No problem! Glad I could help.

[Saitama]

Homework Statement

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 the 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 $t=0$.

Homework Equations

The Attempt at a Solution

Let $\frac{dm}{dt}=\lambda$.
At t=0, momentum of the system, P(0)=0.
At some later time t, $P(t)=(M+m-\lambda t)v$ where v is the velocity of car at time t.

When all the sand falls out, $m=\lambda t \Rightarrow t=m/\lambda$. Hence $P(m/\lambda)=Mv$.
$$F=\frac{P(t)-P(0)}{t}=\frac{P(m/\lambda)}{m/\lambda}=\frac{M \lambda v}{m}$$
Therefore, the speed of car is
$$v=\frac{Fm}{M\lambda}$$
Does this look correct?

Any help is appreciated. Thanks!

 

[TSny]

Your expression for $P(t)$ only includes the momentum of the cart and the sand still left in the cart. But the sand that falls out also has horizontal momentum.

 

[Saitama]

Your expression for $P(t)$ only includes the momentum of the cart and the sand still left in the cart. But the sand that falls out also has horizontal momentum.

Is $P(t)=(M+m-\lambda t)v+\lambda tv= (M+m)v$?

 

[TSny]

Is $P(t)=(M+m-\lambda t)v+\lambda tv= (M+m)v$?

No. The particles of sand that fall out earlier have less horizontal velocity than the particles that fall out later.

 

[Saitama]

No. The particles of sand that fall out earlier have less horizontal velocity than the particles that fall out later.

Do you mean I should evaluate P at time t and t+dt?

 

[TSny]

That approach should work.

 

[Saitama]

That approach should work.

$P(t)=(M+m-\lambda t)v+\lambda tv'$

How do I write the expression for P(t+dt)?
The mass that has already fallen off ($\lambda t$) has still the same horizontal velocity but what about the mass which will fall off at t+dt? What about the mass which is still in the car? I need a few hints to make the equations.

 

[TSny]

$P(t)=(M+m-\lambda t)v+\lambda tv'$

How do I write the expression for P(t+dt)?
The mass that has already fallen off ($\lambda t$) has still the same horizontal velocity but what about the mass which will fall off at t+dt? What about the mass which is still in the car? I need a few hints to make the equations.

You are only interested in the change in momentum of the entire system between t and t + dt. The last term in your expression above apparently represents the momentum of the sand that has already fallen out before time t. Is the momentum of that sand going to change between t and t + dt? If not, you can ignore it since it will not contribute to the change in momentum of the system between t and t + dt.

So, ignoring the sand that has fallen out before time t, the momentum of the system at time t is

$P(t)=(M+m-\lambda t)v(t)$

How would you express the momentum at time t + dt ignoring the sand that fell out before time t but including the sand that falls out between t and t + dt?

 

[Saitama]

You are only interested in the change in momentum of the entire system between t and t + dt. The last term in your expression above apparently represents the momentum of the sand that has already fallen out before time t. Is the momentum of that sand going to change between t and t + dt? If not, you can ignore it since it will not contribute to the change in momentum of the system between t and t + dt.

So, ignoring the sand that has fallen out before time t, the momentum of the system at time t is

$P(t)=(M+m-\lambda t)v(t)$

How would you express the momentum at time t + dt ignoring the sand that fell out before time t but including the sand that falls out between t and t + dt?

After time t, the mass of sand remaining in car is $M+m-\lambda t$. At time (t+dt), mass $\lambda dt$ falls off with velocity v(t) i.e
$P(t+dt)=(M+m-\lambda (t+dt))v(t+dt)+\lambda (dt) v(t)$

Correct?

 

[TSny]

After time t, the mass of sand remaining in car is $M+m-\lambda t$. At time (t+dt), mass $\lambda dt$ falls off with velocity v(t) i.e
$P(t+dt)=(M+m-\lambda (t+dt))v(t+dt)+\lambda (dt) v(t)$

Correct?

Yes.

 

[Saitama]

Yes.

Erm...I end up with a really dirty expression for change in momentum.

$$\Delta P=P(t+dt)-P(t)$$
$$=M(v(t+dt)-v(t))+m(v(t+dt)-v(t))+\lambda(v(t)dt-tv(t)-tv(t+dt)-(dt)v(t+dt))$$

 

[TSny]

I believe you have one sign error in the parentheses multiplying the λ.

Can you express v(t+dt) in terms of v(t) and the acceleration a(t)? (to first-order accuracy in dt)

Then see if you can expand the equation for ΔP keeping terms only up to and including first order in dt.

 

[Saitama]

I believe you have one sign error in the parentheses multiplying the λ.

I cannot find the sign error.

Can you express v(t+dt) in terms of v(t) and the acceleration a(t)? (to first-order accuracy in dt)

Taylor series?
$v(t+dt)=v(t)+a(t)dt$. Looks good?

Then see if you can expand the equation for ΔP keeping terms only up to and including first order in dt.

Do you mean this:
$\Delta P=P(t+dt)-P(t)=P(t)+P'(t)dt-P(t)=P'(t)dt$.

 

[TSny]

I cannot find the sign error.

Check the sign of the 2nd term inside the parentheses multiplying λ.

Taylor series?
$v(t+dt)=v(t)+a(t)dt$. Looks good?

Yes, looks good.

Do you mean this:
$\Delta P=P(t+dt)-P(t)=P(t)+P'(t)dt-P(t)=P'(t)dt$.

Yes, that's good. By expanding the expression for ΔP, I meant to expand out the right hand side of the equation for ΔP after substituting $v(t+dt)=v(t)+a(t)dt$.

 

[Saitama]

Check the sign of the 2nd term inside the parentheses multiplying λ.

Honestly, I still cannot see it. :uhh:

I will write down the expressions again.
$P(t)=(M+m-\lambda t)v(t)$
$P(t+dt)=(M+m-\lambda(t+dt))v(t+dt)+\lambda v(t)(dt)$
See, there is only one term in $\lambda$ which has a plus sign.

I think you are right about the sign. When I evaluate $\Delta P$, I get a term $v(t+dt)+v(t)$ which is undesired. I mean it hinders in simplification.

I will reply to other thread later when I return back home. Thank you for your time TSny! :)

 

[haruspex]

When I evaluate $\Delta P$, I get a term $v(t+dt)+v(t)$

I don't see how that happens. It doesn't when I do it.

 

[ehild]

Just call the mass at time t as M(t). From t to t+Δt, the mass changes by -λΔt. The velocity changes from v(t) to v(t)+Δv.

The momentum of the car and load at time t is P(t)=M(t)v. Δt later, the mass of the car is M(t)-λΔt, its velocity is v+Δv, and the mass of the fallen-out sand is λΔt, its (average) velocity is something between v and v+Δv. Let it be v+Δv'

The change of momentum during t and t+Δt is

(M(t)-λΔt)(v+Δv)+λΔt(v+Δv')-Mv=FΔt.

Expanding and simplifying, MΔv+λΔt(Δv'-Δv)=FΔt
The second term on the left-hand side can be ignored, So you get the equation

M(t)Δv=FΔt -->(M+m-λt)dv/dt=F

A linear de, solve.
ehild

 

[Saitama]

I finally found the sign error.

I plugged in v(t+dt)-v(t)=a(t)dt and simplified the expression further. This is what I got,
$$\Delta P=(M+m)a(t)dt+\lambda(a(t)(dt)^2-ta(t)(dt))$$
I think I have to neglect the term with (dt)^2.
Substituting $\Delta P$ with $P'(t)dt=(M+m-\lambda t)a(t)dt \Rightarrow F=(M+m-\lambda t)dv/dt$ which is the same D.E ehild has shown.

Thanks for the help TSny! :)

The momentum of the car and load at time t is P(t)=M(t)v. Δt later, the mass of the car is M(t)-λΔt, its velocity is v+Δv, and the mass of the fallen-out sand is λΔt, its (average) velocity is something between v and v+Δv. Let it be v+Δv'

Why we deal with average velocity? I am sorry if I miss something obvious, I am not too much familiar with variable mass systems.

 

[arildno]

I finally found the sign error.




Why we deal with average velocity? I am sorry if I miss something obvious, I am not too much familiar with variable mass systems.

Perhaps the treatment I made in the following thread will be helpful to you?
https://www.physicsforums.com/showthread.php?t=72176

 

[ehild]

We wrote the equation for conservation of momentum for the time period Δt. The sand that leaves the car at the beginning of the Δt period of time has the velocity v, the amount leaving at the end of that period has v+Δv. The momentum of the spilt sand is included to the momentum at t+Δt, but it does not have an exactly defined velocity, that is why I used v' for the average velocity for the fallen-out sand.
Anyway, the term proportional to the product ΔtΔv is a second-order small quantity, it can be ignored.

ehild

 

[haruspex]

An interesting thread, but I don't understand why the momentum approach is used at all. Is it not straightforward to write down the acceleration equation and integrate twice?

 

[TSny]

An interesting thread, but I don't understand why the momentum approach is used at all. Is it not straightforward to write down the acceleration equation and integrate twice?

I agree. I first worked the problem starting from the acceleration and integrating.

Since Pranav had started with the momentum approach, I felt like that would be a good way to work it also. It allows you to compare this problem with the momentum approach to the standard rocket problem where the exhaust has a nonzero velocity relative to the rocket.

Also, when writing down the expression for the acceleration, it might not be obvious to some students how to handle the sand that's leaking out.

 

[ehild]

An interesting thread, but I don't understand why the momentum approach is used at all. Is it not straightforward to write down the acceleration equation and integrate twice?

I saw teaching that the momentum equation dp/dt=F is more general as the equation F=ma as the former is valid also for changing mass. And it is true when the added or removed mass has no momentum, the mass just increases while the body is accelerated, like in case of relativistic speeds. Otherwise, it must be handled carefully.
If the mass changes by virtue of added or removed mass which brings in or takes away momentum, the problem has to be handled as a collision and then a short acceleration step with the changed mass. If the added/removed mass has velocity u, F=mdv/dt+(v-u)dm/dt results.

Here, the leaked out mass has the same velocity as the cart, u=v, you get the usual equation F=ma at the end for the varying mass of the car.

In case of a falling raindrop which gains mass during its fall, u=0, and you can not use F=ma.
ehild

 

[WannabeNewton]

An interesting thread, but I don't understand why the momentum approach is used at all. Is it not straightforward to write down the acceleration equation and integrate twice?

The problem is taken from the chapter on momentum in Kleppner and Kolenkow so I'm assuming it's being done in the spirit of the chapter.

 

[Saitama]

I saw teaching that the momentum equation dp/dt=F is more general as the equation F=ma as the former is valid also for changing mass. And it is true when the added or removed mass has no momentum, the mass just increases while the body is accelerated, like in case of relativistic speeds. Otherwise, it must be handled carefully.
If the mass changes by virtue of added or removed mass which brings in or takes away momentum, the problem has to be handled as a collision and then a short acceleration step with the changed mass. If the added/removed mass has velocity u, F=mdv/dt+(v-u)dm/dt results.

Here, the leaked out mass has the same velocity as the cart, u=v, you get the usual equation F=ma at the end for the varying mass of the car.

In case of a falling raindrop which gains mass during its fall, u=0, and you can not use F=ma.

There's a similar example to this in my text which is the reason I started with calculating momentum at t and t+dt instead of directly using F=dP/dt. Thanks ehild! :)

An interesting thread, but I don't understand why the momentum approach is used at all. Is it not straightforward to write down the acceleration equation and integrate twice?

I don't get it. Do you mean to write $F=M(t)a(t)$ and plug in $M(t)=(M+m-\lambda t)$ and $a(t)=dv/dt$?

 

[haruspex]

I don't get it. Do you mean to write $F=M(t)a(t)$ and plug in $M(t)=(M+m-\lambda t)$ and $a(t)=dv/dt$?

Yes.

 

[ovicenzu]

You are only interested in the change in momentum of the entire system between t and t + dt. The last term in your expression above apparently represents the momentum of the sand that has already fallen out before time t. Is the momentum of that sand going to change between t and t + dt? If not, you can ignore it since it will not contribute to the change in momentum of the system between t and t + dt.

So, ignoring the sand that has fallen out before time t, the momentum of the system at time t is

$P(t)=(M+m-\lambda t)v(t)$

How would you express the momentum at time t + dt ignoring the sand that fell out before time t but including the sand that falls out between t and t + dt?

Would it be easiest to treat P(t) at t=0 so that P(t) = 0?

 

[ovicenzu]

Yes.

Why is λ(dt) traveling at v(t) and not v(t+dt)?

 

[TSny]

Would it be easiest to treat P(t) at t=0 so that P(t) = 0?

I'm not sure what you're getting at here. Can you expound on this?

 

[TSny]

Why is λ(dt) traveling at v(t) and not v(t+dt)?

It doesn't matter whether you consider λ(dt) as traveling at v(t) or v(t+dt). The momentum would be λ(dt)v(t) in the first case and λ(dt)v(t+dt) in the second case. But these are the same to order dt. We only need to keep terms of first order in dt in the analysis.

 

[ovicenzu]

I'm not sure what you're getting at here. Can you expound on this?

Does the differential, dP, simplify if we look at the the change from P(0) to P(0+dt)?

 

[TSny]

Does the differential, dP, simplify we look at the the change from P(0) to P(0+dt)?

Yes. What do you get for dP from P(0) to P(0+dt)? How would you use this to solve the problem?

 

[ovicenzu]

Yes. What do you get for dP from P(0) to P(0+dt)? How would you use this to solve the problem?

The same result just quicker thereby expediting the solution process. Thank you very much for your help.