Conserving Momentum in Emptying a Freight Car

[ThEmptyTree]

Homework Statement

A freight car of mass $m_c$ contains sand of mass $m_s$ . At $t = 0$ a constant horizontal force of magnitude $F$ 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 the constant rate $b = \frac{dm_s}{dt}$. Find the speed of the freight car when all the sand is gone (Figure 12.6). Assume that the freight car is at rest at $t = 0$ .

Relevant Equations

$$\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}$$

$$\overrightarrow{p_i}=\overrightarrow{0},~\overrightarrow{p_f}=m_c\overrightarrow{v_f}$$

$$\overrightarrow{p_f}-\overrightarrow{p_i}=\int\limits_{t_i}^{t_f}\overrightarrow{F}dt$$

$$t_i=0,~t_f=\frac{m_s}{b}$$

$$m_c\overrightarrow{v_f}=\overrightarrow{F}\frac{m_s}{b}\Rightarrow v_f=\frac{Fm_s}{bm_c}$$

However, their solution uses differential analysis for states $t$ and $t+\Delta{t}$, yielding

$$v(t)=\frac{F}{b}ln\Big(\frac{m_c+m_s}{m_c+m_s-bt}\Big)$$

For $t=t_f=\frac{m_s}{b}$ according to their solution

$$v_f=\frac{F}{b}ln\Big(1+\frac{m_s}{m_c}\Big)$$

Notice that using approximation $ln(1+x)\approx x$ for small $x$ their solution becomes equivalent to mine.

What am I doing wrong? I am confused. What am I skipping when trying to analyze only the initial and final state?

 

[BvU]

What am I doing wrong?

The sand that falls out also has momentum ... how does your solution account for that ?

$\$

 

[ThEmptyTree]

The sand that falls out also has momentum ... how does your solution account for that ?

$\$

I agree that the sand has speed, but I am only analyzing the cart.

Also, in their solution $\Delta{m_c}=-\Delta{m_s}$ and the sand simplifies.

 

[BvU]

but I am only analyzing the cart

$$\overrightarrow{p_f}-\overrightarrow{p_i}=\int\limits_{t_i}^{t_f}\overrightarrow{F}dt$$ has a righthand side that only applies to the car ?

 

[ThEmptyTree]

$$\overrightarrow{p_f}-\overrightarrow{p_i}=\int\limits_{t_i}^{t_f}\overrightarrow{F}dt$$ has a righthand side that only applies to the car ?

That's what came to my mind, that $\overrightarrow{F}$ acts over the cart + fallen sand, not only the cart. But how is that, if $\overrightarrow{F}$ originates from friction?

 

[BvU]

Also, in their solution $\Delta{m_c}=-\Delta{m_s}$ and the sand simplifies.

$m_s$ and $m_c$ are fixed numbers in their solution.

(but not in $b = \frac{dm_s}{dt}$, so I grant you that their use of the symbols is misleading...)

 

[BvU]

$\vec F$ is the externally applied force (e.g. from a locomotive)

 

[ThEmptyTree]

$\vec F$ is the externally applied force (e.g. from a locomotive)

Still, the locomotive tracts the cart, not the sand, right ?

 

[BvU]

The locomotive is pulling on cart + remaining sand !

$$\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}$$this is one of the (rare) cases where both $m$ and $\vec p$ change ...

$\$

 

[ThEmptyTree]

The locomotive is pulling on cart + remaining sand !

I agree with the remaining sand. The problem is that in their solution they consider the system of cart + fallen sand, and when they write $\overrightarrow{F}=\frac{d\overrightarrow{p_{sys}}}{dt}$ $\overrightarrow{F}$ appears to act on both cart + fallen sand at $t+\Delta{t}$.

When I said I only analyze the cart, that means the cart + the sand left in it.

At $t=0$ even if there is sand in the cart the momentum is null, and at $t=t_f$ there is no sand left.

 

[ThEmptyTree]

@BvU I figured it out. The problem is that $\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}$ only works for constant mass. In my approach I chose the system of cart + remaining sand, but that is not constant mass because sand is constantly falling. I think this is what you were trying to explain, but you did not emphasize it, or I couldn't understand. Thanks for your time!

Case solved.

 

[bob012345]

@BvU I figured it out. The problem is that $\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}$ only works for constant mass. In my approach I chose the system of cart + remaining sand, but that is not constant mass because sand is constantly falling. I think this is what you were trying to explain, but you did not emphasize it, or I couldn't understand. Thanks for your time!

Case solve

Force is the time rate of change of momentum and both mass and velocity can change but applying $\vec F = \large\frac {d \vec P}{dt}$ correctly can be a bit tricky. In general $\large\frac {d \vec P}{dt} = \large\frac {m(t) d \vec V}{dt} + \large\frac { \vec V dm}{dt}$. In a rocket, the second term would be the generator of the force. In this case the force is fixed and external and it produces an increasing acceleration as the total mass of the car decreases.

Starting with the variables in the diagram of post $10$ we can write the system momentum at times $t$ and $t + Δt$;

$$\vec P(t ) = (m_c + m_s -bt)\vec v_c(t)$$ and

$$\vec P(t + Δt) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c) + Δm_s(v_c(t) + Δ\vec v_c)$$

Then we can write $\vec F = \large \frac{Δ\vec P}{Δt} = \large \frac{\vec P(t + Δt) - \vec P(t )}{Δt}$

as $$\vec F Δt = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c) + Δm_s(v_c(t) + Δ\vec v_c) - (m_c + m_s -bt)\vec v_c(t)$$

Then $$\vec F Δt = -bΔt\vec v_c(t) +(m_c + m_s -bt)Δ\vec v_c - bΔtΔ\vec v_c) + Δm_s\vec v_c(t) + Δm_sΔ\vec v_c)$$ Then getting rid of double differentials and noticing that $Δm_s = bΔt$ we get

$$\vec F Δt = (m_c + m_s -bt)Δ\vec v_c~~~ or ~~~ \frac {d\vec v_c}{dt} = \frac { \vec F}{(m_c + m_s -bt)}~~~ as Δt →0$$.

 

[ThEmptyTree]

@bob012345 I appreciate doing all the calculations, but my enigma was why we could not just analyze the initial and final state only.

No, I believe that's not accurate.

I still believe the reason is because during that time interval, the mass of the object (whether it is a rocket, a freight cart, a snowplow gathering snow) changes, and Newton's generalized 2nd law (in the way I applied it) only works for constant mass, either for a single object or a system of objects.

 

[bob012345]

I was just reacting to your statement ;

The problem is that $F=\large \frac{dp}{dt}$ only works for constant mass.

It works for changing mass too in general. You are correct that what doesn't work is not taking changing mass into account not that Newton's formulation only works for constant mass. If you assume constant mass in this problem all you get is that $v(t) = \frac{F}{m_c + m_s} t$ which is just $v(t) = at$.But I did just use the initial and final states to derive the equation that needed to be solved to show it works. The simplest way to do this problem however is to just recognize that $\vec a = \large\frac{\vec F}{m_c+m_s -bt}$ and integrate to find $v(t)$.

 

[ThEmptyTree]

I was just reacting to your statement ;

The problem is that $F=\large \frac{dp}{dt}$ only works for constant mass.

It works for changing mass too in general.

What I said there is indeed misleading.

Now after meditating on it for a while I still don't think I understand what I did wrong in my first approach.

I did not take into account that $$d\overrightarrow{p}=md\overrightarrow{v}+dm\overrightarrow{v}$$
but
$$\int\limits_{t_i}^{t_f} d\overrightarrow{p}=\int\limits_{t_i}^{t_f} \overrightarrow{F}dt$$
should still work right?

 

[haruspex]

It works for changing mass too in general.

$\vec F=\frac{d\vec p}{dt}$ applies to a closed system, and in closed systems mass does not change.
If the mass is changing then you need to consider what momentum the gained (lost) mass is adding (subtracting). To make the equation work you need to treat that as an applied force.
In the example of sand being dropped vertically from a static hopper into a moving cart, you get away with it because the sand arrives with no horizontal momentum. But if sand leaks from a moving cart, it carries its momentum away with it. Overlooking that leads to the conclusion that the cart should accelerate.

 

[bob012345]

$\vec F=\frac{d\vec p}{dt}$ applies to a closed system, and in closed systems mass does not change.
If the mass is changing then you need to consider what momentum the gained (lost) mass is adding (subtracting). To make the equation work you need to treat that as an applied force.
In the example of sand being dropped vertically from a static hopper into a moving cart, you get away with it because the sand arrives with no horizontal momentum. But if sand leaks from a moving cart, it carries its momentum away with it. Overlooking that leads to the conclusion that the cart should accelerate.

In my understanding it is more general than $F = ma$. One just has to be careful how to apply it.

 

[haruspex]

In my understanding it is more general than $F = ma$. One just has to be careful how to apply it.

For clarity, please illustrate its use in the leaking sand model.

 

[bob012345]

For clarity, please illustrate its use in the leaking sand model.

I thought I did that in post #12.

 

[haruspex]

I thought I did that in post #12.

That post was rather confusing.
You wrote
$\vec F = \large\frac {d \vec P}{dt}$
and
$\large\frac {d \vec P}{dt} = \large\frac {m(t) d \vec V}{dt} + \large\frac { \vec V dm}{dt}$
from which one would conclude
$\vec F = \large\frac {m(t) d \vec V}{dt} + \large\frac { \vec V dm}{dt}$.
but followed with

In a rocket, the second term would be the generator of the force.

If the force is $\large\frac { \vec V dm}{dt}$, what does $\vec F$ mean in that equation?
And what is V now, the velocity of the rocket, of the exhaust, or the difference between the two?

It seems to me you did not actually use $\vec F = \large\frac {d \vec P}{dt}$.

 

[bob012345]

That post was rather confusing.
You wrote
$\vec F = \large\frac {d \vec P}{dt}$
and
$\large\frac {d \vec P}{dt} = \large\frac {m(t) d \vec V}{dt} + \large\frac { \vec V dm}{dt}$
from which one would conclude
$\vec F = \large\frac {m(t) d \vec V}{dt} + \large\frac { \vec V dm}{dt}$.
but followed with

If the force is $\large\frac { \vec V dm}{dt}$, what does $\vec F$ mean in that equation?
And what is V now, the velocity of the rocket, of the exhaust, or the difference between the two?

It seems to me you did not actually use $\vec F = \large\frac {d \vec P}{dt}$.

Sorry if I confused you. I indeed fell victim to a common error of assuming $\large\frac {d \vec P}{dt} = \large\frac {m(t) d \vec V}{dt} + \large\frac { \vec V dm}{dt}$
which assumes momentum is always simply $\vec P = m\vec V$ when there can be different contributions. However, that was only in my general remarks and I did not use that directly in the working of the problem. There I used the differential form of $\vec F = \large\frac {Δ \vec P}{Δt}$ to set up the OP problem as recommended in my text An Introduction to Mechanics by Kleppner and Kolenkow to arrive at the result;

$$ \frac {d\vec v_c}{dt} = \frac { \vec F}{(m_c + m_s -bt)}$$.

Which will give $\vec v_c(t)$ when integrated..

 

[haruspex]

which assumes momentum is always simply $\vec P = m\vec V$

Which, for a defined assembly, it is.

we can write the system momentum at times $t$ and $t + Δt$;
$$\vec P(t ) = (m_c + m_s -bt)\vec v_c(t)$$ and
$$\vec P(t + Δt) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c) + Δm_s(v_c(t) + Δ\vec v_c)$$

The second equation does not follow from the first. In fact, if, as you defined, P is the momentum of the cart and its remaining sand then the second equation is not true.
Rather, you would get
$$\vec P(t + Δt) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c) $$.
It's no good countering that the extra term you added comes from the momentum lost in the loss of sand. That term did not appear in the initial equation. You cannot go adding terms just because you know you will otherwise get the wrong answer.

 

[bob012345]

Which, for a defined assembly, it is.

The second equation does not follow from the first. In fact, if, as you defined, P is the momentum of the cart and its remaining sand then the second equation is not true.
Rather, you would get
$$\vec P(t + Δt) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c) $$.
It's no good countering that the extra term you added comes from the momentum lost in the loss of sand. That term did not appear in the initial equation. You cannot go adding terms just because you know you will otherwise get the wrong answer.

I wrote those equations for the system, not just the cart. That is exactly what we are talking about, the system not one part. They seem correct to me. They led to the correct expression for velocity. That extra term is the sand. I used the figure from post #10 as reference.

 

[haruspex]

I wrote those equations for the system, not just the cart.

No, your first equation was

$$\vec P(t ) = (m_c + m_s -bt)\vec v_c(t)$$

which is clearly just for the cart and its remaining sand. If your next equation is for the cart plus all the sand then you changed the meaning of P(t) . The two equations are clearly not consistent algebraically.

Even if we do redefine P that way for the second equation it is wrong:

$$\vec P(t + Δt) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c) + Δm_s(v_c(t) + Δ\vec v_c)$$

Presumably the idea is
"Momentum of whole system at time t+dt = momentum of cart with remaining sand at t+dt + momentum of all lost sand at time t+dt."
Unfortunately the second term on the right is not the momentum of all lost sand at time t+dt; it is only the momentum of the sand lost during time dt. This makes it hard to understand how P(t) is being defined.

 

[bob012345]

No, your first equation was
which is clearly just for the cart and its remaining sand. If your next equation is for the cart plus all the sand then you changed the meaning of P(t) . The two equations are clearly not consistent algebraically.

The cart and remaining sand is the system for the first equation. All previous fallen sand has zero momentum. The second equation is for the cart with $Δm_s = bΔt$ less sand than the first and the same amount of sand moving with $v_c + Δv_c$. We are only concerned with each element of sand just as it falls and looking at one is all we need to do. We are not accounting for all the fallen sand from $t=0$. We start the problem at some arbitrary time.

Even if we do redefine P that way for the second equation it is wrong:

Presumably the idea is
"Momentum of whole system at time t+dt = momentum of cart with remaining sand at t+dt + momentum of all lost sand at time t+dt."
Unfortunately the second term on the right is not the momentum of all lost sand at time t+dt; it is only the momentum of the sand lost during time dt. This makes it hard to understand how P(t) is being defined.

I invite you then to do the problem yourself and see if you get the right answer.

 

[haruspex]

The cart and remaining sand is the system for the first equation.

Which, as I wrote, is what I assumed in the first place:

as you defined, P is the momentum of the cart and its remaining sand

So why did you respond with:

I wrote those equations for the system, not just the cart

So, now you are saying that in
$$\vec P(t + Δt) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c) + Δm_s(v_c(t) + Δ\vec v_c)$$
$\vec P(t + Δt)$ is the momentum of the cart+remaining sand, yes?
Clearly that is wrong.
$\vec P(t + Δt)$ = momentum of cart plus remaining sand at time (t + Δt)
= [mass of cart plus remaining sand at time (t + Δt)][velocity of cart plus remaining sand at time (t + Δt)]
=$(m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c)$

All previous fallen sand has zero momentum.

We do not know the fate of the sand, but it certainly had momentum when it left the cart.

see if you get the right answer.

I have no problem getting the right answer, and whether you got the right answer is not the question I'm posing.
The issue is, what exactly is your method, did it really use $F=dp/dt$ with p being the momentum of a variable mass subsystem, with those variables defined how, and is your method correct?

 

[ThEmptyTree]

I am confused. I don't know what to believe, does $\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}$ work or not for changing mass? If it works, I still can't explain what I did wrong. If it doesn't work, then it contradicts physics?

 

[haruspex]

I am confused. I don't know what to believe, does $\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}$ work or not for changing mass? If it works, I still can't explain what I did wrong. If it doesn't work, then it contradicts physics?

$\frac{d\vec p}{dt}=m\frac{d\vec v}{dt}+\vec v\frac{dm}{dt}$ is certainly valid - that's just algebra. The question is, when can we write $\vec F=\frac{d\vec p}{dt}$, F being the force applied to a system and m being its mass?
The short answer is, only when no mass change is adding momentum to or removing it from the system.
With the current problem, the leaking sand is taking momentum away from the cart+remaining sand system.
One way to deal with that is to treat the resulting change in momentum of the chosen system as another applied force. Since the sand leaves at the current velocity of the cart, the rate at which it adds momentum is $v\dot m$ ($\dot m$ being negative). This makes the force equation (I'll drop the vectors) $F+v\dot m=\dot p$, which reduces to $F=m\dot v$.

 

[jbriggs444]

I am confused. I don't know what to believe, does $\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}$ work or not for changing mass? If it works, I still can't explain what I did wrong. If it doesn't work, then it contradicts physics?

It can work. But it is fraught with opportunities to confuse yourself. If you are contemplating a system with variable mass then you are contemplating a (hopefully continuous) mass flow across the system boundary. That mass flow can be treated as a momentum flow -- i.e. as a force.

It is ever so tempting to differentiate $\frac{dp}{dt}=\frac{d(m(t)v(t))}{dt}$ and obtain $m\frac{dv}{dt} + v\frac{dm}{dt} = ma + v\frac{dm}{dt}$. And it is even correct -- as far as it goes.

But it is easy to forget that this applies to a scenario where the mass being added or removed from the system comes in already moving at the same velocity as the system or leaves still moving at the same velocity as the system.

If you have a stream of sand that is at rest and is added to a moving train, it does not carry in a momentum flow that matches $v\frac{dm}{dt}$. It carries in zero momentum instead.

If you have a rocket that is streaming out an exhaust at relative velocity $v_e$, it does not carry out a momentum flow that matches $v\frac{dm}{dt}$. It carries out a momentum flow given by $(v-v_e)\frac{dm}{dt}$ instead.

In the case at hand, we have a freight car streaming sand out of the hopper at a rate of $\frac{dm}{dt}$ The dropped sand is indeed initially moving at the speed of the car, $v$. The egress momentum flow is indeed given by $v\frac{dm}{dt}$. (One should be aware that $\frac{dm}{dt}$ will be negative).

 

[ThEmptyTree]

If you have a stream of sand that is at rest and
In the case at hand, we have a freight car streaming sand out of the hopper at a rate of $\frac{dm}{dt}$ The dropped sand is indeed initially moving at the speed of the car, $v$. The egress momentum flow is indeed given by $v\frac{dm}{dt}$. (One should be aware that $\frac{dm}{dt}$ will be negative).

So I applied it correctly in the first place but got the wrong answer?

 

[haruspex]

So I applied it correctly in the first place but got the wrong answer?

You overlooked that some of the work done by the force goes into accelerating sand that is later lost.

 

[ThEmptyTree]

You overlooked that some of the work done by the force goes into accelerating sand that is later lost.

OK, so that's what you mean by "fallen sand should be counted as applied force". So instead of $\overrightarrow{F}$ I should have $\overrightarrow{F}+\frac{dm}{dt}\overrightarrow{v}$ as you said in post #28. I think it's clear now.

One thing I didn't understand from the beginning is that $\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}$ is actually the definition of the force, that means total physical force that acts over the system as observed by the chosen frame. I was thinking of $\overrightarrow{F}$ as just the traction force.

Now that I understand the precautions when analyzing the initial and final state only, I consider it's safer to analyze problems at $t$ and $t+\Delta{t}$ since the system has constant mass.

I think now it's case solved.

 

[bob012345]

I am confused. I don't know what to believe, does $\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}$ work or not for changing mass? If it works, I still can't explain what I did wrong. If it doesn't work, then it contradicts physics?

You mentioned you knew the right answer. Was it from a textbook and did they show some steps for the problem?

 

[ThEmptyTree]

You mentioned you knew the right answer. Was it from a textbook and did they show some steps for the problem?

I am currently studying from an MIT OpenCourseWare. In the introductory lecture of chapter Momentum and Impulse they told that $\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}$ is the most general form for Newton's 2nd law, which works even for systems with a change in mass over time. However for changing mass I realized that it contradicts $\overrightarrow{F}=m\overrightarrow{a}$ so I was confused. But after some research I found out $\overrightarrow{F}=m\overrightarrow{a}$ is actually the case of $\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}$ for constant mass.

What was most confusing for me was that people on the internet were saying $\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}$ only works for constant mass, which would have easily explained the mistake in my approach. However the mistake was that I was counting $\overrightarrow{F}$ as traction force, not total force.

I had the right answer with full steps, that was not the problem. The exercise is from the course e-book. My dilemma was why my method was not right.

 

[bob012345]

So, now you are saying that in
$$\vec P(t + Δt) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c) + Δm_s(v_c(t) + Δ\vec v_c)$$
$\vec P(t + Δt)$ is the momentum of the cart+remaining sand, yes?
Clearly that is wrong.

I said it is the momentum of cart + remaining sand in the cart plus the sand that just left the cart at that instant.

$\vec P(t + Δt)$ = momentum of cart plus remaining sand at time (t + Δt)
= [mass of cart plus remaining sand at time (t + Δt)][velocity of cart plus remaining sand at time (t + Δt)]
=$(m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c)$We do not know the fate of the sand, but it certainly had momentum when it left the cart.

And I accounted for that.

I have no problem getting the right answer, and whether you got the right answer is not the question I'm posing.

Please just show your version of the first and second equations so I can see how you want to do it.

The issue is, what exactly is your method, did it really use $F=dp/dt$ with p being the momentum of a variable mass subsystem, with those variables defined how, and is your method correct?

My method was using the figure in post #10 and writing the momentum before and after the mass $Δm_s$ is released. Then I subtracted those two momentum equations and divided by $Δt$ and took the limit at $Δt →0$ to get a differential equation that can be solved for $v_c(t)$.

 

[ThEmptyTree]

https://ocw.mit.edu/courses/physics...2016/readings/MIT8_01F16_chapter12.1_12.3.pdf

Example 12.2

It is basically what you did.

 

[haruspex]

I said it is the momentum of cart + remaining sand in the cart plus the sand that just left the cart at that instant.

What I am trying to get to grips with in your algebra is how you are defining $P(t)$ in general. If it is the momentum of the cart plus remaining sand at time $t$ then necessarily $P(t+\Delta t)$ is the momentum of the cart plus remaining sand at time $t +\Delta t$. You can't arbitrarily define it to be something else.
You probably knew what you meant with your equations but expressed them wrongly.

My method was using the figure in post #10 and writing the momentum before and after the mass $Δm_s$ is released. Then I subtracted those two momentum equations and divided by $Δt$ and took the limit at $Δt →0$ to get a differential equation that can be solved for $v_c(t)$.

Fine, so as I suspected you did not actually use $F=dp/dt$ but instead derived it all from first principles. I see the linked text does the same.

Which still leaves us with the question of whether we can avoid resorting to all this detailed Delta and limit analysis and apply a standard differential equation.
As you saw in post #28 it can be done with F=dp/dt, but really the simplest is to use F=ma, where F is the stated applied force and m is the cart with its remaining sand at time t.
In the case of a rocket or of sand being dropped into a moving cart we can add the actual force (rather than a virtual one to account for momentum brought in or removed by the gained or lost mass) that applies. Much less confusing all round.

I am surprised the MIT text teaches such unnecessarily verbose methods. Life's too short.

 

[bob012345]

What I am trying to get to grips with in your algebra is how you are defining $P(t)$ in general. If it is the momentum of the cart plus remaining sand at time $t$ then necessarily $P(t+\Delta t)$ is the momentum of the cart plus remaining sand at time $t +\Delta t$. You can't arbitrarily define it to be something else.
You probably knew what you meant with your equations but expressed them wrongly.

My equations are exactly what the MIT solution did and I meant the same thing they did. I'll try one more time to explain it by showing you exactly what I said in post #12; I used this diagram from @ThEmptyTree.

Please study this diagram. It is the system at times $t$ and $t +Δt$. In this arbitrary slice of time there is one piece at $t$ and two pieces at $t +Δt$. Nothing is being redefined. Note $Δm_c = -Δm_s$.

Starting with the variables in the diagram of post $10$ we can write the system momentum at times $t$ and $t + Δt$;

$$\vec P(t ) = (m_c + m_s -bt)\vec v_c(t)$$ and

$$\vec P(t + Δt) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c) + Δm_s(v_c(t) + Δ\vec v_c)$$

Then we can write $\vec F = \large \frac{Δ\vec P}{Δt} = \large \frac{\vec P(t + Δt) - \vec P(t )}{Δt}$

Now compare that to what MIT used ;
$$\vec P_{sys} (t) = m_c(t) \vec v_c (t)$$ .
$$\vec P_{sys} (t + Δt) = (Δm_s + m (t) + Δm_c)(\vec v_c(t) + Δ \vec v_c )$$
where $Δm_c = −Δm_s$ , $m_c (t + Δt) = m_c (t) + Δm_c$ and $m_c(t) =m_c + m_s − bt$

substituting their notation and rearranging my equations a bit we get;
$$\vec P(t ) = (m_c + m_s -bt)\vec v_c(t)~~ →~~\vec P_{sys}(t ) = m_c(t)\vec v_c(t)$$ and
$$\vec P(t + Δt) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c) + Δm_s(v_c(t) + Δ\vec v_c)$$
which becomes
$$\vec P_{sys}(t + Δt) = (m_c(t) - bΔt))(\vec v_c(t) + Δ\vec v_c) + Δm_s(v_c(t) + Δ\vec v_c)$$
or
$$\vec P_{sys}(t + Δt) = (Δm_s +m_c(t) +Δm_c))(\vec v_c(t) + Δ\vec v_c)$$
since $-bΔt = -Δm_s = Δm_c$

They are the same and they mean the same.

Fine, so as I suspected you did not actually use $F=dp/dt$ but instead derived it all from first principles. I see the linked text does the same.

I tried to tell you I was using the differential form. I thought it was understood I was deriving the relationship for this case.

Which still leaves us with the question of whether we can avoid resorting to all this detailed Delta and limit analysis and apply a standard differential equation.

Yes. I said that in post #14. Nothing in $\vec F = m \vec a$ says $m$ can't be a function of time.

I am surprised the MIT text teaches such unnecessarily verbose methods. Life's too short.

It is necessary like teaching limits to calculus students.

 

[haruspex]

$$\vec P(t ) = (m_c + m_s -bt)\vec v_c(t)$$ and

$$\vec P(t + Δt) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c) + Δm_s(v_c(t) + Δ\vec v_c)$$

Although the definitions of $m_c$ etc. need a little translation, I agree that your equations match those at the link. And they share the sin of not defining what the 'system' is.
You later supplied a definition, but it doesn't match the equations. If you define p(t) as the momentum of the cart and remaining sand at time t then it is just a matter of algebra that $$\vec P(t + Δt) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c)$$

To get your equation and the one at the link, you have to define P(t) in a very awkward way; so awkward I have failed to find a form of words to describe it.

The whole matter can be resolved by accepting that $\vec P(t + Δt)$ is as I have written it above and then observing that the change in the momentum of this system is due not only to the applied force F but also because we are no longer including a bit of momentum that has gone off with the lost sand:
$$\vec P(t + Δt)-\vec P(t) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c)- (m_c + m_s -bt)\vec v_c(t)+\vec F.Δt+bΔt\vec v_c(t)$$

Edit: should be
$$\vec F.Δt-bΔt\vec v_c(t)=\vec P(t + Δt)-\vec P(t) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c)- (m_c + m_s -bt)\vec v_c(t)$$
The hard part is getting the sign right.

Really it is rather a sloppy and confusing text at that link. I'll try to contact the page owner.
As to the use of limit methods in all the exercises, if it is just to practise such methods it seems like overkill. Do it once then show the workaday method.

 

[bob012345]

Although the definitions of $m_c$ etc. need a little translation, I agree that your equations match those at the link. And they share the sin of not defining what the 'system' is.

I disagree with that assessment . We both did define the system.

You later supplied a definition, but it doesn't match the equations.

It does.

If you define p(t) as the momentum of the cart and remaining sand at time t then it is just a matter of algebra that $$\vec P(t + Δt) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c)$$

That ignores the sand that just left the cart. Why do you want to do that? It is part of the problem. Besides, I'm not defining momentum, I am describing it and how it changes for the whole system as I've said before.

To get your equation and the one at the link, you have to define P(t) in a very awkward way; so awkward I have failed to find a form of words to describe it.

Try it please. Maybe it will help diagnose where your issue really is.

The whole matter can be resolved by accepting that $\vec P(t + Δt)$ is as I have written it above and then observing that the change in the momentum of this system is due not only to the applied force F but also because we are no longer including a bit of momentum that has gone off with the lost sand:
$$\vec P(t + Δt)-\vec P(t) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c)- (m_c + m_s -bt)\vec v_c(t)+\vec F.Δt+bΔt\vec v_c(t)$$

Now that is confusing to me. Is that whole thing equal to $FΔt$? Please carry it through and show what it leads to.

 

[haruspex]

It does.

No, you added a term in your expression for $\vec P(t + Δt)$ to represent the sand lost during (t, t + Δt). That is no longer in the cart. If P(t) is defined generally as "the momentum of cart+remaining sand at time t" then by that definition $\vec P(t + Δt)$ is "the momentum of cart+remaining sand at time t+ Δt". It does not include sand lost during (t, t + Δt).

I'm not defining momentum

I didn't say you were defining momentum. What you need is a clear definition of the function p(t) and equations that match that.

Try it please. Maybe it will help diagnose where your issue really is.

I don't think there's a way. You'd need another parameter, something like p(t,t') = momentum at time t' of cart plus the sand that was in it at time t. What you write as p(t+Δt) is then p(t,t+Δt).

Now that is confusing to me. Is that whole thing equal to FΔt? Please carry it through and show what it leads to.

sorry, typo:
$F.Δt-bΔt\vec v_c(t)=\vec P(t + Δt)-\vec P(t) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c)- (m_c + m_s -bt)\vec v_c(t)$
$F.Δt= (m_c + m_s - bt )Δ\vec v_c$.

 

[bob012345]

No, you added a term in your expression for $\vec P(t + Δt)$ to represent the sand lost during (t, t + Δt). That is no longer in the cart. If P(t) is defined generally as "the momentum of cart+remaining sand at time t" then by that definition $\vec P(t + Δt)$ is "the momentum of cart+remaining sand at time t+ Δt". It does not include sand lost during (t, t + Δt).

First, I am happy to continue this dialog as long as this is a dialog and not a one sided teaching moment from your point of view. You keep saying If P(t) is defined generally as "the momentum of cart+remaining sand at time t" ... but I did not define it as such so that is irrelevant to my argument. I used the definition in my text that if you have a system of particles and an external force acting on the system, you need to keep account of all the particles not just some.

sorry, typo:
$F.Δt-bΔt\vec v_c(t)=\vec P(t + Δt)-\vec P(t) = (m_c + m_s - b(t +Δt))(\vec v_c(t) + Δ\vec v_c)- (m_c + m_s -bt)\vec v_c(t)$
$F.Δt= (m_c + m_s - bt )Δ\vec v_c$.

This is just quibbling. You are looking at the cart only and just adding a fictitious force to cancel out the $bΔt \vec v_c(t)$ term from $\vec P(t + Δt) - \vec P(t)$. Assuming no friction and level ground, if $F=0$ your equation suggests there would be a force acting on the cart to maintain the same speed because of the momentum loss from the leaving sand or as the sand falls otherwise the cart would slow down. There is no such force on the cart or the sand because it already possesses momentum and is moving at speed $v_c(t)$. The cart would continue to roll at the same speed as the sand fell out assuming no friction. We get to the same answer so at the very least you should admit both views are equivalent ways to look at the problem rather than continuing to say I and MIT are somehow wrong.

 

[hutchphd]

First, I am happy to continue this dialog as long as this is a dialog and not a one sided teaching moment from your point of view.

Think of it this way. Consider the mass leaving the system: if it is shoved out (like a rocket) there will be a reaction force (on the LHS) from it on the rocket that is proportional to exhaust velocity. Here the exhaust velocity is zero. The system is the remaining total mass. That works for me.

 

[bob012345]

Think of it this way. Consider the mass leaving the system: if it is shoved out (like a rocket) there will be a reaction force (on the LHS) from it on the rocket that is proportional to exhaust velocity. Here the exhaust velocity is zero. The system is the remaining total mass. That works for me.

In most problems either the applied force is causing all the momentum change or the momentum change causes all the force (rocket). Here, Interestingly or perhaps frustratingly, the change in momentum by falling sand does not cause a real force and the applied force does not cause the change due to the sand falling out so it's muddled. Fortunately most problems are clearly one or the other.

 

[jbriggs444]

Think of it this way. Consider the mass leaving the system: if it is shoved out (like a rocket) there will be a reaction force (on the LHS) from it on the rocket that is proportional to exhaust velocity. Here the exhaust velocity is zero. The system is the remaining total mass. That works for me.

One should draw the system boundaries carefully.

Does one put the system boundary between the rocket and the exhaust stream at the inlet into the combustion chamber? If so, mass transfer takes place at [near] zero relative velocity and the pressure of the exhaust gasses on the nozzle bell is an external force.

Does one put the system boundary at the mouth of the exhaust bell? If so, mass transfer takes place at a high relative velocity and the pressure of the exhaust gasses on the nozzle bell is an internal force.

You get the same result either way, of course. But if one fails to decide then a muddle can result.

 

[haruspex]

You keep saying If P(t) is defined generally as "the momentum of cart+remaining sand at time t" ... but I did not define it as such so that is irrelevant to my argument. I used the definition in my text that if you have a system of particles and an external force acting on the system, you need to keep account of all the particles not just some.

The entire argument is over notation.

You can draw a diagram of a system of particles in two different states and define p1 as its momentum in one state and p2 as its momentum in the other.
You can then write $F.\Delta t=p_2-p_1$, and if the expressions for p1 and p2 involve such as $v$ and $v+\Delta v$ then you may well obtain a valid differential equation.

The confusion arises because you denoted the momenta of these states generically as a function of time, p(t). That implies a definition that can be interpreted at any given time t, but I cannot see a way to do that with your p1 and p2.
In any algebraic development, if you write $f(t)=x(t)y(t)$ then it follows that $f(t+\Delta t)=x(t+\Delta t)y(t+\Delta t)$, but your equations do not satisfy that.

 

[hutchphd]

I used the definition in my text that if you have a system of particles and an external force acting on the system, you need to keep account of all the particles not just some.

So in the end your system consists of $m_s$ spewed along the roadway and $m_c$ trundling down the road. The center of mass of your system is now moving backwards relative to the cart in an unpleasant way and your system has received unknown forces from the ground to bring $m_s$ to a stop. If figured correctly you will be able to work out the result. This doesn't make it the best solution, which indeed it is not.

 

[bob012345]

So in the end your system consists of $m_s$ spewed along the roadway and $m_c$ trundling down the road. The center of mass of your system is now moving backwards relative to the cart in an unpleasant way and your system has received unknown forces from the ground to bring $m_s$ to a stop. If figured correctly you will be able to work out the result. This doesn't make it the best solution, which indeed it is not.

The problem at hand consists only of everything at time $t$ and everything at time $t + Δt$. I kept track of that interval only not all $Δt's$.

 

[haruspex]

So in the end your system

In @bob012345's notation, p(t) does not represent the momentum as a function of time of any definable 'system'. That's why we're all confused. See post #46.

 

[bob012345]

In @bob012345's notation, p(t) does not represent the momentum as a function of time of any definable 'system'. That's why we're all confused. See post #46.

You had an issue with my second equation but now you say my notation is wrong? What notation is that? I merely wrote down the momentum before and after based on the figure. Is the figure wrong?