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!)
Answer: It’s a rocket.
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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.
I’m pretty sure that your explanation is correct. I’m more sure about this one that about either of your earlier brain teasers linked to.
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.
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.
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.)
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.
Thomas Purzycki: Once the train has started moving leftward on a frictionless track, what force causes it to stop?
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.
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.
@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.
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?
@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.
@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.
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.
A frictionless track does not mean that the wheel channel on the rail won’t block a sideways movement of the boxcar.
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.
@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.
db: Thank you. I learned a lot from this.
@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.
@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).
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.
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.
@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.
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…?!
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.
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.
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.
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!
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.
I have a brilliant idea. Why doesn’t someone build a model and see what happens.
@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.
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.
@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…
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?
@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.
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.
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?
@Gordon (35, 10)
I’m not sure my comments will help any, but you are welcome to them.
Your point about pushing someone is a nice solid-body thought-experiment:
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).
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.
@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.
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.
@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.
@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.
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.)
db #37,
Thanks for your thoughts.
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.
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.
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.
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?
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.
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.
@ 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?
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?
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
left during your stroll. What matters is your uncompensated
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.
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..
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.
@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.
@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.
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?
#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.
@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.
@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.
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.
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.
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.
@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.
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
@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.
@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.
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.
@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.
@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.
@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
$$x'(0)=\frac{l w'(t)}{m+w(t)} + \int_0^t \frac{l w'(u)^2}{(m+w(u))^2} du$$
@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)
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…)
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).
(I didn’t read all of the previous comments to see if someone has submitted this already.)
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.
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.
#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
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.
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.
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
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.
@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.
@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.
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.
@ 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.
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.
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).
@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.
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.
@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.
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.
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.
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.
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.
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.
Neil, do you see your comments 81 and 86 are contradictory?
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.
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.
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.
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.
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.
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.
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.
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.
@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?
Neal, did you read no friction? So any momentum counts.
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.
comma after no
sorry, cancel the comma
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.
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.
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?
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.
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.
Ken. It is you who is confused. Your comment on zero friction revealed that. The movement in the track direction is exactly zero.
@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.
@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.
@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.
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.
@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.
@Ken (119)
I look forward to your proof that holes and pipes work differently, and inparticular that all holes work the way you. Suggest.
@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.
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.
Oops a LASER affixed to the left end. The pulse of the laser leaves the boxcar with momentum.
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.
@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).
@db link under 79.
@db
Basically huh? You don’t see a difference between the situation where the exiting particles strike car and when they don’t?
@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.
@kenb I’m curious to know whether you think the hole gets wet.
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
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?
@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)
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.
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.
@KenB (131)
I only require some of the water to touch the glass in order to keep all of the water in the glass.
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.
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.
@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…
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?
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
@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.
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.
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.
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.
#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.
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.
@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… :-)
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).
#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?
@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.
@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.
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?
@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.
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.
@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.
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.
@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.
Ack. In 156 I mean the boxcar is accelerating at a slow rate.
@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.
@Steve: For your next physics puzzle post I suggest something simpler! The three-body problem perhaps.
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.
@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.
@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?
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.
That should be “For every kg of water that leaves the car, a HALF kg of water moves to the right…”
@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.
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).
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?
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.
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.
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.
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
@Harold: I misread the number in 150.
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?
@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.
@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).
Somewhere, the Trainmaster is chewing out the brakeman who neglected to wind up the hand brake and trig the wheels.
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` ).
#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.
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.
Hmm. I think the double-decker remains motionless, but I may have misunderstood the posed problem. Does the track stay dry in this example?
Neil #160 -I am interested to hear about your experiment – was there noticable displacement of the water stream to the right or not?
@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.
#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.
@Harold 184: Correct, that’s my thought experiment. And with the sloshing water we get a shimmying double decker.
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.)
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.
@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…
@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.
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?
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?
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.
@KenB (190)
That’s not what I wrote. But your suggestion (which I christen Ken’s Paradox) might hold in some circumstances.
@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)
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.
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.
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.
#197 – I do not understand your set up -what do you mean by “with the nozzles together”?
@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.
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?