Chewy0087 said:
That's fascinating... any hints as to why? I'm trying to think of the reason...
First off, it should be obvious that if at the end everything is moving to the left we have indeed violated conservation of momentum. At the end, if the bulk of the exhausted fluid is moving to the left, the car can only be moving to the right. So how to get there?The discussion so far has focused on the train car and its contents. What about the liquid that has left the train car? It helps to make the train car + liquid a horizontally-isolated system. One way to do this is to have the liquid leaving the car fall through the rails onto a frictionless surface situated well beneath the rails. With this construction, the train car + liquid system does not interact horizontally with the environment. Moreover, because the surface onto which the liquid falls is well beneath the rails, the train car does not interact with the liquid once it leaves the car.
The momentum of the fluid that has left the car between time
t=0 when the flow first starts to time
t and the momentum of the car plus its fluid contents are given by
\aligned<br />
\vec p_{\mbox{ex}} &= \int_0^t \dot m(\tau)\vec v(\tau)\, d\tau \\[4pt]<br />
\vec p_{\mbox{car}} &= (M+m_f(t))\vec v(t) + \vec p_{\mbox{flow}}<br />
\endaligned
where
- \vec p_{\mbox{ex}} is the momentum of the fluid that has left the car from the persective of an inertial observer,
- \vec p_{\mbox{car}} is the momentum of the car plus its fluid contents from the persective of this inertial observer,
- \vec p_{\mbox{flow}} is the horizontal momentum of fluid flow inside the car from the persective of an observer fixed with respect to the car,
- \dot m(t) is the (positive) rate at which mass flowing from the car at time t,
- \vec v(t) is the velocity of the car at time t,
- M is the (empty) mass of the car, and
- m_f(t) is the mass of the fluid contents of the car.
In steady-state flow conditions, the horizontal momentum flow of the fluid inside the car is given by
\vec p_{\mbox{flow}} = \dot m(t) \vec o
where \vec o is the horizontal component of the vector from the center of the tank to the outlet.
To conserve mass and to conserve horizontal momentum, we must have
\aligned<br />
dm_f(t)/dt &= -\dot m \\[4pt]<br />
\vec p_{\mbox{car}} &= -\vec p_{\mbox{ex}}<br />
\endaligned
Let's start simple: constant flow rate, no transients, and we'll close the outlet valve before all the fluid has left the tank. In this case, while the fluid is flowing from the car, the car will have a constant velocity given by
\vec v = -\frac{\dot m}{M+m_0}\vec o
where m_0 is the initial mass of the fluid content of the car. By ignoring transients we have given the car an impulsive change in momentum of
\delta p_0 = -\dot m \vec o
at time
t=0. How does this arise? It arises from the transients that we have so conveniently ignored. It takes a bit of time for the fluid flow to build up inside the car. There is a transfer of momentum from the car itself to the fluid within the car.
When we close the value (remember that there is still fluid in the car) we will get a reversal of this momentum transfer. The initial and final changes in momentum of the car+contents subsystem have the same magnitudes but opposite directions. However, the car+contents subsystem has lost mass in the interim. The final change in velocity is greater than the initial change in velocity: The vehicle ends up moving to the right.
Now let's make the vehicle drain under its own pressure, and keep draining until it is completely empty. We'll get a similar startup transient as above. Now the momentum flow of the liquid within the car is no longer constant. It is instead decreasing in magnitude as fluid drains from the car. This means momentum is being transferred from the fluid to the car, and this transfer is toward the right. This will make the car slow down, come to a momentary stop, and then start moving to the right.