The departure point is this problem:

A water tank on wheels is moving over an horizontal trail with negligible friction. There is a small opening in one of the walls, at a depth of $h$ below the tank's water level. The cross-section area of the opening is $A$. The initial masses of the tank and the water are $M$ and $m_0$. What is the initial acceleration of the cart?

Can we consider the water at the top of the tank to be stationary? If so, then it is pretty straightforward to find the velocity at which the fluid exits the opening. Then I would guess the acceleration could be estimated by looking at momentum variations. However, this is a varying mass system, so the mass also varies. This ends up being similar to the rocket equation, which involves solving a system of differential equations.

Is there a simpler way to solve this kind of problem (in other words, can you obtain the value of the acceleration without having to solve differential equations)?


Using Torricelli's law (https://en.wikipedia.org/wiki/Torricelli%27s_law) you get $v=\sqrt{2gh}$ irregardless of how big the opening is. Now you can calculate how much mass is leaving the tank at any time by multiplying the volume that is leaving the tank by its density: $$\Delta m=A\Delta s\cdot\rho$$ with $\Delta m$ the mass leaving the tank at any second, $\Delta s$ the distance travelled by the water in a second and $\rho$ the density of the water. Dividing by the timestep and letting the timestep be really small you get: $$\frac{\Delta m}{\Delta t}=A\frac{\Delta s}{\Delta t}\rho$$ $$\frac{dm}{dt}=Av\rho=A\rho \sqrt{2gh}$$ Using the momentum of the stream to calculate the force applied to the cart (assuming the velocity doesn't change in the beginning) $$F=\frac{dp}{dt}=\frac{d(mv)}{dt}=v\frac{dm}{dt}+m\frac{dv}{dt}=A\rho\sqrt{2gh}^2+0$$ $$F=A\rho g h$$ And finally applying the force to the cart: $$a=\frac{F}{m}=\frac{2A\rho gh}{M+m_0}$$

  • $\begingroup$ I think there is an error here: $m_0$ was the initial mass of the water, but the mass of the tank is time dependent as well. See my updated answer. $\endgroup$ – Han-Kwang Nienhuys May 24 '16 at 6:13

For a particular setup, the equations may get very simple: the tank should be massless ($M=0$) and the hole is all the way at the bottom of the container. Then $h$ is proportional to the mass of the water in the container: $m=m_0 h/h_0$, with $h_0$ the initial height. It's straightforward to derive that the force generated by the water jet is $F=2\rho g h A$, where $g$ is the gravitational acceleration and $\rho$ the density of water. The acceleration is then $$\frac{dv}{dt} = \frac Fm = \frac{2\rho g A h_0}{m_0},$$ which is a constant. Technically, this is still a differential equation, but a rather trivial one.

Update With nonzero initial mass $M$, it is $$ \frac{dv}{dt} = \frac{2\rho g A h}{M + m_0 h/h_0}, $$ which is a bit more unpleasant to solve, since $h(t)$ will become an exponential function of time. It's still a trivial differential equation in the sense that the velocity $v$ does not appear in the right-hand side; you just need to take the primitive.


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