Conservation of linar momentum - variable mass

[StefanBU]

If a railroad car traveling down a straight, frictionless track encounters vertical rain that fills it with additional mass, will the velocity decrease in order for the momentum to be conserved? Will the opposite happen if the water is let out from a hole in the bottom?

I cannot see any external forces to affect the system, but the fact that velocity is changing calls for acceleration, which calls for a force - I am having a hard time wrapping my mind around it. How can the velocity of the car change without any work being done?

 

[DrStupid]

If a railroad car traveling down a straight, frictionless track encounters vertical rain that fills it with additional mass, will the velocity decrease in order for the momentum to be conserved?

yes

Will the opposite happen if the water is let out from a hole in the bottom?

no

How can the velocity of the car change without any work being done?

There is work done between the car and the incoming water.

 

[StefanBU]

What would the force against which work is done be? If the incoming water is vertical and the the car motion is horizontal, shouldn't there be a force with a horizontal component that does this?
If there is no such force, what would prevent the car from re-gaining it's original velocity once the mass is gone?

The way I have broken it down is - the incoming water entering the car is an inelastic collision: momentum is conserved but the overall kinetic energy is lower. Hence, to regain it work would need to be done on the car, which leaving water cannot do. Well, if we look at the leaving water, if no external force acted on it it would maintain the velocity of the car (which in reality it won't due to air drag, hitting the ground, etc.). Hence, the leaving water's momentum should be taken into account when establishing the momentum of the system, and velocity needs not to increase (in fact it cannot).

Please let me know if my reasoning is fallacious.

 

[DrStupid]

incoming water is vertical and the the car motion is horizontal, shouldn't there be a force with a horizontal component that does this?

Of course there is a force with a horizontal component. The velocity of the incoming water is vertical and the velocity of the water in the car is horizontal. Therefore the water has been accelerated vertical and horizontal. According to the second law there must be a corresponding force acting on the water and according to the third law an opposite force acting on the car.

 

[andrien]

F=dp/dt, where p is momentum so when there is a change in mass it will also give rise to force and it will be in the direction of velocity or opposite to it.

 

[jbriggs444]

F=dp/dt, where p is momentum so when there is a change in mass it will also give rise to force and it will be in the direction of velocity or opposite to it.

In the usual case we have an otherwise-closed system with an external force being exerted on it. In such a case we can indeed derive F = dp/dt.

However, in the case at hand we have an otherwise-closed system with a mass flow going across the boundary. Or, equivalently, we have a boundary expanding or contracting to include or exclude some mass.

The act of expanding or contracting the boundary of an otherwise-closed system is done with pencil and paper or in the mind of the analyst. It does not impose a physical force on the objects being analyzed.