Yesterday’s puzzle was this: A boxcar filled with water sits on a frictionless train track. A mouse gnaws a small hole in the bottom of the boxcar, near what we’ll call the right-hand end. What happens to the boxcar?

(Spoiler warning!)

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Edited to add: In view of some of the comments below, I’m no longer at all confident of this answer. I’m retaining the rest of the post, including the final paragraph in which I say that I’m pretty sure of this answer, but not as sure as I am of some other things. It looks like my hesitation might have been well justified.

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Several people got this right in comments; let me summarize:

Most of the water coming out of the hole has traveled rightward to get there, and hence, barring the application of another force, will continue traveling rightward forever. (Another force, which we can call “hitting the ground”, does in fact intervene, but the boxcar doesn’t know about that, so it’s irrelevant to the problem.) Since the total momentum of the system is zero, and since this momentum must be conserved, and since the water has acquired rightward momentum, the boxcar must acquire leftward momentum to cancel the momentum of the water. Therefore the boxcar travels leftward forever or until, like the water, it encounters some external force to stop it.

I’m nearly sure that’s right, though I’m less sure about this one than I am about this other one, less sure of that as I am of the existence of conscious beings other than myself, and less sure of that than I am about this one here. I’m pretty sure of all of them, though.

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#### 348 Responses to “About That Boxcar”

1. 1 1 Al

2. 2 2 Doug

The velocity on the X-axis depends on the width and depth of the hole in the boxcar. For a hole infinitely narrow (a perfect point) with a finite thickness (i.e. the hull of the box car has thickness) the water would leave the boxcar with perfect downward motion.

In that case there is no X-axis momentum to the water and therefore the car remains still after the water has left. Though the car would move conserve the center of gravity of the system while the water was leaving. But after vacating the boxcar would remain still.

3. 3 3 J Storrs Hall

Nope. The simple objection is that since all the water comes out sideways from the car (out of the page or down, it doesn’t matter), each bit of it has the x-velocity the car had when it came out — i.e. leftward. You would have us believe that the final result is that the car AND all the water end up with a leftward velocity.

An easier model to understand is to assume that the mousehole is in the center of the car, but someone has put a pipe from it to the right end, with a 90-degree bend in each end. (If there is no bend at the exit end, it *is* a rocket, but that would correspond to putting the mousehole in the back end of the car, not the side; obviously a rocket and no puzzle).

Now as the water gushes into the pipe and goes through the first bend, there is a rocket effect and the car accelerates to the left. But as the stream reaches the exit end and is diverted sideways, the opposite effect occurs and there is a counteracting force.

Note that the counteracting force is greater than the original rocket force at the first bend. The reason is that by the time the water has reached the exit bend, the car is moving, so the change in v_x is now the sum of the v_x it got at the first bend and the speed the car had attained by the time it got to the second. Thus there will be a net deceleration after the stream hits the second bend.

Once the car is in uniform motion, the forces at the bends cancel, so uniform motion is an equilibrium state. Momentum arguments would seem to indicate that this would likely happen at (or asymptotic to) v_x = 0, but who knows. You have to account for the pressure of the water entering the pipe is decreasing to 0 as it runs out, for example.

I wouldn’t be surprised if the overall motion were a damped oscillation.

4. 4 4 Steve Landsburg

J Storrs Hall: I could have worded this better:

*If* the car never moves to the left, *then* the water coming out the hole has righward momentum, which is not offset by any leftward momentum, which is a contradiction. Therefore the car must move to the left.

You are right that once we know the car is moving to the left, we know that the water coming out the hole picks up that leftward momentum. But I think the proof by contradiction still stands, no?

I continue to think this argument is right, though I could still be talked out of it.

(Of course all this shows is that the *initial* motion is leftward and does not rule out your damped oscillation….so I think I need to take your comment very seriously.)

5. 5 5 Thomas Purzycki

Let’s say there are two tanks of water in the boxcar. Tank one is centered at the mouse hole and tank two is the rest of the boxcar. It seems clear to me that draining tank one through the mouse hole will not cause the boxcar to move. If you plug the mouse hole and then pump water from tank two into tank one, the boxcar will move a bit to the left and then stop as its center of mass has moved, but nothing has happened to permanently change momentum. You can repeat the process of draining and pumping until both tanks are empty and the end result is that the boxcar has moved to the left and stopped.

6. 6 6 Steve Landsburg

Thomas Purzycki: Once the train has started moving leftward on a frictionless track, what force causes it to stop?

7. 7 7 Thomas Purzycki

Pumping from tank two to tank one pushes the water to the right, causing the box car to accelerate left. Once it reaches tank one, it hits the the tank wall, decelerating the car back to zero velocity.

That said, depending on how water flows out of the mouse hole in your original formulation, I could see how the boxcar could keep moving if the flow out the hole is not perfectly perpendicular to the tracks. Most of the water has moved to the right to get to the hole, and if it maintains any of that trajectory as it exits, the boxcar needs to go in the opposite direction. If you replaced the mouse hole with a garden hose, I’d expect the boxcar to move opposite where the nozzle is pointed.

8. 8 8 TjD

I am not sure why the initial motion is leftward, the initial motion ought to be downward as the hole is on the bottom.

With some awesome ascii art

WWWWWWWWWWW BB
WWWWWWWWWWWWAWBB
BBBBBBBBBBBB WBB
BBBBBBBBBBBBABBB

The water W has to go through the boxcar B and air has to go up through the boxcar and the water.

I am not sure that the force of water moving left to right has any impact after dealing with air going up and ‘jumping’ over the hole in the boxcar.

Again, not a physicist, but I just dont see it.

9. 9 9 Ken B

@J Storrs Hall:
Work in the boxcar frame. Imagine a slow stroboscope over the hole. It lets a small amount of water through, which is deflected right by the bend in the pipe, deflecting the boxcar left thereby since the pipe is affixed to the box car. Now the boxcar is in motion left, and after the water has left the pipe it is moving at constant velocity relative to the ground and is again a valid intertial frame. Repeat. You never see a rightward force on the boxcar in the boxcar frame, and so not in any inertial frame. So there can be no oscillation.
In the limit of the stroboscope the math gets messier but the result is the same, as the arguments about inertia/centre of mass show.

10. 10 10 Gordon / Brooks

Steve,

I got it wrong, but let me ask if the following is essentially the same account of what happens, but with different words.

First, if you and I were on/in the boxcar (with no water), and I’m standing directly to the left of you (as one would view your diagram), and I push you rightward (but you remain in the boxcar), does the boxcar move leftward?

If so, perhaps the following way of thinking makes sense: In your puzzle, as the water flows out near the right end, there is a rightward flow of water (as you note). This rightward flow is due to water pressure. Prior to the creation of the hole the water (at corresponding heights) throughout the boxcar is at equal pressure, which I suppose means the molecules throughout (i.e., all the way from left to right, as well as cross-wise), at equal depths, are equally compressed, and are trying to push out to restore their density at ambient pressure (i.e., at the top). After the hole is created the water flows out the hole, reducing the water pressure on the right side, causing the water molecules on the left to push the molecules to their right rightward, much like my pushing you rightward in my example. If my pushing you rightward would cause the boxcar to move leftward, then I suppose it’s analogous that molecules on the left pushing molecules on their right rightward would similarly move the boxcar leftward.

Does that make sense? Is it essentially the dynamic you’re describing, just expressed differently?

11. 11 11 Ken B

@Thomas Purzycki
That’s my explanation from comment 24 on the post. What you are missing is that the *force* on the boxcar ends, so its *accelearation* ends, but once set in motion to the left it will continue left.

12. 12 12 Thomas Purzycki

@Ken B

I agree that the forces stop, but there are two equal and opposite forces. The water accelerates right, travels right toward the mouse hole, then accelerates left to stop at the hole. The (force * time) causing the water to accelerate and decelerate are equal and opposite, with the boxcar being at velocity zero at the end of the move.

Another way to think about it is if it were a passenger car instead of a tanker. If I started on the left end of the car and walked to the exit (mouse hole) on the right end of the car, the car would move left on the tracks under me, and then stop once I stop at the exit. As long as I jump out of the exit at a velocity exactly perpendicular to the tracks, the car would have ending velocity zero, but will have moved slightly to the left.

13. 13 13 Neil

The two compartment idea is a good way to see the why the boxcar must move left. Let one compartment be a tall narrow one directly over the hole-to-be. When the mouse chews the hole, the water flows out and the cart stays put. Now punch a hole in the wall separating the compartments. The remaining water now jets to the right from the full compartment into the empty compartment pushing the boxcar to the left.

14. 14 14 khodge

A frictionless track does not mean that the wheel channel on the rail won’t block a sideways movement of the boxcar.

15. 15 15 db

I did promise myself I wasn’t going to spend more time on this…but since the published answer appears to need correcting…

The position of the hole (deliberately shown to be exiting perpendicular to the dimension of possible motion. The hole is small (mouse) so we can assume that any left-right momentum of the water is not permitted on exit: the water exits left-right stationary in the frame of the boxcar. (Or this is a really dull problem).

So it’s not a rocket. Rockets expel momentum by ejecting matter backwards relative to their frame. It’s important: the boxcar exhibits leftwards motion because of the movement of water *STILL IN THE BOXCAR* not that which has been ejected.

Despite J STorrs Hall’s analysis above (with which I broadly agree), I find the momentum argument more compelling that the force analysis argument (they are both equally valid ways of approaching a dynamics situation, but considerably easier to resolve momentum post transient behaviour and to be confident that one has not over-valued a small effect)

There are three components of momentum (my proof yesterday simplified to a two-part divided body problem) — the box car, the water still in it, and the water that’s left it.

At the initial point, the water exiting the boxcar does so with no l-r momentum (the box car is at rest, it exits stationary relative to the boxcar), the rest of the water shifts right and so the boxcar shifts left. This is the initial impulse on opening (gnawing) the hole which sets up the initial leftwards travel of the boxcar.

With the boxcar moving left and the water exiting stationary in its frame, the water exiting the system does so with leftwards (observer frame) momentum and takes it out of the boxcar forever. The boxcar must gain rightwards momentum from this. ie it acts to slow and reverse the motion.

On the issue of whether this system them oscillates or runs off to infinity, I think we need to consider that the flow of water is exponentially decreasing so the first pass of the oscillation carries a disproportionate momentum compared with the second pass. The third with the fourth, etc. So at the end of the process there is net leftwards momentum lost to the water and the box car travels off to the right infinity. There is probably a transient phase of oscillations prior to that.

16. 16 16 db

@Ken B (11)

The boxcar frame is a poor choice of frame as it is not an inertial frame: it accelerates at the start of the action, and so momentum is not preserved in that frame. It’s a dangerous place to be.

A better choice of inertial frame is that of the observer (in which system momentum totals zero). In this frame, it is clearly not possible that the car moves leftwards forever whilst dumping water leftwards forever. There’s too much leftwards momentum in the system at the end.

@Thomas (12)
It’s tempting to think that the force to stop the water is the same as the force to start the water. But
i) it’s moving inside a box which is now moving left, so actually you’re not stopping the water, you’re trying to move it left with the box.
ii) there’s less water to stop, as some of it has leaked out.

This means that the car does more than stop — it accelerates rightwards.

This complexity is a big reason why I prefer the momentum view rather than the force view in order to analyse the problem.

17. 17 17 Steve Landsburg

db: Thank you. I learned a lot from this.

18. 18 18 Ken B

@db 16: I was very careful to note in 9 that in my stroboscope example the boxcar is only an inertial frame between “pulses”. My answer in 11 is quite separate.

Again, imagine that the water is on an inclined plane sliding out the right side. It pushes the boxcar left. Or look at J Storrs Hall’s tube construction. There is the applied force is gravity. The water comes down and is deflected by the bend in the pipe (an inclined plane). The pipe exerts a normal force, which has a rightward component. So the falling water pushes the boxcar left.

19. 19 19 Thomas Purzycki

@db

You’ve got me convinced I was wrong. I now realize why my pump and dump and passenger car models are not equivalent to the original which has a constant flow out of the boxcar (even when moving).

20. 20 20 JohnW

db:

Your assumption that the water jet exiting the boxcar has no horizontal momentum component is a poor one in a problem where it was specifically stated that we have a frictionless track. Assuming something that is small is exactly zero may be reasonable when there is friction, but not when there is zero friction.

But let us modify the problem so that instead of a mousehole, we have a hose connected to a valve on the right side of the boxcar, the hose is immensely long, and the end of the hose is directed straight up or down.

Also assume that, instead of a track, the ground is an infinite plane that is frictionless in the left-right direction, but not in the perpendicular direction.

Let us look at the boxcar from the inertial frame corresponding to the ground.

Now we can look for errors in your analysis.

You write: “the rest of the water shifts right and so the boxcar shifts left. This is the initial impulse on opening (gnawing) the hole which sets up the initial leftwards travel of the boxcar.”

I thought you stated that you would be choosing an inertial frame and looking at momentum, so I do not follow your reasoning here.

I expected reasoning like this: examine a mass of water dm, the first water to exit the boxcar through the hose. The dm mass of water exits the end of the hose with zero left-right velocity component, and so zero L-R momentum. Therefore the boxcar also has zero L-R momentum and zero L-R velocity. This remains true as we let dm approach zero. So at no point does the boxcar acquire L-R velocity.

If the the problem had started with the boxcar moving at a constant L-R velocity and the valve to the hose was opened, after a sufficient time had elapsed all of the water would have exited the boxcar, and both the water and the boxcar would still be moving with the same L-R velocity, even though the water is no longer in the boxcar.

With conservation of momentum, I do not see how you can obtain any other result.

Bottom line for the original problem is that if the jet of water is allowed to have a non-zero rightward momentum component (which it should, since there is a slight pressure gradient across the hole from left to right), the boxcar ends moving left, with the “pool” of water moving right (L-R momentum equal and opposite to that of the boxcar) if we assume the water also has a frictionless path in the L-R direction. If the exiting jet of water is for some reason assumed to have zero L-R momentum component, then the final state is that the boxcar velocity will be the same after the water has exited the hole.

21. 21 21 AMTbuff

I’m with db’s post number 2. That’s the cleanest version of the problem and it has a clean solution: preservation of center of gravity and preservation of zero net left-right momentum.

This is not rocket science.

22. 22 22 Ken B

@db 15: ” we can assume that any left-right momentum of the water is not permitted on exit:”

This is not so. Think for a moment of it not as water but as sand — easier to visualize. The hole is on the right end, and a small bit of sand falls straight down. Now there is a gradient in the remaining sand, the heap is higher just left of the hole, right? So we have a pressure gradient.

23. 23 23 db

Wow. This place is just the politest debate on the internet. Godwin is possibly divergent here.

@JohnW(20)
I’m happy to agree that my assumption is poor. I only put it in place to make the problem interesting, and because I thought it was what the framer intended. I’m happy to substitute with the long hose: the result remains that we may assume water exits the car with no l-r component in the frame (not inertial!) of the car.

I do also apologise that I’ve sung the praises of the momentum view and then given an impulse summary in my comment above. All the momentum work is in yesterday’s comments where you can see me blundering through the problem step-by-step. I thoroughly entertain the idea that I might not yet have blundered all the way to the right answer. I am enjoying the journey.

The summary of the momentum argument is:-
Always zero momentum in observers inertial frame so
i) centre of mass of system can’t move — so when water starts exiting and places the centre of mass of the water+boxcar to the right of the centre of the box car, the boxcar needs to move to the left so that this point does not move in the observer’s frame.

I find this a considerably more powerful argument that worrying about water sloshing, forces etc. The initial deflection must be leftwards. I hope the sloshing argument helped motivate the system argument*.

ii) the leftwards motion of the car is matched by the exiting water (recall my earlier assumption that it exits with no l-r component in the frame of the box car), so the car is dumping left-momentum in the observer’s frame. The right momentum has to go somewhere so it lives in the (car + remaining water) which must therefore travel rightwards (and the position reverses)

I recognise the possibility of transient oscillations that J Storrs Hall identified earlier and suggested damp to a stop, but I’m going to assert that the exponential reducing mass flow rate means that the initial leftwards momentum dump always trumps the subsequent rightwards momentum dump so when the water is all gone there is net leftwards momentum loss so the box car must have net rightwards momentum. It sails off in to the (rightwards) sunset. (This conclusion is virtually impossible to achieve in a point-wise force-view of the world).

I might be wrong about that assertion — the only way to be sure would be to write down the equations and solve them. That is left as an exercise to the reader.

* system arguments are traditionally powerful, but hard eg — light refracts to picks a path through materials so as to minimise its total journey time … how on earth does it know…?!

24. 24 24 Ken B

One more try.

The hole is at the right edge. Virtually all of the water is left of the hole (the rest is above it). Eventually this water leaves via the hole.

How did it get over the hole to fall through?

If the hole stayed put and the water just on its own moved right we’d have a violation of conservation of momentum. Similarly if the hole (and boxcar) moved left and the water didn’t move, likewise.

So that means the water moved right and the hole moved left. This is true at every moment. So it’s a rocket.

25. 25 25 JohnW

db:

Your analysis makes no sense to me because you assume zero L-R momentum and then you say the boxcar moves L-R. That is a contradiction.

I already gave a much simpler analysis. Consider the first mass dm of water that exits the boxcar. If dm has zero L-R momentum, than so does the boxcar by conservation of momentum. Now let dm approach zero. So even at time dt from the opening of the hose valve, the momentum of the boxcar is zero. And it remains zero, since this analysis is true for any mass dm.

26. 26 26 Ken B

db: “Wow. This place is just the politest debate on the internet. Godwin is possibly divergent here. ”

Not on the arithmetic threads my friend! Some of those get nasty. This is rocket science: it brings out the hesitant and tentative in all of us.

27. 27 27 db

Steve – thanks for posting this question. I should have been doing other things for the last couple of days, but this has been much more entertaining. If I could figure out a way to be emailed when the next question comes up, I would subscribe.

I took a look at the other two questions about which you were more certain. I’m afraid that I disagree with your Google answer!

28. 28 28 Ken B

One last last try to convince Steve that Steve is right. (Usually a redundant effort.)

Do we agree that the internal processes, fluid dynamics, pumps, etc do not matter for the final state? It really all comes down to conservation of momentum. The actual details of the motion will depend on the process but not the final state: if the water was given momentum to the right then the boxcar was given it to the left.

Imagine the water as hole shaped ICE tubes standing on end in the wagon. As each one falls out I slide then next over the whole and it falls out. What happened? I had to push against the boxcar to get the ice over the hole. The ice moved right, so I had to push it rightwards, so I had to push leftwards against the box car.

29. 29 29 Neil

I have a brilliant idea. Why doesn’t someone build a model and see what happens.

30. 30 30 db

@KenB (28, 26, 24)

The bit of your argument where I struggle is “if the water was given momentum to the right, then the boxcar was given it to the left”. That’s fine as far as it goes, but the water gives that momentum back before exiting (it has to go out stationary relative to the boxcar to get out of the hole). Something must stop the water from bursting out of the back of the boxcar: it must stop and match speed with the boxcar and then exit.

That tends to stop the boxcar, and since it is lighter when it stops the column of water/ice than when it started it (having lost some water in the meantime), the force not only stops the car but more than stops it.

Which is just as well, as the car can’t keep going left, dumping leftwards moving water and also conserve system momentum.

31. 31 31 Steve Landsburg

Ken B (#28): Yes, this is the argument that initially convinced me that the boxcar moves left forever. It still convinces me that the boxcar *initially* moves left.

But I now think things could get more complicated after that, largely for reasons well expressed by db. After all, the leftward moving boxcar imparts leftward momentum to the water, so while we know that the *first few drops* of water have net rightward momentum when they exit, I’m not sure we know that all, or even most, of the water, has net rightward momentum when it exits.

32. 32 32 db

@neil (29)

I was wondering about the size of this effect.

The biggest liquid transport car is given by Wikipedia:-
…rated at 50,000 US gallons (190 m3; 42,000 imp gal). It first hit the rails in 1963, remained in service for over twenty years, and is now on display at the Museum of Transportation in Saint Louis, Missouri. This behemoth is 89 feet (27.1 m) in length, weighs 175,000 lb (79,400 kg) empty.

Full that is 269T. If I drain a tonne of water out of one end, then that displaces the CoG by about 4 inches relative to the centreline.

You’re going to need a lot of axle grease to make that frictionless…

33. 33 33 iceman

Anyone less troubled than they initially thought yet?

For a non-physicist, where are we at in terms of the ‘preservation of center of mass’ principle? Clearly that’s not a ‘rocket’ situation no?

34. 34 34 db

@Steve (31)

I’m troubled by this idea – “…so while we know that the *first few drops* of water have net rightward momentum when they exit, I’m not sure we know that all, or even most, of the water, has net rightward momentum when it exits.”

If the hole is such that the flow is out sideways (rather than being angled forward or back) then I think the first few drops have no momentum in the observer’s frame (since they always come out at the same speed as the boxcar which is initially at rest).

The boxcar moves left to keep the centre of mass in the same place.

Where has the rightwards momentum gone? It is tempting to give it to the exited water, but that has no momentum (and increasingly has left-momentum as the boxcar moves off). The rightwards momentum is in the internal water. It is moving relative to the boxcar. Slosh.

35. 35 35 Gordon / Brooks

db,

If you don’t mind, could you tell me if you think my #10 makes sense (at least with regard to initial motion upon the onset of outward flow of water out the hole) or not? I’m wondering if my way of thinking about it reflects the same dynamic you (and some others) are talking about.

Thanks.

36. 36 36 Max

It’s tempting to reason that since water is moving right, the car should move left. But the water is only moving right WITHIN the car. To make a rocket, the water has to LEAVE the car moving right. If the water is assumed to exit straight down (unrealistic, but so is a frictionless track), then the car doesn’t move.

Right?

37. 37 37 db

@Gordon (35, 10)
I’m not sure my comments will help any, but you are welcome to them.

You push Steve rightwards and the box car moves leftwards until he stops himself (or the wall does) and the force he uses to stop himself stops the boxcar. The centre of mass of the combined system of (boxcar + Steve) does not move.

If we continue this to say that Steve jumps out of the hole, then there’s a slight nuance over whether Steve stops himself relative to the boxcar and then drops out of the hole (which I prefer), or whether he is allowed to take a running sideways leap through the hole without stopping himself relative to the car (which I really do not like).

I think what you’ve written about molecules and water pressure passes muster but it’s not needed. We don’t actually need to think about the underlying mechanism for the movement of the water at all. All you need to know is that the centre of mass of the water is moving rightwards and since it is a uniform fluid, you just need to know the shape of the fluid (which is intuitive: it starts nicely symmetrically sat in the boxcar, and it distorts to be asymmetrically biased towards on the right side).

38. 38 38 Robert Ferguson

A slight change in the problem makes things clear and is equivalent to the posed problem. Put the hole in the middle of the rightward half of the boxcar. Put a divider in the middle of the car, so that half the water is to the left and half the water is to the right. The divider is water proof.

Step 1: Open the hole. The water in the rightward part of the boxcar drains out. Since the hole is centered in the right half of the boxcar, none of the arguments to date, right or wrong, lead to motion of the boxcar.

Step 2: Close the hole.

Step 3: Remove the divider. The water in the leftward part of the boxcar runs to the right moving the cg to the right relative to the boxcar. The boxcar moves to the left by an equal amount of the cg move, to retain the cg position of the boxcar system.

We are now back to a position equivalent to the initial position, but with half the water and a finite move to the left.

Put in the divider and repeat indefinitely.

This will produce an infinite series of finite and decreasing moves that converge to a fixed finite position of the boxcar to the left of its initial starting point.

39. 39 39 db

@Robert (38)

This is a lovely thought experiment, but differs from the original problem in which water leaves the box car whilst it is in motion. In the thought experiment the step is completed, the car stops and then more water is drained.

The distinction is important for two reasons
— firstly the mass of the boxcar system changes whilst the water is still in motion so when it stops the rightward moving water it does so on different terms than when it starts it moving rightwards and
— secondly the water leaving a leftwards moving boxcar is leftwards moving and takes momentum out of the boxcar.

Just following the thought-experiment through, you make quite a strong assertion that the infinite series converges at a finite point. For that to happen, the jumps need not just to be getting smaller, but getting smaller at a specific rate. As it happens, I think it does converge as the amount of water halves each time so the shift is probably a bit like 2^-n which is a convergent sum.

40. 40 40 Neil

Max @36

The water has to leave the car, but not to the right. Imagine a boxcar with water jetting out of the right hand side and falling to the ground. We agree the boxcar moves left. Now, around the jet, put 4 walls (3 sides and a ceiling without a floor). Other than the box car being heavier, nothing changes and it still moves to the left. Now put in a floor with a hole so the water drains out downwards at the same rate as it is coming out of the jet. Nothing changes. But we can now consider the whole contraption as a boxcar, and it will still move left even though the water is jetting to the right within the boxcar.

41. 41 41 Ken B

@Steve 31:
How about my ice example then.

Its pretty easy to see that under SOME circumstances the motion is purely to the left. My stroboscope, or ice examples show that I believe. So can internal operations, which might be sloshing etc, affect the final state? If so the problem is under-determined.

Imagine the whole is minute. I think you get such low forces it clearly goes left. Imagine the hole is the sudden disappearance of the right half of the boxcar. You clearly get a hug impetus left for the car. So the extremes match too.

More to the point, in the inertial frame where the boxcar is *instantaneously* at rest I think my argument and yours applies at that instant.

42. 42 42 Ken B

@38: Nice but you have not shown the boxcar comes to rest. I don’t see how it can in fact, as no rightward force is applied to it. A force is applied to send it left but none to stop it.

43. 43 43 Steve Landsburg

Ken B: I certainly don’t want an answer that depends on things like like the micro-properties of sloshing. But my problem with the situation where the hole is minute is this: A tiny bit of water comes out, the boxcar starts moving left, and (according to you, and according to me 24 hours ago), the boxcar continues moving left. Which means that almost all of the water emerges from a left-moving boxcar and hence has leftward momentum (at least partly offset by the rightward momentum it gains by moving from one end of the boxcar to the other). That seems like an awful lot of leftward momentum in the system. Can it *all* be offset by the rightward trip across the boxcar?

(Edited to add: An earlier comment of db’s calls this reasoning into question. I need to digest it.)

44. 44 44 Gordon / Brooks

db #37,

I realize my discussion of the molecules and water pressure as the force moving the water is (apparently) unnecessary for those more familiar with physics concepts involving “center of mass” of the water, etc., but, if valid, it’s a way I can conceptualize what’s going on. Something on the left pushes something to its right rightward within the boxcar, creating a force on the boxcar in the opposite direction (if that’s correct). If indeed my pushing Steve within the boxcar has this effect (until/unless he stops within the boxcar), then it seems the water molecules pushing other water molecules rightward would have this effect, too. If all that makes sense, it’s easier for me to conceptualize than discussing “center of mass”, etc.

Thanks again.

45. 45 45 Pieter G

Yes, it’s a rocket. Remember that the right-hand side of the box car must raise slightly, angling the stream to the right. This is due to the water not dripping out of the hole, but being squeezed out the hole by the water pressure inside the box car.

Therefore, the car accelerates leftward until the water is all gone, then continues to travel at constant speed leftwards. Center of mass arguments are of no value because the bottom of the box car does not remain level.

46. 46 46 Pieter G

The force squeezing the water out of the hole is that exerted by the column of water above it, tilting the RHS of the box car up slightly. This force decreases as the water level drops, but does not reduce to zero until the car is empty.

47. 47 47 Mike H

Imagine a bathtub, floating on a still, calm lake. The plug hole is at one end of the bathtub – the north end, in fact.

Now, I yank out the plug, and water starts gushing into the tub.

Does the bathtub move north?

48. 48 48 Ron

If this boxcar is empty, then if you walk left to right, the boxcar
will start to move left and will stop that motion when you stop.
It’s the unbalanced internal acceleration at both ends of the walk
that causes this motion and later completely cancels the motion. As
long as all motion is kept within the boxcar, no permanent motion is
generated; the Dean Drive never worked. You can’t sneak back and
repeat it. Move back to your original position, and the boxcar
moves back to its original position.

In order to impart continuing motion, the system needs to dump
momentum externally. It does so via some of the right-moving
water exiting the hole before its momentum is stopped by the
boxcar itself. This constitutes a (small) rocket thrust. You
must use the frame of reference of the boxcar to visualize
this, or you get silly results[1]. Some of the counter arguments in
this thread are the equivalent of saying a rocket can’t accelerate
past the velocity of its exhaust. No, that’s not the way it works.

The boxcar accelerates left, and it continues to accelerate as more
right momentum (relative to the boxcar) gets dumped out the hole.
Once the water stops flowing, all acceleration stops and the
velocity of the system is fixed at its current value.

NOTES:
[1] It actually works with any frame of reference, but the added
complication of a different frame makes things much harder to see.

49. 49 49 db

Worked this morning on the Google problem with a pen and paper this time. Developed the one family ratio, but am full of respect for Douglas Zare’s weighted solution across all families.

So given Douglas’ solution is mathematically correct, at what point do we have to consider taking on Steve’s bet that (1/2 – 1/4k) is closer than 0.5 to the observed proportion of girls in a k-family simulation test run n-times?

Given I could even find the middle, I’m hard pressed to find the variance of the distribution, but I’m prepared to consider that when n/k is less than about 1/4, this is better odds than playing roulette, although it’s always worse than tossing a coin.

50. 50 50 suckmydictum

@ db

Please don’t start that up again.

I originally posted a bogus argument why the car would not move, which I quickly regretted, and then this bugged me so much I had to go ask a physicist friend. He said the COM is preserved by Noether’s theorem and the reason that the cart does not go forever is because the draining water does impart an instantaneous force, but the time averaged force is zero. Does this make sense?

51. 51 51 Harold

db said “If we continue this to say that Steve jumps out of the hole, then there’s a slight nuance over whether Steve stops himself relative to the boxcar and then drops out of the hole (which I prefer), or whether he is allowed to take a running sideways leap through the hole without stopping himself relative to the car (which I really do not like).”

I imagine a line of people in the boxcar. A trapdoor opens and the end person falls out. The people all run towards the hole and drop out one by one.

We have several situations: 1) the people fall through the hole without touching the sides. 2) the people all stop then one falls through the hole. 3) Most of the people keep running, and only the one who falls through the hole stops then falls.

1) I think this is just a rocket. It is the same as people running off the back off the trolley. The velocity of the trolley can never be greater than the rightwards velocity of the people running and jumping off, so we never have any left falling people.

2) – All stop then one falls through hole. Obviously, the first person falling through the hole does not affect COG of the system, although it does of the car. There is no need for anything to move sideways.

After he has fallen through, a trapdoor closes, and all the people walk to the right until the next person is over the hole. The COG of the system has moved to the right, the boxcar moves to the left. When they all stop moving (with the next person over the trapdoor, I think the boxcar stops moving, because of the “stopping” force they all exert. Is this correct? The trapdoor opens and the next person drops down. The whole thing repeats, and the boxcar moves a fixed distance to the left in a series of jerks, then stops. (Is this a correct analysis?)

Situation 3) Simplify further and assume 3 people only. The first person drops through, then the others walk to the right as before. The boxcar starts to move left as before with constant velocity. Person 2 stops, but person 3 keeps moving. The car slows as he stops, but keeps keeps moving to the left because person 3 is moving to the right. The trapdoor opens, and person 2 falls out with velocity to the left. We now have both the boxcar and the man moving to the left, and person 3 moving to the right to keep things even. If person 3 stops what happens to the boxcar? It must move to the right to balance the person moving to the left. Where does this force comes from?

Perhaps if we compare it to what happens if person 2 stayed in the boxcar. Both people move to the right, the car moves to the left. Person 2 stops. Then person 3 stops. We know (?) the “stopping force” equals the “starting force”, so the car stops. Person 3 stooping is just enough to stop the car. In our example, person 2 has left the car making it a bit lighter, so the stopping force now exceeds that needed to stop the car, and the car moves to the right with just enough velocity to match the momentum of person 2 moving to the left. If the trap door now opens, we have both person 3 and the boxcar continuing to move to the right.

If we have lots of people, I would guess that the car moves to the left, slows down, and at the end moves to the right. I guess there would be no oscillation.

This is mostly re-phrasing db for my own understanding. Is the water case different in a significant way from the 3 person case?

52. 52 52 Ron

Steve, you had it right, and now you’ve backed away from it.

That seems like an awful lot of leftward momentum in the system.
Can it *all* be offset by the rightward trip across the boxcar?

The answer is yes. Let’s simplify this. Our railcar is now a very
light flatbed car. You and I are standing on the left side. I run
right as fast as I can and leave the car before I reach the far end.
Without question, the car is now in continuous motion to the left,
agreed? Of course, you are moving now at the same speed as the car,
leftwards.

Now, you stroll at a constant speed and trace my path. Despite all
the leftward momentum of the system, you still leave the car with a
rightward velocity with respect to the car. The momentum
within the system is immaterial. The car has higher resultant
leftward velocity because you left it, just as is did for me. The
speed of the car down the tracks is also immaterial. It doesn’t
matter that an observer standing next to the tracks sees you moving
rightward acceleration, again, with respect to the car.

In order for the car to slow down, it needs some momentum soaked up
from outside its own closed system. What possible external force do
you see doing this? The only one I can see is air resistance, which
will eventually stop the car.

To extend this to the water problem, put up a wall. As long as the
opening in the wall lets us exit at our unchanged rightward speed
without hitting its right edge, it doesn’t affect the result. The
opening has to be much bigger for people than for water molecules,
of course.

53. 53 53 TjD

It seems most of the explanations, especially the sloshing ones assume instinctively that the hole is the width of the car. Whereas it is a tiny hole, so a vortex effect would appear.

Apparently a large part of this experiment has been done https://en.wikipedia.org/wiki/Coriolis_effect#Draining_in_bathtubs_and_toilets and a counter clock wise vortex would appear in the boxcar due to earth rotation.

I am not sure what that means for the boxcar though..

54. 54 54 Ron

Harold:
Ooh, nice example for situation 3. That’s where the
outside momentum comes from. Let me simplify it and put numbers
to it.

Situation 3 needs only 2 people. One is standing on the trapdoor.
The other starts walking. During the walk, we drop person 1. What
has happened?

Let’s say the boxcar has mass 98 and each person has mass 1. Person
2 starts to walk at speed 1. Lets solve the speed of the overall
system at this time. Since v= f/m, the car is moving left at speed
1/100 or .01. Person 1 drops. Subsequently, person 2 stops. It
still takes the same force for person 1 to stop. Again solving for
v= f/m, it becomes v= 1/99. The change in speed is approximately 1%
more than at the first change. The car ends up with constant
rightward motion.

Here’s why I think it doesn’t affect the water problem. This
momentum dump relies on dumping mass while there’s uncompensated
motion during the dump. Water is incompressible. During the water
exit, the only uncompensated momentum is going out the hole. Any
other right-hand current is instantly compensated for by the right
wall.

55. 55 55 Ken B

@Steve 43: That’s why you need to look at the inertial frame where the car is instantaneously at rest. If the water moves right *relative to the car* then the car is given a leftward impulse.

56. 56 56 db

@Ron (54) [@Harold (51)]

I agree with much of this post (and Harold’s earlier post) — it elegantly provides a discrete world explanation of the effect of reduced mass.

However, I can’t agree that the water isn’t moving rightwards from the point of view of the tank (your final conclusion is that the rightwards current is stopped by incompressibility). I could argue that the sink of the hole means that the condition on incompressibility isn’t met, but far simpler to just ask you to look at the shape of the water, part in the tank, and part poured out, and it’s clear that it’s moved rightwards.

* * *

I’ve been thinking more about the condition that the water must exit the boxcar perpendicular to the track and at the same speed as the hole / boxcar.

Obviously this choice is the choice of the framer of the question: he can tell us that the mouse gnaweded a hole that points rightwards and we’re all rocket scientists. However I prefer to believe that our framer wants us to have an interesting question, so I’d rather that the water exits perpendicular to the tracks.

So there’s a residual question about whether the water is allowed to carry momentum rightwards out of the hole relative to the boxcar. In the discrete thought-experiments above, the long-suffering Steve has found himself running backwards in the carriage and then doing a mixture of stopping and diving, not stopping and diving sideways out of the hole and — in one example — leaping clean off the back of the carriage without stopping.

In the case of the water trying to leap sideways out of the hole whilst running backwards, I think it’s a reasonable to work with the assumption that the hole is small so as the water travels through the hole the side of the hole act on it so as to bring the water to an identical speed as the hole/boxcar. I increasingly like JohnW (20) idea of sticking a hosepipe out there, or even a short straw to make this point clearer.

Again, it’s the choice of the framer whether the mouse gnawed a deep small diameter hole (straw), or if it was able to blow a massive gash in the side of the tank, allowing water to spurt backwards and forwards. I prefer to think that the framer might give us the solvable and interesting problem of the straw, but am in his hands, of course.

57. 57 57 Ken B

I have to stop BUT this is interesting.

Imagine a Cylinder shaped boxcar with the flat end on the sides, so the profile we see in Steve’s picture is a circle.

1) hole at equator on right side. Clearly a leftwards rocket.
2) hole at south pole. Stationary.
3) hole at equator on left side. Clearly a rightwards rocket.

Now imagine a hole on the right just below the equator. I claim it is clearly a leftwards rocket. Now just a bit lower. Again I claim its a rocket. Repeat. At each stage don’t we have a rocket?

Integrate the volumes above the hole and see you have lost more from the left of the hole than from the right of the hole.

If you think this will osciallate, when do the oscillations start? If you think the thing will come to a halt, where does your halting force come from?

58. 58 58 Harold

#57. If the hole is not pointing vertically down then we have a rocket. In all the cylinder examples it is not pointing down until we get to the very bottom. In the boxcar it is pointing down.

If we fix the boxcar to the rail, where will the water end up? Imagine a series of tubes beneath the boxcar connected to measuring cylinders. If all the water ends up in the one directly beneath the hole, we do not have a rocket. If some or all of it travels to the right and gets into the other tubes (and none to the left), we do have a rocket. Steves solution seems to indicate that some water would travel to the tubes on the right. That is one problem. It becomes a different one if all the water ends up in the tube directly below the hole. I guess a real mousehole in a thin boxcar would allow the water to continue to travel to the right, as it was doing in the boxcar. If we imagine a small enough hole and a thick enough skin to the boxcar, then all the leftward velocity could be removed from the water.

59. 59 59 Ken B

@58:
What matters is the direction of the water flow relative to the boxcar, not the hole being down. Water can leave the hole at an angle.

Imagine a chute inside the boxcar and water running down the chute to the downward facing hole. It shoots out the hole at an angle.

To come back to a point made by several above, what are the forces? Gravity and normal forces from the ground are the ONLY applied forces. Internally water pushes the walls and the walls push the water. Aslong as the net force applied by the walls to the water is rightward the box will accelerate leftwards. Now I agree that if there is a lot of sloshing you can get shimmying etc. But a mouse hole si small. We get a reasonably steday diminishing jet of water coming out the hole. It is water that mostly came from left of the hole so has rightward momentum.

60. 60 60 Ron

@db (56)

Yes, the thickness of the wall with respect to the hole is critical.
You’ve postulated a very thick wall with a small hole. In that
case, you’re correct; no thrust because the horizontal movement is
damped within the hole. I postulated a wall as an ideal plane (zero
thickness). In that case, there is thrust. I would point to the
frictionless track as implying an ideal planar wall, but that’s not
explicit in the statement of the problem.

61. 61 61 Ken B

OY. Another example, with a downward hole.

I am rolling bowling balls from the left end to the right. When I set the ball going the car moves left. This motion will cease when the ball hits the far wall ans sticks, and even reverse a bit if the ball bounces. But what is there is a hole in the floor and the ball drops through the (downward facing) hole before hitting the opposite wall? Then we have a rocket.

This is the whole debate. If the exiting object carries x momentum wrt the car then the car must carry momentum in the opposite direction. If the exiting object leaves carrying no x momentum then the car does not. True whether the object is a ball, water, or a person.
So when the water moves right inside the car, does it give up all its momentum before leaving the hole or not? I think clearly not for water draining out.

62. 62 62 db

I think that pressure in a fluid acts in all directions, and that fluid accelerates in the direction where that pressure is not met with normal reaction. The fluid accelerates out of the hole in the direction of the hole – ie perpendicular to the surface in which the hole lives.

@Ron (60) – I’m not sure the thickness of the hole is that relevant because of the pressure argument., although a nice straw-like hole does help the mind think about this.

@KenB (61) — your bowling ball bounces off the back edge of the hole and wobbles around a bit as it exits, transferring all its relative momentum to the car.

Fortunately it isn’t the whole debate about whether the exiting water carries rightwards momentum relative to the car — the leftwards displacement of the car can be explained by the rightwards motion of the water still inside the car, even without needing momentum to be carried away from the car.

63. 63 63 Bernard

I’d argue that the boxcar will start moving to the left and then invert its motion to end up moving to the right.

We write that the center of mass (projected on the x axis) stays constant.
Notations:
m = mass of the boxcar
w(t) = mass of the water in the boxcar
x(t) = position of the boxcar (x(0)=0)
l = position of the hole (l>0 since it’s on the right)

we have: 0=[m+w(t)]x(t) – \int_0^t [x(u)+l+x'(u)(t-u)]w’(u)du

if we differentiate once with respect to time and choose t=0:

[m+w(0)]x’(0) = l w’(0) 0 which is pretty reasonable, the boxcar will be moving to the right by the end.

64. 64 64 Ken B

@db 62: I think that depends on the hole. As Ron notes, a plane is usual assumption in idealized problems. But just make the hole large compared to the ball. Like say a mouse hole compared to a water molecule!

You are simply wrong about what the whole debate is about. If the water on net *leaves* with a rightward momentum then the car ends with leftward momentum. If the water does not on net carry away rightward momentum then there is no rocket, there is the shimmying that some predict. If I am in a boxcar with balls and no holes I cannot make a rocket. If I am on a flat car with balls and no walls I can make a rocket.

The “on net” means doing a sum. You can do this in any inertial frame or over any set of inertial frames.

65. 65 65 Bernard

looks like the comment section didn’t like latex notations. And since I don’t want to write it again, I’ll just give the method:
(1) center of mass doesn’t move
(2) differentiate once when time=0
(3) differentiate once and integrate by parts once with position of the box car on the left and position of the water on the right.
(4) assume that the flow intensity decreases and gets sufficiently close to 0

66. 66 66 Scott H.

@Ken B 63 & 61. I think what db is saying is that the water could be leaving with zero horizontal momentum relative to the car, but — for a while at least — the car/water would have imparted momentum relative to a stationary observer. This imparted momentum leak of the falling water is part of the debate.

btw… thanks a lot Steve. I needed to work today.

67. 67 67 Ken B

@Scott H 66:
db is assuming that but saying otherwise. Why do I say he says otherwise? Because db wrote this: ” it isn’t the whole debate about whether the exiting water carries rightwards momentum relative to the car”. But since momentum is conserved where it goes is the whole issue.

IF there is no momentum leak THEN there is no rocket. There will be oscillations I think.
IF there is a momentum leak THEN there will be a rocket. Maybe a really slow pathetic shuddering one that accelerates and decelerates but one that at the end rolls off forever.

Lets rig up a hose under the hole. It slopes down and to the right. All the water exits via this hose. So with the hose in place all the exiting water, all of it, has rightward momentum relative to the car. And it is clearly a rocket.

68. 68 68 neil

I do not think it is a rocket. That answer can be crossed out. The jet leaving the boxcar perpendicular to the allowed direction of motion cannot accelerate the boxcar, we all agree on that I believe. I do not think that the water moving within the box car can accelerate either, ignoring turbulence. It is no different than if the water jetted out from a hole in the center or bottom of the boxcar. Water flowing to the right has to “stop” and turn to exit the hole, which offsets reaction of the right flow action.

69. 69 69 Ken B

@Bernard:
I’d like to see your derivation of the equation. I cannot frankly make it out but it looks like you are doingf something with the com, and so are assuming the the expelled water has no x momentum. Which is the whole question.

70. 70 70 db

@Bernard — the inclusion of the x’(t-u) term in the integral suggests that the water exits the boxcar with no relative l-r momentum.

I’m fine with that, but you probably need to state it as an assumption, assert it as a universal fact, or appeal to the framing of the question.

71. 71 71 Bernard

@kenB,@db
Yes I am assuming zero l-r momentum of the water relative to the boxcar (‘steady flow’).
Not sure why my latex code didn’t go through,
the 2 results are:

$$x’(0)=\frac{l w’(0)}{m+w(0)}$$

and

$$x’(0)=\frac{l w’(t)}{m+w(t)} + \int_0^t \frac{l w’(u)^2}{(m+w(u))^2} du$$

72. 72 72 db

@KenB (passim)

Indulge me for a moment and imagine that we have a straw, but it’s one that we can choose the angle back or forth if we please.

If the straw points straight out then we have the problem that Bernard’s equations describe. If it points a little rightwards then we have a bit of a rocket added on top (and conversely if it is pointed leftwards).

Indeed I could allow you a straw pointing rightwards and still have the boxcar deviate initially left and then roll off to infinity.

Bernard’s equation becomes:-

0 = (m+w)x – int[0-t] {x + l + (t-u)(x’-CSw’)w’ du}

Where
– 1/C is the density of the fluid x cross-section of the hole
– S is sine(angle of straw)

73. 73 73 db

Sorry I have a bracket in the wrong place, but hopefully point is clear that the two problems live in a space together.

I’m sure Bernard is off solving this chap as we speak (it’s a little trickier…)

74. 74 74 Bennett Haselton

I think it all depends on the exact shape of the mousehole gnawed in the side of the boxcar and whether the boxcar has nonzero thickness.

This is because it all depends on what the water does after it shoots out of the hole (or what it *would* do if it could travel forever in infinite space) — then the boxcar has to do the opposite, along the axis of the train track.

The original drawing made it look like the hole is in the side of the boxcar, through a nonzero-thickness side. In that formulation, the water could all shoot out perfectly sideways. That would cause the boxcar to move a finite distance in the opposite direction, then stop (because the water has moved a finite distance in the opposite direction along the axis of the train track).

On the other hand, if the side of the boxcar has zero thickness, then most of the water flowing out of the hole will flow to the side and back a bit at an angle, which will cause the boxcar to move in the opposite direction (and continue forever, on a frictionless track).

75. 75 75 Harold

So where are we with this one? There are two different possibitities. If the water leaves with some velocity sideways relative to the boxcar then we have the rocket type force. This is entirely feasible given a thin wall to the boxcar.

Another possibility is that the water leaves with no sideways motion relative to the boxcar. Initially the car moves left. Water leaving with no velocity relative to the boxcar will have left velocity relative to the track (observer). If water leaves to the left, then something must move to the right to keep COG of the system in the same place. This could be water that leaves to the right, and the boxcar wobbles back and forth spraying water to left and right. Or it could be the boxcar itself, which would move to the left, stop, then move to the right.

Is it correct that Bernard’s equations suggest it moves to the left, stops then moves off to the right and carries on? Are we pretty sure they are right? This seems a simpler and more elegant solution than an oscillation, since we would need complicated terms to define the frequency etc.

76. 76 76 Ken B

Re Harold 75: The boxcar cannot sail off in either direction if the water in toto does not carry off momentum in the opposite direction. This is why I find oscillations implausible. Permanent osciallations are impossible. If it moves left then right then left etc then there must be alternating forces applied to the car; these can only come from the water and so must halt. Now dampened oscillations are conceivable but strike me as wildly implausible.

One problem with ANY equation like Bernard’s is that it makes assumptions about the water. If the water moves all in one direction then its a rocket. If the water moves in a complex fashion with sloshing and breaking up into blobs going each direction etc then it’s hard to believe a simple equation and hard to tell which leftward momentum is water and which is boxcar.

There are in the end only 4 possibilities:
The boxcar never moves.
The box does not move after a period of dampened oscillations.
The boxcar goes right forever and the COM of the water goes left.
The boxcar goes left forever and the COM of the water moves right.

77. 77 77 Harold

#76 Clearly the boxcar can sail off in one direction if the COM of the water moves in the other direction.

I think we can reject never moves. The initial leaving of water surely must create a left force on the trolley because the COM of the system has moved to the right.

Once we have a left movement, we have both the water that leaves the trolley and the trolley moving to the left. The trolley may eventually be stopped by movement of the water inside, but the water that has exited cannot be stopped (by anything within the system). This clearly cannot persist, so something must move continually to the right at some point to counter the water moving to the left.

The options appear to me:
1) Trolley slows, and then finally moves to the right. Obviously any water leaving the trolley now is moving to the right.

The water leaving the trolley as it starts to move to the right then acts in the same way as the leftward moving water. As soon as the momentum of water moving right balances the momentum of that moving left, the trolley must move to the left again. This could end up with the trolley wobbling back and forth with water leaving in left and right directions. This give us either:

2) The trolley eventually comes to rest and only the water continues to move off in left and right directions.

3) The trolley eventually moves off to either the left or right

78. 78 78 Harold

Quick thought – what happens if it is a pressurised gas inside? The gas will simply expand to fill the space – no possibility of sloshing. Gas leaving the car expands equally in both left and right directions, so from the boxcar perspective it has COM at the hole. Does this make either the equations or visualisation easier?

Imagine mercry gas to give some weight to he issue.

79. 79 79 Ken B

Here is an answer I found for a similar problem, where we explicitly assume the water flows straight down wrt the boxcar: note the pipe. As I noted above the boxcar in that case must come to rest.

http://www.phys.unsw.edu.au/~jw/1131/Tank.pdf

Whether this applies to the given problem is another question. The right side of the pipe is where the water applies the force to stop the car, and there is no such in a hole in a plane.

80. 80 80 iceman

I agree with all of you. Actually you had me at relativity (#52).
Now it seems the question comes down to whether the hole takes away the momentum.
[Harold this seems to be the basis for “once we have a left movement, we have both the water that leaves the trolley and the trolley moving to the left”? BTW thanks for helping me reconcile rockets with COM arguments – I take it any momentum of exiting water is presumed to continue indefinitely as well]

Assuming an ideal plane was intended (rather than being left to speculate over the thickness), there would still seem to be a turn involved — but does this convert the momentum to downward, or just remove the floor as barrier to the normal gravitational pull?

PS what about tilting that leads #45 to say “COM arguments are of no value” – is this cheating the problem? And the bathtub story #47?

Happy holiday all

81. 81 81 Neil

I am at a loss to understand why people think, at least in the problem as it was originally posed, that the center of mass of the boxcar moves laterally. Water seeks its own level. The center of mass is lowered as the water drains out, and that is it

If you add pipes or compartments (as I did in an earlier “proof” that boxcar moves)you can contrive a situation in which the COM moves right and convince yourself that the boxcar must move, but in the original problem the COM does not move laterally and there is no need to consider conservation of momentum. The boxcar stays put.

82. 82 82 Ken B

@Neil 81:
Not remotely true. Your argument applies even if the hole is in the end of the boxcar. But then the thing is plainly a rocket. Precisely because of the conservation of momentum.

83. 83 83 Neil

@Ken B 82

Momentum is ALWAYS conserved. The issue is acceleration and mass. Momentum is equal to MASS times velocity where velocity is directional (vector valued). When momentum is in the direction of the track, the mass is that of the boxcar which we assume is low enough that its velocity can be appreciably affected by the departing water. When the momentum is perpendicular to the track, as in the problem, the mass is that of the earth, because the boxcar cannot move in that direction, so the acceleration is effectively zero.

When I walk on a sidewalk, my forward momentum is matched by a backward momentum that is sunk into the enormous mass of the earth. The effect on the earth’s spin is so small we treat it as zero.
When I walk on a treadmill, it is a different matter. The mass of the treadmill is small and my forward momentum results in a backward velocity to the treadmill.

84. 84 84 db

I had the joy of an international flight this weekend which afforded me time to solve Bernar’s stated equations. It becomes clear that with no or little relative motion of the stream wrt the boxcar the solution is a step impulse leftwards which gradually reduces to rightwards motion forever. There is no Oscillation if the hole remains constamt.

For a small hole, I cannot believe that the water does anything other than acccelerate in the direction of the missing normal force and the relative motion Is accordingly small.

I’d prefer the question were framed with a tap/ dump valve, but a mouse is small enough for me to be happy with the assumption: it’s not like we are talking about a hole in the side of high velocity pipeline: almost all of the water moves very little.

85. 85 85 Guy

@ Neil 83

On its journey to the hole in the boxcar the water has to acquire some momentum parallel to the tracks. This momentum has to be balanced by the momentum of the boxcar (initially) moving in the opposite direction.

86. 86 86 Neil

Guy@85

No. This goes back to my point @81 which is that the center of mass of the boxcar-cum-water moves DOWN not to the right, which again is perpendicular to the direction of allowed movement. Think of the water molecules as layers of marbles. After the water has drained out, there is one layer of marbles left covering the floor of boxcar–that is why the floor of the boxcar will be wet. That layer does not leave the boxcar through the hole, it evaporates. There is no net right movement of center of mass.

87. 87 87 Guy

Neil@86

But the centre of mass of the water does move to the right. If the boxcar didn’t move, all of the water would be under the hole. Given that there are no external horizontal forces, something else has to move for the horizontal centre of mass to be preserved (the vertical centre of mass can change because there’s a vertical force acting on it).

88. 88 88 Guy

@db

I’m disappointed there is no oscillation. After initially thinking it was a crazy idea, I had convinced myself it was possible (perhaps depending on the relative mass of water/boxcar and size of hole), but the sums are beyond me, so I’ll believe you.

89. 89 89 Neil

Guy@87

The location of the water mass that has left the boxcar is irrelevant. The only center of mass that matters is that of the boxcar and the water remaining in it. That COM goes down only and there is no accelerative force in the direction of allowed motion.

90. 90 90 Ken B

@89
Neil, if the Com of the water moved right the the com of the boxcar moved left, because the total com does not move.

There is movement of water and car in the x direction because of normal forces between parts of the system. Imagine the water is a block of ice and the boxcar is a wedge. As the block slides down the wedge the wedge also moves. So your notion that only what remains in the car is obviously mistaken. Every part of the system contributes momentum which must be counted.

91. 91 91 Guy

Neil@89

I disagree that the water mass that has left the boxcar is irrelevant. It is part of the (boxcar+water) system under consideration. You cannot account for its change in (horizontal) COM without external (horizontal) forces or a corresponding COM change (in the boxcar) such that the whole system’s COM is preserved. But I’m repeating myself now. I think we’ll have to agree to disagree.

92. 92 92 Neil

The claim that the location of the water mass matters after it has exited the boxcar is prima facie absurd. The boxcar system has no way of knowing where the water goes after it exits–it only knows that the water exited with momentum perpendicular to the track. The land on which the water lands could slope any which way and the water could end up anywhere. Come on–this is physics 101.

93. 93 93 Ken B

Neil, it’s not the position of the water, it’s the momentum it carries at it leaves the car. And whether the momentum is perpendicular to the track is just what is being discussed.

94. 94 94 Neil

If the walls are perfectly perpendicular and the hole is perfectly engineered, the average momenta of the water molecules leaving the boxcar must be perpendicular to the wall that has the hole.

Imagine the water as layers of slipperly marbles. All of the outside marbles are pressing equally against the four walls. Think of the marble in the bottom layer in the right hand corner. Create a hole at that corner just large enough for a marble to pass through. If you do it in the wall shown in the problem, the marble will pop out perpendicular to the wall and the track. (The marble has no knowledge of the lateral location the hole. It does not know whether it is pressing against the right wall of the boxcar or another marble to its left.) It could be in the center of the boxcar for all it knows.

95. 95 95 Ken B

94
Mouse holes are larger than water molecules. Water “gushes” out in the problem statement.
We have discussed the nature of the hole endlessly.

But even in your marbles example Neil, does a marble from the far end have to travel to exit the hole? Yes. So you need to count its momentum. Your claim is we can just ignore the exiting marbles. We can’t.

Finally are you saying a marble that shoots straight out the hole so has its trajectory perp to the surface of the hole as it travels means perp to the rails? Because that isn’t so if the car is in motion.

96. 96 96 Ken B

The com of the whole system moves down. The com of individual parts of the system can move left or right. When a falling bomb explodes the com just keeps going down, but parts can go any which way.

97. 97 97 Guy

Neil,

I’m going to give this one more go:

What happens to the water after it has left the boxcar (including – importantly for thinking about it – nothing) is – of course – irrelevant (as long as it doesn’t splash back against the boxcar). I’m not proposing some spooky action at a distance. I’m not claiming either that the water exiting the boxcar at t=0 carries any (horizontal) momentum.

I am claiming that the position of the water after it has left the boxcar can be used to reason about the position of the boxcar + remaining water – because it is part of the same system (to which no external horizontal forces were applied).

You know that velocity and momentum are vector quantities (I know you know this because you’ve mentioned it yourself). These vector quantities can be separated out into (for instance) horizontal and vertical components.

You’ve also apparently heard of conservation of momentum. Have you considered that the preservation of center of mass is just as fundamental (follows directly from) conservation of momentum? Horizontal momentum (and center of mass) and vertical momentum (and center of mass) must be conserved independently (they are orthogonal).

Imagine that the boxcar, the water, frictionless tracks and a constant gravitational field are the only things in the universe. After the water leaves the boxcar, it falls forever.

If – as you propose – the boxcar (and therefore the hole) doesn’t move, all the water will have moved to the right by approx. half the length of the boxcar. The (separable) horizontal COM of the system will have moved to the right. The principle of preservation of COM (and therefore the preservation of momentum) will have been violated.

The movement of the boxcar can be explained if you acknowledge that the contents of the boxcar (whether water or your marbles) has a velocity (and therefore momentum) different from the velocity of the boxcar.

98. 98 98 Neil

Ken@95

I repeat, water molecules in the car cannot know where they are located. They cannot tell whether they are pressing against a wall or pressing each other. When a hole is opened, water molecules flow out not knowing whether they and the hole are in the right hand corner, the center, or the left hand corner of the car. Explain to me how the can know and then I will listen to your argument.

Since the water molecules cannot know their lateral position, they must flow out of the car in the same way wherever the hole is. Would you argue that they flow to the left if the hole is in the left hand corner and perpendicular only when the hole is in the center? Then I repeat, how do the water molecules know what to do?

And whether the water molecules flow perpendicular to the track is a red herring. Only the car matters.

If you have a rocket sled pushing perpendicularly against a wall, do you somehow expect the sled to move sideways along the wall? of course not. Instead the momentum is in the opposite direction of the exhaust of the rocket and sunk into the mass of the earth.

99. 99 99 Neil

Guy@97

There is no more violation of the conservation of momentum than there is in the fact that when I begin to walk and gain forward momentum the earth doesn’t accelerate in the opposite direction. For the last time, you have to consider mass.

100. 100 100 Harold

db #84. This is what I get by extending the number of people using Ron’s momentum calculation. Each person walks towards the trapdoor, bumps into the wall at the end and stops, then falls through the trap in the floor at one end whilst the rest keep walking. As each person drops through the total weight reduces, so the effect of the next person on the wall at the end is greater, slowing the motion. By the last person this effect is enough to reverse the motion of the car which moves off to the right.

101. 101 101 Ken B

Neil
Of course the world moves when you walk. Equal and opposite reactions.
Water responds to pressure differentials.
And for the nth time, momentum not position. Momentum not position.

102. 102 102 Neil

BTW, to forestall yet another red herring. My description of the water molecules above as stationary is a simplification. Of course they are in constant brownian motion equally likely to move one direction as another. Doesn’t change a thing.

103. 103 103 Neil

Ken. Are you reading what I say? The movement is neglible and we treat it as zero, just as we do when the rocket sled is pressing against the wall or the boxcar is pressing perpendicularly against the tracks. In the limit, as mass approaches infinitely, dv approaches zero. The mass of the earth is close enough to infinity for the purpose of this problem.

104. 104 104 Ken B

@harold 100:
Imagine the walkers do it one at a time. Each waits until after the previous guy has dropped out of sight before moving. Each person hitting the right wall does brings everything to a standstill. Then he exits, with no x momentum. Car and contents do not move as he drops. Repeat for each person. Are we agreed the car is at rest when the last one exits? At rest left of where it started?

Now imagine he drops through the hole before hitting the right wall. Are we agreed the carriage moves left at a constant speed after he exits and before the next guy starts?

Now if when he hits the wall he pushes away as he drops, having then momentum to the left the contraption will accelerate to the right?

I have follow ups, but do we agree on these?

105. 105 105 Ken B

Neal, did you read no friction? So any momentum counts.

106. 106 106 neil

Ken,

No friction simply means that if the boxcar is accelerated along the track, it will never slow down. Mass determines how much the velocity will change for a given force. In fact, a jet of water (in the direction of the track) over a finite period of time (the time needed to drain the car) would accelerate a large mass boxcar by an insignificant amount just as a small rocket can barely accelerate a massive asteroid before it runs out of fuel. However, whatever acceleration can be achieved in the direction of the track, only an infinitesimal (i.e., approximately zero) acceleration can occur perpendicular to the track because then the jet has to accelerate the earth.

107. 107 107 neil

comma after no

108. 108 108 neil

sorry, cancel the comma

109. 109 109 Guy

Neil@99

(definitely my last comment on this question)

“There is no more violation of the conservation of momentum than there is in the fact that when I begin to walk and gain forward momentum the earth doesn’t accelerate in the opposite direction. For the last time, you have to consider mass.”

For goodness sake. When you’re walking on the Earth you are part of a system which includes the Earth.

“when I begin to walk and gain forward momentum the earth doesn’t accelerate in the opposite direction”

Really?

Screw you guys, I’m going home.

110. 110 110 neil

Okay, smart guy. Tell me how much you think you accelerate the earth when you accelerate your body to 5 mph. Prepare to give me many zeros after the decimal point or use scientific notation.

111. 111 111 Guy

Ok – firstly: sorry (everyone) for the exasperated tone of my last comment (and for the lie that it was my last comment).

To (not) answer your question: I don’t know (can’t be bothered to work out) how much, but the answer isn’t zero.

Do you think the answer is zero?

More importantly: what momentum from the system in question do you think gets transferred to the Earth?

112. 112 112 neil

All of the momentum of course–that is what conservation means. But we are talking velocity here. Since I figure your mass is no more than 100 kg but the earth’s is 6*10^24 kg, when you accelerate to 5 mph you will accelerate the earth in the opposite direction by about 8*10^-23 or .000000000000000000000008 mph. In any physics problem I have ever solved in my life that would be treated as effectively zero.

113. 113 113 Ken B

Neil, no one but you is talking about movement or acceleration of the car in anything but the x direction. That is the whole damn question: what happens in the x direction.

You are simply confused.

@guy: Sympathies.

114. 114 114 Neil

Ken. It is you who is confused. Your comment on zero friction revealed that. The movement in the track direction is exactly zero.

115. 115 115 Ken B

@114: “ok smart guy”, as you put it. How does water at the left end of the car exit a hole at the right end of the car if there is no movement in the x direction? Either the water moves or the car does. That’s movement. And if the water does, and exits the hole as you describe, then the car must, to keep the x component of the COM unchanged, because in your description the COM of the water ends up at the x coordinate of the hole.

116. 116 116 db

@kenB (104)
Imagine your walkers set off when the previous one is halfway to the hole.

They are organised in weight order, heaviest first. The move slowly and mostly stop at the hole to drop out.

The cart accelerates rightwards until there are no more walkers.

117. 117 117 db

@Neil (passim)
I’m finding it a little hard to understand your exact argument – but consider if you will that the tracks are on a bridge and the water may fall a long way before hitting the ground.

Consider the solution that the car remains stationary.

All the water exits in a column under the hole. The boxcar is empty.
The CoM of the boxcar is clearly where it was at the start. The CoM of the water has moved from the centre of the car to the hole (and a long way dOwn, but I’m considering l-r dimension only). So the CoM of the boxcar + water system has moved rightwards towards the hole.

There is no lateral force acting on this system, so we’ve violated Newton’s First Law. So my solution that the car remains stationary must be wrong.

118. 118 118 Ron

Neil @98:

You seem to have defined away any possibility that water currents
exist. This would surprise a lot of people, including most SCUBA
divers.

Also, @112, “negligible” is not zero, and there are effects you
can’t handwave away. You’re arguing engineering, not science.
Nobel prizes have been awarded to people who have found “negligible”
differences that meant current theories are wrong or have failed to
account for a previously-unknown effect.

119. 119 119 Ken B

@db 116
Consider when guy 1 hits the wall. Does he come to rest wrt the car? I assume you mean he does. So he exits relative to the ground moving LEFTWARD. After all, the car is moving left under the effect of runner 2. Now guy 2is going right and everything else left. We have a recursive problem. In that case I think you may be right. But again that’s not how holes work, it’s how pipes work. If things drop through the hole without giving back their momentum the men exit moving right not left.

120. 120 120 db

@Ken (119)
I look forward to your proof that holes and pipes work differently, and inparticular that all holes work the way you. Suggest.

121. 121 121 Ken B

@bd
I gave it! You fall through a hole before hitting the far side of it so carry with you your momentum. With a pipe/wall you collide and then there are 3 cases as in 104. Imagine the hole is wide the walker falls through with no contact to the right side it or any wall. Easier visualize with rolling balls.

Sloshing seems like it could be the equivalent of a man pushing off. You need that to end up with the empty car going right. But that requires collision between the car and the exiting object.

The reason a hole in the right end of the car would certainly be a rocket: the water carrying momentum right.

I think we either have a rocket or a car shifted left but motionless at the end. However you could be right if there is sloshing.
The only answer that *cannot* be right is Neil’s.

122. 122 122 Ken B

Imagine the car is empty except for some mirrors and a mirror affixed to the left end. Fire a pulse.

1. The hole is wide enough for the laser beam to pass through the hole without touching. Rocket leftward.

2. The beam is reflected off a mirror and exits in the y direction. Car ends up translated left, stationary.

3. Mirror angled so beam passes through hole angled left. Rocket to right.

So which seems to best describe water? I dunno. I thought 1 at first, and it still sems most likely given this problem, but cannot reject 2, or even 3. With a pipe I favor 2.

123. 123 123 Ken B

Oops a LASER affixed to the left end. The pulse of the laser leaves the boxcar with momentum.

124. 124 124 Peter Tennenbaum

I have not followed all the posts and threads; perhaps someone else pointed out the following, often seen in the so-called “real world”:

The water will splash upon hitting the ground and some water might well hit a wheel, thus imparting a force. Further analysis of this scenario yields a host of additional variables to consider.

125. 125 125 db

@KenB (121)

I’m afraid that our standards of proof are different. That’s fine, but you will struggle to persuade me of the truth of what you say with such loose reasoning.

I cannot see how there is a solution where the car is displaced left and is stationary without the water making discrete journies (I don’t think it does).

126. 126 126 Ken B

127. 127 127 Ken B

@db
Basically huh? You don’t see a difference between the situation where the exiting particles strike car and when they don’t?

128. 128 128 db

@ken (126)
I’m travelling in France at the moment so writing on a phone that can’t read a pdf, I’m afraid.

What does it say? I’m convinced the solution to the water leaving with zero relative velocity (pipe / straw) is that it moves off to the left and then accelerates to the right and goes rightwards forever as the water runs out. Bernard’s equation shows that without much doubt.

I’m not a fluid dynamicist. I know that stationary fluids exert a force in all directions which is countered by a normal reaction at the surface and a hole will result in a nornally directed acceleration. I don’t know the effect of a drift in the fluid. Particularly one where the fluid drains in from both sides. I was hoping for a definitive statement or proof so that I can learn more.

129. 129 129 db

@kenb I’m curious to know whether you think the hole gets wet.

130. 130 130 Keshav Srinivasan

db, if you’re not able to view PDF’s, are you at least able to view images? Try this image version of the PDF:

http://tinypic.com/r/dbsdu8/5

131. 131 131 Ken B

Keshav, thanks.
Db, the hole can get wet from the edge of the stream can’t it? Must all the water in your glass touch the lip when you pour it out?

132. 132 132 db

@Keshav – thank-you that is extremely kind.

The solution is wrong for two reasons, but mostly he gets it wrong on the first line.

The displacement he calculates supposes that the tank and the water are not moving (which is great as t approaches 0 from below, but just no longer holds once t>0. He is writing a CoM equation, but as soon as the tank moves, the stream of water is moving. So the CoM equation must contain a term for the position of the stream of water.

Having got the wrong differential equation he does then solve it, but he is quite sloppy about the initial t=0 discontinuity. X is not smooth – it’s not even differentiable at t=0. So the author needs to take greater care at that point (same for m). One might also object to his ‘physicists formality’ in deriving the differential, dividing by the infinitesimal, but it’s broadly right…where functions aresmooth.h

But he’s already made a fatal error in line 1, so this makes no material difference.

I am assuming the author is solving the same problem that we are (in his case with a handy zero-relative-velocity pipe and tap)

133. 133 133 Mike H

The determinant is the average horizontal momentum of the water coming out. This, in turn, depends on the geometry of the hole and the boxcar.

Clearly, we can get the water’s momentum to be in any horizontal direction we like by attaching a bent pipe to the bottom of the hole.

If we ban pipes (mice don’t have plumber’s licenses, usually) we can still get direction we like by placing a little pipe or similar above the hole :

+===== — walls of “pipe”
| O — hole
+=====

this will make the boxcar into a rocket, moving to the right, no matter where it is in the boxcar.

If one idealises the boxcar, making the floor flat, etc, the answer still depends on the fluid dynamics of the water filling it.

134. 134 134 Mike H

Now for a little thought experiment.

I shall assume the boxcar floor is a rectangle, length 1.

I shall assume that the water molecules slide along the floor of the boxcar until they reach a hole, then fall out. They collide perfectly elastically with the sides of the boxcar. Also, that they don’t otherwise interact – so water is a 2-dimensional ideal gas. Also that consumers are rational. Close enough.

The horizontal velocity of the molecules follows some probability distrubution, with mean zero. Yes, E(V)=0.

Now, I ask the mouse to instantaneously gnaw a trench all the way across the boxcar, x units from one end.

Water molecules from either side of the trench start to fall into it – but only those heading towards the trench. The mean velocity is E(|V|)=K for those on the left, and E(-|V|)=-K for those on the right.

since x of the molecules are on the left, and 1-x are on the right, the net change in momentum is m(Kx-K(1-x)) = mK(2x-1), where m is the mass of my water.

This is nonzero, so for this type of hole, the boxcar is a rocket. It follows that there exist holes for which the boxcar is a rocket, even in this idealised case. In fact, to make the boxcar not a rocket requires careful fine-tuning of x.

Conjecture : the boxcar is almost always a rocket; that is, if the mosue nibbles a random hole, the probability of the boxcar becoming a rocket is 1.

135. 135 135 db

@KenB (131)

I only require some of the water to touch the glass in order to keep all of the water in the glass.

136. 136 136 JohnW

The water tank solution seems more likely to be correct than db’s solution.

db assumes that “as soon as the tank moves, the stream of water is moving”. The solver of the water tank question assumes that the exiting stream of water is not moving (relative to the initial at-rest frame). Neither db nor the water tank solver provides justification for their assumption.

If we assume away all the L-R momentum of the exiting water stream in the boxcar question (a poor assumption, but since we already know the answer in that case, let us change the problem to one with a long hose or pipe that makes the L-R momentum of the exiting water stream zero relative to the boxcar), then it is not difficult to imagine a model where the water exits the boxcar in units of mass dm where each dm is stationary relative to the initial at rest frame. For example, assume the water is actually ice cubes, and after the first ice cube drops, the next ice cube slides (or is pushed) into place above the hole. The boxcar moves while the ice cube slides into place, but stops moving once the ice cube stops above the hole. The next ice cube drops, and the process repeats. If you then let dm approach zero, in the limit this is very similar to the boxcar problem.

137. 137 137 db

There’s been a lot of debate about the direction which the water flows out of the boxcar. In fact this makes exactly no difference to the initial movement of the car.

If you continue with Bernard’s equation (63), you can alter these to reflect the extreme case rocket — perhaps with the nozzle pointing flat back to the left (ie tending to create a rightwards moving rocket).

The most obvious alteration being an additional Cw’ term inside the integral — where C is a constant (a function of density, hole cross-section, orientation of pipe and pipe drag) so that Cw’ is the velocity of flow out of the pipe.

On differentiating Bernard’s equation and solving at t=0+, you can see that the initial velocity of the boxcar is *unaltered* by pointing the nozzle of the jet of water in any direction. This makes intuitive sense: at t=0+ there isn’t enough water that has exited the system to have any rocket effect yet, but he water inside the car has been all set in motion towards the nozzle with a consequent shift in momentum.

What happens next depends on where the nozzle points and how the water exits, but fascinatingly, the initial motion depends solely on the location of the hole and the rate of flow relative to the total system mass.

138. 138 138 db

@JohnW (136)

I’m sorry I wasn’t clearer in the argument against that “tank” solution. I thought it was obvious that as the tank moves, so does the hole. Let’s tackle it more directly.

The solver’s first line is an infinitesimal equation, beloved of engineers and formally tricky, but can work provided you are careful:-

(m+M)dx – Ldm = 0

This is possibly true for the very first drop of water, but there’s a number of things not considered here:-

He doesn’t really define x. His diagram is unclear, but his use of x in his explanation shows that he means x to be the position of the centre of the tank. This means that the position of the tap is x + L, not L (it starts at L: x starts at 0). This is what I mean by the “hole moves with the boxcar”.

He also doesn’t consider the movement of the previous dm’s of water — if they also move (before hitting the ground) — then they must also be taken into account in the motion of the centre of mass of the system, but they are neglected.

Effectively he has assumed that all dm fall from the same place and at rest in the lab frame. True for the first dm…

139. 139 139 nivedita

db, I tried taking a simplified one-dimensional boxcar with the water flowing at a constant speed to the right and exiting with zero lateral velocity with respect to the car. Conservation of momentum implies that the boxcar initially moves to the left, but after all the water has drained out, moves to the right at a constant terminal velocity. I can get that by just solving the differential equation, but I am still confused by one thing: at some point in the process, the boxcar is at rest, with the water inside the boxcar moving to the right (its rightward momentum balancing out the leftward momentum of the water that has previously drained out); and then the boxcar moves to the right. What force causes it to move to the right?

140. 140 140 nivedita

Oh, I get it now, if the boxcar is at rest, the water hitting it as it comes to rest just before exiting will give it an impulse to the right

141. 141 141 db

@nivedita
If you’ve solved the differential equation, then you will already have the answer. The car is subjected to a constant rightwards force because w” (to use Bernard’s notation) is always positive (the flow of water is slowing out of the tank — ie the momentum of the water relative to the tank is decreasing: this was the momentum that set-up the initial leftwards impulse in a stepfunction at t=0. It is gradually unwound over time.

How to motivate this in terms of physical objects? When the boxcar is stationary it is dumping water straight down so it’s tempting to think there can be no forces. But it’s not a rocket! The water is moving in the boxcar relative to its frame with a momentum of -lw’. But that flow is slowing (as the level drops the flow slows. You need to explain how the water’s momentum can slow if the boxcar and jet of water do not: clearly the answer is that the boxcar is where this momentum goes and off it trundles rightwards from rest again.

142. 142 142 db

St to clarify something I just read in your 139: if the exiting water is at a constant speed then you get a different result. You need the flow to be slowing (as it does in the real world: typically as an exponential decay.

Apologies: typing on phone is a nightmate.

143. 143 143 Al D P

Perhaps I’m the only person to agree with Neil. There would be no movement in the boxcar in either direction because the mass of water and dripping speed would have to overcome the boxcar’s mass, which wouldn’t be possible with a small hole.

If I stand next to an empty boxcar on frictionless tracks I would be unable to push or pull it in either direction because I would not be strong enough to exert a force greater than the mass of the boxcar.

144. 144 144 Harold

Ken B #104- a bit late, but yes, I agree with all your statements. It is interesting that if they go one at a time the boxcar ends up at rest.

145. 145 145 Harold

#139. I can explain the force moving it to the right in the case of individual discrete chunks – people in my example. If all the people move to the right, then stop (as in pumping water from a tank on one side to a tank on the other), then the trolley moves to the left, then stops. We know this from the conservation of momentum, and we can explain it in terms of forces also. As the people accelerate to the right, they exert the same force to the trolley to the left, so it moves to the left. When they stop, they exert the same force again in the opposite direction. This exactly cancels out the initial force, so the trolley stops.

In the case of the people, each one stops, then falls out. The sum total of the forces to the right is the same. However, after each person falls out, the weight of the trolley is lower, so the same force gives greater acceleration. This means the trolley must end up moving to the right.

#142 db. It is interesting that a reducing flow is required for the equation. I had assumed my person example was a crude approximation to the smooth integrated equations, yet the flow in my case is (or can be) constant.

146. 146 146 Harold

To clarify, the car moves to the right if the starting force is exerted on a heavier car than the stopping force. For this to happen, someone (or some water) must leave whilst the person in the car is in motion. This means the person (or water) leaves from a moving car, since the car must move to balance the person (water) moving. The consequence is that if someone (thing) leaves with leftward velocity, then the car will end up moving right. Thus the “momentum” and “force” approaches agree, as they must.

147. 147 147 db

@Harold (144) -
In your example you run out of people: this represents a decreasing flow. Suddenly and just as the last person leaves the car.

Ideally you’d have all the people arranged in order of decreasing weight… :-)

148. 148 148 db

Interestingly (and I appreciate that by comment 147, I might have set the bar pretty low, and I do fear that we may have stayed at this party after the host has gone to bed), the phenomenon that is exhibited is very similar to waterhammer heard in creaky old piping.

When the tap is turned on, the entire column of water in the pipe starts moving creating an instantaneous large impulse on the pipework. If this manages to interrupt the flow then the impulse is reversed and the pipe oscillates loudly.

That initial impulse from the standing column of water suddenly starting to move is exactly the thing that kicks the boxcar into motion. An impulse which doesn’t really depend on the orientation of the tap (my 137).

149. 149 149 Harold

#146 db – thanks for the clarification – the exponential decrease is a realistic depiction, but it works also with a sudden decrease. On a side note, waterhammer can occur in new piping too! Plumbers hopefully install to avoid it.

Are we all sorted now- has agreement broken out? Can the guests go to bed too?

150. 150 150 Ken B

@147, 144:
db, if they go one at a time it does not matter what their masses are.

@138: the equation with dx and dm is correct. Imagine the instantaneous inertial frame in which the boxcar is at rest. These are the first order terms.

Overall I agree with 135. Maybe a leftward rocket but maybe just a shift. I lean still to rocket, but wouldn’t bet heavily against shift. However depending on geometry and the innards of the box, which we cannot see, the final state could be boxcar moving right or left. I think you need very special things to happen to get it going right though.

151. 151 151 Ken B

@Al DP:
You are stronger than you think!
You are confused about mass and exerting a force ‘greater than the mass’. These are meaningless statements; force is not mass, mass is not force. Forces cause masses to accelerate. even small forces, even for big masses.

The boxcar will move on the fictionless surface if you push it sideways *no matter how weakly you push it*. This is Newton’s law. F=ma. The only force to counter yours would be friction, and there is no friction.

152. 152 152 Harold

Ken B #150 – is it 135 you are agreeing with?

For any real hole made by a mouse in any real boxcar, there must be more molecules leaving to the right than leaving to the left if the hole is wider than 1 molecule. Thus for almost any real hole there will be some “rocket” type forces.

If we revert to marbles, then it is quite easy to remove all the rightward motion form the falling marbles by having the hole one marble wide and a length of tube. If we have such a boxcar filled with marbles, it will initially move to the left, then to the right.

We have postulated a theoretical hole, where all the left-right momentum were removed from the water before leaving the car. In this case the car moves left, then right.

Do you agree with this?

153. 153 153 db

@KenB (138)
“Instantaneous inertial frame” makes no sense. It’s either inertial or it is not. If it is not, then it’s a dangerous place to be doing momentum calculations. I thought we’d covered that before. The reason it is dangerous is because it lets you write the wrong equation down.

They are the first order terms of what, precisely? Not the CoM equation.

The right starting point is to write down the CoM of the system and make it stationary for all time (since that’s the constraint we use)

That equation isn’t it. It’s nearly d(CoM), but it’s wrong. It’s formally nicer to look at (CoM)’ in any case, and in this instance more revealing.

154. 154 154 Dave

Change the metaphor:

If you took a big bucketloader, and removed all the water from the “right” hand side (of the picture), the boxcar would move left (due to the potential energy release of the water as it flows in to fill the empty space.)

Then, when the water hit the “right-hand” wall, the movement would stop. You could say it was cancelled by the force against the wall, but what really happened was that the mass reached equilibrium.

This revised metaphor works, because the force of the water jet (perpendicular to the tracks) can be ignored.

155. 155 155 Ken B

@153: Huh? Not only does it make sense but it’s a perfectly common thing to do with problems. There is an inertial frame going at every single fixed velocity relative to any other inertial frame. So if the vehicle is moving left at 4 m/s relative to eartch, pick the frame going 4 m/s left relative to earth. In that frame the car is stationary. And if the vehicle is being accelerated, pick that frame only for that instant. And look at the forces being applied at that instant. If you want to use Newton’s Laws directly you need to pick an inertial frame; if you want to look at differentials you can.

156. 156 156 Al V.

I’ve been away for a few days, but reading the comments, I am convinced that Ken B is correct. Assuming a zero thickness hole, the water leaves the boxcar with rightward momentum. Assuming that the boxcar is resting on a frictionless plane, the water will continue to travel rightward forever. Also, notice that initially the water will have a much higher velocity to the right than the boxcar has to the left. If the boxcar has mass independent of the water, then even after all of the water has been exhausted, the boxcar will continue to travel to the left, while the water continues to travel to the right.

Thus, momentum is preserved, as the boxcar’s momentum to the left is balanced by the water’s momentum to the right. And the center of mass will not move, as the water continues rightwards while the boxcar continues leftwards.

157. 157 157 Ken B

@152: Yes Harold I agree with my own argument! :)

The complexity is fluid flow and sloshing. If water can pile up at the right side of the whole it can exit leftward as it sloshes back. I find this implausible because the boxcar is big, it moves slowly compared to the water rushing out. But I cannot for all holes prove it cannot happen. But if it can happen then it’s not so easy to see where all the water goes.

A word to those relying on pressure arguments. The water pressure is NOT equal in all directions when water is draining out a hole.

158. 158 158 Ken B

Ack. In 156 I mean the boxcar is accelerating at a slow rate.

159. 159 159 Ken B

@Harold 152: No, 135 is clearly wrong. I can pour water out of a glass despite the fact some of the water is touching the glass.
And only some of the water leaving the glass touches the rim.

160. 160 160 Ken B

@Steve: For your next physics puzzle post I suggest something simpler! The three-body problem perhaps.

161. 161 161 Neil

Obtain a small rectangular plastic box, preferably clear. Drill a small hole in the right hand bottom corner. File away any irregularities and seal the hole with tape. Fill the box with water and place on the edge of a table. Leave it for several minutes to ensure the water reaches stationary equilibrium. Carefully remove the tape without disturbing the box, and observe the direction of the water stream leaving the box from above. If you are interested in the hydrodynamics of the water in the box, add a drop of food dye.

I’ve done this already, so I know what happens. Keep on arguing.

162. 162 162 Ken B

@Neil 160:
It could leave perfectly perpendicular to the box to and you’d still be wrong that the boxcar never moves. Has the centre of mass of the water shifted sideways?

Actually this raises an interesting point I wanted to make. When the boxcar moves the stream may look like it is pointing left or right, thethe orientation of the stream is not what mattters. What matters is the momentum of the atoms as they leave the boxcar. Throw balls perpendicularly out of a moving car and it looks like a stream going backwards, but that’s a red herring.

163. 163 163 db

@kenb -
I don’t understand how you can simultaneously defend the differential equation with no term in it for the water that has left the box, at the same time as saying’what matters is the momentum of the atoms as they leave the boxcar’

I’m somewhat at a loss as to what will persuade you, since mathematics seems to have failed as well as broader physics arguments. Can you give me a clue or write an eQuation?

164. 164 164 Neil

Ken,

As I have said before, I know the center of mass of the water in the box does not move left or right because the surface of the water remains level. Since the boxcar is level, the water would need to tilt in order to move the COM right or left, and that does not happen because water has so little viscosity. For every kg of water that leaves the car, a kg of water moves to the right keeping the COM in the center.

Now if I did the experiment with molasses the surface would tilt– downwards on the right side of the car shifting the COM left. Since water has non-zero viscosity, it must do the same very slightly, but with a pure physics problem like this one assumes an ideal fluid, just as we are assuming zero friction and a perfectly engineered boxcar and hole.

165. 165 165 Neil

That should be “For every kg of water that leaves the car, a HALF kg of water moves to the right…”

166. 166 166 db

@Neil

The kg of water that has left the box still hass mass and should be included in your calculation of the CoM. I agree that it’s reasonable tO assume fLat water and CoM of water in the box + box remains at centre of box.

167. 167 167 db

I’m still not a fluid dynamicist, but I think that a small hole (cross sectional area) regardless of thickness of wall, has a very high Reynold’s number, so the drag on the water flow is huge and the flow is determined solely by the static pressure and not the motion within the tank, and is normal to the surface of the tank.

We are told it’s a mouse (the SI unit of ‘small’), and that the fluid is water (viscose for these purposes).

168. 168 168 Neil

My former professor would pose physics problems like this. When we students started worrying about details, like the thickness of walls or irregularities of any kind, he’d say “that is an engineering problem, this is physics!”

I guess that is my approach here.

I would like to pose a question free of hydraulics, which was always my weakest subject.

Since the subject is rockets, if we were attach a standard chemical reaction rocket exactly perpendicular to the wall of the box car where SL drew the hole, would we agree that the box car would not move any track direction? If not, why?

169. 169 169 Peter Tennenbaum

Assuming that the boxcar material has mass, the appearance of a hole means a loss of mass and, therefore, a shift in the center of mass and the center of gravity of the boxcar.

Note that this is INDEPENDENT of the water moving towards the hole and exiting.

170. 170 170 db

If Steve hasn’t used this as an opportunity to get back in touch with Prof Goodwillie, then this problem will have failed to have been as useful as it could have been.

171. 171 171 Harold

Ken B 158 and Ken B 150 are contradictory. When you said you agreed with 135 I presume you meant some other number.

As far as I can see, everyone agrees with the theory except Neil. IF the water leaves with rightward momentum we have a rocket. IF the water leaves straight down then the trolley moves to the left then right. There is some disagreement about what would happen in the real world, but for the sake of the puzzle we can stipulate either condition. Where does the sloshing argument come from in #156? I don’t think anyone has said this is necessary.

Neil for some reason no longer considers the water that has left the boxcar for COM considerations. I think he is confusing force arguments with COM arguments. As I think db said earlier, you can use either.

In the force argument, the water that has left the boxcar has no further part to play, but you must consider pressure gradients within the boxcar. There must be differences in pressure or there can be no flow. Water flows from the left of the car to the right, so there must be higher pressure on the left of the car. This exerts a force on the left, so the car will move to the left initially.

In the COM argument, the water that has left the boxcar must be included as it is part of the system. The COM of the water part of the system has moved to the right. Since there is no external left / right force to move the COM of the system, the boxcar part of the system must move to the left.

This is repeating what others have said – I can’t see any other way to convince Neil.

The force argument becomes more difficult to visualise if the hole is small enough and the fluid viscous enough to remove all the rightward momentum form the fluid in the boxcar before it leaves. We must visualise another pressure gradient in the hole the other direction, exerting a force to the right. It seems quite possible that the hole could be small enough that surface tension causes the water to form a drop which leaves perpendicular to the tracks – i.e. no left – right momentum wrt the boxcar.

172. 172 172 Al D P

The only benefit frictionless tracks gives the boxcar is that it can now move along the tracks without the need for wheels turning. The boxcar itself can be slid along them. Provided that enough of a force is applied that moves the mass of the boxcar and the water remaining inside. Frictionless does not mean the boxcar doesn’t weigh anything. I can push against a boxcar all day and it won’t move even with frictionless tracks. Neil is right

173. 173 173 Ken B

@Harold: I misread the number in 150.

174. 174 174 Keshav Srinivasan

Al DP, are you really disagreeing with Newton’s second law of motion? If the boxcar weighs 100kg and a little kid pushes it with a force of 1 Newton, then it will undergo an acceleration of .01 m/s^2. Are you really disputing that?

175. 175 175 db

@Al DP (171)
The benefit of the frictionless track is that it means we can forget about the impact of the track and the Earth on the l-r direction momentum calculation.

If a small fly bumped into the right side of the car, then it would sail off to the left forever. Very slowly.

If you leaned against it, it would go a bit quicker, but go it would. If you do intend to try this, then be careful you don’t slip on the frictionless track.

More rigorously:-
Newton’s Third Law means that if the car didn’t accelerate when you pushed it, it must be exerting an equal and opposite force on you. Where does that force come from?

Normally this would be a resistive static frictional force, given by a maximum of (mu)N where (mu) is the coefficition of friction and N is the normal reaction force of the surface (in this horizontal case, equal to the weight of the car). In this instance (mu) is 0 so the maximum frictional force is zero, so it is overwhelmed by your push.

176. 176 176 db

@Harold (170)
Whilst I like to broadly agree with what you have written, it is possible to choose variables so that the nozzle (where fitted) is pointing rightwards, but that the overall motion is dominated by the leftwards impulse and rightwards acceleration of the perpendicular nozzle.

If you develop the equations that I developed in my 137 — the variant of Bernard’s equation with an additional Cw’ term inside the integral, then you can follow the same calculation through and derive a constraint on C so that the motion remains rightwards forever DESPITE the rightwards nozzle.

But you have to do the maths to believe that (although you ought to be able to intuit it from taking deviations from the perpendicular nozzle case which you seem to already believe).

177. 177 177 Stephen Karlson

Somewhere, the Trainmaster is chewing out the brakeman who neglected to wind up the hand brake and trig the wheels.

178. 178 178 TjD

Re : Math formulas, if a math formula does not imply a vortex movement in the water, then the formula is wrong ( does not match reality ).

179. 179 179 Harold

#175. I think I already believe! I have said that there will be “rocket forces” if some of the water moves right, rather than a rocket. There will also be rightward forces, as long as some of the water exits straight down. I assumed that a small rocket force could be overcome by the rightward force.

180. 180 180 Ken B

For fun imagine the boxcar is frictionless inside too, and that we have two identical ones. Stack them one above the other. Connect the identical holes by a frictionless pipe. So now the water flows frictionlessly from the upper car to the lower car.

I suggest the double decker car shimmies back and forth about its original location. More or less forever.

The COM of the contraption is fixed so it does not drift off. No wter is lost. But it must move as the ater drains and the COM of the water moves. When the last drop has drained I suggest the shimmying continues. After all we turned ptential energy into kinetic. The water could I suppose be slohing back and forth in the Y direction causing no movement, but that seems implausible. So there must be x direction sloshing. That moves the cars back and forth.
This lasts until the energy dissipates in noise or air resistance etc. Not heating the box: no friction!

I choose the word shimmying carefully; I do not see that this needs to be nicely periodic like a pendulum.

181. 181 181 db

Hmm. I think the double-decker remains motionless, but I may have misunderstood the posed problem. Does the track stay dry in this example?

182. 182 182 Harold

Neil #160 -I am interested to hear about your experiment – was there noticable displacement of the water stream to the right or not?

183. 183 183 Ken B

@181:
Yes The water flows from the upper to the lower chamber. You can have the holes at the same x coordinate, or different. The water will spread differently in the lower one (I think!) than it drains from the upper. For example a drop enters the pipe. At the end it is moving faster than when it entered the pipe. If the pipe slopes that will give the drop x speed. Plus water will spread on the foor etc. So I think the COM of the water moves as the water flows, so the box must move.

Since the water falls there must be kinetic energy after the draining. So there are only two possibilities: the water sloshes in a way that does not move the COM of the water, which seems unlikely, or it sloshes in a way that wiggles its COM, shimmying the cars. Think of a wabbling ballon of water on some ice. The COM of the rubber part probably wiggles a bit as does the com of the water. the whole COM being constant.

184. 184 184 Harold

#180. I assume ther is no hole in the bottom car? Is this it: The water runs through a tube from the right of the top car to the right of the bottom car. The water then flows along the bottom car.

When the flow starts, the water fills the pipe, so the COM moves to the right, the car moves to the left. There will never be the rightward force we discussed earlier, since the truck combo does not get lighter. At the end of the flow, the COM is back where it started. I would guess the movement would be a small one to the left, then a small one back to the right at the end as the level in the lower car returns to horizontal.

However, as we have done away with friction, the energy in the moving water has nowhere to go – it cannot dissipate. I think this probably does mean we would get the water in the bottom truck sloshing endlessly.

I am not sure I have got your set-up correct.

185. 185 185 Ken B

@Harold 184: Correct, that’s my thought experiment. And with the sloshing water we get a shimmying double decker.

186. 186 186 Ken B

db: “Newton’s Third Law means that if the car didn’t accelerate when you pushed it, it must be exerting an equal and opposite force on you. Where does this force come from?”

This isn’t right. If I push the car it pushes me with an equal and opposite force. Always. Even if another force keeps the car from accelerating. (In fact the world attached to the car accelerates.)

187. 187 187 Neil

Harold,

When the hole was first opened, the water left in what appeared to be a perpendicular direction, as far as I could eyeball it. The flow was fairly laminar for the first inch or so of the jet, and then broke up. As the water level receded the flow started to droop like it does with a Pissing bucket. Also a vortex formed as in draining a bath tub. When the water level got low enough it started dribbling in any which direction and I couldn’t tell what was happening.

When I put a drop of food dye in the container, I could see the general drift of water down and toward the hole.

188. 188 188 db

@KenB (186)
You’re right: it was sloppy to cite Newton III. Newton I is superior there. I was amazed to have to cite anything, to be honest…

189. 189 189 Ken B

@db: Mostly I was pointing out you got the law wrong. Normally I wouldn’t but after some of the comments here it’s clear some posters do not understand Newton’s Laws of motion, so the error seemed worth pointing out.

190. 190 190 Ken B

db in 196 writes:
“If you develop the equations that I developed in my 137… the motion remains rightwards forever DESPITE the rightwards nozzle.”

I want to be sure I understand your claim. You claim that if instead of a hole on the side of the boxcar we have a nozzle, and that nozzle is directed to the right, say parallel to the ground, that the car will end up moving to the right?

191. 191 191 iceman

Harold I forget, do you think you can get the rightward move with a continuous water flow, or just a discrete process like people stepping off?

192. 192 192 Al V.

By my calculation, the velocity of the box car is proportional to 1/ln(W+B-Ft) – 1/ln(W+B) where W is the initial mass of the water in the boxcar, B is the mass of the boxcar, F is the rate of flow of water mass out of the boxcar, and t is the elapsed time. This assumes a constant flow of water out the hole.

Thus, if I have a 10 meter 1000 kg boxcar containing 1000 kg of water, flowing out at 10 kg per second, then it will take 100 seconds for all the water to flow out. In that 100 seconds, the boxcar will move 2.85 meters total, and achieve a final velocity of 66 cm/sec.

193. 193 193 db

@KenB (190)
That’s not what I wrote. But your suggestion (which I christen Ken’s Paradox) might hold in some circumstances.

194. 194 194 db

@iceman (191)
You can get the final rightwards motion with both the continuous case and the discrete case provided in the discrete case the people keep moving when others drop out.

The final rightwards motion is proven with flow perpendicular to tracks, and conjectured even in cases of Ken’s Paradox (a rightward directed flow)

195. 195 195 db

I just wrote out the constraint for Ken’s Paradox, and it’s surprisingly simple and peasing. Is there a way to write maths in here, or is it best to handwrite and scan. Would feel happier if checked by someone who is neither on a train, nor several beers to the good.

196. 196 196 Harold

Iceman: I agree with db #194 – there will be a rightward force if there is movement in the car as the car is getting lighter, either flow or discrete lumps jumping off.

Can’t wait for the solution to Ken’s Paradox! I am hoping it has meaning in language other than maths.

197. 197 197 Ken B

Ken’s Paradox! I like it. I asked because I was unsure about what db was saying, and think I have a reductio. I think I can prove that if the nozzle is as described in 190 that we have a rocket.

Consider a mirror image of the boxcar, with a nozzle on the far (hidden) side also parallel to the ground and also point to the right side of the picture. Now link these two cars together. It makes no difference physically but let’s do it with the nozzles together. This is like a boxcar with a nozzle sticking out the right hand end. (If you are worried about the gap between the boxcars, imagine a thrid boxcar joined in the middle with a backward pointing nozzle to fill in the gap.) Does anyone really believe that is anything but a rocket? Therefore, if there is a rightward nozzle we have a rocket.

198. 198 198 Harold

#197 – I do not understand your set up -what do you mean by “with the nozzles together”?

199. 199 199 Ken B

@198

Boxcar A has nozzle on north side, box car B has nozzle on south side, B is north of A. The nozzles of A and B are adjacent, parallel, and pointing east. A and B are joined together rigidly.

200. 200 200 iceman

Sorry I know we’ve moved on to Ken’s Paradox but just to clarify, the ultimate rightward move with continuous flow requires that all the water momentum be converted to perpendicular upon exit?

201. 201 201 Harold

Are these on different tracks? Initially I assumed they were on the same track.

Are you postulating two boxcars linked together, where in one the hole is at the right of the car (as in original diagram), and the other the hole is at the left of the car? The water in car A moves to the right, the water in car B moves to the left. In this case the forces generated by the internal movements will cancel out, so the only force is generated by the water leaving the nozzle. We have a rocket. Or to put it another way, the COM of the combined water does not change until it leaves the two cars.

202. 202 202 Ken B

@201:
I am responding to db’s claim about “even if the nozzle points right.” I am am using a reductio to show that if the nozzle points right it is a rocket for sure.

203. 203 203 Harold

I am not sure. For the rightward force to occur there must be some water that moves left. If the nozzle were very small, so the water dribbles out with only a very small rightward momentum (or it is pointing only slightly to the right), is it not possible that the car could move left faster than the water leaves the nozzle relative to the hole? This would mean that the water is leaving to the right wrt the hole, but the hole is movinbg left faster, so the water actually leaves to the left wrt the observer?

We have forces in two directions, how do we know which is the larger?

204. 204 204 Ken B

@203:
Harold, add up the momemtum. If the big car moves left the small drop must move right at a greater speed. Rememebr, the car gets more momentum all the time as it is pushed.

There is really no question at all what happens if the water leaves the car through a pipe facing right. Rockets work. Rockets do not reverse course when they exhaust their fuel. Voyager is not returning to earth.

205. 205 205 Harold

I will have to think about people falling through the hole with no right momentum and throwing a pea to the right off the back of the car before they fall out. The key is that some of the force on the car is the movement of the water still in the car. With most rockets I think this is too small to make any difference.

206. 206 206 db

@KenB (197 etc)
I don’t find your reductio compelling. Your 204 “there really is no question at all…” I why this is called Ken’s Paradox. It seems impossible that a rightwards facing nozzle could allow rightwards terminal movement.

I’m not sure it’s proven, but belive for long carriages with small holes, it might be true.

But I really need to write out last night’s drunken scrawl and check it. I can certainly achieve a infinite (time) rightwards acceleration, but I need to check that it overwhelms the initial leftwards impulse (or can do so)

Harold (205) has one intuitive explanation, which certainly helps for the nozzle not fully flat right. I need to work on the fully flat non-rocket.

@iceman (200) – the condition that the water exits with no relative motion is sufficient for the leftwards shove followed by rightwards drift forever. It is not necessary – indeed that’s what has me thinking about Ken’s Paradox: what is the necessary constraint on the nozzle?

207. 207 207 Ken B

“The key is that some of the force on the car is the movement of the water still in the car.”
Which must balance out unless the water LEAVES the car with momentum. You can have sloshing and the car shimmying, or sloshing and periodic motion back and forth of the car but the car cannot move on net unless water leaves with net momentum.

208. 208 208 Ken B

@206: So at the end, does the last bit of water leave the right facing nozzle? Because if it does then that water’s x velocity (to the right is positive) must be HIGHER than the car’s. Otherwise it wouldn’t go throught the nozzle. But you say the car’s velocity is rightward, ie > 0. So at some point before the end you have the whole contraption, the car and the remaining water, moving *right* already (velocity > 0). And how did that happen? What applied a rightward force to the water and the car *together*?

209. 209 209 db

…Unless you have a lot of water leaving leftwards and fast early on…

210. 210 210 Harold

Ken B “Which must balance out unless the water LEAVES the car with momentum” Of course, but water leaving with no movement relative to the boxcar does have momentum if the boxcar is moving. db can correct me if I am wrong, but what I think must happen is this. The initial water leaves the rightward facing nozzle, which accelerates the car to the left. The water inside the car moves to the right to get to to the hole, which accelerates the car to the left also. If the acceleration due to the water inside the car is sufficient, then the initial water will be moving left to the observer, even though it leaves from a right facing nozzle. The water hammer effect. if the car is long and thin, the whole column of water moving inside the pipe could exert a large force. This could only happen with an incompressible fluid, not a gas.

We await the results of db’s analysis. Intuitively I couldn’t say if such a thing were possible or not.

211. 211 211 db

@harold – broadly yes, although I think the initial acceleration is all due to the motion of water inside the car, not the as-yet trivial amount of exited water-bourne momentum.

I’m not sure where the condition for incompressibility comes from. It’s not obviously a problem, but then not that interesting, either… :-)

In a long car. The initial impulse can be (arbitrarily) large. That’s a lot of left to play with.

Analysis is the only way. Today’s flight didn’t yield the answer, I’m afraid.

212. 212 212 Harold

The incopmpressibility thing is not important. Earlier I had speculated what would happen if the fluid were a gas.

213. 213 213 Ken B

OK, 210 and 211 are simply confused. You are assuming that water inside the the car can (not must but can)
1) move right wrt to the car
2) smash into the car causing it to accelerate in the positive x direction
3) leave the car throught the right facing nozzle moving right wrt to the car
4) thus cause the car to move right *wrt to the ground* even though the water has left the car moving right

1, 2 and 3 are correct but they are incompatible with 4. It’s just not possible because just before the impact of the water on the right end of the car the whole kit and kaboodle are moving left, so have leftward momentum. If the car and the water both end up moving right that momentum has disappeared.

My guess is you guys are not being careful with what is and is not an inertial frame, or assuming that accelerating the car rightwards causes it to *move* rightwards.

214. 214 214 Harold

Ken B #213 – I think you are correct. Thinking again, for the car to move right, we must have at least some water moving left. If the car is moving to the left wrt the ground, the COM of the water must be moving right wrt the ground. When the water leaves the car, it must still be moving right, unless we stop its movement wrt the car.

From a forces perspective, I also can’t see where the rightward force comes from unless we stop the water wrt the car before it leaves.

So I can’t see how a rightward nozzle could be other than a rocket, assuming no friction / surface tension / viscous drag at the hole.

If we are allowed to slow the water as it moves through the hole, then I think it possible that the car could end up moving right.

215. 215 215 Ken B

Harold: “So I can’t see how a rightward nozzle could be other than a rocket, assuming no friction / surface tension / viscous drag at the hole.” Almost there Harold! But the momentum argument I presented works even with those. Those can cause impact with the car, transferring momentum, but the argument allows for impacts with walls anyway.

216. 216 216 Harold

#215. if we have drag at the hole we can end up with no right momentum in the water, and we are back with the same situation as with the hole at the bottom.

I await db’s analysis, but are we in agreement up to:
1) If the water leaves straight down the car moves left then right
2) If all the water leaves from a right pointing nozzle without friction it is a rocket.

If so, we still have intermediate situations where the water is slowed or pointed partially down. In this case, the water could end up leaving with left momentum wrt the ground, even if it leaves from a right pointing nozzle.

217. 217 217 db

Careless to assume no viscous drag at the hole!

The initial leftwards velocity is independent of the nozzle direction.
We can make it arbitrarily large with large l.

All we need to final rightwards movement is that more waterborne momentum exits leftwards than rightwards.

The flow of water is strongly decreasing — so that initial velocity has a huge impact on the net waterborne exited momentum.

It’s a relatively simple derivation to add in the Cw’ element and creat an inequality on C to maKe always rightwards acceleration of the car.

I don’t find anything significant challenges to the idea in Ken’s four-point plan. But then it is called Ken’s paradox…

Boxcars are nothing. Try having people shoot laser guns at each other in a relativistic context (i.e. approaching lightspeed) and you end up with people being dead in one frame of reference and alive in another.

219. 219 219 Harold

db “The initial leftwards velocity is independent of the nozzle direction.”

Ok, lets see if I can figure this out. It would be great if the paradox is real!

Thinking of people on a very long rail car, all lined up behind one-another. Simultaneously they all start to run to the right. This exerts a large force on the car, which moves to the left. It can only move to the left if the COM of all the people move to the right, so the people must have some velocity to the right. This will depend on the weight of the car and people. Assume quite a light car, and the car moves quickly left, but the people must move slowly right. Whilst everyone is on the car it is not possible for both car and people to be moving left no matter how fast the car is going in that direction.

Now the first person falls off the back of the car. One extreme is that he slows to a stop before falling – that is the vertical hole scenario. He will leave with leftward velocity. The other extreme is he slows not at all. In the latter case he must leave with rightward velocity. I don’t see how at this extreme we can have anything other than a rocket.

However, if he slows even a little bit, then he could leave with left velocity, and the car could end up travelling to the right.

I assume this will also happen with the water in the car. If the flow is slowed at all as it goes through the hole, the eventual rightward movement of the car is feasible. The force that causes this is the force that slows the water.

So we are back with assumptions. If we can have a frictionless track, then why not a frictionless hole that doesn’t slow the water? Alternatively, if the rest of the system is “real” except the frictionless track, then we will have slowing of the water as it flows through the hole.

Is this accurate? Does the people experiment translate, or is there some essential part missing?

220. 220 220 Harold

#218. I recently read of a problem relating this and quantum cryptography. A station on Earth sends two entangled photons to two satellites. The first measures the photon, causing the entangled photon’s wave function to collapse also as measured by the second satellite. The twist is that the two satellites are travelling towards each other. From the POV of Earth, the photons arrive simultaneously, but each sees the other’s clock running slow, so each sees the others measurement to happen first. Which measurement caused the wave functions to collapse?

221. 221 221 Ken B

“so each sees the others measurement to happen first. ”

I double.

Neither can SEE the event until ANOTHER photon travels from the event to the observer. This is NOT a minor objection. There is a big big difference between inferring/calculating the time value of an event in some frame or other and SEEING it.

222. 222 222 Harold

#221 -yes I phrased that badly. To an observer on sat 1 the measurement on sat 2 will appear to happen first. To an observer on sat 1, the measurement at sat 2 appears to happen first. I am not certain the is any better, but it is how it is put in the original article. The nature of cause and effect in uncertain.

It was from New Scientist
- diagram is here, but article is paywalled.
http://www.sciencedirect.com/science/article/pii/S026240791361009X

223. 223 223 Neil

I found this.

http://www.hep.princeton.edu/~mcdonald/examples/tankcar.pdf

Apparently the car does move, although not as a rocket.

224. 224 224 Harold

Neil – good find! I like the way they specify the problem by saying the water leaves the car vertically in the rest frame of the car. Later on they say “To guarantee the latter condition, the drain might have to be placed in a sump whose vertical walls can assist in absorbing the horizontal momentum of the water flow”

225. 225 225 Guy

Thanks Neil. That looks like a pretty thorough treatment of the “water leaving vertically w.r.t the boxcar” scenario. I think I’m ready to move on now (and forget all about nozzles etc.)

226. 226 226 Guy

Also; well done db – I think you nailed it!

227. 227 227 Ken B

The link under 223 is wise enough to specify explicitly a simplifying assumption about the water! As has been noted endlessly above, yes if the water carries no net x momentum when it leaves the car we get motion not a rocket. If it leaves with momentum we do get a rocket.

228. 228 228 Neil

I wonder if that air track experiment he describes at the end of the article has ever been done. I’d like to know because I still have doubts that the car moves, even on the frictionless track.

229. 229 229 Keshav Srinivasan
230. 230 230 Neil

If you haven’t had enough of this by now (I have), another good site for the leaking tank wagon problem is

http://physics.stackexchange.com/questions/1683/mechanics-around-a-rail-tank-wagon

231. 231 231 Ken B

We’re missing the big picture. What does Goedel’s Theorem say about the boxcar, and does Farnsworth McCrankypants agree?

232. 232 232 Ken Arromdee

Any bit of water that reaches the hole will cause thrust in the opposite direction that the water is moving when it’s leaving the hole.

The wider the hole is relative to the width of the boxcar the greater the thrust is because if the hole is wider, the greater the chance that the water will flow over the hole and fall out while still flowing rather than hitting something, coming to a rest, and then falling out of the hole without carrying all that momentum.

The wider the hole is relative to its thickness, the greater the thrust, because if the hole is n a thick surface the water can lose momentum by hitting the sides of the hole. With a small enough hole this effect can be zero (and the first effect moot).

If the water does not provide thrust because the hole is small eniugh that the flowing water always hits the sides of the hole before falling out, then the boxcar will shift to the left, keeping the center of mass of the boxcar+water constant.

If the water is providing thrust and the boxcar starts moving, the water will continue to provide thrust since it will be carrying rightward momentum relative to the boxcar, so the boxcar will continue to speed up until the water is gone. It is true that the water is being carried along by the boxcar and may actually be moving left relative to the ground even though it is moving right relative to the boxcar, but water that is moving left relative to the ground can still provide thrust which accelerates the boxcar left–an observer on the ground would see that as a boxcar that moves left and splits into a faster left object (less full boxcar) and a slower left object (falling water).

233. 233 233 db

@Neil (223) – that’s a good paper — much more clearly lays out the route through Bernard’s equations than I was able to manage.

I’m still pondering Ken’s Paradox. I remain convinced it is provable.

234. 234 234 Harold

db – any news on the proof?

Is this a good summary:
If the water exits vertically, the car moves to the left, then stops, reverses and ends up heading for the sunset to the right. Exactly when it turns round depends on the way the flow reduces.

If the water exits without friction to the right, we have a rocket. There is no way to produce the rightward force.

Intermediate situations are of two types: water exits to the right, but slowed down, or water exits with some angle to the right. I think these are more or less equivalent.

If the water leaves at 0.01° off the vertical towards the right, I would guess that the force that produces the rightward travel in the first case would (for most boxcars and holes) overcome the tiny rocket type force, and the car would still end up travelling to the right.

If the water leaves at 0.01° to the horizontal, there will be only a tiny force to the right, and I would expect the rocket forces to be larger.

However it may be possible to stipulate starting conditions (length of car, size of hole etc) where it would only take a small deviation from the horizontal to produce a rightward motion.

235. 235 235 Ken B

In the paper, which I have not read in detail, equation 9 clearly ends up 0. This is what you should get if, as per the problem, the water exists straight down.

You cannot write a sensible equation without *assuming* something about how the water moves as it leaves the hole.

236. 236 236 iceman

Fyi Farnsworth does not like the idea of wagons leaking water all over. He does not like it at all.

Still don’t quite get how we end up going rightward with a continuous water flow (as opposed to people stopping and jumping off).

Btw what’s the all-time record for # of post comments?

237. 237 237 Harold

#235 I presume a hole at 45° is the same as one at 90° and one at 0°. We can calculate the COM changes for the horizontal (rocket) hole using half the flow and appropriate fomula for velocity of water leaving a hole. The COM changes for water inside the tank will be the same as calculated for the tank with the vertical hole (using flow through both holes). We then just combine the two.

We adjust the rocket forces depending on the angle of the nozzle. the other forces always stay the same.

238. 238 238 Ken B

I data-mined Farnsworth’s cell calls and he spent a lot of time talking to the railway safety inspectors. I am having the IRS audit him.

239. 239 239 Brian

Steve,

Thanks for re-opening the comments. I have what I think you will agree is the definitive solution to the problem. The short version is this:

The boxcar is not a rocket.
There is no net change in the center of mass.
Therefore, the boxcar does not move at all.

Since this solution is contrary to what everyone has said so far, except for Neil (and even he surrendered on the strength of the McDonald paper), I will lay out the detailed reasoning.

First, we can agree that if the water flows straight down, there is no net momentum change in the horizontal direction, so the boxcar cannot be a rocket under those conditions (this has been stated by many in the comments). But what does the water do?

There are a number of ways to see that the water must flow out vertically, i.e., along the gravitational field direction.

First, the water shoots out because of the pressure (force per area on the hole), which is due solely to the weight of the water above the hole. Even for a hole on the right wall (which WOULD make a rocket), the water motion depends only on the height of the water and not on the length of the boxcar. The force on the water effectively comes from the molecules bouncing off the far wall (the piece opposite the hole), which is unbalanced because the hole can’t bounce them back. In the case of the hole in the bottom, the opposite “wall” is the water surface lying along the gravitational equipotential, and the water flows along the gravitational field.

Second, water can only flow rightward out of the hole if it is moving rightward along the boxcar bottom right where the hole is. But this is impossible. Friction and surface tension create the “no-slip” condition of hydrodynamics, where the velocity at a surface is identically zero. It’s true, of course, that the water must move to the right to fill the vacated space as water flows out, but this happens primarily near the top where the velocity can be maximum. The no-slip condition combined with water viscosity prevents horizontal motion near the boxcar bottom. Water can only reach the hole, then, by flowing straight down (along the gravitational field). Horizontal motion near the top is lost by collisions long before the water reaches the hole.

Third, these ideas are easily tested experimentally (as Neil apparently did). Take a clear-plastic Q-tip container, punch a hole in the far-right bottom, trim off the excess, fill with water, and observe. The water flows straight down, as expected.

THEREFORE, THE BOXCAR IS NOT A ROCKET!

But what about the center-of-mass condition. Water clearly has to move right to exit the hole on the right. Since the horizontal C.o.M. of the combined system must stay constant, shouldn’t the boxcar first move left and then right again, as in the McDonald paper? Shouldn’t there be, at least, a transient displacement?

The answer is NO. Here’s why.

The rightward motion of the water will certainly try to move the boxcar to the left via the reaction force. But here’s the subtlety: the centripetal gravitational field is not perpendicular to the hole but points slightly to the left. The water therefore flows out SLIGHTLY TO THE LEFT(along the gravitational field), creating a reaction force on the boxcar to the right that cancels the leftward reaction force.

The claim of exact cancellation might seem implausible, but is easy to see from a C.o.M. argument. The water and boxcar begin with their C.o.M. over the gravitational center of Earth. As far as the boxcar knows, the falling water travels all the way to the gravitational center and continues up again to China (say), before falling down through the center again and returning to the boxcar. This motion continues forever with the water C.o.M. at the gravitational center of Earth. That is, because Earth’s gravity is centripetal, there is ultimately no net change in the water’s C.o.M. and the boxcar never needs to adjust itself.

THEREFORE, THE BOXCAR EXPERIENCES NO MOTION AT ALL.

240. 240 240 Scott F

I don’t know if any of you have ever watched water drain from a tub, but *spoiler* sooner or later it starts to form a whirlpool-like phenomenon. YouTube 2 liter vortex, or just take a bath, and watch it for a while when you let the water drain. If that happens, I’d imagine the only force exerted would be that which the water leaving the hole exerts (again I site the 2 liter bottle experiment). What happens once the boxcar moves and how that changes the creation/stability of the whirlpool, is far beyond my understanding of fluid dynamics and physics. Just thought I’d throw in another line of thought.

241. 241 241 Ken B

@Scott F: Yes, this is part of my thinking in 235. Without simplifying stipulations you cannot write a sensible equation. The cited paper has such a stipluation, that the water leaves perpendicular to the car. That may not be quite realistic, especially as the equations also assume uniform flow ie no gurgling.

242. 242 242 Scott F

@Ken B: Got it. I also failed to realize (this might have been mentioned or implicit in a comment above), that with time all flow would stop unless the mouse’s sister also chews a hole at the top to allow air to come in. Therefore, the no gurgling bit is probably a stretch. Thanks for the clarification

243. 243 243 JohnC

Ah, the memories. See Kirk T. McDonald’s “Motion of a Leaky Tank Car”; and cf. his “Wagon in the Rain” (and similar problems).

244. 244 244 Neil

Brian@239

Yes! Your centripetal argument is exactly right. I knew the water-boxcar system could not be affected by the water that already left the car, but I didn’t pick up on the fact that the water stream had to be slightly left of the perpendicular because the hole was right offset. That was the missing piece I wish I had known before I “surrendered”.

245. 245 245 Guy

@Brian

Sorry, I’m not buying it. The water has it’s C.o.M shifted right (to reach the hole). It now falls towards the center of the Earth. The C.o.M. of the Earth+water does not change as it does so (or as it makes its way back from China). The Earth is falling towards the water at the same time as the water is falling towards the Earth.

246. 246 246 Brian

Guy,

You say ” The C.o.M. of the Earth+water does not change as it does so (or as it makes its way back from China). The Earth is falling towards the water at the same time as the water is falling towards the Earth.”

Yes, this is true, but the C.o.M. of the combined system doesn’t change as the water moves right either. You are biasing yourself by applying the argument only after the water leaves the hole. Apply the argument over the entire motion (before the water comes out) and you will see that the water by itself has no horizontal change in its center of mass. Therefore, neither the boxcar nor the Earth has a change.

247. 247 247 Guy

Brian,

“the C.o.M of the combined system doesn’t change as the water moves right”. Yes – because the boxcar moves left. The Earth doesn’t move to compensate for this movement because of the friction-less tracks and the movement is perpendicular to the gravitational forces.

If you prefer we have a change in angular momentum of the water (around the center of the Earth). The Earth doesn’t compensate for it (because of the friction-less tracks), so the boxcar must.

248. 248 248 db

@Brian (239) I’m afraid I don’t fully buy yOur argument that the water flows left exactly enough to cAncel the shift from the centre of the truck.
If you prefer not to simplify to a cartesian analysis of the momentum, then you can consider the polar analysis. The angular mmomentum is conserved so the boxcar must move

249. 249 249 Brian

Guy and db,

You both want to make angular momentum arguments. I’ve never thought about it that way, so I applaud your use of yet another perspective. Great minds think alike. :)

The angular momentum does not change, however. Prior to flowing the water and boxcar are not moving, so they have no angular momentum (in the Earth’s frame). Immediately upon leaving the boxcar the water is falling along the gravitational field toward Earth’s center. This motion has no angular momentum since r || p and r x p = 0. Consequently, as long as the falling motion is centripetal, any angular momentum gained when moving right (while still in the boxcar) is lost as soon as it falls out. Since the water angular momentum has no net change, neither does the boxcar’s angular momentum.

This perspective confirms once again that centripetal motion is the key to preventing the boxcar from moving at all!

NOW are you convinced?

250. 250 250 db

“any angular momentum gained when moving right (while still in the boxcar) is lost as soon as it falls out”

Where does it get the angular momentum from whilst it moves rightwards? It can’t just borrow it from God and give it back later so that momentum is preserved at the start and the end — it must be preserved for all time.

It comes from the boxcar — which therefore moves the other way.

Your argument is a classic error in qualitative discussion which are seen in all sorts of arguments (and vaguely blog on-topic — is extremely common in economics). Someone points out a thing which does one thing and a second thing which opposes and makes an argument that they balance out without observing that the first effect is a billion times larger than the second…

If you’d like to prove this out, then you can run the equations with your leftwards deviation in the stream and see what effect it has on the boxcar. The answer is very little indeed — the deflection is of the order L/R where R is the radius of the Earth (about 6 million metres). Wonderfully, these equations are a special case of the equations to solve Ken’s Paradox — since you are effectively proposing a (very very very) mildy deviated nozzle.

251. 251 251 db

I abandoned the proof of Ken’s Paradox after the comments were closed, but I picked it back up again this afternoon and wrote it out after seeing they were open again last night.

To recall Ken’s Paradox: We consider a boxcar with a nozzle instead of a simple hole. We can choose to point the nozzle in any direction. The nozzle directs the flow so that it removes any doubt about the direction of the flow out of the boxcar that the simple hole creates. The straight down solution is provided by McDonald’s paper — the boxcar moves off left and then rightwards forever.

Ken’s Paradox is that for a sufficiently long carriage:-
– even if the nozzle is pointing the jet leftwards, the carriage initially moves left
– even if the nozzle is pointing the jet rightwards, the carriage eventually moves right forever.
This isn’t the behaviour that one would expect in a rocket, but it is exhibited in the boxcar.

To do this I stood on McDonald’s shoulders and added in the nozzle effect to his equations. In addition to the standard assumptions and generalised flow, I needed to take a specific exponential flow rate associated with a viscous nozzle. A better mathematician might be able to force an inequality out of my [4] without such a strong condition. I chose units to avoid death-by-ink but a lost no more generality than insisting the flow rate was proportional to the height of water.

Appreciate any comments or if I need to clarify anything.

252. 252 252 Brian

db,

You say “Your argument is a classic error…. Someone points out a thing which does one thing and a second thing which opposes and makes an argument that they balance out without observing that the first effect is a billion times larger than the second…”

It’s clear you’ve not yet processed my arguments. I stated up front that exact cancellation seems implausible, but then showed rigorously that they do in fact cancel. Both in terms of C.o.M. and in terms of angular momentum, the beginning value is the same as the ending value. That guarantees the two effects offset. The offsetting effect does not just occur for all the water taken as a whole; it occurs for every individual packet of water one can define. In the case of angular momentum, every definable packet begins with L = 0 (no motion) and ends with L = 0 (motion along the gravitational field), so no packet of water gives a net angular momentum to the boxcar.

Your dismissal of this reasoning is not valid. You haven’t shown a flaw. Mathematics doesn’t add any light to the matter–it’s just more difficult to do than the qualitative analysis. This is a blog after all, and more amenable to qualitative discussion than quantitative exposition. But I understand your complaint–you want a microscopic explanation rather than a macroscopic one. There’s nothing wrong with that preference. But you can’t reject macroscopic arguments as being invalid just because you don’t prefer them.

1) The leftward motion due to gravity is NOT identical to a slightly leftward nozzle. The latter would cause the boxcar to move right because the water keeps going forever. The effect of gravity is to distribute the water around Earth’s center of mass, which gives no net motion.

2) The net momentum change (angular or horizontal) at the hole is indeed very small, but so is the net change to fill in the vacated space (which never actually vacates–all the motion occurs at once). Most of the water moves vertically. The immediate horizontal motion occurs from both sides. Now it’s true that the longer left side has to replenish the right side, but it does this by sloshing over to the right side and hitting the right wall. The horizontal motion cancels, as it would with no hole.

253. 253 253 Neil

Just to add to Brian’s already excellent exposition, realize that in his hole-in-the-world example the CoM of the car and water is changed, and momentum is not conserved, if the car does move. This alone proves that the slight leftward angle at which the water exits must be enough to compensate for the impulse of the right-ward flow of water in the car while it is draining. Of couse there does not have to be a hole in the world. It is simpley that, once the water has left, the car has, and cannot have, any knowledge of what happened to it.

254. 254 254 Guy

Brian,

I think both your momentum argument and your C.o.M argument are flawed.

Momentum first. You say that the beginning value and the end value is the same. That might be true, but, as db says, you cannot borrow momentum and give it back later – it must be conserved at all times. A very short time after the hole is opened the water leaving the hole reaches its maximum velocity. This water is replaced by water traveling from the center of the boxcar (it might help to conceptually “isolate” this water by imagining a horizontal pipe running from the middle of the boxcar, along the floor to the hole so that the bulk of the boxcar water has no horizontal momentum). So the boxcar water reaches its maximum horizontal (or clockwise) momentum very quickly. This momentum cannot be cancelled by the small amount of water which has left the boxcar in this time.

Now C.o.M.: You say that the water that has left the boxcar ends up with its C.o.M. over the center of the Earth (and therefore centered under the boxcar). I would argue that it ends up not at the center of the Earth, but at the C.o.M of the Earth-plus-water-as-it-left-the-boxcar. It’s true that this is immeasurably close to the original C.o.M. of the Earth, but I hope you agree that doesn’t matter.

255. 255 255 db

Brian (252)
“Your dismissal of this reasoning is not valid. You haven’t shown a flaw. Mathematics doesn’t add any light to the matter–it’s just more difficult to do than the qualitative analysis.”

I’m afraid that I can’t really deal a fatal blow to your “reasoning” because it’s just not rigorous enough to argue with. It’s like nailing jelly (jello if you’re in the US) to a wall. Mathematics is harder than qualitative reasoning because it forces you to be rigorous to be right or wrong. It structures thought and exposes it to the bracing wind of criticism.

In that sense I don’t really disagree with you at all: you haven’t said anything yet. There’s nothing to disagree with. That said – almost every sentence you’ve written is flawed, so I find it hard to believe your argument.

It’s clear you’ve not yet processed my arguments.
– Give me something to process

I stated up front that exact cancellation seems implausible, but then showed rigorously that they do in fact cancel.
– You and I have different definitions of rigour.

Both in terms of C.o.M. and in terms of angular momentum, the beginning value is the same as the ending value. That guarantees the two effects offset.
– As it is with Newton’s cradle through one oscillation, but noone doubts the second ball moves. If you assume that the boxcar doesn’t move then the boxcar doesn’t move, although it doesn’t explain how the water does.
- Circular arguments are key because the key to arguments is circular reasoning.

The offsetting effect does not just occur for all the water taken as a whole; it occurs for every individual packet of water one can define. In the case of angular momentum, every definable packet begins with L = 0 (no motion) and ends with L = 0 (motion along the gravitational field), so no packet of water gives a net angular momentum to the boxcar.
– Rubbish. Once the boxcar starts moving the packets have angular momentum and then they leave with angular momentum which they forever take away from the boxcar, making it move.

* * *

Let me ask you a question and try to put some rigour into your thinking that we might all learn something — what would persuade you that you are wrong?

256. 256 256 Harold

Just when you think it is safe to come back to the boxcar! The circularity of the Earth does indeed add an interesting twist to the problem.

If we assume no air resistance and a hole throught the center of a cool, non rotating Earth, the water will fall through the Earth and back out to the surface, then fall back again endlessly, oscillating for all time. (I think there was a post about the how long it would take a while ago.) The COM of the water thus oscillates around the COM of the original boxcar/water system.

Because this is oscillating in this direction, something must compensate to keep the COM the same. The boxcar has no connection with the water any more, so it cannot be the boxcar. Initially I concluded that it must be the Earth. Therefore, we must consider the COM of the baxcar/water/Earth system, not just the boxcar/water.

However, the boxcar is now on a circular track, which goes round the Earth and back to where it started. Therefore the boxcar if moving in one direction will also oscillate around the COM. Is it possible that just as the water gets to the other surface and starts falling back to the center of the Earth, the boxcar has got 1/4 way round and is now heading back towards the COM of the Earth also? It sounds implausibly fast.

257. 257 257 db

@Harold (256)

Usually we would be happy with a locally flat Earth and do our work in Cartesian coordinates — thinking in terms of “left/right” “up/down” etc.

If we prefer, we can embed this one-dimensional problem into three dimensions and consider it that way. In this instance we would look at angular momentum in the 3-system and reduce it to prove that it is the same as “linear” mommentum in the 1-system.

Probably the easiest way to proceed is with an energy argument — insisting that the track lies along an isopotential. We can then use Hamilton’s equations to derive the constraint on the momentum and show that it obeys a similar solution to the Flat Earth solution. I haven’t written this out, but believe that’s the way it will play through. All you need to do is show that momentum is conserved along the line of the track and you are back to the original problem.

If you think about this from a problem-solving approach, the goal in this construct (and all constructs) is to reduce this three-body, 3-dimension problem to a 2-body 1-dimension problem prior to hitting it with the maths. What is essential is that whatever solution we come up with, it must match the flat Earth solution when D/R is small — just as Relativity and Quantum mechanics must reduce to Newtonian mechanics at everyday scale and pace.

The false reasoning of the Round Earth solution can just as easily be applied to the Flat Earth problem where it is more obviously nonsense. Either way you can’t displace stuff without moving it — whether altering its l-r coordinate or its latitude.

To be honest I thought I was done when I solved Ken’s Paradox. Disappointed noone has corrected the obvious error in my working.

258. 258 258 Ken B

@256 and others. The COM can move if there is a net applied force. There is in the hypothetical here you are discussing (gravity). And once the water starts its way down to tunnel to China that force keeps changing. So the analysis of the tunnel to China tells you nothing at all about the boxcar.

259. 259 259 Ken B

Here’s a new worked out example. I will play with a couple variations.
db, Ken, and Harold are in theboxcar, which has a frictionless floor, holding handles.
The riders leavee perpendicular to the car, instantaneously.

Each has mass m, and the boxcar has mass m.

First one at a time, as I did earlier.
Harold pushes off and slides towards the hole. He smacks the right side, comes to rest wrt to the car,
and exits the hole perpendicular to the wall.
What has happened so far?
The boxcar together with its remaining occupants has moved a distance left and stopped.
Repeat for Ken and then db. Same conclusion.
This confirm my earlier analysis of stroboscopes.

OK, now let’s do what db did, and have them start moving *before* the previous guy exits.
Harrold pushes off. He has a certain velocity, call it 3h rightward.
The car and its occupants now have velocity h leftward.
(Convince yourself that this is happening with a brief accelration in an inertial frame
so the equations apply.)
Ken pushes off. Wrt to the car he pushes off with speed 2k.
In the ground frame of reference
the car and db are now moving left h+k, Ken is moving right 2k-h, Harold is moving right 3h,

Let’s balance:
left: 2m(h+k)
right: m(2k-h) + m3h

Now Harold hits the wall and stops wrt to the car.
This impact will not affect Ken, who is in motion on the frictionless floor.
So we have a leftward momentum 2m(k+h) meeting rightward 3mh.
In the original problem 3h and 2k represent forces from water pressure, which should diminish.
So its reasonable to let k=h for our example. Let’s do that and see.
That leaves leftward momentum of mh, and the moving mass is
the car, Harold, and db. It is therefore moving h/3 leftward.
Ken is sliding (note he has the correct mh rightward).
Harold exits.

We could put db in motion here but first let’s see what happens if we do not.

Now Ken hits the car. mh rightward BUT (2/3)mh leftward. Harold is gone.
When Ken exists the car and db will be moving right, as per db’s analysis.

So let’s have db push off before Ken’s collision.

The car was going h/3 left and now gains d. Let d = h. The only affected masses are
db and the car.
SO: Ken h right
car 4/3 h left
db 2/3 right
Ken hits and we have for the Ken-car mass net momentum (mh/3) left.

Finally db hits. That leaves us with net right momentum for db-car.
So the car ends up moving right (per db).

Note that this example DEPENDS on the frictionless floor.
No frictionless floor and we can no longer ignore the guy in motion.
I don’t know how well that fits a box of water; pressure is transmitted.

FINALLY FINALLY
I doubt that the water leaves with NO x momentum.
Imagine that suddenly the entire right end half of the side or floor of the boxcar vanished.
The water would clearly come out with a lot of x momentum.
Imagine the right quarter. A lesser effect but still there.
Continue to mousehole size. The x momentum never goes away entirely.
That would be a rocket.

260. 260 260 db

I can assure you, I weigh considerably more than m.

261. 261 261 Guy

db – “Disappointed noone has corrected the obvious error in my working.”

I think you are using a meaning of the word “obvious” of which I was previously unaware.

262. 262 262 Neil

@259

This example fails to model the water in the tankcar for two reasons. First, water molecules have kinetic energy and are in constant brownian motion because they are not at absolute zero. They do not need to “shove off”. Second, water molecules below are pressed on by those above them because of gravity. It is gravity that creates the water pressure within the car that evacuates the water. That is why we need to put the hole at the bottom of car, not the top.

If we had one layer of water molecules covering the floor of the car, there would be no jet. The molecules would leave randomly.

263. 263 263 Harold

#261 – I second that! Please enlighten us db, or was that a tease?
#259. I don’t think friction alters the result – we just have friction to get started instead of handles and friction to stop us instead of hitting the wall. It would make the instantaneous claculation difficult, but surely overall the forces are all the same if we ignore heating?
257 & 258 – thanks for clarifying the situation.

264. 264 264 Ken B

@Neil 262:
It’s a pressure differential that “pushes off”. Water moves from the left end of the car to the right, due to pressure differentials. The applied forces are gravity and normal forces from the boxcar walls and floor.

265. 265 265 db

@263 there’s a sign missign and a reversed ine

266. 266 266 db

Early post….

@263 there’s a sign missing and a reversed inequality in Lemma 2. I still assert the right answer. Nice thing about Lemmas – they are Machiavellian.

267. 267 267 Capt. J Parker

Ok, I’ve gone from “doesn’t move” to moves then stops (based on lumped masses exiting the boxcar) and now I’m back to the boxcar moves left like Steve Landsburg said. I think the tank problem is a good way of thinking about this BUT, the author of that PDF is messing with you. The tank keeps moving left. In that problem solution m=0 at a defined x but dx/dt is not zero at that same x. I think you can prove this if you assume m=mo-ct where c is a constant (dm/dt?) Differentiate x WRT t and solve for dx/dt @ x=L ln(M/mo+M)
db, Ken B, Harold. Do you guys have the definitive solution? My brain got too fried to read all those posts.

268. 268 268 Harold

#267 It is true, there are an awful lot of posts! I personally think db’s analysis is close to definative, if the water exits vertically then the boxcar moves left then right forever. I think Ken B, db and I agree on this, with some dispute from Brian and possibly Neil. I believe that the discussion has moved on to whether this can happen if the nozzle points right.

269. 269 269 db

@Harold (268)
In terms of the discussion, I think I’ve proven Ken’s Paradox in my proof ( http://www.turl.st/maths/KensParadox.pdf ) I’ve not heard any criticisms of it here so far. I remain all ears, however.

In other words: With well-behaved nozzle flow, regardless of the direction of the nozzle, a sufficiently long carriage dominates the behaviour: the car moves initially left despite a leftwards jet and moves rightwards forever despite a rightwards jet.

@Capt J Parker (267)
There are several pdfs (one of which is mine) so need more clarity. Mine is an expansion of the McDonald paper. I think the Wolfe paper is wrong — its treatment of the hole is flawed meaning that his statement of the constraint on the system is wrong.

270. 270 270 Neil

FYI–the McDonald paper is not just a pdf. It was published in the American Journal of Physics in 1991. That doesn’t mean it is not wrong. But it was peer reviewed I suppose.

271. 271 271 Capt. J Parker

@db, In (267) The tank problem I was refering to was from the Ken B (79) post http://www.phys.unsw.edu.au/~jw/1131/Tank.pdf
But, it looks like I’m way behind the 8-ball and I need to read McDonald and your analysis. Thanks
Thanks Ken B and Harold too. Good luck on the bet with Steve Landsburg ;)

272. 272 272 Brian

db (255),

You say “In that sense I don’t really disagree with you at all: you haven’t said anything yet.”

Thanks for confirming that you haven’t really grasped my argument.

“what would persuade you that you are wrong?”

Well, I can think of lot’s of things regarding my own argument, but since you like the concreteness of calculations, let’s look at McDonald’s approach to the problem (of which you have done a variation). I will propose my own variation and I’d like to hear your analysis (it doesn’t require any extra math) of what happens.

Let’s suppose that we make the water drain at a constant rate (as McDonald does as an example), so Mdot = constant. Furthermore, let’s hold the boxcar still (by some external force) until the constant flow is established, then remove the force at t = 0. We follow the motion from that point on.

What do you think the boxcar does? Does it move? If so, how? Most importantly, if the boxcar moves, what are the forces and where do they act to cause such motion?

If you are right, this discussion would go a long way to convincing me that I’m wrong. Fair enough?

273. 273 273 Ken B

@Harold 268: Yeah I am convinced that IF we stipulate that the water leaves the car perp to the car in the frame of the car, that it ends up moving right. I doubt that is realistic in the given problem. So does MacDonald who writes
“the water can leave the tank
with zero relative velocity. To guarantee the latter condition, the drain might have to be
placed in a sump whose vertical walls can assist in absorbing the horizontal momentum of
the water flow.”
Which is what several of us noted above. A mouse hole probably won’t work.

An early poster mentioned making assumptions to get a “good problem”. When you do that you get this stipulation.

274. 274 274 Brian

Ken B,

You say ” IF we stipulate that the water leaves the car perp to the car in the frame of the car…I doubt that is realistic in the given problem. A mouse hole probably won’t work.”

It’s easy to verify for yourself that the water comes out vertically as long as the hole is clean. There’s no need for a sump. As I said in my original post (239), it’s easy to test with a boxcar-like shape, such as a Q-tip container. My post also highlighted the theoretical reasons the water has to exit that way. This point is not really debatable. Test it for yourself if you don’t believe it.

275. 275 275 Brian

Harold (#268),

You say ” if the water exits vertically then the boxcar moves left then right forever.”

This result from the McDonald paper appears to depend on faulty initial conditions, which is the point of my scenario in #272. The vertical flowing water never exerts a horizontal force on the tank car, so how can it ever start moving? How can it ever reverse direction?

276. 276 276 db

@Brian (272)

If the boxcar is held still for the “gnawing” and the flow of water established before we release the boxcar and then is constant then there is no motion. That can be seen by McDonald’s [5].

However unless the boxcar is being topped up, there will be an impulse when the water runs out (or the tap is turned off). M” become large and positive so the car accelerates rightwards with this impulse and goes off to the right horizon. If the water never ends then it does not move.

If the boxcar is held still for the initial impulse and then released, but there is an angled nozzle, the motion is determined by my [4].

However in the originally posed problem, the boxcar isn’t held, and so the impulse which allows the rightwards flow of water to be set-up does move the boxcar, and that moves the hole, and that moves the stream of water, and that is why the motion is complicated.

277. 277 277 db

@KenB (273)
McDonald’s constraint is sufficient for his work, but is unnecessarily strong for the conclusion. The water does not need to exit exactly perpendicular to the boxcar — it can be directed by a nozzle and still exhibit the same motion as with the perpendicular flow.

Even if that does seem like a paradox (to you in particular!)

I don’t know what the weakest possible constraint is, but I suspect it is that the hole is small enough so that the flow is largely determined by internal static pressure rather than internal dynamic pressure. I’m not a fluid dynamicist. I wish I knew one to ask.

* * * *

MORE GENERALLY:-
The larger the boxcar and the smaller the hole, the more the motion is dominated by the internal flow of the water rather than the external flow of the water.

278. 278 278 db

@Brian (275)
It’s not the water falling vertically that exerts the force on the boxcar, it’s the water moving rightwards within the box car that does.

In the moment before the hole is opened, the water inside the boxcar is stationary. In the very instant that the hole is opened, the entire mass of water starts moving right. The size of this step impulse is M’(0+)D (the initial rate at which mass is suddenly appearing at the hole instead of at the centre, times the distance from the centre) — so if the hole is a long way from the centre, or the flow is suddenly fast, then that’s a huge amount of momentum that has just appeared in the system.

That’s what kicks the boxcar leftwards.

279. 279 279 Brian

db (276),

You said “If … the flow of water established before we release the boxcar and then is constant then there is no motion.

However unless the boxcar is being topped up, there will be an impulse when the water runs out (or the tap is turned off).”

You almost got it right. Look at McDonald’s eq. 26. The rightward kick doesn’t happen if v(0) = 0.

In other words, if the water flows at a constant rate before the tank car is released, the tsnk car NEVER moves!

Now how is that possible? Doesn’t all that water still flow rightward while all the water is draining out at a steady pace? Isn’t that rightward motion still arrested by the sump? How can everything depend on these transient effects? How can the endpoints matter but the huge bulk of water in between doesn’t?

The answer is that the transient effects identified by McDonald are not real. They are an artifact of his calculation using COM.

You’ve said you want a calculation. Here it is. Let’s use conservation of momentum instead of COM. From conservation of momentum we get

MtankVtank + MwaterVtank + integral[(-Mwaterdot)Vwater dt'] = 0.

But if the water flows out vertically, Vwater = Vtank, giving

(Mtank+Mwater)Vtank – integral[(-Mwaterdot)Vtank dt'] = 0.

But by inspection it’s obvious that a solution to this equation is
Vtank = 0, so the tank car never moves.

Taking the derivative of this equation wrt time gives the force equation

(Mtank+Mwater)Vtankdot = 0,

which verifies that the acceleration of the tank car Vtankdot = 0 for all time.

In other words, the transient effects don’t exist.

What do you say to that?

280. 280 280 db

@Brian (279)

You’re missing two terms from your momentum calculation.
A – The water is moving relative to the tank and you need to add that in to your calculation of the momentum of the system. It’s M’D (the rate at which water appears at the hole instead of the centre). If this is a topped-up system then the location of the top-up might be important too — I’ve not really thought about that yet.

B – The momentum of the system you describe is not 0 so the RHS of your momentum equation is wrong. It is M’(0+)D.

The issue you mention is discussed at length by McDonald in his paper in and around the [26] you quote — so if you carefully follow what he’s written then you should develop a better understanding of what is going on.

281. 281 281 Harold

I thought I had it, then this comes along: ““If … the flow of water established before we release the boxcar and then is constant then there is no motion.”

Surely, the COM of the water is still moving to the right, so the car must still move to the left?

In the forces model, the pressure on the left is higher than the right, so the force on the left wall is higher, and the boxcar moves left.

I don’t quite understant the constrain “the flow is constant” as this cannot happen if the water is draining from the car unless the hole is getting bigger.

282. 282 282 Ken B

@274
I know it comes out almost perpendicular. That argument requires *perfectly* perpendicular. You cannot measure that accurately with your eyeball. Try a big hole. a quater of the container length.

283. 283 283 Ken B

@db 277:
Well if the car moves right with a perp flow then by continuity there should be a small angle at which the nozzle can be and still get that effect. But it will be very slight. A rightward facing nozzle parallel to the gound, as I stipulated, is a rocket.

284. 284 284 db

@KenB (283)
“A rightward facing nozzle parallel to the gound, as I stipulated, is a rocket.”

As I proved: not for a sufficiently long boxcar it isn’t.

285. 285 285 db

@Harold (281)

Sorry to lay this on you on top of everything…

The motion of the water rightwards doesn’t move the boxcar if it is held whilst that flow is being established (the holding agent produces a force on the system which creates the rightwards momentum — contained within the water). Once the flow is established, to move the boxcar, we need a *change* in the water’s flow. (Impulse = change in momentum). Think waterhammer again. It’s the tap/flow going on or off that creates the force on the pipe.

If we assume that the water flow doesn’t change, then the momentum does’t change and so there’s no more motion. I agree it’s a bit odd to have constant flow water, but a topped-up tank works in this regards (although the top-up stream then needs to be included in the calculation).

The fact that intuition based on some force explanation doesn’t give the same answer shows a weakness in both intuition and the forces model.

286. 286 286 Guy

Harold(281)

In case you’re still worried about the water’s COM moving right (I was after I read your comment): Imagine instead of holding the boxcar still as the water flow begins, the boxcar pushes something leftwards down the track with enough force to keep the boxcar stationary. The leftward motion of this counterweight will balance the rightward motion of the water, keeping the original COM preserved (I think).

287. 287 287 Harold

#285 #286 Thanks again. That does make sense. I come back to my incompressible thing again – I think my intuition is mixing up an ideal gas (perfectly compressible) with “ideal” water (perfectly incompressible). If we had an ideal gas in the car, we can establish a pressure and density gradient accross the car. Each gas molecule is independant, and assumed to be going at the same speed. The force on the left is because there are more molecules hitting the left wall than the right, because some have left from the right side through the hole. Oveall, the COM argument is I think exactly the same (assuming a sump with baffles to direct the gas vertically). In this case there is no initial large impulse, as the entire mass of gas is not moved at t=0. (In practice this is true as you do not get water hammer in gas lines I think). However, the COM arguments still must predict left movement then right. How would we model this? Surely the end result would have to be the same, but db’s equations would not apply.

In the water case, we are assuming total incompressibility – the volume does not change with pressure. When a molecule leaves the hole, all molecules move over to fill the gap. Am I talking nonesense?

288. 288 288 Neil

db@284

This would seem to be a problem for your proof, then. Would not a right-facing jet parallel to the ground make the boxcar a rocket by definition? If this is not a rocket, what is?

289. 289 289 Ken B

@Neil 288
Indeed. This was the point of my reductio. Rockets do work! Voyager is not coming back.

A little googling found this http://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation

290. 290 290 Brian

Neil (284),

Exactly right. db just hasn’t caught on yet. But what he will try to argue, I think, is that it matters whether the flow comes out the bottom and is bent to the right or flows out a hole in the right side. This makes no difference, of course. The net change of flow momentum is still the same.

291. 291 291 Brian

db (280 and 285),

You haven’t responded clearly yet to my point regarding eq. 26.

Do you now accept that an initially held boxcar with steady flow never moves? Please answer yes or no.

Regarding the McDonald paper and eq. 26, I am very familiar with what he wrote. I also know it’s nonsense. Peer reviewed or not, the hand-waving arguments about transient forces are wrong.

Look, this is pretty basic physics. Water flowing rightward cannot impart a net momentum change to the boxcar because it doesn’t have a net change in momentum itself.

Before the hole is opened, the water has zero net momentum (it’s not moving), therefore (Px)water = 0. After bouncing around the boxcar, it eventually comes out in the vertical direction, so again (Px)water = 0. With no net change in horizontal momentum, it never imparts a net momentum to the boxcar. This is true for water starting ANYWHERE in the boxcar. Claims of transient forces can’t get around this basic constraint.

Let’s look specifically where McDonald goes wrong around eq. 26. He refers to the last bits of water (which by assumption have their COM at the tankcar’s COM) carrying all the momentum -MdotD. Then he says “all this momentum must suddenly be transferred to the tank, which leads to an impulse and hence the final [rightward: my add] velocity found above.” Basically he’s saying the last remaining drops leap from the center of the tankcar to the hole at D, bash into the sump, and give the tankcar a large rightward kick. But he’s ignoring that the initial jump has to give a compensating leftward kick to move from the center to the hole.

Of course, he’s also ignoring that it’s IMPOSSIBLE for the water to have a momentum MdotD RELATIVE TO THE TANKCAR and still maintain its COM with the tankcar COM (recall that McDonald assumed this from the beginning).

292. 292 292 Brian

Ken B (282),

You say “I know it comes out almost perpendicular. That argument requires *perfectly* perpendicular. You cannot measure that accurately with your eyeball. Try a big hole. a quater of the container length.”

You can see it plainly enough. Just let the water fall for a while. It hits the spot on the ground directly below the hole. If there’s ANY sideways motion, no matter how small, you’ll see a deviation if you let it fall far enough. There is no deviation.

Also, I can’t imagine why the size of the hole relative to the container matters. The mouse is not going to bite through 1/4 of the boxcar. All that potentially matters is the ratio of hole diameter to hole depth (material thickness). I had a ratio of at least 10:1, and probably 20:1. No sump there!

293. 293 293 Ken B

@Brian 279
What is waterdot? For that matter, what is tankdot?

294. 294 294 Ken B

Brian: “With no net change in horizontal momentum, it never imparts a net momentum to the boxcar. This is true for water starting ANYWHERE in the boxcar. Claims of transient forces can’t get around this basic constraint.”

Correct. But that does not mean the car never moves. There can be a transient period where the water and the car are both moving. If the water leaves with no net momentum then the cart picks up no net momentum. But the car could have still been displaced. I discussed this at length in several places including my recent example of my trip with Harold and db.

Before:
$===== After$
=====

all motionless in the x direction.

295. 295 295 Ken B

@294: dratted white space formatting. With , for a balnk space

,,,$,,,====== ,,,,$
,=====

That should work!

296. 296 296 db

@Neil (288) and @Ken (289)

A rocket is something which is dominated by the effect of the momentum leaving the nozzle — ie the rocket propulsion.

A sufficiently long boxcar is dominated by the impulse from the water flow within the car rather than the rocket propulsion outside it.

That the nozzle can point in a way that your (rocket) intuition suggests would cause movement in the other direction to that which it occurs is why it is a Paradox. That Ken particularly can’t see it is why it is called Ken’s Paradox.

I don’t fall over when I pee. My motion is dominated by my weight on my feet rather than the expulsion of mass. I wouldn’t describe myself as a rocket in those circumstances, much thought a man might dream a little.

297. 297 297 db

@Brian (291)

I think I’ve covered McDonald’s [26] quite comprehensively in my 276 and pointed out the errors in your equations in my 280, which you haven’t addressed.

A held tank with constant flow does not move until the flow changes (ie ends).

I think you would do well to examine in what way you think McDonald is hand-waving with the transients. His equation for CoM seems sound unless you can see an error in it?

* * *

A simple question, then:

Scenario: The tank isn’t being topped up, but the brakes are on. The hole has opened and a flow just started.

Consider all the water — both that still inside and that which has left the carriage.
– Is the centre of mass of this total body of water moving rightwards at all?
– How fast?

Saying that the centre of mass of a body is moving rightwards is equivalent to saying that body has rightwards momentum.

The stream of water outside the tank clearly has no rightwards momentum, so the body of water inside the tank must have that momentum.

Where does it come from?

298. 298 298 Neil

db@297

So you think there is a thing that looks exactly like a long rocket with an exhaust and all but it doesn’t go anywhere. I didn’t know there was such a thing. And I am talking about a rocket in frictionless space, not you peeing the toilet.

299. 299 299 Brian

db (297),

I admire your tenacity, even if it is misguided. You say “A held tank with constant flow does not move until the flow changes (ie ends).”

Well, that’s not what eq. 26 says. It says the boxcar doesn’t move even after the flow changes. According to eq. 26, it doesn’t move ever. That’s what I’m looking to hear you admit. So no, you haven’t addressed eq. 26 comprehensively.

McDonald’s equation for COM is NOT a sound starting point because it only holds for closed systems not subjected to external forces. In this problem the only system for which this holds is the water-boxcar-Earth system. The problem is that the Earth can shift infinitesimally and still maintain the COM without the boxcar moving relative to it. There’s no easy way to account for this in McDonald’s COM equation, which is why it’s not a good starting point.

My original argument showed how by considering Earth as part of the system, the boxcar can remain still while the COM stays the same. It’s because the water ends up at the COM instead of being offset by a distance D.

Conservation of momentum is a more fundamental principle without limitations on its use, which is why it’s a better starting point. My equations are not wrong. Let me explain each one and indicate why it’s correct.

MtankVtank is the mass and velocity of the boxcar. The boxcar is rigid and moves at Vtank by assumption. It’s the motion of the COM of the boxcar alone. No debate there.

MwaterVtank is the mass of the water still in the tank and Vtank is the velocity of the boxcar. This is the term you would dispute. Again, the velocity/momentum of the bulk is the velocity/momentum of the COM. But McDonald specifies that the COM of the water in the boxcar coincides with the COM of the boxcar. Therefore, the two have the same velocity. If they didn’t, the water would build up on what side of the boxcar. We know that doesn’t happen without an acceleration. In any case, McDonald ASSUMES that it doesn’t build up on one side. That’s a fundamental logical contradiction in his work.

Integral[etc] is the integral of the mass of each water element times the velocity at which it moves. Since the water leaves perpendicular to the boxcar (by assumption), it follows that Vwater = Vtank. I assume you have no problem with that part of the equation–it seems self-explanatory. (Note for Ken B: the dot designation indicates the time derivative of that term. Vdot would be an acceleration).

db, here’s a task for you. If you think the MdotD term represents a real momentum (I say it doesn’t), please derive it WITHOUT using McDonald’s COM equation (which I don’t accept). Use whatever you like–conservation of momentum or angular momentum or energy. I don’t think you can because it’s an artifact of the COM equation. But if you CAN derive it by another means, I would consider that much more convincing.

Finally, with regard to your simple question, I claim that the boxcar and water don’t move, so the COM of all the water (in and out) would move slightly to the LEFT due to the centripetal direction of the falling water along Earth’s gravity. This is compensated by a very slight shift of the Earth to maintain a constant COM.

300. 300 300 Brian

db,

Just in case I wasn’t clear in my previous post, McDonald’s COM equation is incorrect in part because it is internally inconsistent. He assumes that the boxcar water remains flat, thereby keeping the boxcar water COM aligned with the boxcar COM. He then derives a result that implies a rightward acceleration MdoubledotD. But such an acceleration would incline the water and shift its COM relative to the boxcar. His failure to account for this makes his equation invalid (and there may be other reasons, such as not taking into account the Earth).

301. 301 301 db

@Brian (298, 299)

Let me be clear about your answer to my simple question. You think that a shape (the body of water) that is shifting to the right (it’s growing a stream out of the hole) has a centre of mass that is moving to the left?

Can you clarify how that can be the case?

Can I also ask if you use physics in any kind of job or profession that involves public structures as I’d very much like to ensure that I avoid driving over any bridges that you have designed.

302. 302 302 db

@Brian (298)

I’ve derived the momentum of the water for you by figuring out the mass of the object times its velocity. I am hopeful that we can agree that is a valid derivation of momentum.

I’ve even done a picture. I hope that helps.

Perhaps if you think I’m wrong you’d annotate my statement or draw your own picture. If you think the answer is wrong because of some small variation such as surface level, then perhaps you’d like to quantify that and show how it results in zero internal momentum.

http://www.turl.st/maths/BriansQuestion.jpg

303. 303 303 Ken B

db in 296:
“A rocket is something which is dominated by the effect of the momentum leaving the nozzle — ie the rocket propulsion.

A sufficiently long boxcar is dominated by the impulse from the water flow within the car rather than the rocket propulsion outside it.”

db, I confess I am floored that you could say something so confused. If momentum is ejected rightward the remainder must move leftward. Period. This is conservation of momentum. Now if the left moving thing is a container with water in it the water might slosh back and forth, so the container will shimmy/osciallate as the whole mass moves left. But since none of the water can exit anywhere but the right end of the container, moving right wrt the container, the container cannot end up moving right.

Imagine two balls on a spring oscillating back and forth in the x direction, as the pair slowly move off to the left. Then at some moment either ball can be moving right but will shortly move left faster. Neither ball can sail off to the right if they remain conjoined.

304. 304 304 db

@KenB (303)

I think you mean confusing rather than confused. It’s your Paradox, not mine.

If you have a leftwards moving thing and it ejects matter rightwards insufficiently fast to make that ejected matter move rightwards in the lab frame then you have motion that isn’t what you might think of as a rocket, despite a nozzle that looks like it would be a rocket in some direction.

If the leftwards momentum of that rightwards (boxcar frame) ejected matter is sufficiently large then the boxcar will go forever rightwards spewing the last dribble of water ahead of it until it runs out of water and sails off into the rightwards sunset.

With a long boxcar that describes the motion (see http://turl.st/maths/KensParadox.pdf).

I have imagined your balls, but can’t see what they have to do with it.

305. 305 305 Harold

“I have imagined your balls, but can’t see what they have to do with it.” Best not to spend too much time on that.

306. 306 306 Ken B

@db 304:
“If the leftwards momentum of that rightwards (boxcar frame) ejected matter”

With a horizontal right pointing nozzle, can water leave the car with a greater leftward velocity than the car? No.

Water sloshing inside the car is like one of the balls, and the car is like the other. There can be oscillation around the COM as the COM moves off leftward.

It’s not a paradox, it’s a reductio.

307. 307 307 db

@Harold (305)
I need all the fun I can get now that the hard bit of this problem is cracked. It’s just sweeping up.

308. 308 308 Neil

db claims here that there exists a sufficiently short boxcar that is a standard rocket traveling to the left ejecting water to its right, and that there exists a sufficiently long boxcar traveling to the right ejecting water to the right that only looks like a rocket. This implies there exists (in a frictionless world) a boxcar of a particular intermediate length such that the boxcar is stationary in the lab frame while ejecting water to its right.

That seems to be a violation of Newton’s third law to me.

309. 309 309 db

@KenB (306)

You have three balls. One of them heads off leftwards out of a rightwards pointing nozzle.

The other two you are free to play with as much as you like as they head off rightwards.

310. 310 310 Ken B

309: “One of them heads off leftwards out of a rightwards pointing nozzle.” Huh? You can only exit a right facing nozzle by moving right wrt to the nozzle. So if in my frame the ball leaving the nozzle is moving left then the the nozzle is moving left even faster.

308: Nice.

311. 311 311 Guy

Neil(308)

No. The boxcar always starts off to the left. There should be one that comes to rest just as the water runs out.

312. 312 312 Ken B

@311:
So what is left moving left? Not water expelled from the car. It came out the right facing nozzle. That’s the contraint.
You need the momentum to balance. You have NOTHING moving left faster than the car. If the car is at rest then you have nothing that moves left; impossible.

313. 313 313 Guy

Ken(312)

Some of the water that left the nozzle will be moving left, and some will be moving right.

314. 314 314 Ken B

For pity’s sake Guy draw a picture. For the car to move right there must be water moving left with respect to the car. Water that exited the nozzle. How can water that leaves the car moving *right* wrt to the car also be moving *left* wrt to the car. Water only gets out the nozzle if it is moving right wrt to the nozzle, correct? And the nozzle if fixed to the car.

315. 315 315 Guy

But the velocity of the car isn’t constant. The car always has leftward motion at some point. Do you think it’s impossible for water leaving the car rightwards wrt to the car to be moving left wrt the world? If the car slows down and starts moving to the right, then that water is now moving left wrt car.

316. 316 316 Neil

Guy@311

Thanks for the correction. But even then there is a moment at which the car is stationary in the lab frame (when it changes from moving left to right) and water is spewing out to the right.

317. 317 317 db

Guy has the answer to 308 exactly right.

318. 318 318 Ken B

“Do you think it’s impossible for water leaving the car rightwards wrt to the car to be moving left wrt the world?”
More than anyone else on this thread I have been vaery careful about frames of reference Guy. Read what I said very carefully, and you will see I have answered this question.

“If the car slows down and starts moving to the right”
Big if. And what causes this? Not water leaving the nozzle. And water that stays in the car can only cause oscillations not permanent reversals, as the sloshing water will hit the other end of the car. So what?

319. 319 319 Neil

I retract 316. Velocity is changing at that moment of course.

320. 320 320 Harold

AT the start, the boxcar is stationary, but there will always be an impulse to the left initially. As the amount of water in the boxcar is much greater than the water that has left (initially) the boxcar must always start off left. I think that water in the boxcar must be moving to the right, else the COM could not remain stationary. If the nozzle is not exactly horizontal, then is it possible for the right velocity to be less than the left velocity of the boxcar. For an incompressible fluid, is it possible to slow it down before it leves the nozzle? If not, then the water leaving initially must be moving to the right at the same velocity as the water in the boxcar.

321. 321 321 Mike Sproul

I hung a 1 gallon jug filled with water from 2 strings suspended from a shelf about 7′ high in my garage. The jug hung horizontally about 1 foot from the floor. I used a hot 10-penny nail to melt a hole about 1/8″ diameter into the bottom of the left side of the jug, then covered the hole with tape and filled the jug. I made some marks to record where the string hung, sighted to the string and the marks from across the garage, and then removed the tape from the hole, allowing the water to flow out of the left side of the jug. There was an air hole on top of the jug.

The jug didn’t move. More accurately, I was only able to dampen the swinging of the jug to about 1/16th inch, and as near as I could tell, the jug was swinging over the same 1/16th arc with the water flowing as it had with the tape covering the hole.

If I had run a 1/8″ pipe up that hole, put a 90 degree elbow (A) bending right, continued the pipe to the middle of the jug, and then put another 90 degree elbow (B) bending up, then the water passing through elbow B would have pushed the elbow (and the jug) to my right. But as the water entered elbow A it would have exerted an equal force to the left. That’s why the jug didn’t move.

322. 322 322 Guy

Mike,

Was the hole in the side of the jug or the bottom? You wrote “bottom of the left side”, but your discussion of hypothetical pipes implies the hole was in the bottom. Also could you tell us the dimensions of the jug?

323. 323 323 Guy

Harold,

I don’t think the velocity of the water leaving the nozzle has much to do with the velocity of the water in the car. It will be determined by the pressure at the nozzle’s opening inside the car which will be mainly a function of the water depth (although there will be a bit of extra pressure from the decelerating water).

324. 324 324 Guy

Ken B,

I have read your comments very carefully. I took your comment at 312 to mean that no water in the world can be moving left. Have I misunderstood you?

At 318 you say “Big if”. I agree that it seems improbable, but the only person to have done the maths seems to have proved that it is possible. Nobody has pointed out any errors (apart from db himself). Brian challenged the basis of db’s equations, but I don’t think his (Brian’s) reasoning about the problem is sound.

“And what causes this? Not water leaving the nozzle.” Agreed. It is the rightward momentum of the water in the boxcar being transferred to the boxcar.

325. 325 325 Mike Sproul

Guy:

It was a 1 gallon jug about 12″ high and 6″ wide. I turned it on its side and put the hole in the ‘bottom’, which was formerly the side.

326. 326 326 Ken B

Guy
Look at 303 or any of my earlier comments. In 312 I was explicitly replying to 311. In 311 you have the car at rest. Any water leaving the car at rest must be moving right wrt to the world.

“It is the rightward momentum of the water in the boxcar being transferred to the boxcar.”
How? Look at the total momentum here. The car and its contents at this time are moving left wrt earth, and there is net leftward momentum. Suddenly you have it ALL reverse direction. After all, the car is in your mind moving right, and so is the water that left the car. Nothing moving left is left :) (It cannot be water left in the car as then it will smack the left end of the car and (work it out) send the car left again. There can be osciallation about a left moving COM, but the COM cannot move left.)

At this point frankly I give up. Rockets work, no matter what you say. db’s notion that there is a limit to how long rockets can be is equivalent to arguing that is the nozzle is narrow enough the rocket travels backwards. (Scale up or down.) It’s wrong.

327. 327 327 Neil

I believe you are right, Ken. Rockets do not take off in one direction and then later start backing up simply because they have certain dimensions. The challenge is to find the error in db’s math. And since he is using McDonald’s math, it casts doubt on his solution regarding the perpendicular jet as well.

328. 328 328 Neil

Actually, there is a boomerang rocket, but I don’t think it works on the principle described here.

http://www.npr.org/blogs/krulwich/2013/04/30/180057507/the-boomerang-rocket-ship-shoot-it-up-back-it-comes

329. 329 329 Ken B

@neil 237:
I gave a link earlier today with a derivation of the correct equation.

What’s wrong with db’s math? MacDonald’s derivation does not apply as he assumes “As previously noted, because the water leaves the tank with zero relative velocity, …”
With a rightward facing nozzle the water does NOT leave with zero relative velocity. (Much less a negative one!) (I am not vouching for McD’s math, just noting it does not apply to the given situation.)

330. 330 330 Guy

Mike,

Thanks for the details. I forgot to ask about the rate of flow out of the hole. To show some willing I performed the following experiment: I punched a hole in the bottom of a Pringles tube roughly 1/8 inch in diameter and filled it half way with water. I collected the water in a measuring jug for ten seconds. I collected 100ml (sorry for the switch to metric). Does that sound about right?

I calculate the change in velocity of your jug should be about 0.0015 m/s

331. 331 331 db

I look forward to reviewing any errors in the maths I’ve posted. I am not perfect but I’ve laid out my reasoning as clearly as I can so that it can be challenged and reliably built upon.

Rockets typically have considerably more forceful ejection of matter than water dribbling out of a nozzle under gravity.

The motion of LOX down a rocket will have a trivial effect compared with the thrust generated by the burning of the fuel. That’s some proper trhust.

Rockets work alright — even big ones — when the rocket effect dominates the effect of the internal water flow. When it’s the other way round it’s not a rocket. Simples.

332. 332 332 db

@KenB (329)
McDonald’s scenario explicitly assumes perpendicular flow. I follow his method and adapt it for non-perpendicular flow. It should be no surprise that McDonald uses a note that the jet leaves with perpendicular velocity.
I don’t depend on that at all.

333. 333 333 Mike Sproul

Guy:

The water flowed out of the jug at a rate that filled 1 cup in 45 seconds. I thought about enlarging the hole to 1/4″, and then 3/8, etc, and I thought about punching the hole in the side of the jug to see what a true rocket would do, but my initial results, and the thought experiment of the 1/8″ pipe with 2 elbows, convinced me that any hole that squirted the water downward would cause no movement at all.

334. 334 334 Brian

I will be gone for a few days and unable to post. I didn’t want anyone to think I had run away. :)

A few parting comments for now.

db–I looked at your “derivation.” Unless I am missing something, you didn’t derive anything. In any case, everything you wrote involves a COM equation and I explicitly challenged you to derive the MdotD term without using COM.

Also, regarding your bridge comment–tsk, tsk. No need to get nasty. I don’t build structures, no, but for what it’s worth, my dad was a civil engineer and nothing he built ever fell down. :)

And you’ve still never responded clearly to my question about eq. 26. Do you agree that the boxcar never moves, even after the water stops flowing, if it is held while establishing constant flow? Why are you avoiding stating this clearly?

Regarding db, Harold, and Guy…well, you guys are very confused at this point. Everything I’ve argued up to this point is correct. The boxcar never moves. There’s no initial leftward motion and no rightward force. The COM is fixed for the boxcar-water-Earth system. Any bulk motion of the boxcar would violate conservation of momentum. In principle, Ken B’s sloshing could occur, but it doesn’t in practice because the water fills in empty spaces long before the heavy boxcar has a chance to move measureably.

I have no problem with db’s calculation–it’s just a good example og GIGO. McDonald’s (and therefore db’s) conclusions are faulty because the starting point is wrong. McDonald’s assumption of a horizontal water surface in the boxcar is not physically consistent with his later conclusion that the boxcar accelerates. The equation can’t give correct results if it’s not internally consistent. The one case where it IS consistent, for the held boxcar with constant flow, it predicts no motion.

Finally, I have a question for Steve. Are you going to comment at some point on your current view of this problem? Have our latest discussions sparked any thoughts? Just curious.

335. 335 335 Guy

Mike

Your flow was less than suggested by my (rather slapdash) experiment, so the expected change in velocity would be even smaller – around 1mm/s. I think that would be very hard to detect. A bigger hole would help.

I think your pipe argument (I’ve been using exactly that arrangement to help me think about the problem) holds for a steady flow. The force applied by the left elbow to accelerate the water at the left elbow will exactly cancel the force at the right elbow. When the hole opens and the water starts to flow, however, the entire cylinder of water in the pipe must accelerate. Thereafter, the flow of water decreases, as the water level falls requiring a force from the right elbow.

336. 336 336 db

@Brian (334)
My derivation of momentum is from first principles. I take a definition that the momentum of a body is the velocity of its centre of mass. While I suppose it might be fun to try to derive momentum from some sort of round-the-houses method — perhaps using Hamilton’s equations — it really does seem pointless.

After all — if someone asked you to figure out the momentum of a 1kg sphere moving at 1m/s – you probably wouldn’t start by calculating the energy of the system and taking the partial wrt position and integrating wrt to time. That would be loopy.

I’m afraid I see little merit in your other arguments:
– The issue of the level of the water you should run through McDonald’s calculations with a variation factor for that effect to show that for most reasonable situation the effect is trivial and has no effect on the overall motion of the boxcar. Or if you prefer to show the threshold at which that tiny effect overcomes the dominant physics. I’ll happily read your proof.

– The issue of the off-centre centripetal flow is similarly a trivial effect, although I’d delight in reading a proof that it is otherwise.

– Your thinking about the momentum of the water inside the boxcar is simply wrong. It’s moving and it has mass. It has momentum. I really can’t contort my mind into a position where I can properly empathise with what you are thinking so I’m afraid I am helpless to dig you out of that particular hole.

337. 337 337 db

@Guy
I am full of respect for your experimental approach and in particular your calculations of significance. Excellent work.

I do like your pipe simplification. It chimes with the “water hammer” explanation I was clumsily making previously, and it makes the momentum of the water inside the boxcar obviously the M’ flow in a pipe D long.

338. 338 338 Harold

Guy #323. I realise I was sloppy with my language. The water will usually have extra velocity because of the pressure, but I am talking about the *minimum* it must have. For the car to move left then right, some water must move left to begin with. From a right pointing nozzle, we know it has right velocity wrt the boxcar. This means it must leave the nozzle with a lower right velocity than the boxcar has left velocity. I am suggesting that the minimum right velocity the water has *must* be larger than the boxcar’s because the water *must* be moving to the right (wrt observer) in the boxcar. Unless we can reduce right velocity in the nozzle to be travelling slower than the water in the boxcar. We can clearly do this by pointing the nozzle at an angle, but can we do it if the nozzle points horizontally? If we cannot get water to leave to the left, we cannot have the car moving to the right.

Brian to db: “Do you agree that the boxcar never moves, even after the water stops flowing, if it is held while establishing constant flow? Why are you avoiding stating this clearly?” As I understand it, we have a motionless boxcar. The hole is instantaneously opened. This causes all the water to start moving right, and the boxcar to move left. After this motion has been started, the right movement of the water in the boxcar has no further effect. All the momentum to the car is transfered instananeously at the beginning (for constant flow). So there will be no left movement of the car if it is clamped initially. If the water is then stopped from moving right, then this will have an effect. If the water exits down, then stopping the water will cause the boxcar to move to the right. If the water is not stopped and flows out of a right pointing nozze, it depends on the velocity of the water leaving. If it leaves with greater velocity than the water is moving in the boxcar, then the water has to accelerate. A force must be exerted, presumably from the boxcar. The boxcar moves left. This is the rocket, and is what we would expect to happen for largeish holes.

If the water can leave with slower right velocity than the water in the boxcar is moving, then the water has negative acceleration. the force must come from the boxcar, so the boxcar moves right.

Comparisons with normal rockets are not too helpful. Incompressible fluids cannot store energy by applying pressure. Nobody building a rocket would use such a fluid.

339. 339 339 Guy

Harold,

Sorry, I think I see your point now. I think immediately after the hole opens, the water leaving the nozzle probably must be moving right wrt an observer. Is it possible that at some later time, when the pressure has dropped and the boxcar is moving left that some of the water leaves to the left (wrt observer)? I need to think about this a bit more (actually I really, really don’t).

340. 340 340 Harold

#339. For the boxcar to end up moving lefdt faster than the water is moving right, there must be rocket forces. If the initial movement were -x for the car and x for the water. The boxcar may end up moving left at -2x, the water at -x and the exiting water at 10x. As the flow reduces the water flow may slow down, so it moves with less than 2x, and will travel left. We will already have sent a lot of fast water right though.

341. 341 341 db

@Harold (338)
That question about whether the rightwards flow could be overcome even if the nozzle were pointing flat right was exactly the reason that I worked through the maths of Ken’s Paradox.

For the intially held boxcar, I laid out the answer to Brian’s questions on McDonald’s [26] in an earlier post. The M’D is transferred back to the box car on cessation of the flow. Until then the boxcar remains stationary as the force is based on M”. If the boxcar has a rightwards nozzle then the final motion depends on the geometry of the nozzle and the density of the fluid. It can always be overcome by a long enough boxcar.

342. 342 342 db

@Harold (340)

The speed of the boxcar leftwards is initially M’(0)D / (M(0)+m)

The speed of the water out of a rightwards nozzle is initially
M’(0)/Ap where A is the cross-sectional area of the muzzle and p is the density of the fluid.

It’s far from clear to me that either of these should necessarily be bigger than one or the other. It’s a question about the geometry of the box vs the geometry of the nozzle.

Repeat after me:- For a sufficiently long boxcar…

343. 343 343 Harold

#341 Bear with me bit here please to see if I can get this. I am trying to reconcile the math with my weak understanding of the processes occuring.

Velocity from the nozzle v = (-Mdot)/Ap. (Since the angle is horizontal we can dispense with the sintheta’s I think). Mdot is the rate of mass transfer? This gives us units of v of distance/time, so that is OK.

The velocity of the water in the boxcar ( call it v’) is the same mass flow, but with a larger area, so the velocity of water in the car is less than v.

Initially v’ must be positive (right), since the boxcar moves left. The COM change can only be balanced with a positive v’

An immensely long car gives a big impulse left because there is a lot of water moving right. But however long the car the COM of the water must be moving right.

Unless we can make v < v', I don't see how we can ever get any water travelling left. I don't see how we can make v < v' unless we tilt the nozzle. (Obviously for a vertical nozzle v = 0).

If no water travels left, the boxcar cannot move right.

Have I misunderstood something?

344. 344 344 Ken B

@343
Correct. Pressure is forcing water out the nozzle. This water carries momentum. Consider for a moment an inertial frame in which the tanker is instantaneously at rest. If the hole were coverd the COM of the tanker and the water inside would be moving uniformly. With the hole open some of the potential energy is converted to kinetic energy, resulting in two opposite bits of momentum. The squirting water dm coming out the nozzle is ACCELERATING right in that frame, so the tanker must be ACCELERATING left (in this frame).

Now what can we say about this frame compared to the earth? it is moving left. We know that because
1. we all agree intially the tanker moves left
2. we just saw that the tanker will accelerate left wrt to a specific inertial, and therefore wrt to all inertial frames. So the tanker+remainingwater is never subjected to a net rightward force. It always accelerates leftward.

345. 345 345 db

@Harold (343)
The fact that you are missing is that the hole is moving at rate x’ since you seem to be working in the labframe (although you aren’t specific on that saying that the boxcar moves left suggests that you are).

Your equation for v should be
v = x’-M’/Ap

That might well be leftwards for large negative x’ — such as in the case of a long boxcar.

I dislike your use of v’ for the velocity of the water relative to the boxcar as that is confusing with dv/dt but if you must… then

v’ = -M’D/M

So the velocity of the internal water in the labframe is
v’ = x’-M’D/M

So (v-v’) = -M’(1/Ap – D/M)

So it depends on the geometry in my 342 as to whether the boxcar water is moving faster than the jet. Although I’m really not sure how that is interesting now I come to conclude it, but it was what you were looking for!

More interesting is to integrate M’v to get the location of the CoM of the exited water…

346. 346 346 bigjeff5

So, I didn’t read all 345 responses, but I did read the first 40 or so, and I didn’t see anybody point out what seems blatantly obvious to me:

100% of the water that moves left to right (thereby imparting leftward motion to the boxcar) must hit the right wall of the boxcar with an exactly equal amount of force, thereby imparting rightward motion exactly equal to the previous leftward motion, BEFORE it can exit the mouse hole. If the water does not do this, it does not leave the boxcar in a different direction that it had previously been traveling (i.e. through the mouse hole), and physics has somehow gone horribly horribly wrong.

The major point I think that people miss is that every single molecule of water that travels any amount from left to right has to transfer its left to right motion back to the boxcar, cancelling that motion completely, before it can gain the correct motion to travel out the mouse hole.

347. 347 347 Harold

#345. My nomenclature is unclear and I see that I have mixed up my frames as well – apologies. The velocity of the water in the boxcar was intended to be in the lab frame, but I see that by using the velocity of the water leaving the car it must be in the boxcar frame. However, having anothergo, I still come to the same conclusion.

This time initially all in the lab frame.

The initial movement is the water to the right as the hole opens. The boxcar moves left in the lab frame. If we take at or just after t-0 the water having left the car can be ignored, so we have boxcar moving left and water in the boxcar moving right.

If we have a very long light car, then the car may be moving very fast to the left, and the water very slowly to the right, but the water must always be moving to the right in the lab frame to keep COM balanced. The velocity of the water leaving the nozzle must be greater than the velocity of the water in the boxcar in some frame of reference.

Just picking numbers out of my rear to get my thoughts in order. Say the car is moving at -9 and the water at +1 in the lab frame.

The water accelerates through the nozzle, so must be leaving at more than +1 in the lab frame. The key points are that if the boxcar is moving left in the lab frame the water in the boxcar must initially be moving right. AND the velocity of water leaving the boxcar must be greater than the velocity of water in the boxcar (in any consistent frame of reference) since it has the same mass flow rate, but is travelling in a smaller cross section.

I cannot see how we can ever get the water to initially leave to the left unless we can slow it down in the nozzle so it leaves with positive velocity in boxcar frame but -ve velocity in the lab frame. I do not see how we can slow it down in the nozzle without slowing down the flow in the boxcar.

AFter a time, the water and the boxcar could all be moving left, acclerated by the water moving. But then the water can not exert any rightward momentum.

It seems we have a different conclusion from the math and my contempations. I am prepared to say the math is right, but it must be possible to reconcile the math with an understanding of what is happening.

#346 bigjeff5. One way to look at this is to think that water has set off to the right and imparted an impulse to the boxcar to the left. By the time it arrives at the right wall and imparts the same impulse to the right, the boxcar has got lighter because some water has left. The same force therefore gives greater acceleration to the right.

348. 348 348 Peter Tennenbaum

A close friend of mine–a genuine mechanical genius–claims to have a clear-cut solution. Last night he told me that he is writing it up and would post it soon. Roughly speaking, the essence can be stated in three sentences–or so he says–but he also says that he will post all requisite details. If possible, please do not close this thread as I would love to read Prof. Landsburg’s response(s) (if any) to my friend’s so-called solution (of which I know absolutely nothing about as I never thought about the technical aspects of the problem–only various unstated assumptions and their consequences).

My friend may well be wrong, but I am quite sure that what he writes will be as interesting as anything else posted on the subject (errors can often be more “stimulating” than solutions).