Mass leaving a cart pushes nothing unless it leaves with a speed of its own

The mesh that makes its own mass took the Reynolds transport theorem and asked what goes wrong when the region moves and the arithmetic forgets to account for how. An identity that should be exact produced mass from nothing, on a moving cell, because the volume the cell swept and the flux through its faces were counted by different rules. That was a question about a calculation.

The same identity decides a question about objects, and a much more familiar one. A cart on a smooth track has a hole in its floor, and it is full of water. It rolls along at a steady speed, leaking. What does the speed do?

It is a rewarding question to pose, and many will reach for the form of Newton’s second law they were taught as the general one, $F=d(mv)/dt$, and say that with no force acting and $m$ falling, $v$ must rise. That was Newton’s own wording, more or less. It is wrong here, and the cart’s speed does not change, and the way it fails shows what a control volume is for.

What the theorem says #

The theorem is about the rate of change of a quantity inside a region that moves, and it splits into what is changing inside and what is crossing the surface. Draw a control volume around the cart, moving with it. It contains a mass $m(t)$, all at velocity $v$ at this level of approximation. Fluid leaves at a rate $q=-\dot m$, and at a velocity relative to the cart of $-w$, so $w>0$ is backward. The momentum in the volume at time $t+dt$, plus what left during $dt$, minus the momentum at $t$, equals the force times $dt$:

$$(m-q\,dt)(v+dv)+q\,dt\,(v-w)-mv=F\,dt ,$$

and all terms in $v$ except the first cancel, to give

$$m\frac{dv}{dt}=F+qw .$$

That is Meshchersky’s equation of 1897. With $F=0$ and $w$ fixed it integrates to the rocket equation, $v=v_0+w\ln(m_0/m)$, which is the middle curve in the figure. With $w=0$ it says the cart’s acceleration is zero. The mass that leaves has taken away its own momentum, $q\,v\,dt$ of it, and what’s left has the same velocity as before. Nothing was pushing the water out of the cart along the track, and nothing pushes the cart.

The wrong form, read literally, contains an error of exactly $qv$. It drops the second term of the expansion, the momentum of what left. If the control volume is treated as a body and the leaving fluid forgotten, its momentum seems to fall with no force acting and the law seems to demand a compensating rise in $v$. It treats a body as having lost mass and kept its momentum. But the momentum went with the mass. Newton’s quantity of motion is a statement about a fixed collection of matter, and the leaking cart is not one.

A rule no force could obey #

There is a one-line test that needs no derivation at all. A force is what two parts of the world do to each other, and it cannot depend on the velocity of someone watching at constant speed. Apply the form $d(mv)/dt=F$ to the cart in the frame of an observer moving at $V$, where its velocity is $v+V$, and with $F=0$:

$$\frac{dv}{dt} = \frac{q\,(v+V)}{m}.$$

So its acceleration can be set to anything by choosing a passing train to do the calculation from. A cart at rest in the observer’s frame gets none. That is immediately absurd, and that is why the rule is not merely an approximation that fails at large $v$. Meshchersky’s equation has only $w$ in it, a velocity difference, which is the same to every observer, so there is no way for it to fail this test. Plastino and Muzzio, reviewing the abuse in 1992, set out why $F=d(mv)/dt$ cannot be the law of motion for a system whose mass is variable.

Rain does the opposite #

Now change what crosses the boundary. Rain falls vertically into an open cart moving at $v$ along a frictionless track. It enters at velocity zero along the track, so relative to the cart it arrives moving backward at $v$; in the theorem’s terms, something enters at a relative speed. The equation is

$$m\frac{dv}{dt} = -r\,v ,$$

and $mv$ is conserved. Here, and only here, the naive form gets it right, because what crossed the boundary had no momentum in this frame; that’s the coincidence that keeps it alive. In a different frame even this form fails.

Add a leak straight down at the same rate, and something remarkable happens. The cart’s mass stays exactly constant, so the naive form predicts constant velocity. Yet the cart slows exponentially, at a rate $r/m$. Each second, a fresh portion of still water arrives and must be accelerated, and an equal portion at the cart’s own speed leaves, taking its momentum. The cart with a steady mass is losing momentum steadily to the water passing through it. Mass in and out is balanced and momentum is not, and any rule framed in terms of the cart’s total mass can never see this at all.

The force $rv$ has a name in engineering. A jet engine takes in still air at rate $r$ and throws it out faster. Its net thrust is $r(V_e-V)$ and the $-rV$ is ram drag: exactly the rain on the cart. A steam locomotive taking water at speed from a trough between the rails with a scoop, a practice of Ramsbottom’s from 1860, was doing the same. A wing’s boundary-layer suction pays it too, and it is the first term in the accounting that the actuator disc keeps implicit by taking everything in the frame of the disc.

And there is an energy lesson in it. A cart being kept at a steady speed against rain has to be pushed with force $rv$, so the motor delivers power $rv^2$. The kinetic energy that the water gains per second is only $\frac12 rv^2$. Half the power goes into the violence of accelerating each raindrop abruptly, as a perfectly inelastic collision between a drop and the cart. Sand falling onto a moving conveyor belt is the classic form of this result, and it holds whatever the material’s properties, because it comes from the kinematics of the velocity jump, like the loss with no viscosity in it at a sudden expansion. No friction coefficient appears, and the same factor of one half is there. The mechanism has to be a viscosity or a friction, and its value does not matter.

A chain shows the same accounting running the other way. Lift a chain from a heap on the floor at a steady speed $v$, and each link has to be brought from rest to $v$ in turn: the force needed is the weight lifted so far plus $\lambda v^2$, with $\lambda$ the chain’s mass per unit length. That second term is the rain’s again, and half the work it does is lost to the jerk at each link, a result Cayley had in 1857. Let the same chain fall off a table’s edge instead, and the links leave with the speed they already have, which is the leak: no such term appears, and the part hanging simply falls. Hoist a chain and it is a cart in the rain; drop it, and it is a cart with a hole in the floor.

A real leak is a thruster #

A hole in the bottom of a cart is not the practical case. A hole in the back wall of an open tank is, and it brings pressure into the problem and produces a famous factor of two.

Torricelli’s result gives the exit speed from a depth $h$, $u=\sqrt{2gh}$. So a small hole of area $a$ passes $\rho au$ per second at $u$, for a thrust of

$$\rho a u^2 = 2\rho g h a .$$

That is exactly twice the force that seems to be available. If the hole were plugged, the plug would be pushed by $\rho g h a$, and the pressure on the opposite wall facing it would balance. Open the plug, and the imbalance on the walls ought to be the missing patch, $\rho g h a$ again. So where does the second half come from?

The missing half comes from the fluid’s motion near the hole. Water approaching the hole accelerates, so by Bernoulli the pressure on the back wall near the hole falls below hydrostatic. The opposite wall, where the fluid is still, keeps its full pressure. So the unbalanced region on the wall is more than the area of the hole: it is an extra patch, around the hole, of total deficit exactly $\rho g h a$. The thrust can only be found by taking the momentum flux at the exit. Trying to get it by integrating the wall pressures needs the whole near-hole flow, which is exactly the free-streamline problem, and the integrated result must come out equal.

The cart’s final speed shows one more thing the rocket equation hides. The exit speed falls as the tank empties, so this is a rocket with a falling exhaust velocity. Using the mean exit speed, two-thirds of the initial one, and the mass ratio gives 7.08 metres a second. The true figure is 5.31. The difference is that the highest speeds of exit come when the cart is heaviest and least able to respond, and the low speeds when it’s light. So a tank pushed along by its own gravity is a poor rocket by comparison with a pressurised one.

Borda’s tube #

Borda’s reasoning of 1766 took this to its elegant conclusion, and the theorem is what makes it possible. Fit the hole with a tube that reaches well inside the tank, a re-entrant mouthpiece. Now the flow near the tube’s inner end is well away from any wall, so all the tank’s walls keep hydrostatic pressure, up to and including the wall around the tube. Then the thrust is exactly the missing patch: $\rho g h a$ where $a$ is the area of the tube. That’s a statement about pressures, and it needs no flow calculation at all.

But the thrust must also be the jet’s momentum flux. The jet has speed $\sqrt{2gh}$ and some area $a_j$, so its momentum flux is $2\rho g h a_j$. The two are equal only if

$$a_j=\tfrac12 a .$$

A jet in that tube must contract to half the tube’s area, and the argument is complete with no streamline ever having been drawn. Measured contractions for a well-made Borda mouthpiece come out close to 0.5, and the free-streamline theory agrees exactly, as it should. The usual sharp-edged hole has a contraction of 0.61, whose thrust of 1.22 sits between; a free-streamline calculation gives $\pi/(\pi+2)=0.611$ in two dimensions, and no such simple momentum argument can be made because the wall near a sharp edge has a real pressure deficit. So the contraction, which had been a measured nuisance, turns out to be the wall’s pressure account, seen from the jet.

Where it matters, and where it does not #

The same accounting recurs in one form or another wherever fluid crosses a boundary.

A rocket is the case where the theorem’s two terms separate most cleanly, since its propellant is carried. A jet engine carries only its fuel, and so has the ram-drag term. A propeller carries nothing, and the theorem must be applied to the stream rather than to the vehicle, which is the whole method of a rotor’s momentum theory.

The error’s consequences are generally small, and that is part of why it persists: in a problem where the exit relative speed is large compared with $v$, the rule $d(mv)/dt$ is off by a small amount of order $v/w$, which is what makes it look harmless. Where it does grave harm is in teaching, because it says that force and momentum belong to a quantity of mass rather than to the collection of parcels. Following the parcel is the discipline the theorem exists to support. It is also the discipline behind the momentum with no value, which has the same shape of mistake one level up, in a region without a boundary at all, and behind a blob of fluid whose push reaches everywhere because no surface encloses what was moved.

What the picture cannot show #

Internal motion. The cart’s contents are treated as all moving at the cart’s speed. Real water sloshes, and while a cart accelerates its fluid develops a tilt and transfers momentum back and forth. It is a small effect on the mean, and over a short time it dominates the force. The momentum balance in the first equation counted the water as rigid, and a surge in a tank at a steep acceleration is a different problem.

A jet that does not stay attached. A hole in a real wall has a viscous boundary layer and the flow rate depends on the Reynolds number of the hole, though for a hole of any practical size in water the correction is a few per cent. The exit speed here is Torricelli’s, with no discharge coefficient other than the contraction.

Friction and rolling. The frictionless cart is what lets a single equation stand. A real cart’s rolling resistance would make the difference between the three speeds in the first figure into a difference in how soon each one stops, and it does not depend on the frame either.

Relativity. An entirely separate reason why $F=d(mv)/dt$ is not the general form is that in special relativity mass depends on the frame. The correct relativistic rule has its own invariance and contains a term for mass-energy radiated, and that problem has the same structure as the leak. It is not used here.

The convention #

Exit speed $w$ is the fluid’s velocity relative to the cart, positive backward. The first three figures use units in which the starting cart mass and the starting speed are one, with mass rate 0.05 per unit time. The tank uses SI units with water of density 1000 and gravity 9.81. The tank’s outflow is taken as quasi-steady in a small hole, so its depth falls by Torricelli’s rule. The thrust in the bar figure is a multiple of $\rho g h a$, with $a$ the hole’s area and not the jet’s. The 0.61 and 0.98 are measured contraction coefficients for those holes; the ½ is derived.

How it was checked #

The ledger’s check with most content is the tank’s: marching the cart’s speed from the thrust alone, then summing the momentum of everything shed at its own velocity in the ground frame, gives a total for the whole system that must be zero, since no outside force acted. It comes out at 0.84 against the cart’s 532, or 0.16 per cent, which is the march’s error at 4000 steps. The model refuses a final mass greater than the starting mass, a hole as large as the tank, and rain that does not fall.

Who found it, and when #

The first variable-mass equations are Buquoy’s, published between 1812 and 1815, and Poisson treated them too. William Moore derived the rocket equation in 1813, and Tsiolkovsky’s better-known 1903 statement came independently. Meshchersky’s general equation dates from his dissertation of 1897. Torricelli’s speed is from 1643; Borda’s tube from 1766. Reynolds’s transport theorem was stated in his 1903 book on the sub-mechanics of the universe. The clean statement of when $d(mv)/dt$ applies is in Plastino and Muzzio’s paper of 1992.

Still open: the slosh that stops a leak-driven cart #

The water here is a rigid mass until it leaves. The next calculation would give the tank a free surface carrying its lowest sloshing mode, couple that to the cart through the mode’s own momentum, and open the hole suddenly, so that thrust’s onset excites the mode. It would ask how large the speed oscillation is relative to the steady speed gain, whether the exit speed’s modulation by the head at the back wall feeds back on the mode and pumps it or damps it, and at what tank length the outflow’s frequency matches the mode’s and a cart with an open hole becomes, for a while, a self-excited oscillator.