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.