# Freight cars and sand

I have come across a few situations involving leaking and loading freight cars being pushed with a constant force. Here, I present two situations that leave me in doubt.

1) A car of mass $M_0$( sand of mass $m$ included ) at rest experiences a constant force $F$ at $t=0$ and starts to leak from the bottom at a constant rate $b$ at the same time. I am to calculate its speed when all the sand leaks out.

Here, I am concerned with the system's horizontal momentum.

I have, $P(0) = 0$ and $P(t) = (M_0 - m)v + x$, where $t$ is the time taken for all the sand to leak out, $v$ is the speed attained by the car at $t$ and $x$ is the momentum of the escaped sand subsystem.

Since the mass of the system remains unchanged, I have, $\Delta P = (M_0 - m)v + x = \int_{0}^{t} F \mathrm dt$

I do not see myself in a position to calculate $x$, and that is why I'm not sure if this approach is correct. While I do know the solution to this question( a different approach ), I would like to know if there are any flaws with my reasoning here. Can I expect the right answer if $x$ is correctly known?

The next question concerns the loading of a freight car.

2) A 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 a steady rate $b$ from a hopper at rest along the track. I am to calculate its speed when a mass $m$ of sand has been transferred.

Once again, I'm concerned with the system's$( M + m )$ horizontal momentum. I am given, $\frac{\mathrm d M_{\mathrm car}}{\mathrm dt} =b$. Mass of the freight car at any time $t$ is then found out to be $M + bt$.

At some time $t$ I have, $P(t) = (M + bt) v$, where $v$ is the speed at that time.

Sometime later, $P(t+\Delta t) = (M + bt + b\Delta t) (v + \Delta v)$

The average rate of change of momentum is given by, $\frac{\Delta P}{\Delta t} = ( M + bt) \frac{\Delta v}{\Delta t} + b( v + \Delta v)$

When I take $\Delta t \rightarrow 0$, I see that the net external force depends on the speed of the car. This is an absurd result because if I shift my origin from my current inertial frame to another inertial frame( moving with a different speed ), the force acting on the car will be different. This is not possible. Where could I have gone wrong with my reasoning? The solution to this question is simple. However, I am curious to know where I went wrong.

• Could you explain: from the last equation of second problem, when $\Delta{t}$ goes to zero, how do you gather dependency on velocity ? – npojo Jan 17 '18 at 5:50
• @npojo $\Delta P$ gives me $(M+bt) \Delta v + b\Delta t(v + \Delta v)$. On dividing by $\Delta t$ we get the last equation. – R004 Jan 17 '18 at 5:54
• I understand that. I am talking about the next step. You obtain a complex equation involving both v and acceleration which should be equated to a constant force and then resolved for v and a. Sort of the other way around from your statement. – npojo Jan 17 '18 at 6:00
• @npojo Correct me if I'm wrong. I get $F=(M + bt)\frac{\mathrm d v}{\mathrm dt}+ bv$. According to the solution( which makes sense unlike mine ), $bv$ should not be there in the equation. – R004 Jan 17 '18 at 6:21
• bv is a reminiscence of b kg of sand which instantly gained velocity of v when reaching the car. Why shouldn't it be there? – npojo Jan 17 '18 at 6:30

There is an asymmetry your two problems (as well as an error in an answer by Mark):

When sand with initial zero horizontal velocity lands on the car already moving with a speed $v$, there is a case of inelastic collision in which a mass of sand $dm=b \, dt$ instantaneously acquires speed $v$, so there should also be a force, corresponding to the momentum we need to impart to accelerate this inflowing sand. The magnitude of this additional force is $$|F_1|=\frac{dp}{dt} = \frac{v\, dm }{dt}=\frac{v\, b dt}{dt}=b v,$$ and the direction is opposite $v$. This additional force is absent in your first case since the sand leaves with the same horizontal speed as the car.

If you consider this situation in a different inertial frame moving horizontally with a speed $u$, then the sand dumped into the car would have a nonzero impulse $dp = u \, dm$, and so its speed would need to be changed by $v-u$ to move with the speed of the car. So, you are not wrong: the net force in the second case does depend on the speed of a car.

And so, if while the car is moving under the hopper we dump into it total mass $m$ of sand the speed at the end is calculated by solving the differential equation $$\frac{dv(t)}{dt}=\frac{F_1+ F}{M(t)}=\frac{-b\, v(t)+F}{M+b\,t},$$ subject to initial condition $v(0)=0$.

• This clears it for me. +1 – R004 Jan 17 '18 at 7:23

There are times when momentum calculations are a handy shortcut for solving physics problems. This isn't one of them.

To calculate the change in velocity of an object after a given period of acceleration, you integrate the acceleration from $t_0$ to $t_1$: $v = \int_{t_0}^{t_1} a(t) dt$. Now, the problem doesn't give you acceleration, it gives you force. But you can use Newton's laws to get acceleration from force by $F = ma$, or solving for $a$, $a = F/m$.

In both problems, the mass isn't constant. It's increasing or increasing at a rate $b$, from which you can derive a formula for mass over time: $M(t) = M_0 + bt$.

The final pieces you need are the time $t$ over which the force is applied and the initial velocity (which shows up as the constant of integration). You can get the time by dividing the mass $m$ added or lost by the rate $b$ of change, and the initial velocity is given to you as $0$.

Substitute the formula for M(t) into the acceleration formula, substitute that into the velocity integral, set $t_0$ to 0 and $t_1$ to the end time you found, and solve.