# Impulse-momentum theorem problem

• Haorong Wu
In summary: 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.
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.

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.

PeroK
Haorong Wu said:
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##.

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=

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.

Haorong Wu said:
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.

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?

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.)

hutchphd, PeroK and Haorong Wu
ergospherical said:
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?

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.

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.

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

PeroK
Haorong Wu said:
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.)

Haorong Wu
Haorong Wu said:
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.

Haorong Wu

## 1. What is the impulse-momentum theorem?

The impulse-momentum theorem is a fundamental principle in physics that relates the change in momentum of an object to the impulse applied to it. It states that the impulse, which is the product of the force applied to an object and the time it acts, is equal to the change in momentum of the object.

## 2. How is the impulse-momentum theorem used to solve problems?

To solve problems using the impulse-momentum theorem, you need to identify the initial and final momentum of the object, as well as the impulse applied to it. Then, you can use the equation impulse = change in momentum to calculate the unknown quantity.

## 3. What are some real-life applications of the impulse-momentum theorem?

The impulse-momentum theorem is used in many real-life situations, such as car crashes, sports, and rocket propulsion. It helps engineers design safety features in cars by calculating the force and time needed to stop a car in a collision. In sports, it is used to analyze the impact force of athletes and equipment. In rocket propulsion, it is used to determine the amount of force needed to launch a rocket into space.

## 4. Can the impulse-momentum theorem be applied to systems of multiple objects?

Yes, the impulse-momentum theorem can be applied to systems of multiple objects. In this case, the total impulse applied to the system is equal to the total change in momentum of the system. This allows for the analysis of collisions between multiple objects.

## 5. How does the impulse-momentum theorem relate to the conservation of momentum?

The impulse-momentum theorem is closely related to the conservation of momentum, which states that the total momentum of a closed system remains constant. The impulse-momentum theorem can be derived from the conservation of momentum by considering a specific time interval and assuming that the net external force acting on the system is constant during that time interval.

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