of momentum, which will total to a resultant pointing due left. All this will contribute to the leftward momentum of the box, causing it to slow down. And the momentum of the molecules that are entering, and therefore increasing the momentum of the system will be exactly the same as that of the molecules that would have hit the side if there weren't a hole there.

Force isn't just provided by static pressure, as your argument suggests - a dynamic force is applied to the inside of this box from the molecules entering it, equal to the sum of the rate of change of momentum of these molecules. Question is - is this force the same as it would have been if the hole hadn't been cut?

However, something I haven't thought about so far is that the molecules entering will give ALL their momentum to the box (taking, if you like, those that have entered, and subtracting those that have left again, of which there will be fewer). However those that bounce off the outside wall actually impart TWICE their momentum to the box, if they leave on average at an equal and opposite velocity. Still with me? If that's the case, then the box will definitely be left moving to the right, with a momentum equal and opposite to that of the molecules entering it.

If that's true, I'll change my mind and say if goes right.
I'll go away and think about it some more.
Please talk amongst yourselves.

It will not move. I liken this example to when you see cartoons grab the doorknob and then try to pull the door open while their feet are on the door.

My guess is that thrust is created but the pressure of it hitting the back of the can cancels it out.

Thrust occurs when something is pushed out a small opening under pressure. So in this case the rest of the world is the "combustion chamber" and the inside of the can is actually the "outside" of the nozzle. Unfortunately the gases rushing in to the can push on the bottom of the can.

This little msg board is driving me crazy :).

The can moves to the right.

We can not assume friction and/or particular levels of pressurization, they are not specified in the problem. The only thing we have is a container and an imbalance of gas pressure. In the second case, if it's more useful, imagine the vacuum container as being inside another highly pressurized container. Atmospheric pressure was never specified. The net effect should be the same.

The people saying the can is stationary due to conservation of momentum are missing a few crucial pieces.

The people arguing about "throwing masses" or "pulling masses" are applying an imperfect analogy. It would be more apt to say you were throwing masses *at* a plate to impart momentum on that plate. If you were inside, and in the center of, a container, and you threw masses in opposite directions (equally), those masses would strike either wall each imparting equal and opposite force on the walls (assuming elastic collison with those walls). If you repeat the steps, but one side is missing, then the container will move because only one mass strikes a wall, imparting a force on that wall (the other mass flies out of the container, not interacting with anything).

A rocket moves because there is gas escaping but that escaping gas is applying *no* force on the rocket. In a pressurized container there is *equal and opposite* forces applying all along the inside (kinetic theory of gasses, or the ideal gas law, much like the mass example above). The force on the rocket is from the gas striking the *inside* of the rocket not being balanced by an opposing force. The momentum of the gas escaping is important, but it's a proxy for the momentum of the gas particles striking the inside of the rocket opposite the hole (where the escaping gas is applying no force). This means there is now momentum being imparted to the inside of the container opposite the hole but not in the other direction. It's these particles, *striking the inside* that are imparting an acceleration to the rocket.

In the vacuum case:

For those saying the "energy on the gas entering the hole", the vacuum imparts no additional energy to the gas molecules. In fact they are the same as the outside air pressure. It's these "normally energized" gas molecules passing into container that make it seem like there is an energized flow.

Now the key to the problem is imagining a free body diagram and the forces applied to it. Imagine a box with big dark arrows pointing in on it in all directions (outside pressure). These arrows are caused by little gas molecules hitting the outside of the container in a random fashion. To simplify this description, lets assume it's not a perfect vacuum inside, therefore inside the container there are teeny-tiny arrows pointing outward (inside pressure). For every arrow, dark or light, there is an opposite arrow. These arrows all balance, therefore no force, and no acceleration.

At time t=0 (the hole is made), the teeny arrow and the big arrow that were applied at that surface vanish (where the hole was). However, there is still a teeny arrow and a big arrow on the opposite side of the container. If you sum all the forces, now the only remaining force is a big arrow, opposite the hole (from which we've subtracted the magnitude of the teeny arrow).

Now lets advance time... air molecules start entering the hole, the net effect is that (due to the kinetic theory of gasses) that the inner arrows start to get bigger (the inside pressure starts to go up). This may be the part that's causing so much confusion, but the force due to the entering gas molecules, is *applied equally in all directions along the inside of the container*. With the quantity of molecules and their rapid motion, any intial directionality should be "smeared out" as they bounce against each other. The jet really has nothing to do with it and is more of side effect.

At some point, time t2 before pressure has equalized, we compare the forces again. We sum the forces and we once again arrive at a big arrow, opposite the hole that's now opposed by a slightly larger internal pressure. This means we're still left with a net force pushing *towards* the hole reduced by a larger amount.

At the last point, t3, the pressure has equalized. If you look at the diagram there are now big arrows pointing in from all directions on the outside, but also equal and opposite forces pointing from the inside out. This results in *no net force applied*. However, no net force means *no acceleration*. No acceleration can be translated also as no *de*celeration. In other words, for the entire time between t0 and t3 there was a net force applied, and therfore an acceleration. This acceleration translates to movement. At t3, there is no net force, no acceleration, therefore no *change in velocity* and the container continues to travel along to the right.

Like all physics problems of this nature I'm making some simplfications, assuming ideal gas, eleastic collisons, and disregarding the pressure gradient (incoming gas doesn't equalize forces *instantly*), but the *net effect* is the same.

This is all very curious.

Seems I'm striking out alone for sanity and truth, like a solo warrior setting out for distant lands, ignored or dismissed by the world as crazy... all very romantic and glorious. I'm flying the flag for the school of "can moves to the left, slows as the can fills, and ends up with a small velocity to the right".

It's a small school, it seems. Only me.

I've set out some rather lovely reasoning in posts 103 and 205.

It's understandable that most of the posts here are from people who just turn up, splurge whatever they've thought of on the hoof and then leave without any reference to previous posts that argue exactly the same or exactly the opposite - but I can see there are plenty of thoughtful folk out there and I'm interested to see if there's anyone who's actually reading anything that's been written, who has at least a little grasp of gas physics, and wants to take me up on my reasoning. If not then I'll doff my cap, take my leave and go fly a little flag for something obscure (but correct) somewhere else. There'll be violins and stuff...

So, jump right, sidle left, anyone?

It won't move. Once it's filled, the pressure on the inside and outside is equalized.

The can will not move. The vaccum theory described below (the vaccum hose in the middle of the floor) follows my logic that there is no propulsion in the vaccum filling like there is in the air expanding and pushing.

The can will be moving to the right even after the pressures are equal. It will not be stopped by anything other than friction. Many arguements using the conservation of momentum here are flawed. Conservation of momentum actually tells us it will continue to move once it is in motion. When the pressures equalize, it will stop having the unbalanced forces that caused movement in the first place, but the balancing pressures do not stop the motion of can (eventually friction will).
I think law of conservation of energy solves this problem. Conservation of energy tells us the potential energy of the pressure differential will be turned into the kinetic energy of the can moving. When pressures equalize the can will still be moving to right per Newtons 2nd law. (neglecting heat losses, friction and other nitpicky stuff)

Can will move to the right. The air won't move into the can, the can will move to envelop the air that sits to the right of it. Then, it will soon stop , no friction, but not much kinetic energy either.

My inclination is to exclude friction from consideration here. There seems to be agreement that the difference in air pressure creates a force on the can pushing to the right. Of course, this force would not be sufficient to overcome the static friction between the can and the surface up which it is resting. But if that friction is ignored, the can will move to the right.

Still ignoring friction, the can would continue to move at a constant speed after the pressure has equalized. The right-ward acceleration is then zero, but without friction there is no force acting toward the left that would cause the can to stop.

c. - It might be easier to think of a piston in a sealed cylinder - If you were to draw back the cylinder, creating a vacuum, and then release it, the system as a whole would not move - in the case of the can, when you puncture it (let's say it's floating in the air) it would move towards the center of mass of the can AND the mass of the air, in this case to the right but "After the vacuum is filled" the momentums (momenti??) of the air and the can would cancel out and the system would not move.

The question asked is "what will happen after the vacuum is filled?"

Once the vacuum is filled the can will not move, as there will be no foce acting on or in the can.

However, if the question were "what will happen as the vacuum is filling?"

While it is filling there will be no movement by the can, unless there is a solid object nearby for the vacuum to act on. It would be like turning on a vacuum cleaner and leaving the hose lying on the middle of the floor.

Can will move to the right - briefly, until the vaccum inside is filled with air. The air pressure surrounding the vaccum tries to fill the space. Most gets filled by pushing air in from the right, but some is filled by pushing the can to the left to meet it. At that point it will stop.

Unpunctured, the can has equal, opposing forces on both sides (Fl and Fr), which keep the can from accelerating in either direction (stationary). Once punctured on the right side, Fl would move the can to the right until Fr impacted the inside face of the left side of the can. The can would then stop moving as equilibrium would be restored (assuming that dynamic friction would apply to the side of the can).
Picture your 2 hands holding a can sideways in mid air by applying opposing horizontal pressure on the top lid (right hand) and bottom (left hand). If the lid then burst by the pressure, your left hand would push the can toward the right until your right hand hit the inside of the bottom of the can, when the can would stop.
I AM WICKED SMART!

The vaccuum creates suction, like the aptly named vacuum cleaner. We tend to imagine that a vacuum pulls objects towards itself, but Newton tells us that the both the vacuum and whatever is being sucked up are being "pulled" toward each other (they are actually being pushed due to the decreased air pressure in the space between the vacuum and the object). Since the items being sucked up by a vacuum are typically far less massive than the vacuum itself, we only really see this force applied to dust particles and cookie crumbs, while the vacuum cleaner remains fixed in our hands. But on a frictionless surface, the force causing the crumbs to move toward the vacuum will also cause the vacuum cleaner to move toward the crumbs, albeit at a much slower speed.

In "poof", the can is analogous to the vacuum cleaner and the air outside the puncture corresponds to the crumbs.

it won't move as the air sucked in will come in from one direction (as the air pushed out is) but rather from a 180 degree arc. The force will not be sufficiently directed to move it. This is a variation (I think) of the problem described in the book Genius about Feynman about whether a circular-rotating sprinkler (that turns because water is forced out) would go the opposite way if placed under water and a vacuum was attached to suck water in. It didn't move, same reason, the water doesn't come in from a single direction.

Nope, I'm entirely wrong.

Before the vacuum can is punctured the pressure of the gas is equal on all sides. When the can is punctured, the pressure on the opposite side from the hole will be briefly higher than the pressure next to the hole. This pressure difference will push the can to the right.

Ok, I didn't really mean "space" but just that the can is not reasting on anything.

If we are talking about a can floating in space, the vacuum can should move to the left as air fills it. The air rushing in will exert a force on the inside walls of the can, forcing it backward from the hole.

The can will not move for the same reason that a said boat will not move if you attach a propeller to it to blow on the sail. The force of the air outside the can is the same force that is pushing into the can. The equilibrium of forces is achieved when the can is emptied of vacuum. There is no extra force available to move the can one iota.

I have realized that my earlier question is incomplete.

So there is in either case, a decreasing unbalanced force upon the can, causing it to move. F=MA. Now in the second (vacuum can) case*, there is air resistance force F(sub r) in the direction opposite to movement. However it is approximately proportional to movement squared and in the opposite direction. It will slow the can down, but won't stop it before the pressure differential becomes zero.

*actually, in a fixed size universe there is air resistance from the increasing external air pressure as well.

The can will not move, but the earth will shift a fraction to the left. Duh!

It moves to the right. You total up the forces. Both cases are merely special cases of the same equation. The forces on any part of the cans surface are the force of air pressure on the inside and air pressure on the outside plus the forces holding the can togther from ajacent portions of can. If you total up the forces you see what happens.

If you look at any section of the sides of the can, the forces are exactly ballanced by the forces on the section of can directly opposite. The same is true for the ends, EXCEPT for the the small section of end that is opposite the hole. If there is higher pressure INSIDE the can then the net pressure pushes the can leftward. If there is higher pressure OUTSIDE the can than the net pressure pushes the can to the right.

The can will not move. Consider the case of the sealed can. the can is stationary because the average forces on the can in every direction are balanced. The CoM of the system is at the center of the can( assuming a symetric can and an infinite atmosphere. Once pressure has equalized the center of mass will be at the center of the can once again. Unlike in the jet case where at the start CoM is in the can, and afterward the CoM is not at that location. There may be a transient velocity imparted to the can durring the transition between them, but this should be rapidly disipated via friction.

The important thing to remember here is the exact wording of the question. It asks whether the can will still be moving AFTER the vacuum is filled. My physics professor husband and I argued about this last night. There will be a slight move in the direction of the puncture (provided the friction forces holding the can in place are not too great), his answer. It will not provide enough of an acceleration to cause the can to continue moving after it is full (my answer).

I think that between us both we have the correct solution for the whole system, although the wording of the question means that the answer is c) not be moving. My husband pointed out that this is a great example of why multiple choice science tests are so troublesome, the people who say that the can moves to the right understand the system well, but aren't reading the question carefully.

Hick Ninja - who says the gas has zero momentum afterwards? A box just moved in it. It'll be moving. See my comments below.

Ok, I'm labouring the point now, perhaps.

Assuming that the cans are in some kind of enclosure (a bounded universe) with double the volume of the can then we can treat the two systems as analogous. In one the can has merely been inverted.

So, the jet of air in the upper can is rushing out of the can and pushes the can to the left.

In the lower can the jet of air is rushing out of the universe and pushes the can to the right.

Newtonian physics would suggest that the jet of air would cause a reaction in the opposite direction.

Z.

Danno: it will stop because of friction with the air not because the air will rush in.

The question is not "will the can move", rather "What will the can be doing after the vacuum is filled".

Assuming no friction and you manage to puncture the can without actually moving it (good luck with that!), the can should not be moving.

The vacuum will initially draw the can to the right, and then the force of the air hitting the end of the can will cause it to stop (conservation of momentum).

The can will move briefly to the right then stop moving. Consider conservation of momentum in both cases: In the first case, the air keeps moving to the right, and the can continues moving to the left, with velocities inversely proportional to their respective masses. In the second case, some of the air moves to the left until the pressure balances and the can stops the inward/leftward movement, and thus can will move right until this happens at which point the momentum will cancel again and the can will stop. Therefore, the can will have a brief rightward twitch, but no continued movement. Looking over the other recent responses, K. Becker in /m236 has it spot on, and probably said it better than I did.

moves to the right.

the air rushing in is irrelevant. The hole IS relevant.

The hole's creation removes the balancing force that existed before its creation to the opposite force that continues from the left side. The left side therefore pushes the can to the right.

Think of the whole system as your frame of reference, and it's a little easier. In the top picture, the air is forced out of the can to the right from the pressure difference. After the system reaches equilibrium, the air is moving to the right, and the can must be moving to the left to preserve 0 momentum in the whole system. In the bottom picture, the air has 0 average momentum before the puncture, and also 0 average momentum after equilibrium. Therefore, the can must also have 0 momentum after the system reaches equilibrium, due to conservation of momentum. So it would not be moving. However, the can will move slightly to the right until it reaches equilibrium. So it moves to the right until the pressure is the same inside and outside of the can, at which point the can stops moving.

To update my answer a bit, the can moves right when the air enters on the right and the can carries that momentum after the can is filled, so it will still be moving right AFTER it is filled until the friction of the air slows it to rest.

Initially the can will move to left slightly and hardly be noticeable. Then jump back to center then to the right before making a can balancing noise. Maybe spin then hit you in the foot then finally stop.

After the can is filled it won't be moving at all. Equal pressure inside and outside of the can will keep it stationary.

Before the vacuum can is punctured air is pressing in on it from all sides. When the can is punctured on the right side, the air pressure causes the air to enter the can, which lessens the net pressure on the right side, so the can will be moving right due to the greater net pressure on the left side.

The can will hardly move in the second picture because the air is being drawn from all directions into the hole. It may move a little bit...but nowhere near as much as the first pic.

The can will not move. The movement that occurs when a pressurized can is punctured occurs because of the release of energy from the can. When a non-pressurized can is punctured there is no release of energy from the can, rather there is probably an increase in the entropy of the system because of the equlization of pressure, hence no movement.

Jon's got it- if it moves at all (depending on friction) it should go to the right. Think literally of a little vacuum sucking in air or a squid moving slightly backwards when it siphons in water before shooting forward.

After a good night's sleep, I'm staying out on my limb and saying the can moves to the left. There is no wording in the puzzle that indicates that the can starts out in a vaccum but for the sake of argument, lets place it on a frictionless surface. Find an empty pop can, lay it on its side on a sheet of ice with the opening facing you. Now blow in the hole and watch what happens... Sure there's less air pressure on your side, but the can doesn't come racing towards you. The inrush of air carries the can away from you because the momentum created by the influx carries the can in the same direction, and continues even after the influx stops. The can goes left.

Ben (224) - Doesn't work like that, or else rockets wouldn't be able to work in outer space. It's just Newton's third law, action and reaction.

Edited by author 01-09-2007 09:03 AM
Again, I agree with K. Becker, but see my comments at 205 and 103. I think there will be some bulk movement of air to the right as a result of the initial brief movement of the box. This will continue after the box has filled, as the moving air is outside the box. Because of this, and conservation of momentum, the box will be moving left.

I agree that the can will move to the right in order to conserve momentum . . . but it feels like it should continue moving even after the can has been filled due to simple inertia . . . I guess I can see how that might violate conservation of momentum, though. I just have a hard time believing that the can will come back to a complete stop.

Actually, it's easier than one might think.
Forget the accelerations and such. Remember the critical point: Momentum is conserved. That is, Sum(m sub i * v sub i) over all masses, velocities, doesn't change.

Net momentum of the whole system before the can is opened is zero. (Yes, the air molecules are bouncing around, but their momentum sum is zero).

Now, imagine that the air and the can are in as big a box as you desire. How about a mile on a side?

Open up the can. Air has to move from right to left to fill the can - there's no other way in.

While the air is moving in the can has to move from left to right in order to conserve momentum.

When the air has finished moving, so, too, will the can.

The can will not move. Once the vacuum has been filled, the air inside the can will be the same pressure as the air outside the can. A normal can when breached would move only because it has been pressurized.

Hey, this is easy, isn't it?
The can will move slightly to the right as long as air is streaming into it,
but only if there is no friction to the can itself.
Air pressure will be slightly larger on the left as long as air streams to fill up the vacuum.
Might also depend on mass, weight and size of the can and hole.
If the can is large enough the streaming might cause enough motion in air to generate a noticable stormy influence and pushback.

The can doesn't move, though there might be an initial "jump". The air entering the canmust fill the can, so it hits the back of the can, but must also hit the indie front of the can.

It will move to the right: on puncturing, outside of can air pressure @ rt face will drop, a.p. @ left face of can will still be at atmospheric. it will be "blown to the right".
Assumed no resistance to can movement, ie weight of can, air resistance etc. Problem becomes more subtle if these come into play.

It will not move. When filled the can holds many times more pressure than the atmospheric pressure. Releasing that pushes a lot of stuff out. When it's a vacuum, the change in pressure is only equal to atmospheric pressure, which is comparatively not much and it doesn't pull that much stuff in. Besides that, what does come in hits the back of the can which counteracts any pull to the right.

I still stand by correct answer being (c) due imply to the fact that we are supposed to be determining the movement "after" the vacuum is filled.

Again, this is less of a physics problem and more of a word problem.

A force is a force, and an impulse is an impulse. Air resistance is a force, and it provides an impulse (change of momentum), decelerating you. If an impulse is then received that was equal and opposite to the one that started you moving, it will be overkill (because air resistance has slowed you), and it certainly will send you backwards.

This doesn't happen when you stop running, because when you stop running you don't do it by applying an equal and opposite momentum - you merely apply what you need to stop running.

Bob (217), air resistance slows you down, and as you become slower, air resistance becomes less. As soon as you stop, there is no more air resistance. When you run, it does _not_ push you "back again", it merely slows you down.

JJ, I wonder why you came to the conclusion that I ever said the can could possibly move to the left (I specifically stated the opposite), and why you missed that I clearly stated that "for all practical purposes" it wouldn't move (I did so on the very first line of my original post)? Is it because you're an idiot?

The can will be moving to the right. I am assuming a frictionless environment. Consider the problem using a triangular can, perfectly balanced on one of its points -- it would have to fall to the right. Now consider a spherical can -- it would have to roll to the right.

I am not sure how one would open the air hole without affecting the outcome, but that is part of what makes this stuff fun.

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

If the air outside the can is still, C will happen.
If the air outside the can is moving in either direction, pick A or B, depending upon the weight of the can, relative to the force of the wind.

Since the question clearly states AFTER the can has been FILLED with air - there is no more expenditure of force that could create directional movement.

OK...I reckon that when the can is punctured, the air starts rushing in, making less pressure on the right-hand side, so there will be a net rightwards force - the can will start moving to the right. It takes a little time for the air to cross the can - not much, but a bit. When the air hits the left-hand side inside the can, that will start to reduce this force, up until there is equal pressure inside and outside the can. So we start with a net rightwards force, and end up with no net force. So, assuming that the can is in a frictionless environment, after the vacuum is filled the can will be moving to the right.

Edited by author 01-09-2007 06:39 AM
The air escaping from the pressurised can will push against the air outside the can, creating thrust to move the can forward - so it moves. However, the air rushing into the can is simply filling a vacuum, so will not create any thrust and not move.

It will not be moving simply because the pressure on the outside of the can is greater than on the inside. When a hole is punctured, air fills the can and it will move very slightly, then stop.

it wont be moving. the air stream pushes the can to the right. at the same time it hits the bottom of the can an therefore pushes it to the left. both forces are equal. no movement.

Assuming that the can is resting on the ground meaning any movement will be owing to air activity.

It will not move after the vacuum is filled just as any empty open can (e.g. a can of Cola) doesn't move if left on a table.

All forces (air pressure in and out) are in equlibrium

I'm not thinking it should move at all. In the case of the air moving out, the puncture causes a displacement because it's under pressure and the equalizing force is being channeled directionally and in THAT direction. However, in the case of a punctured vacuum can, the air, presumably at something like normal air pressure, rushes in simply to fill the space within the can. Something inside the can might be jostled quite a bit (I'm thinking about the coffee grounds dancing when I open a fresh can of ground New Orleans blend) but the air moving in is just swirling around inside the container. And I don't really know it, but I suspect that it's probably not nearly as much air to fill the vacuum as was likely compressed into the first one. So, I'd guess the vacuum can stays put.

Nature abhors a vacuum.

Velocity is achieved by air molecules being compressed, bumping and compressing other air molecules.

When atmosphere rushes into the container, a balance with inner and outer pressure is the result. No propelling force occurs. The container 'doesn't move'.

Maybe obiwan (94) was right.

Yes, nex (215), I agree it sounds silly. But if you run forward air resistance does push you back again; and if the force that made you run forwards is balanced a split second later by an identical force in the opposite direction, you will indeed find yourself moving backwards.

When there's any resistance to motion, the impulse (change in momentum) required to stop you will always be less than the impulse required to start you. If the impulse stopping you is the same, it'll send you backwards.

It's just a little subtler than it looks. The way running works is counter-intuitive anyway - when you run, you're actually pushing the ground backwards, and it's friction (which is a resistance to motion) that accelerates you forwards. Would you dismiss that on the grounds that it looks ridiculous? A force is a force: you need the big picture, and to work out where an imbalance of forces might be.

so nex,
it goes right and left? and then it doesn't move at all.

wow, since you covered every answer, you must be right.
dig your thong out of your ass and make a choice.

it moves right!
see? its not so hard.

/m205 :
"I'm still saying it does a quick jump to the right as the air rushes in, and then slowly sidles off stage left. I seem to be alone in this, but I'll persist, as nobody's followed up on anything like my point 3 yet (see post 103)."

Please excue my cheekyness, but I dare say that it's just not worth following up on. The reasoning in that point is like "if I run forward, then air resistance will push me back again; therefore, if I run forward, I end up moving backwards."

/m199 :
LOL! My post ( /m194 ) became so long because (A) there are so many people here who consider only a _subset_ of all forces involved, think they have _everything_ covered, and therefore think they have a solution. I tried to show them that there is more to consider. And (B) I didn't have the time to make it shorter.

Do you think "someone told me I should get a girlfriend because mine was the longest" would work as a pick-up line?

The can will not move. The air pressure on the outside of the can will always be higher than the pressure inside the can until both can and external environment have equalised. As such, any pressure (force) exerted inside the can (to potentially move it) will always be less than the pressure outside (keeping it in the same place).

Well,...i guess it wun move at all...it's just air filling up "empty space"...NO MOVEMENT DUDES

The can will move to the right.

Anything that occurs inside the can will not have any significant effect on the forces and acting on the can as a whole and therefore its motion.

As air enters the can it creates a partial vacumm to its right. Since the pressure to the right of the can is less that that to the can's left, there will be net force exerted on the can to the right as equilibrium is restored.

The can moves to the right.

I see a few people are talking about forces being negligable, that they're so small they don't count. I'm afraid the question doesn't allow us to think like this. It doesn't mention the magnitude of movement and only specifies that movement exist. It may only move the width of an atom, but that is undeniably movement though admittedly tiny. Movement is what the question is about.

Contrary to people's thought then is the fact that air does in fact have mass. More than you might think. A largish hot air balloon will hold 40 to 50 tonnes fo it.

There are no sizes specified in the puzzle so I'll try to think big to show the forces aren't negligable.

Imagine a large container, the size of a hot air balloon. It's empty at the moment but to fill it we know will take about 50 tonnes of air. We puntcure on side of the container and wait. The container fills with air sucked from one side of it. We know that when the pressure is equalised 50 tonnes of air has been moved from the side of the container which was punctured, to the centre of the container. You can't just move 50 tonnes without creating a bit of force. We can plainly see that if the container doesn't move then a lot of force has been applied to the air to move it from the side of the container to the middle of it and that has to come from somewhere!

The container is caused to move to the right by the extra pressure on the left (or the lack of pressure on the right depending on how you look at it). It gains momentum, however small, and this keeps it moving once the pressure is stablised.

Yes it's the obvious answer, but you have to be able to justify it. I just can't see how all that air can move without any forces involved (it can't). Once any force is involved on the container, it has to move. Simple.

Edited by author 01-09-2007 05:04 AM
Actually, I changed my mind. The can will be moving to the left because of the amount of energy applied to the air molecules as they enter the vacuume. Also because of the poster that said he broke flouresent tubes and they went left (only real scientific data here).

To Neil: I don't beleive in your airflow on the left internal side of the can. The analogy with blowing in a cup doesn't work, because your cup is initialy full of air, thus by blowing in it you increase pressure on one side. But here, the air enters an empty area. Now with the speed of air particles (much closer to the speed of ligth than to the speed of an airflow), we can safly talk of the "pression" in the inside of the can (quasi-static hypethesis) (if the hole is small enough).

I think it moves right then stops.

peejaybee (160) has the best analogy. Imagine pulling a weight in space with a massless rope. You move toward each other until you collide, then you both stop. The can moves to the right as it is filling (cancelling the momentum of the air entering). Then, the air hits the other side, stopping the molecules and stopping the can. So, *after* the can fills (which is the question!!!!) the can stops moving. (if the can also moves because of pressure drop on the right, that motion is stopped by inrushing of surrounding air to equalize the pressure drop).

The counter example, opening the pressurized can, is analogous to throwing a weight or pushing off a weight in space. Both objects keep moving away from each other.

Think about how the problem would be different if the left side of the can was a 1 way permeable membrane that would let air molecules pass from right to left. Then it would be like pulling that weight in space, but not colliding.

Hope that helps.

We have a squillion people saying 'it will move to the right' - very good! - now how about answering the question?

Will the can still be moving after the vacuum is filled?

(has anyone had a look in the book?)

The can moves to the right.

The area of the vertical surface on the left is larger than the area of the vertical surface on the right. (since the surface on the right has a hole in it.)

Since the pressure outside on the can is uniform, then the force pressing the can on the left is greater than the force on the right. That means that the can has to move right.

I'm still saying it does a quick jump to the right as the air rushes in, and then slowly sidles off stage left. I seem to be alone in this, but I'll persist, as nobody's followed up on anything like my point 3 yet (see post 103).

There's a lot of talk about conservation of momentum implying that the box must be stationary once it's filled - but all you can say with conservation of momentum is that the box (and air inside) will end up with equal and opposite momentum to the air outside. So it tells us nothing unless we know the final state of the air outside.

If you agree that the box will initially move to the right (which most people seem to), then you'll probably agree that this would cause the external air to be pushed to the right too. My argument (please see post 103 for reasoning) is that, after the box is filled, momentum will have been passed from the box to the air pushing the air to the right, and the box will end up moving left.

This will slow down gradually by air resistance (there's no friction, but you can't say there's no air resistance), which I'm assuming takes place over a longer timescale than the filling of the can and the initial jump to the right.

It's like the song from the Rocky Horror Show. Only the other way around.

NB there are other posts by a Bob (hello Bob) but that's not me - I've added a D this time so I know who I am.

(Talk of conservation of energy is also pointless - there's loads of energy released when a vacuum is punctured. All we can say is that some of that will move the box, some will move the air and some will heat it all up, and it'll all add up to the energy released.)

This is another force balance question (like the airliner on a treadmill).

The puncture in the can will create an area of low pressure to the right of the can, air from all around will flow into the low pressure area. As the can is blocking some of this airflow a small force to the right will be applied to the can by air moving into the low pressure area.

However we also have the force supplied by the air flow hitting the back of the can (think blowing into a cup).

I think (although I havn't done the math) that somewhat counterintuitively the can would still move slowly to the left i.e. the force of the air 'blowing' into the can will be greater than the force applied by the air flow to the right into the low pressure area.

NOTE: this assumes that the static friction between the can and the surface on which it is standing is lower than the force imbablance.

Edited by author 01-09-2007 04:02 AM
The can doesn't move.

The question is set up to make moving to the right the obvious answer. But if the obvious answer was correct, it wouldn't have been worth bothering to link to from Boing Boing.

And the can doesn't move to the left, because that would be too silly.

So it doesn't move.

Quod Erat Demonstrandum.

This seems fairly straightforward to me.

there will still exists similar gas pressure on the left side.
there will be decreased gas pressure on the right side as it enters the vaacuum.
ergo, the pressure from the left would nudge it to the right.
this is not a political statement.

IMHO, the can does not move. In example one the can moves left because air "bumps" into other air. But in example two, there is nothing, just a vacuum, so the air goes in. Sure it bumps into walls, but it bumps into all walls at the same time. It just does not move. Then again, I never studied physics.

Edited by author 01-09-2007 02:44 AM
Mmmkay. There are two approaches to this problem, one is the tendency for the center of a mass to remain in a fixed position, the other based on forces of pressure as a gas achieves equilibrium.

The center of mass solution, which I firmly believe in but don't want to dig into a textbook to give a formal detailed proof.

The fixed position for the center of mass explains why rockets work in a vacuum, As the mass expands, the container (or rocket) must move for the center of mass to remain fixed. This is the explanation that the professor from Los Alamos gave us.

Applying this to the vacuum container becomes more difficult to explain. Imagine out in space a larger container filled with gas, In the exact center of this larger container is the vacuum container positioned so that the center of the vacuum is also the center of mass for the gas in the larger container. It could actually be anywhere but it's easier to picture when everything is lined up this way.

The vacuum container is punctured on the right side and the gas enters, unevenly filling the container. In order for the center of mass to remain fixed in the center, where the diminishing vacuum exists, the vacuum container must move to the right. Once filled, the mass is equal throughout the closed system and momentum causes the once vacuumed container to continue moving to the right until either friction or the wall of the larger container stops it.


The pressure and equilibrium solution, which I'm a bit sketchy on but still want to put out there.

For the pressurized container, and ignoring the force of the expelled gas against the external environment, the gas is expanding in basically all directions to reach equilibrium with the external environmental and applies a degree of force against the container in the area where the gas is exiting.

The same is happening inside of the vacuum container at the point of entry. The gas is expanding to reach some equilibrium and exerting force on the inside wall of the container predominantly where the gas is entering the container. It could also be argued that there is a negative pressure on the outside of the container at the point of entry pulling the container as well.

These forces move the vacuum container to the right. Once the pressure on the inside matches the outside, there is still momentum acting on the once vacuumed container so that it continues to move to the right until friction or some other external force causes it to stop.


In the best of all theoretical worlds for problems like these there is no friction to be accounted for.

@nex message#194
I think you need to find a girlfriend/boyfriend at a composition class or stop doing speed.

no discernible movement assuming the local pressure differential created at the side of the puncture and mass of the can aren't taken to ridiculous extremes. Stop trying to conjure up a war of analogies and just draw a vector diagram including the statics. I know I am no fun at all. I just love all the analogies meant as illumination. It reminds me of all the different ways people try to give directions - get a map and construct your own representation.

It will be moving to the right.

Once gas starts to vent into the can, a pressure differential
will be in effect to the right of the can (next to the hole) compared
to the left side of the can. This pressure differential essentially
"sucks" the can toward the right (more proper is to view that it is
pushed from the left). The pressure differential will subside quickly,
but before it subsides, the can will accelerate. Note that the can
will start to slow down due to drag once the pressure differential
subsides, but that is a separate issue.

no movement. Stop trying to conjure uo a war of analogies and just draw a vector diagram including the statics. I know I am no fun at all.

I didn't see anyone mention this, but forgive me if I missed it.

When the air rushes out of the pressurized can, the can moves because the air rushing out pushes against the air around it (action/reaction.)

When the air rushes in, it's rushing into a vacuum. Nothing to push against, therefore no movement.

For all practical purposes, it will not move, and I can prove it.

Before you read my lengthy comment (or not), here's the most important part: These experiments are actually quite easily conducted in your own home on a lazy sunday afternoon. You don't need a zero gravity environment; suspending the can on some string is enough to see if it would move or not. The answer is that in theory, you might be able to make the vacuum can move a tiny little bit, but this force is not even remotely in the order of magnitude as the force moving the pressurised can, so in practice, the vacuum can won't move at any significant velocity; you might as well say (in the context of this question) that it stays at rest. If you don't believe this, again: You can try it out and see for yourself.

STP ( /m102 ): Your reply is my favourite one, you get a gold star for this.

So here's the answer to the problem and to some misstaken comments here (henceforth TAP&SMC): The "opposite effect" would not be observed, there is no opposite effect. There will be some stuff moving around, and some of that stuff will consist of the can, but whether the movement of the can would be measureable in a real-world experiment is another question. This question is significant! See, you can't say, "whatever movement to the right, however miniscule, is enough to check answer B." Because gas molecules are always moving, by definition. (Otherwise the temperature would be so low that the gas would at least become liquid; some gasses in the air can even become solid.) So there are always molecules bouncing against the can, moving it a tiny, tiny bit.

On with TAP&SMC: The can will, with absolute certainty, not move to the left. (A reminder: Left is direction in which the _pressurised_ can moves! If you don't see the contradiction here, I have this perpetuum mobile to sell you ...) The idea that the can will move while it is filling with air, but come to rest in the instance the pressure is equalized, is extremely silly; this is comic book physics, not real-world physics. The idea that the vacuum can will move to the right, but never come to rest as it is inself in a vacuum ... err, well ... I don't have words for this :-)

Want to see explanations for these assertions? OK, but I'm not a physics teacher, so this will not be concise:

The "poof" effect does not work in reverse. This has been quite well researched and is widely known. A similar experiment is a lawn sprinkler (of the type that uses the water pressure to spin the top part around) submerged in a swimming pool. It will still turn under water, but if you switch it from "blow" to "suck" (Spaceballs style), it won't move at all. If you want to see an explanation for that, you can find it easily with your favourite web search engine.

A vacuum cleaner sticks to a surface because the air pressure above the sucking head is higher than below it. This is because the seal between the sucking head and the floor is sufficient to significantly delay the equalisation of air pressure.

In the can experiment, there is no seal of quite this type. Mind you, if the vacuum can was fitted snugly into a sealed tube, it will move upon being opened on one side. Without the tube, there is no pressure difference that big.

When you puncture one side, its area becomes smaller than the area of the opposite side, with less gas molecules exerting presure on it. But all in all, the surface area of all left-facing surfaces is still the same as that of all right-facing surfaces! It just takes a while until the pressure on all of these is equalized. As with the vacuum cleaner, this equalisation is delayed by a leaky seal, which in this case is the punctured side of the can. During that time, there is in fact a miniscule pressure differential and a tiny, tiny force acting on the can. The outside air/gas pressure to the right of the can is a tiny, tiny bit lower on the punctured side, compared to the intact side. The pressure within the can is a tiny, tiny bit higher on the punctured side. This adds up to something that is pretty darn insignificant, but not zero.

But at this point, there's still something we haven't considered yet: The moment the can reacts to this rightwards force, it moves towards gas molecules to the right, so these get to exert a greater leftwards force upon the can. The left side of the can will move away from gas molecules to the left, so these get to exert a smaller rightwards force, which on the bottom line is again a net leftwards force. If this miniscule leftwards force doesn't cancel out the miniscule rightwards force, it will at least make it pretty darn insignificant by one further order of magnitude.

None of this, by the way, has anything to do with "sucking yourself forward", by analogy to "blowing yourself forward". You cannot suck yourself forward. This has been conclusively demonstrated and is widely known; see above. Therefore, the vaccum can experiment will work _nothing_ like a turbine. You cannot use air like a rope on which you can pull yourself forward. (Try it ;-)

Also consider: the case with the pressurised can and the one with the vacuum can are two entirely separate experiments. There is no causal link between them. Therefore, conservation of momentum is no valid argument. The vacuum can is not influenced by a fictitous can from an unrelated, hypothetical scenario.

And there's yet another thing I want to mention: Everyone who says that the can will move, but must stop moving as soon as the pressure is equalized and no more force is acting on it, is wrong! If you throw a stone, it won't stop dead as soon as it leaves your hand. Yes, if an object is at rest, and no force acts upon it, it will stay at rest. But conversely, if an object is moving, and no force acts upon it, it will keep on moving.

Yet another thing to consider: The air/gas flowing in will hit the left inside surface, pushing the can to the left; but it will also hit the right inside surface, pushing the can to the right. The left inside surface is a wee bit larger, but while the air still moves in, pressure is a wee bit larger on the right side.

Oh, and I have yet another very important announcement to make. It is _not_ true that conservation of momentum says the can must be at rest after the pressure is equalized! "BUT", I head peterc ( /m47 ) say, BUT if there is no momentum before the air moves, than there can't be any momentum once it's done moving! This is wrong. What you really mean is that all momentum ought to add up to zero, but you forgot that gas molecules are always moving. If upon completion of pressure equalisation the can moves to the right, it will continue to move for quite a while, until friction has stopped it. It will therefore penetrate a _large_ volume of gas-filled space. (Tee hee, he said 'penetrate'.) It will therefore displace a large amount of gas molecules. On average, these will be displaced to the left, because the can moves to the right. So you have one solid, relatively heavy, but relatively small, object moving to the right, and many very light air molecules moving to the left. Thus, momentum _is_ conserved, even though the can still moves.

And Nate ( /m57 ): The essence of the poof is that air escapes out of the previously closed system in question. You have constructed a new closed system, out of which no air escapes, therefore you did not succeed in your efforts to connect a poof to a foop. So the bad news is that your thought experiment is invalid, but the good news is that your intuition was correct.

Someone (not calling names in this case) said: "The idea that the first can will move is (at least) questionable." Err, so you never inflated a balloon as a child, let it go and saw what happened? You must have had such a sad childhood; I feel sorry for you :-(

Many people see, in their mind, air rushing to the left. In a way, you are right; around the hole, air rushes to the left. But as a result, elsewhere air gently flows to the right. These molecules move slower, but they are much, much more numerous. Overall, air is flowing from outside the can to the inside of the can, and the left/right forces cancel out.

Whee, that was fun. I'm looking forward to next week's queef and feeuq puzzle!

Why does anyone think the vacuum can will go anywhere? The can moves from depressurising because the force of the gas is expanding from the opening point along the axis of movement which is defined by the opening and the direction it is most open towards relative to the centre of the cavity, and the shape of the nozzle also will have an effect.

The vacuum can does not move anywhere because the force of inrushing gas equalises by being mostly directed towards the wall of the can. In actual fact the main site of force in this scenario is the walls beside the opening will tend to want to collapse inwards.

Explosions cause a generation of movement away from the centre of the centre of the expanding force, implosions cause motion towards the centre of the expanding force - in other words force is being absorbed, not projected.

here's a nice visual... imagine one side being open and allowing them to escape... rocket. It's not a stretch to imagine the opposite case.

http://www.7stones.com/Homepage/Publisher/Thermo1.html

The can does not move.

The air rushes in, you may assume that the hole into which it rushes becomes like a jet (equal and opposite reaction etc etc).

But the air rushing in also streams against the back wall of the can pushing it thus cancelling out any chance of movement.

I don't know.. common sense? Maybe i need to get my vaccume cleaner an air tight box and a can and test this out.

No movement:

OK, I buy the 'it shifts to the right while filling and comes to rest again once full' argument.

Opening the can to the vacuum has the same effect as moving one of the walls instantaneously. The momentum that the can would have received from the gas pressure on the outside face still arrives on the inner wall, just a bit later, and there is no net force.

Oh and because there are collisions on the outside, oposite the hole, that are not counteracted, their is a net force to the right. The can moves to the right.

It's moving to the right afterwards.

I'd like to suggest that everyone who replies have their answer as the first line of their post... makes it easier to parse the hundreds of replies.

When the can is filling it moves right because there are more air molecules pounding on the left side than the right side (because the right side has a hole). Assuming the sides are symmetrical except for the hole helps to visualize this.

Once the can has filled there is no force on it, so it keeps moving until some frictional force stops it. The problem statement didn't include any frictional forces to stop it besides the air.

ok here goes another attempt.

Imagine the first scenario with the gas molecules being little billiard balls bouncing off each other and the insides of the can. Due to the random motion, for every bounce off the left side of the inside there will be a bounce off the right side of the inside in an equal and opposite direction so the two "bounces" cancel each other out. This is a simplified view of "pressure" the force acting equally on all inner sides of the container because these little molecules are randomly bouncing off of them. Now you puncture the container. When the molecules reach the hole they escape and don't bounce (don't impart any force on the can) but there are still random bounces on the inner opposite surface, it's *these* bounces that impart momentum on the can. Hence you get the rocket effect.

Now reversing the situation. The higher pressure on the *outside* of the container can be imagined as little billiard balls again, but they're battering the can equally from all sides in a random fashion (external pressure). You now open a hole on one side, now the billiard balls are battering one side, but not the other (there's a hole) so they start passing through the hole into the inside of the container.

As the billiard balls enter the can they start interacting with each other (but there are still fewer than outside), and start bouncing around in random fashion on the inside of the can (the pressure inside starts going up). While this is going on, there is still a steady battering on the outside of the container opposite the hole that is not counteracted. These collisions on the outside impart momentum on the opposite side from the hole. As the relative number of "billiard balls" on the inside reaches the same number as the outside the random collisions inside and out now cancel.

the can is traveling to the right.
once the hole is punctured on the right side, as air enters the can, there will be slightly less pressure on the right side of the can, thus causing the can to travel to the right were it encounters less resistance than the left side..

the force that stops the can is the air squirting in from the right

pretty obvious

Mike-
Thanks. i was wondering if anyone would catch that.

OK, so you agree there is a force acting while the can fills - the can accelerates while it fills.

there is no force to stop the can. so the can continues moving after it is filled.

at the beginning there is a positive force on the left hand side of the can which is greater than the force on the right side of the can. (Same pressure, different surface areas). The can therefore begins to move to the right.

conservation of momentum: after the vacuum is filled the can is stationary

explanation: the air entering from the right has momentum and slows the can as it hits the left hand wall of the can. The up/down components balance out.

the difference in potential energy between the initial state and the end state (an increase in entropy) is accounted for by the higher velocity of the air particles inside the can (higher temp of the air) in the end state.

You are right, G!

The box will not be moving after it is filled. When the air begins to fill, the box will move to the right. But, once everything settles the box will not be moving.

CDKS: I am sorry, but mrscienceteach is right.

By conservation of momentum, the initial total momentum is zero,so the final total momentum will be zero. When the box is pressurized and punctured, the total momentum is still zero, but since the gas has rightward momentum, the box moves to the left. When the box is evacuated and punctured, since the gas has no net momentum after things settle, the box must also not have any momentum.

Edited by author 01-09-2007 12:40 AM
The first step to this problem should be identifying the body in question. Only external forces cause change to the body (or something like that).

In the first case, if we draw an imaginary boundary around the box, it can be clearly seen that the escaping air is an external force to the right. Therefore, an equal and opposite reaction will cause the box to move to the left.

In the second case, drawing another imaginary boundary around the box, no external forces are present; therefore, no change is done to the system.


.... right?

You are correct Bob, the question says "after the vacuum" is filled.
However you are incorrect to assume when it will stop. It will stop moving when friction exceeds it's momentum. In a zero G environment, the can will move for a longer period.

In the first case (pressurized can), a LOT more air is moving,
because the can is pressurized. Plus, the air surrounding the
can offers resistance to the escaping air, it's essentially
pushing against it.

In the second case, air enters the can only until the pressure
equalizes, then it stops. As it enters, it does not "push"
against air that's already entered in the same way it "pushes"
while escaping.

So I say the can does nto move.

The question says "After the vacuum is filled," not "while the vacuum is being filled." The can will experience a movement to the right while the can is being filled, but because of the viscosity of its environment (it is not in a vacuum) it will stop moving before the pressures inside and out are at equillibium.

thompat: there's an Air and Space Museum

Edited by author 01-09-2007 12:14 AM
Again, the can will move because of a lower pressure on it's right side rather than it's left. I'm not sure how you are missing that. It is a minute, but it is still there.

Rocket propulsion is based on the conservation of total momentum of a system. When the pressurized vessel is punctured, thus causing the the gas to be evacuated, the gas within the vessel must move in a fashion to balance the momentum conveyed by the exhaust; it must move in the opposite direction and of equal magnitude. The vessel itself moves because it encapsulates the pressurized gas, which, in turn, is moving in the opposite direction of the exhaust.

When the pressurized gas surrounds the vessel and the vessel is subsequesntly punctured, the same physics applies. However, since the vessel does not encapsulate the volume of gas that is reacting in the opposite direction of the exhaust, the vessel will not move.

The can will definitely will be moving to the left after the vacuum is filled, but only IF the hole is large enough to admit air fast enough for the moving air's mass momentum to overcome any incipient friction holding the can in place.
I used to work at a retail store, and we collected the old flourescent light tubes. These were quite long, about 8 feet in length, and still had a partial vacuum inside. For fun, we would line them up on a table, take a screwdriver, and whack off the end. If you hit them hard enough, the glass would break cleanly and knock off one end while the rest of tube remained intact.
The air would whoosh into the tube, and initially there was no movement. But a few fractions of second later, the mass of air entering the tube would impact the far left end, and the entire tube would take off like a rocket, to the LEFT according to the original problem definition.
Try it! It was sure fun for us! But be safe and wear safety glasses, and be ready to clean up a big broken glass mess afterwards.

To the people who assert that the can does *not* move, I have these 2 questions:

How do you account for the net leftward movement of air?

Has the center of gravity moved?

No, CDKS, momentum must be conserved. With the positively pressurised can, the momentum of the can is balanced by the momentum of the air rushing the other way. When the air is all gone from the can, it's still rushing away from the can. Total momentum of the system is zero.

When the can is negatively pressurised, the air rushes in to fill the vacuum in the can but is stopped by the can's walls. If the can is moving at the end then it must be balanced by some air moving in the opposite direction. But there's nowhere for the air to go: it's stuck in the can. Ergo, the only way to conserve momentum is for the can to stay still.

Forget pressure: think of it one particle at a time. The reason it moves to the left in the first case is that the particles all move at random, and there are more particles knocking it to the left than there are to the right (because of the hole). In the latter case, it moves to the right: particles hit it from every direction, but fewer hit it from the left because there aren't as many places to hit.

Correct observation that is added to the pressure change.

It will move to the right as the can is filling with air. The reason for this is the momentum of the gas moving to the left has to be balanced by something moving to the right. The only thing that can move to the right is the can. When the can is completely filled, the can will stop moving.

"Not moving
Conservation of momentum. Zero total momentum in the system before rupture = zero total momentum in the system after rupture."-mrscienceteach

Wrong. Flawed logic. The pressurized can had no momemtum and it moves...once punctured. It's called potential energy. Cans with positive an negative pressures both have potential energy.

I hope you aren't teaching my kid.

correct! it will move right till the point of fullness. the air that is "gobbled" and will create negative pressure - causing it to move forwards - "pulled" towards the right.

Not moving

Conservation of momentum. Zero total momentum in the system before rupture = zero total momentum in the system after rupture.

Edited by author 01-08-2007 11:35 PM
Wrong. Like I said, the lowered pressure on the outside of the fan will "pull" it in that direction. The air entering the can will not be uniform on the left wall of the can. Geez people, go to school.

To quote that provided site: "....so the can will be moving very slightly".

For year physics teachers say that the neutron has no mass because it is so small it is almost nothing. Well, something isn't nothing. Mass is mass. Movement is movement.

Here's another way to look at it. I'ts just the reverse of using a fan mounted on a sailboat to try to blow air into the sails to make it move forward.

The can will not be moving.

An air filled can acts like a rocket, tossing air out the back and flying forward due to conservation of momentum. Imagine tossing a ball in space.

The vacuum-filled can sucks in air, which makes you think that it will be like an inverse rocket, except that the air that is sucked in will slam into the opposite wall of the can, resulting in a net change in momentum of zero. Imagine pulling a ball with a rope in space, and what will happen when you slam into the ball.

Just to add some more fuel to the fire; it is only one source, and an internet one at that, but may help:
http://van.physics.uiuc.edu/qa/listing.php?id=335

The can will move to the right,because the pressure change will cause a lower pressure to the right and will continue to move to the right until friction stops it. Period.

As air to the right rushes into the can, it produces a semi-vacuum to the right of the can, causing it to wobble to the right as it is 'pulled' in that direction.

The can does not move at all.

The momentum of the air particles rushing in to the can are exactly balanced with the momentum of air particles around the outside of the can rushing to replace them. Air pressure on the back side of the can is unimportant and virtually entrely counterbalanced by the area of the front of the can which does not have a hole.

if you accept

a) that the pressurized can moves to the left when opened in a vacuum
b) that a can filled to ambient pressure and opened would remain stationary

then you have to accept c) that the evacuated can, when opened to the air moves to the right.

the forces in case b) are the sum of a) and c), so the vacuum filled can will move to the right with the same momentum that a pressurised can would move to the left.

All forces act between the parts of the system, so total momentum is conserved. Beforehand, everything is at rest (neglecting little air particle wiggles), so total momentum is zero. As air is rushing into the can, this is momentum to the left, so the can must move right. If the can is much more massive than the air (probably) this may be small. Once filled, when no more air is flowing, the can must again stop.

Will not move. The pressure surrounding the can stays constant.

The can will move to the right.

There is an unbalanced force on the left side of the can. This will push it to the right.

The can will not move, as the air moving into to the empty can will not do so with enough velocity so as to overcome the friction.

Not that hard. A vacuum cleaner sucks to a surface - i.e., the vacuum wand pulls toward where the air is entering. Try it and see. So, the can will move right, which makes sense when compared to the air expulsion example.

Another way to look at it: let's say the can is pressurized. You open the can, air goes right, can goes left, net momentum is zero. Fair enough.

Can is a vacuum. You open the can, air goes LEFT... if the can goes left too, we've just magically made everything in the whole system go left without ANY corresponding rightward movement. Can't happen. So the can has got to go right.

The intake of air into the right side of the can will create a low-pressure zone on that side. Therefore, the higher pressure on the left side of the can will be the only force not in equilibrium.

The can will move right, and continue to move right until brought to rest by friction.

For a comparison, consider the "caterpillar drive" submarine propulsion system. Water is drawn into the submarine, creating a low-pressure area ahead into which the submarine "slides".

The can feels a force of air pressure. Make a hole in the can on one side and the force decreases as the area decreases. Hence the can moves right. After equilibrium it is still moving right.

Simply the propulsion of the air entering at the right side is offset by the force of the air hitting the left wall of the canister.....no movement.

"After" the vacuum is filled the can will....

The can will not move!

The question is not "during" the vacuum being filled but "After".

The air entering the cannister will disperse to all sides of the container and not effect a movement of the cannister. The question is will there be a low pressure situation on the right strong enough to drag the can to the right and I vote no, that any low pressure will be more easily brought to equilibrium by movement of air only. Yhe cannister never moves.

Yeah. It's going to move to the right. The important concepts in this problem are conservation of energy and momentum. There's a pressure differential between the inside and outside of the can before the puncture, which means there is potential energy in the system. When the pressure is equalized, the potential energy in the system is gone, which means the energy is distributed between heat energy in the gas and kinetic energy of the can. Thus the can cant be standing still when the pressure has equalized. (note, the can may stop moving due to friction eventually, but immediately after the equalization of pressure, it will be moving)

Which way the can is moving can be determined by a little conservation of momentum analysis and some free body force analysis. Since we know the momentum of an object is proportional to its velocity, and that momentum is just the integral of a force over time, we can determine the velocity and its direction by looking at the forces on the can over time.

The can is the free body we are interested in, and the only forces that could possibly move the can are the forces on the inside and outside of the can's skin caused by the pressure of the gasses. We can assume that the hole is perfectly centered for this problem. If this is the case, the forces around the circumference of the can will balance out, since the pressures outside the can are symmetrical around the cans axis. In this picture, this means that the force on the top and bottom edges of the can will always be equal and opposite. Thus the only forces we need to worry about are the forces on the left and right sides of the can.

To calculate the force caused by pressure on a surface, you just multiply the pressure by the area of the surface. At the beginning, there is no pressure inside the can, so the force pushing from the inside of the can is negligible. The forces from the outside are the only ones you need to look at immediately after the can is punctured. Obviously the surface area of the edge of the can with the hole is less than that of the edge without. Since the pressure at the beginning is constant throughout the gas, the force on the right side pushing left will be less then the force on the left side pushing right. By this analysis, it has to accelerate towards the right when you puncture it.

Even as the pressure inside the can increases, The force on the right edge will always be leftward, and the force on the left edge will always point right. But the force to the left will always be less than the force to the right, since the pressure on the inside is always less than or equal to the pressure outside.

So since we know the can is moving, and since we know the net force on the can is always to the right, the momentum (ie the integral of the force over time) has to be finite and directed to the right.

Word I just solved this problem instead of going to the gym.

Assuming a frictionless enviroment and little mass to the can, it will move to the right as it fills. When the pressure becomes equal it will be moving to the right. The opening will "scoop" more air, causing the pressure in the can to be higher, therefore the last movement will be to the left.

Edited by author 01-08-2007 09:44 PM
Won't it move to the right, since, by sucking vapor in, it's decreasing the pressure on its right side.

I'm assuming it's in a frictionless environment.

I like the unicorn theory, too.

Mark, we need an answer here.
I think that everything has been covered outside of taking the can into space and puncturing the can with a Unicorn Horn-- Which changes the entire experiment.

Everyone knows that a unicorn couldn't live in space.
There's no air in space....

Assuming that the particles enterring the can become assimilated into it's mass, similar electrons surrounding a nucleus, in that they aren't physically built into the can but are included for the sake of experimentation as part of the can, the can will move to the left. Physics tells us that an object will always move to a less densely populated area. The particles rushing in would not be replaced quickly enough to ballance off the change in pressure and the can would move to the right until it had been filled and the particles surrounding it are dispersed evenly again.

won't the can just collapse in on itself? it won't move.

The can will move to the right at first, and then stop when it hits equilibrium. Conservation of momentum shows that when the gas particles stop moving, the can must have stopped as well. While the air is flowing into the can, the can will move to the right, but as the particles hit the back wall it will slow down, coming to a stop at equlibrium.
Before the gas particles hit the back wall, you can say that the can is moving right due to the pressure differential. Air pressure acts on all parts of the can, however there is a void on the right side where the pressure does not act, so the can will move to the right.
You can also look at the center of mass of the system. By conservation of momentum it must be stationary at all times. While gas flows to the left, the can must move to the right, but when the gas stops moving, so does the can, slowed by the impact of the gas against the back wall. The can will end to the right of its starting place, but it will be motionless

Again it will move to the left (at least if the canon-shooting aspect is true).
The first example is comparable with a recoil.
The second one would be an impact.

The can will not move, but the earth will. The earth will move toward the can and the amount of movement can be determined by the size of the hole in the can and the size of the hole in your head.

I think the can will move to the left again.

In the original case, the air coming out of the can was pushing against the air already there, thus moving the can left.
In the second case, air entering the can pushes against the back of the can, moving it left.
I think. :)

In a real system, there might be a weak area of low pressure near the puncture that might cause a weak movement to the side of the hole (the right). Think of a vacuum end underwater next to your hand. It could close the gap as water rushed into the can.

The force would be very weak, and in an ideal system (where pressures equalize throughout the system instantly), I'm not sure there would be any at all, as the "thrust" would be against all internal surfaces of the can such that they cancel each other out.

Primary (ideal system) answer: No movement

Fudging (for real systems) answer: VERY small movement to the right....

Covering all my bases (even though they're wrong) answer: Definitely movement to the left...

The can will move to the right because the molecules entering the vacuum will create a lower pressure to the right of the can. The can will shift to the right towards the area of decreased pressure. maybe.

When a can of compressed air is punctured, the air is directed in one direction as a jet. This will force the can in the opposite direction given that the force of the jet is powerful enough to overcome the can's inertia and any friction on any surface it may rest on.

Consider a fan. Place your hand on the exhaust side and you feel force. Place your hand on the intake side and there is very little force, and little cooling effect. The air is being taken from every direction other then the direction the air is being forced.

So in the case of a can containing a vacuum that has been punctured, air is drawn in from every direction from the outside. There is not enough directional force to move the can in any one direction. In a weightless condition, the incoming air's force to the left against the inside of the can, will cancel any force to the right caused by any suction. The can will not move. When the can is full. It will continue to not move.

c) The can will not move

This type of problem is analogous to the "Reverse Sprinkler" problem.

yeah, it's a word problem. after the pressure is equalized, the can is not moving.

I'd further point out that there is pressure pushing on *all* sides of the can, it just so happens that at that one point, the foop, it is actually entering the can. It is not creating thrust, or lift. So it's not moving.

bk

bretorium.com

It would move to the right as it displace the vacuum inside.
Visualize a floating cylinder in which the right end disappears, I would expect it to jump to the right to fill with air. Simplest way to collapse the vacuum, what with pressure applied to all the sides of the cylinder except the missing end.

The can will move to the right.

This follows from first accepting that in "poof" the can will move to the left. Then, we invert the universe. The inside of the can becomes the outside, and the outside the inside. As the compressed air now moves left, with the same logic as in "poof" the can will move to the right.

It is perhaps easier to see this while keeping in mind the intermediate phases while the universe inverts itself, when for example the vacuum and non-vacuum spaces are of comparable size.

This problem is analogous to a person standing on the right end of a raft and moving to the left side of the raft. The raft will move to the right. We can apply the same logic to this: the can will move to the right. This follows from conservation of momentum.

Ned (114) has the right idea here. The CV method precisely shows what will happen.

The key bit here is to look at the state after the pressure is equalized.

In the rocket case, after the can is empty there is mass of the can moving to the left, and a mass of air moving to the right. The momentum of the two masses is identical/opposite, the center of mass of the can + air system hasn't changed.

This is true in the 2nd case as well. There is no external input, so there can be no change in the overall center of mass. For the can to be moving after the pressures have equalized, there would have to be some mass of air also moving in the opposite direction. But any motion of air the left while filling the can can't continue because the can is closed, the air is stopped by the can itself. The can may move a bit while pressures equalize, but the motion cannot continue once the pressure inside the can matches the pressure outside.

Now, what happens if there's a microscopic black hole in the can that keeps sucking in more air? ;-)

Another good example of why the can will be at rest after it has filled with air would be to think of if the pressurized can did not empty itself into a lower pressure environment (or a vacuum), but instead emptied itself into the vacuum can. Imagine the two are taped together and sealed such that no air can escape. You start with the high pressure air all inside the can on the right, then create a puncture allowing the air to travel to the vacuum can on the left. Momentarily there will be a net momentum of air particles to the left, which will move the two cans to the right, but after the pressures have equalized the two cans must remain stationary. It would be like an astronaut with a medicine ball tethered to her leg. She can throw the ball and propel herself backwards, but when the tether becomes taught she (and the ball) will stop moving.

If this were not the way it worked you would be able to build devices made from high pressure/low pressure can combinations that could propel themselves without any countering force on the environment. A bit like pulling ones self up by ones bootstraps.

Consider a vacuum cleaner: as you apply the nozzle to a surface, what makes it stick? Consider an airplane wing: what gives it lift?

The pressure on the left end of the can will exceed the (temporarily lowered) pressure at the right. Net force is to the right.

thompat(108,112) and AM (88) are misinterpretting a fundamental law of physics. It is not the case that when the pressures are equal that the motion stops. Once the pressures in this example equalize, then the momentum of the object will no longer be changing. This is not the same thing as saying it has stopped. Consider an object falling toward the moon. Because space is a vacuum, the object will experienc no pressures anywhere, yet it will move (indeed accelerate at approximately 1/6 earth gravity) toward the moon.

A bullet, when fired from a gun in the vacuum of space, will not stop when it leaves the barrel because it's "pressures have equalized". Rather, it will stop accelerating and continue at the velocity it had when it left the barrel, in the absence of gravitational fields, induced currents due to slicing through a non-uniform magnetic field, blocks of wood, or etc.
The same thing happens to our can. The only difference here is that the velocity ofthe can and it's contained air at the end of the collision is zero, for the reasons I have mentioned in post 70.

This *is* a physics problem. It's even a simple one. But it is a *subtle* one. The solution really does come from the principle of impulse and momentum. The can experiences a net force that is the difference between the areas of the two sides (holed and hole-free) for the duration of the experiment. The air crossing into the can experiences the same exact force for the same exact amount of time, but in the opposite direction. Newton's Second Law, the time rate of change of momentum is proportional to the applied force, integrated with respect to time yields the Impulse and Momentum equation. Simply, Force x Time = The total change in momentum. Therefore, the momentum change of the can and the momentum change of the air inside the can at the end of the experiment are equal in magnitude but opposite in sign. Opposites may not attract, but they do cancel each other out when the meet. The can is definitely at rest at the end of the experiment.
Interestingly, the can will lurch when the hole is initally opened, and then slow to a stop on exactly the time scale of the experiment.

I remember reading somewhere that it will not move, reason being that in the first case the water is going in one direction (more or less). But in the second case, the water is coming into the can in all directions, thus reducing the thrust.

Which makes me wonder if it actually should move, if this practical problem is overcome to make the incoming stream of water come in one direction.

Conservation of momentum says that after the can has filled there will be no motion.

In the converse example with the punctured pressurized can the can is moving to the left after all the air has exited the can because the air that has left the can is moving to the right. The momentum of the air exactly matches that of the can but in the opposite direction.

While the can is filling it will move slightly to the right because of the pressure difference between the right and left sides. During this time the momentum of the air traveling to the left into the can will match the momentum of the can traveling to the right. Once the air gets into the can it will bounce around inside the can and eventually and come to a state where the cumulative momentum of the air will be zero and all of the leftwards momentum will have been transfered to the can, exactly canceling out the rightwards momentum the can temporarily gained.

Benjamin-
Ya, it is silly response, but a viable one given the wording of the problem.
Space has nothing to do with the problem. The way the problem reads deals with the state of the can after the vacuum is filled with air, not while the transfer is taking place.
This particular problem could theoreticly be taking place three weeks after the vacuum is filled. Every other variable is pretty much covered.

I agreee- It is a silly answer, but a wholly viable one.

I'm sticking with the can sitting still.
Anyone else?

The can won't budge.

But I can't explain why.

I think of this scenario, but I'm probably wrong.

Try blowing a peice of paper on a table.
Now try inhaling at the peice of paper.

The can will move to the left. Newton says object rest until acted upon. The The stationary can will be affected by the movement of the air rushing into the vacuum. Because the air is moving from right to left [per the diagram], its momentum will act upon the can and move it similarly. The can, being stationary, will not rush towards the influx, because it is inert.

thompat, post 112, I believe your reasoning is... rather silly...

"When the vacuum is filled, there will be equal pressure on all sides. It will not be in motion at all."

Just because there is equal pressure does not mean there is no motion. If that same vacuum can is thrown into space, will it somehow stop moving because the pressure on the inside and outside of the can is equally zero?

The can moves right.

For every action, there is an equal and opposite reaction. (Conservation of momentum, if you prefer.)

Since the gas rushing into the can is moving left, the can must (remarkably) be driven right.

Alternately, consider an argument from relativity. Imagine from the point of view of an observer in a different frame of reference. The gas rushing left to fill the can is IDENTICAL to the can moving right to collect the particles.

The can moves right.

It will move to the right (towards the hole, opposite the direction of the airflow) due to conservation of momentum.

Make a control volume (CV) around the inside of the can. Within this volume, we're going to keep track of anything that enters and leaves.

When the can is punctured, air will start rushing into our CV, moving left. As mass with momentum pointing to the left enters the control volume, there will thus be momentum flow to the left inside the CV. This MUST be balanced by "momentum flow" to the right, and it will manifest itself as a reaction force on the can to the right.

It is easy to demonstrate the relation between momentum flow and force:
F = ma = m dv/dt
momentum = p = mv
momentum flow into CV = rate of momentum change of CV = dp/dt = d(mv)/dt = m dv/dt (assuming mass is constant, which it is for the can)

Another way to look at it is that you have pressure on all sides of the can except for where the air comes in (there is pressure on the vaccuum here, but not on the can), so there is a force imbalance to the right. The force at any given instant will = (ambient pressure - internal pressure)*area of hole

You can apply both of these approaches to the more intuitive rocket example and you will see that the results are consistent.

Intuitively, the air doesn't suck into the can and bang against the back wall, the can sucks itself towards the air.

To verify this experimentally, take an empty soda can and submerge it upside down deep in a tub of water. Quickly turn the can sideways, and dense water rushes in and light air rushes out, resulting in net momentum flow into the can. The can will tug TOWARDS the hole into which the water is moving. You can do the same for quickly filling a cup in tub. It's only a slight pull, but it's there.

Edited by author 01-08-2007 07:48 PM
Many things could happen.
I would integrate the normal stresses on the outer surface of the can, according to this, initially the pressure unbalance would tend to move the can towards the puncture side.

Afterwards, under certain circumstances the can might bounce a bit left and right, because of air compresibility, and can inertia, imagine a pneumatic shock absorber.

In the end all macroscopic motion will end because of frictional losses.

So, my veredict is that the problem statement does not contain enough information to predict the motion of the can.

Sorry--- I should have elaborated.

The problem states this:
"After the vacuum is filled the can will..."
then we are given the three choices.
When the vacuum is filled, there will be equal pressure on all sides. It will not be in motion at all.

This is a word problem and not a physics problem.

Simple answer: C

To the right. For every action, there is an equal and opposite reaction.

It will move to the right.

Inrushing air will lower the outside air pressure to the right of the can, and the can will be pushed by the correspondingly higher pressure on the left.

The can will remain stationary, and the universe moves right.... Heh.

If you could find a can that could support the air pressure on the outside of the can, then I believe the can will remain stationary, as air rushing into the can would create force on the interior of the can, but would be almost instantly equalled by the pressure from the outside of the can. Unless the air pressure on the outside was of a great magnitude, I doubt it would move.

Refer to Post #88.
It is absolutely correct. It seems that this is more of a word problem than an actual Physics Problem.

Surely it won't move.
When it's full the pressure inside will be the same as that outside, therefore no forces in any direction.

I could, of course, be entirely wrong.

Just to cover my ass I'll add that perhaps AS it fills it moves to the right, once filled it'll continue in this direction.

It won't move. Look at the picture. There is equal pressure all around the can.

The inital condition is a vacuum can surrounded by air, with nothing moving. This may lead some to believe that after the can is punctured and the pressures equialize, the can must again be at rest to conserve momentum. The problem in this reasoning is that the existance of a difference in pressure is the existance of potential energy. When the hole is created in the can so the air moves to fill the vacuum, potential energy is being converted to kinetic energy, which will cause the can to move. So, in the end, the can must be moving (conservation of energy). Because the motion of the air is from right to left to fill the can, I believe the final motion of the can will be to the right.

Depends on the size of the hole.

With a pinhole, the air will filter in slowly. The outside air won't be disturbed much, and the air inside will probably hit all the walls at about the same time, kinda neutral.

Now take the opposite. If you remove an entire side instantly the air will rush in. There will be no "Side" on the right to counteract the rush, so as the air slaps against the left side of the can it will move leftwards. (actually this will probably just counter the initial movement that will be caused for the ms that the left-hand side of the can has air on one side and vacuum on the other, so again we're kinda neutral.

So I guess I'd say no significant movement.

Richard (post 81) is most of the way there, but I think I've got a cute third point that swings it:

1. yes, the can will start moving to the right as the can is filling, because there are more molecules hitting its external surfaces from the left that the right (as there's no hole in the left side)

2. yes, after a short time this impulse will be exactly balanced by the fact that all the molecules that would have hit the missing piece at the front will instead hit the wall at the back (either directly, or indirectly by imparting their momentum to other gas molecules within). So with this balancing leftward impulse you'd expect the can to come to rest again once it's full. But...

3. while the can was moving right, it will have been hit more by external gas molecules on its right side (due to its motion), and this will add to the leftward impulse, tipping the balance slightly so that once full, the can will find itself moving to the left and will continue to do so until the resistance to motion of the gas brings it (exponentially) to a halt.

4. There may be a point 4 to do with the fact that the air molecules entering the box will tend to be faster moving than average (faster moving molecules are simply more likely to enter). So the gas in the box will be warmer and faster moving than the gas outside, but I can't get my head fully round that one. I'll leave that for someone else to ponder. The faster molecules within are also more likely to leave once they're in. My intuition says this won't result a net impulse in any particular direction, but I'm not sure, so I'll stick with my first three points unless someone can convince me otherwise.

Edited by author 01-08-2007 07:11 PM
Y'all should read your Feynman! He wrote about this one (or an equivalent version) in Surely You're Joking Mr. Feynman:

     I once did an experiment in the cyclotron laboratory at Princeton that had some startling results. There was a problem in a hydrodynamics book that was being discussed by all the physics students. The problem is this: You have an S-shaped lawn sprinkler -- an S-shaped pipe on a pivot -- and the water squirts out at right angles to the axis and makes it spin in a certain direction. Everybody knows which way it goes around; it backs away from the outgoing water. Now the question is this: If you had a lake, or swimming pool -- a big supply of water -- and you put the sprinkler completely under water, and sucked the water in, instead of squirting it out, which way would it turn? Would it turn the same way as it does when you squirt water out into the air, or would it turn the other way?
     The answer is perfectly clear at first sight. The trouble was, some guy would think it was perfectly clear one way, and another guy would think it was perfectly clear the other way. So everybody was discussing it. I remember at one particular seminar, or tea, somebody went up to Prof. John Wheeler and said, "Which way do you think it goes around?"
     Wheeler said, "Yesterday, Feynman convinced me that it went backwards. Today, he's convinced me equally well that it goes around the other way. I don't know what he'll convince me of tomorrow!"
     I'll tell you an argument that will make you think it's one way, and another argument that will make you think it's the other way, OK?
     One argument is that when you're sucking water in, you're sort of pulling the water with the nozzle, so it will go forward, towards the incoming water.
     But then another guy comes along and says, "Suppose we hold it still and ask what kind of a torque we need to hold it still. In the case of the water going out, we all know you have to hold it on the outside of the curve, because of the centrifugal force of the water going around the curve. Now, when the water goes around the same curve the other way, it still makes the same centrifugal force toward the outside of the curve. Therefore the two cases are the same, and the sprinkler will go around the same way, whether you're squirting water out or sucking it in."
     After some thought, I finally made up my mind what the answer was, and in order to demonstrate it, I wanted to do an experiment.
     In the Princeton cyclotron lab they had a big carboy -- a monster bottle of water. I thought this was just great for the experiment. I got a piece of copper tubing and bent it into an S-shape. Then in the middle I drilled a hole, stuck in a piece of rubber hose, and led it up through a hole in a cork I had put in the top of the bottle. The cork had another hole, into which I put another piece of rubber hose, and connected it to the air pressure supply of the lab. By blowing air into the bottle, I could force water into the copper tubing exactly as if I were sucking it in. Now, the S-shaped tubing wouldn't turn around, but it would twist (because of the flexible rubber hose), and I was going to measure the speed of the water flow by measuring how far it squirted out of the top of the bottle.
     I got it all set up, turned on the air supply, and it went "Puup!" The air pressure blew the cork out of the bottle. I wired it in very well, so it wouldn't jump out. Now the experiment was going pretty good. The water was coming out, and the hose was twisting, so I put a little more pressure on it, because with a higher speed, the measurements would be more accurate. I measured the angle very carefully, and measured the distance, and increased the pressure again, and suddenly the whole thing just blew glass and water in all directions throughout the laboratory. A guy who had come to watch got all wet and had to go home and change his clothes (it's a miracle he didn't get cut by the glass), and lots of cloud chamber pictures that had been taken patiently using the cyclotron were all wet, but for some reason I was far enough away, or in some such position that I didn't get very wet. But I'll always remember how the great Professor Del Sasso, who was in charge of the cyclotron, came over to me and said sternly, "The freshman experiments should be done in the freshman laboratory!"

Feynamn doesn't say how it ends, but James Gleick does in his biography of Feynamn: Feynman was banned from the lab. And the sprinker doesn't turn.

The can won't be moving. When the air from the vacuum starts rushing in, the can will still be moving to the left from the air that had rushed out of the can when it was punctured. As the can fills up with air it will create momentum towards the right which will subtract from the momentum to the left. Since it's a vacuum, the air rushing out will have the same amount of force as the air rushing in, so the total momentum will be zero when it is filled and the can will have been brought to a stop. This is assuming that the can was somehow punctured in such a way as to have not created any momentum in either direction and, since this is a thought experiment, I've assumed that it was done in such a manner.

A number of people seem to be under the mistaken impression that the can containing the vacuum also resides in a vacuum. If that was the scenario, there wouldn't be any air rushing into the punctured can, now would there?

I think it will move to the right: the inflow of air means lower air pressure on the punctured side while the air pressure on the opposite side remains unchanged.

tried it in a vacuum chamber? did you A) suck all of the air out of the can with your vacuum B) reseal it with the vacuum C) open it without any opposing force touching the can in a 1 atm environment - doesnt sound like it.

also to the "the air pressure decreases locally outside the puncture so it will move to the right" people - why would the can move to fill this space and not the air surrounding the can, which is in much more abundance and would move much easier (i.e. the atmosphere 'moves' to fill in the empty space)

I think I'm done arguing this is getting silly - where are the physics profs to tell us who is right

The question has basically the same wording as this site: http://van.physics.uiuc.edu/qa/listing.php?id=335

it won't move. my understanding of momentum tells me that a vacuum (mass=0) will not move, period. the air will move inside it, however. this much i gather from my AP physics class 6 years ago...

The can will move to the left. Remember, when you poke an object, you impart a certain amount of force to puncture the can, and anything strong enough to keep air out against a good vacuum is going to take a LOT of poking. So the can will end up moving left, although it may slow down a bit from the air rushing in reducing the leftward momentum.

d. A fish!

Can't see it moving. 1 atm moving into a 0 atm vacuum, is a lot different than a can pressurized to 10-20 atm going into a 1 atm room. Simple volume of air. That said, if this was a frictionless, gravity-less environment... it would probably move to the right as air rushes into the hole... but the impact of that air against the back wall of the canister is going to counteract that a bit. So very little motion I would imagine in a perfect situation. Real life, nothing, just sit there.