Freight car problems

Question

I am confused with the way we need to use different approach in some similar questions from Kleppner and Kolenkow mechanics like those given below:

1. 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 a mass of sand m has been transferred.

$P(0) =0$

$ P(t) = (M + bt)v $

$impulse = $$\int P = (M + bt)v = $$\int Fdt = F $$\int dt = Ft$

$ v = Ft/ (M + bt)$.................................(1)

But I got the following result which as easily seen comes from the integration of

$1/[(M+bt)/M]*dt$,

$v=F/b*ln[(M+ bt)/M] $................(2)

On the other hand we get the same result given by eq(2) for the following question:

2. A freight car of mass M contains a mass of sand m. At t = 0 a constant horizontal 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.

I couldn't really understand why we use different approaches to deal with these two question.(Actually,the only difference is in the last step i.e. while solving the integral). But why the twist all of a sudden?

I would be very thankful if someone could help me understanding the difference between these two question, maybe the difference in the physical context these two questions hold that can justify why we solve them differently.

PS: I am not asking for anyone to solve them cause I know the solution.It's just that I wonder if I've missed out certain underlying physics which creates the difference.

Answer 1

The integral allows for the fact that the mass gets heavier so a given impulse produces less velocity, but not for the fact that the mass of sand being poured from the hopper needs to be brought up to speed.

Consider a similar set up with no F but a nonzero initial velocity. The freight car in the first question would slow down, whereas the one in the second question would not change speed. That's the difference.

Answer 2

You will have noted that before Newton's second law is used a system of constant mass has been defined.

In your first example the system of constant mass is the freight car and all the mass which is added to the freight car.
How can that be?
Consider a time $t'$ when the freight car and all the sand inside it is travelling at a speed $v'$.
During a short time interval a mass of sand $\Delta m$ is added to the freight car.
The mass of sand $\Delta m$ starts with a speed of zero and is accelerated to as current speed of the freight car $v'$ (ignoring second order terms) and then over a period of time that sand is further accelerated to the final speed of the freight car $v$.
So overall the change in momentum of the sand is $\Delta m (v-0)$ and that is true for all the sand of mass $bt$ which is added to the freight car althogh the "intermediate" speed will be different.
In this case it does not matter about the time sequence of the acceleration of sand from rest to speed $v$ rather all that mattered was that all the sand started at rest and finished up at speed $v$ so the total change of momentum of all of the sand was $\sum\limits_{\text {all sand}}\Delta m (v-0) = bt v$.

In your second example although the speed of the sand at the start is well defined as being zero the change of momentum of each "grain" of snad is different and dictated by the speed of the freight car at the instant the grain of sand falls from the fright car.
In this case the system is defined at a time when a mass of sand $\Delta m$ falls from the freight car and Newton's second law is applied over a short interval $\Delta t$ which in the end is made "vanishingly small".
An integration is then done to in effect apply Newton's second law an infinite number of times to a number of infinitesimally small amounts of sand to evaluate the total change of momentum of the freight train and the sand.

There is no reason why you cannot use this approach for your first example but you must include the fact that the acceleration of each grain of sand occurs in two parts - getting to the current speed of the freight car and then reaching the final speed of the freight car during the rest of the journey.