Impulse-momentum theorem problem

[Haorong Wu]

Homework Statement

A cart full of sand starts to move by a total force $f$, which is parallel to the cart movement direction. The sand is lost at a rate of $\rho$ per second. What is the velocity of the cart?

Relevant Equations

$fdt=d(mv)=mdv+vdm$

The solution is from $fdt=d(mv)=mdv+vdm$ and separate the variables and then integrate them.

But at first I tried this method. At time $t$, suppose the mass of the cart is $m$, and its velocity is $v$. And suppose at time $t+dt$, its mass will be $m-\rho dt$, and its velocity becomes $v+dv$. Now I use the impulse-momentum theorem, I have $$fdt=(m-\rho dt)(v+dv)+\rho dt v-mv=mdv.$$ This is clearly wrong, but I could not figure out the mistakes.

 

[ergospherical]

Rather, the momentum-balance approach is correct and the provided solution is wrong.

The general equation for (one-dimensional) variable mass systems is
$f + v_{\mathrm{rel}} \dot{m} = m \dot{v}$
The sand is ejected at $v_{\mathrm{rel}} = 0$ and then $f = m\dot{v}$,
in agreement with your result.

 

[hutchphd]

For me this is easiest to see in the case of f=0. Then what the heck does the first "Relevant Equation" say? Nothing good.

 

[haruspex]

Relevant Equations:: $fdt=d(mv)=mdv+vdm$

That equation is for a closed system. If m is just the remaining sand at any instant then that is not closed.

If we try to use the equation with m being the cart plus all the sand it started with then dm=0, but now there is no well-defined v for m: different grains of sand are moving at different speeds.

Instead, we can note that the system m=cart+remaining sand changes momentum both because of the applied force and because of momentum carried away with lost sand: $d(mv)=fdt+vdm$, simplifying to $fdt=mdv$.

 

[phystro]

It seems to me that what you derive is Newton's second law F=Ma. Why not start from there and set M (the mass of the cart) as M(t)= Μ-ρt, you can then integrate to find the velocity. (You need to double check the following calculations as they might be wrong):
https://www.wolframalpha.com/input/?i=int+A/(B-cx)+dx=

 

[Haorong Wu]

Thanks, @ergospherical, @hutchphd, @haruspex, @phystro.

Also, from my equations, $fdt=(m-\rho dt)(v+dv)+\rho dt v-mv=mdv$, in order to obtain the relavent equation, $fdt=d(mv)=mdv+vdm$, it seems the given solution assume that the lost sand suddenly lost its velocity so the term $\rho dt v$ vanishes. This looks strange to me.

Anyway, I decided to tell the student to ignore this problem. It is not worth spending more time in it.

Thanks, again.

 

[PeroK]

Thanks, @ergospherical, @hutchphd, @haruspex, @phystro.

Also, from my equations, $fdt=(m-\rho dt)(v+dv)+\rho dt v-mv=mdv$, in order to obtain the relavent equation, $fdt=d(mv)=mdv+vdm$, it seems the given solution assume that the lost sand suddenly lost its velocity so the term $\rho dt v$ vanishes. This looks strange to me.

Anyway, I decided to tell the student to ignore this problem. It is not worth spending more time in it.

Thanks, again.

You're missing the point that mass cannot be created or destroyed (in Newtonian physics)! The concept of the mass of a "system" changing is fundamentally impossible. Instead, the mass being lost is being transferred out of the system. Therefore, you do not have a closed system and do not have conservation of momentum.

You can, of course, have a sub-system changing its mass, as long as you fully account for the mass that is being transferred out of the sub-system.

That's why a blind application of Newton's second law with a variable mass can go badly wrong.

 

[Haorong Wu]

Thanks, @PeroK. I am not sure which sentence I have written is related to the creation or destruction of mass. Could you point it out?

I understand the mass is changing and the conservation of momentum is not applied here. I modeled that the mass at time $t$ is $m$, and at time $t+dt$, a small fraction $dm$ is dropped, so its momentum will not change while the rest mass will take up some momentum due to the external force $f$. In total, I treat the whole mass $m$ including the dropped part $dm$ as a whole system, and apply the impulse-momentum theorem to it. Is that correct?

 

[ergospherical]

The only thing you really need to keep in mind is that $f=\dot{p}$ applies to a system which is not gaining or losing any matter.

(This usually means extending the system to include the mass which is ejected in a small time interval.)

 

[phystro]

The only thing you really need to keep in mind is that $f=\dot{p}$ applies to a system which is not gaining or losing any matter.

Why not?

 

[ergospherical]

If the system is gaining or losing matter then $\dot p$ has a contribution due to momentum flux across the boundary, in addition to the forcing term $f$, r.e. the Reynolds transport theorem.

 

[phystro]

Well, certainly that's true, that's why you write the momentum as p=m(t)u, and you set m(t)=M-ρt , which gets rid of that contribution.

 

[ergospherical]

It’s not that simple, as this problem demonstrates.

 

[haruspex]

Thanks, @ergospherical, @hutchphd, @haruspex, @phystro.

Also, from my equations, $fdt=(m-\rho dt)(v+dv)+\rho dt v-mv=mdv$, in order to obtain the relavent equation, $fdt=d(mv)=mdv+vdm$, it seems the given solution assume that the lost sand suddenly lost its velocity so the term $\rho dt v$ vanishes. This looks strange to me.

Anyway, I decided to tell the student to ignore this problem. It is not worth spending more time in it.

Thanks, again.

Perhaps @PeroK was misled by your punctuation. I think you meant:

"from my equations, $fdt=(m-\rho dt)(v+dv)+\rho dt v-mv=mdv$."
(Which is correct. Then, new sentence...)

"In order to obtain the equation, $fdt=d(mv)=mdv+vdm$, it seems the given solution assume that the lost sand suddenly lost its velocity so the term $\rho dt v$ vanishes."
(Yes, that seems to be the blunder made.)

 

[PeroK]

Thanks, @PeroK. I am not sure which sentence I have written is related to the creation or destruction of mass. Could you point it out?

My point was simply the one that @ergospherical has made, but I was wanted to emphasise the conceptual side that the lost mass has to go somewhere. Before you dive into any equations, it's important to realize that a closed system cannot lose mass. You can only redefine which units of mass are allocated to which sub-systems.

When we say a rocket loses mass, it's perhaps more precise to say that some mass that was previously considered part of the rocket is no longer considered part of the rocket.

If you understand that conceptually, then you're not going to make mistakes with the mass of an object being a function of time.