Why Do Sand Flow Rates Affect Freight Car Momentum Differently?

[Order]

Homework Statement

Just have to ask one more question . I have two problems which I don't understand in Kleppner Kolenkow. They seem to contradict each other in my view. The first is 3.9 and the second is 3.10.

3.9 A freight car of mass M contains a mass of sand m. At t = 0 a constant horisontal force F is applied in the direction of rolling and at the same time a port in the bottom is opened to let the sand flow out at constant rate dm/dt.
Find the speed of the freight car when all the sand is gone. Assume the freight car is at rest at t = 0.

3.10 An empty freight car of mass M starts from rest under an applied force F. At the same time, sand begins to run into the car at steady rate b from a hopper at rest along the track.
Find the speed when the mass of sand, m has been transferred.

Homework Equations

F = dP/dt

The Attempt at a Solution

I try to find the differentials for these two problems.

3.9 P(t) = (M + m - dm/dt*t)v
P(t + dt) = (M + m - dm/dt*(t + dt))(v + dv) + dm/dt*dt*v
The last term dm/dt*dt*v is what I don't understand. The physical meaning of it being that the car gets an extra "boost" when mass flows out. The full differnatial equation becomes:
dP/dt = dv/dt*(M + m - dm/dt*t)

Lets look at the differentials of the other problem.

3.10 P(t) = Mv + vbt
P(t + dt) = M(v + dv) + (v + dv)b(t + dt)
In the same spirit as of the last problem I want to add a term b*dt*v, because the car is slowed down when sand at rest hits the car of speed v. The full differential equation becomes:
dP/dt = dv/dt*(M + bt) +bv
which is correct from check.

My question is, why add an extra term in 3.9 and not in 3.10?

 

[hikaru1221]

For 3.9, the phenomenon is like this: M+(m-dm/dt*t) is running, and the whole thing drops a mass of dm/dt*dt. If you exclude the "extra" term, the system in the first equation will be different from the system in the 2nd one: while it's the whole thing M+(m-dm/dt*t) in the 1st, it would be just M+(m-dm/dt*t) - dm/dt*dt in the 2nd. That's totally fine until you write this equation: P(t+dt) - P(t) = Fdt. You know why?

The same reason applies for 3.10. However, the "extra" term actually hides itself in this case.

 

[Order]

The same reason applies for 3.10. However, the "extra" term actually hides itself in this case.

Yes I see! The whole system includes the mass that is dropped in 3.9, even if it is not on the car.

In 3.10 the mass hides as a term b*dt*0 in the P(t) equation because the hopper is at rest. Thanks Hikaru!

I think i can see why. I'm a little confused about if there are internal or external forces missing, but subtracting two different systems (P(t + dt) - P(t)) cannot be right.

 

[hikaru1221]

Subtracting those two is still okay (though quite pointless), but equating the difference with Fdt is the problem.

 

[gneill]

Is there some particular reason why you want to take a momentum approach to this problem?

 

[Order]

Is there some particular reason why you want to take a momentum approach to this problem?

Well, the force is constant, so I guess it could be solved before the university level, when you learn that Work=Force*distance, but I was a little confused about momentum conservation and wanted to give air to the paradoxes in my head. Now I can apply it much better and I feel more confident too.