I Change in Momentum from mass contribution to the moving object

[DaTario]

TL;DR Summary

How to use Newton's second law to treat differently the two cases below in which a system has its mass increased?

Hi All,

in the figure below we see two cases in which a wheelbarrow moves with speed v, containing a certain mass of water. At a given time t0 a drop of water is added to the system by a dripper. In case 1, the drop enters the system, which increases in mass, having the same speed as the cart. In case 2 the drop is at rest relative to the ground on which the cart moves. Considering Newton's second law expressed in terms of the derivative of linear momentum, how do we treat each case to understand the respective consequences of the increase in mass?
Obs.: consider possible existence of an external and constant force acting on this system.

Best wishes

 

[PeroK]

You need to show us your best attempt at this problem before we can help.

 

[DaTario]

You need to show us your best attempt at this problem before we can help.

Hi Perok, I have just assembled these figures (plane, dripper, etc) to mount the exercise. I have a general doubt concerning the way mass enters or leaves a system. It seems that elements of mass can enter a system with some initial velocity, for instance. My question has to do with this. How do we manipulate the formalism to make it manifest the entrance of mass with or without some given initial velocity. Specifically the formalism implied by the second law, with the derivative of the linear momentum included.

 

[PeroK]

Hi Perok, I have just assembled these figures (plane, dripper, etc) to mount the exercise. I have a general doubt concerning the way mass enters or leaves a system. It seems that elements of mass can enter a system with some initial velocity, for instance. My question has to do with this. How do we manipulate the formalism to make it manifest the entrance of mass with or without some given initial velocity.

Let me make a general observation. All over the Internet, including in some otherwise reputable sources, you will see Newton's second law expressed as:$$F = \frac d {dt}(mv) = \frac{dm}{dt}v + m\frac {dv}{dt}$$This is not correct. The changing mass, which can only be achieved by mass physically entering or leaving the system. This problem may have been set to illustrate the problem with that equation.

 

[DaTario]

I would like you to believe that I have just mounted this problem, based on a doubt I have had for years. In which sense you say that this very famous way to express Newton's second law is not correct?

 

[PeroK]

I would like you to believe that I have just mounted this problem, based on a doubt I have had for years. In which sense you say that this very famous way to express Newton's second law is not correct?

Wikipedia has a good analysis of this:

https://en.m.wikipedia.org/wiki/Variable-mass_system

 

[DaTario]

Thank you. It seems to be a nice material. Is it correct to say that the concept associated with the verb 'accrete' is somewhat related to the concepts involved in inelastic collisions?

 

[Ibix]

In this context, one could say that your droplet of water is accreting to your cart, yes, because the two things join together.

In my experience it is not a word that is used except as a technical term in some scientific contexts.

 

[DaTario]

Thank you, Ibix and wrobel. I confess now I am interested in Perok's explanation on why presenting the often called 'the more complete and comprehensive version of Newton's second law' is not correct.

Why is it not correct to tell students that $\vec F_{ext} = \frac{d}{dt}\vec p$ ?

 

[PeroK]

Thank you, Ibix and wrobel. I confess now I am interested in Perok's explanation on why presenting the often called 'the more complete and comprehensive version of Newton's second law' is not correct.

There are several threads on here dealing with this. The main issue is that you then have to define a force in cases where there is no external force using the normal definition $F = ma$.

As an extreme example, consider a large number of balls with a blue light inside. When the light is switched off, the ball is white.

Now, we consider the momentum of the system of blue balls and the system of white balls. Each time a light is switched on or off the mass of the two systems change, and by the erroneous definition of force each system exerts a force on the other. But, with the usual definition, there is no force applied to any ball nor to either system.

 

[DaTario]

Are you taking into consideration the light as a content of energy that can be translated into mass content by Einstein's equation?

 

[DaTario]

In the context of Einstein's Relativity this equation seems to be useful when we try to explain why a constant force does not make a particle reach velocities higher than c.

 

[Ibix]

Are you taking into consideration the light as a content of energy that can be translated into mass content by Einstein's equation?

No. He's just arguing that there is a system of white balls and a system of blue balls. The mass of one system is $m$ and the mass of the other system is 0, and he can choose which system has the non-zero mass by turning a light on or off and hence changing the balls' colour. Thus he is changing the systems' boundaries and moving mass between them without actually moving anything, just changing the light. Changing the masses of the systems like that would imply that the system of blue balls (which doesn't even really exist when the light is off) exerts a force on the system of white balls if you take $\frac{d}{dt}(mv)$ as your definition of force.

 

[Ibix]

In the context of Einstein's Relativity this equation seems to be useful when we try to explain why a constant force does not make a particle reach velocities higher than c.

No - the so-called "relativistic mass" dropped out of favour decades ago because it causes even more confusion. It clings on in popular science versions of relativity, unfortunately.

Mass (the concept that used to be called "rest mass") is an invariant in relativity. It only changes for the same reasons it does in Newtonian physics - some bit of a system was removed and put somewhere else (possibly in some other form). Relativity just gives you more options for how that can happen and what forms it can take and the extra complexity that masses aren't additive.

 

[DaTario]

Ok, I got it. But in this case we would be just redefining the frontiers of our system. Interesting this purely 'linguistic' way to use the formalism so that a phenomenon seems to have occurred.

 

[Ibix]

Ok, I got it. But in this case we would be just redefining the frontiers of our system. Interesting this purely 'linguistic' way to use the formalism so that a phenomenon seems to have occurred.

I'd say "conceptual" rather than "linguistic". There's nothing mathematically wrong with defining a quantity $\frac{d}{dt}(mv)$. It's just that you can construct experiments (like the balls and lights one) where that quantity does not behave at all like our "if you kick something it starts to move, if you don't it doesn't" concept of force, so it's probably better not to call it force.

 

[DaTario]

No - the so-called "relativistic mass" dropped out of favour decades ago because it causes even more confusion. It clings on in popular science versions of relativity, unfortunately.

Mass (the concept that used to be called "rest mass") is an invariant in relativity. It only changes for the same reasons it does in Newtonian physics - some bit of a system was removed and put somewhere else (possibly in some other form). Relativity just gives you more options for how that can happen and what forms it can take and the extra complexity that masses aren't additive.

Ok, I have heard and read about this rejection of the relativistic mass increase, but formally the relativistic factor is still there multiplying the rest mass. I would also expect gravitational effects related to such 'mass increase' to take place when the particle accelerates. Of course one could also explain that in terms of the energy density at the position of the particle. Although I will not argue in favour of the variable mass, I will just note that the equation $F = \frac{dm}{dt}v + m\frac {dv}{dt}$ had the advantage of showing two possible consequences of the application of a force on a particle: 1) to produce a variation in its velocity and 2) to produce a variation in its mass.

 

[A.T.]

Ok, I got it. But in this case we would be just redefining the frontiers of our system.

Just like you can arbitrarily define which drops become part of the wagon, you also arbitrarily define which balls change from one system to the other.

 

[DaTario]

I'd say "conceptual" rather than "linguistic". There's nothing mathematically wrong with defining a quantity $\frac{d}{dt}(mv)$. It's just that you can construct experiments (like the balls and lights one) where that quantity does not behave at all like our "if you kick something it starts to move, if you don't it doesn't" concept of force, so it's probably better not to call it force.

This example of PeroK reminds me a bit of a notion related to the Einstein's relativity. If two very tall trees fall as in the figure below, their interception can eventually travel faster than light, which does not violate Einstein's postulates.

 

[A.T.]

Why is it not correct to tell students that $\vec F_{ext} = \frac{d}{dt}\vec p$ ?

In principle you can call any term with the right dimensions a 'force' (see inertial forces), but conceptually there is difference to forces that for example can cause deformation.

 

[Ibix]

formally the relativistic factor is still there multiplying the rest mass

Sometimes. Sometimes it's $\gamma^3$, or somewhere in between the two. Or just something different. So the problem of keeping track of which "relativistic mass" you should use in which circumstance is one reason to dislike the concept.

I would also expect gravitational effects related to such 'mass increase' to take place when the particle accelerates.

Not even close, I'm afraid. You need the sixteen components of the stress-energy tensor to describe the source of gravity in relativity. And people making the assumption you did is one of the reasons relativistic mass was deprecated.

2) to produce a variation in its mass.

But applying a force doesn't change the mass. Not in Newtonian physics nor relativistic physics.

And I think relativity is way off topic here - perhaps we should stick to Newtonian dynamics.

 

[DaTario]

Not even close, I'm afraid. You need the sixteen components of the stress-energy tensor to describe the source of gravity in relativity. And people making the assumption you did is one of the reasons relativistic mass was deprecated.

And I think relativity is way off topic here - perhaps we should stick to Newtonian dynamics.

Perhaps you are right with respect to the gravitational events I alluded. With respect to stick to Newtonian physics I also agree. But most of the texts that are in favour of this more general second law equation defend this position based on the fact that in the context of relativity, this equation remains valid, while $F_{ext} = m \cdot a$ don't.

 

[PeroK]

I don't know how relativity came into the discussion. Surely the idea of an object changing colour is within the realms of classical physics.

You could have a football player running around and turning his shirt inside out to change its colour!

The other issue with the erroneous definition of Newton's second law is that force is no longer invariant across inertial reference frames:

$$F = \frac d {dt}(mv) = \frac{dm}{dt}v + m\frac {dv}{dt}$$

Note that the $v$, being the instantaneous velocity of the changing mass, is a frame dependent quantity. Hence force becomes frame dependent.

One important point about $F= ma$ is that it is independent of the choice of inertial reference frame.

 

[DaTario]

I don't know how relativity came into the discussion. Surely the idea of an object changing colour is within the realms of classical physics.

You could have a football player running around and turning his shirt inside out to change its colour!

The other issue with the erroneous definition of Newton's second law is that force is no longer invariant across inertial reference frames:

Note that the $v$, being the instantaneous velocity of the changing mass, is a frame dependent quantity. Hence force becomes frame dependent.

One important point about $F= ma$ is that it is independent of the choice of inertial reference frame.

Ok for this three observations. But I think you agree that the last one is just a disadvantage, not an intrinsic error of this definition. I would sumarize my doubts asking you: Is the resultant force on a particle the rate of change of the linear momentum of this particle?

 

[PeroK]

Ok for this three observations. But I think you agree that the last one is just a disadvantage, not an intrinsic error of this definition.

I would say it is an intrinsic error because with that definition you are no longer doing Newtonian physics. Unless that was the intention, then it's wrong.

I would sumarize my doubts asking you: Is the resultant force on a particle the rate of change of the linear momentum of this particle?

Yes. Because under Newtonian physics a particle cannot change its mass.

 

[Juanda]

That took a strange turn into photons and relativity. Just to add what I think is a simpler example with classical physics of why this is not a force.
$$\frac d {dt}(mv) = \frac{dm}{dt}v + m\frac {dv}{dt} \neq F$$

Imagine a rocket moving at a nonzero speed that's expelling fuel symmetrically all around itself. We know the rocket would keep moving at a constant speed because of the conservation of momentum (fuel expelled one way is counteracted by its symmetric equivalent). However, according to the previous equation, the rocket should be under a force that would accelerate it because the mass of the system is changing.

To solve the conundrum the easiest thing is to consider the rocket hull to be a system with constant mass and the fuel inside a different body that will apply a force on the rocket when it accelerates because it needs to work on the fuel's inertia that's still in the rocket too.

As @PeroK said, in Newtonian physics a particle is not allowed to change its mass so the change in linear momentum will always be
$$\frac d {dt}(mv) = m\frac {dv}{dt} = F$$

 

[DaTario]

Thanks also to A.T. and Juanda. Is there any textbook on Newtonial physics that presents this simple but impactful idea: 'in Newtonian physics a particle is not allowed to change its mass' ?

and a possible continuation could well be the following: 'it also applies to the relativity of Einstein.'

I don't remember to have seen this ideia in Halliday & Resnick & Walker, Alonso & Finn, Serway & Jewit, Tipler & Mosca, Keller & Gettys & Skove, Feynman & Leighton & Sands and Paul Hewitt among few others. I honestly think the defense of this idea here by some of you was very convincing. It somewhat reminds me a paper on Am. J. of Phys. (if I can remember well) in which the authors defend the idea that an additional zero-th law of Newton is necessary for the students. "A force ceases to produce acceleration at the very moment it ceases to be applied".

 

[PeroK]

Thanks also to A.T. and Juanda. Is there any textbook on Newtonial physics that presents this simple but impactful idea: 'in Newtonian physics a particle is not allowed to change its mass' ?

Newtonian physics is built on three conservation laws (for a closed system): conservation of mass, momentum and energy.

A corollary of which is that a particle cannot change its mass. Note that, in particular, it cannot "decay".

and a possible continuation could well be the following: 'it also applies to the relativity of Einstein.'

In SR, these three conservation laws are unified in the conservation of the energy-momentum (four-vector). The energy being the zeroth (time) component, momentum the three spatial components; and the mass being its magnitude - which is also an invariant. This is represented by one of the most important equations in physics:$$E^2 = p^2c^2 + m^2c^4$$Where $m$ is the rest mass of particle. This is invariant, but not conserved. In particular, a particle can decay into two or more particles with less total rest mass.

 

[DaTario]

This is represented by one of the most important equations in physics:$$E^2 = p^2c^2 + m^2c^4$$Where $m$ is the rest mass of particle. This is invariant, but not conserved.

Just as an attempt at a little provocation, doesn't this equation for energy in SR also involve a frame-dependent term?

 

[Juanda]

Thanks also to A.T. and Juanda. Is there any textbook on Newtonial physics that presents this simple but impactful idea: 'in Newtonian physics a particle is not allowed to change its mass' ?

I wish I could. Here in this forum, there are people way more capable of explaining the theoretical background and providing renowned sources. In fact, I feel you have a deeper background than myself already from all the books you mentioned.
I'm just a ME but I happen to have battled against this particular concept in the past because I couldn't make sense of it during my time in university. If you allow $\frac{dm}{dt}v$ to contribute to the force then you run into situations that are not possible as the one I described with the rocket expelling fuel symmetrically.
This of course applies to classical mechanics. How things work when Quantum Physics and Relativity are involved is beyond me so I'm gonna sit this one out and read your conversation that looks interesting.

 

[PeroK]

Just as an attempt at a little provocation, doesn't this equation for energy in SR also involve a frame-dependent term?

We can rearrange the equation to get:
$$m^2c^4 = E^2 - p^2c^2$$The invariance of $m$ then depends on the invariance of the RHS, which we recognise as the magnitude squared of the four-vector $(E, \vec p c)$ in Minkowski space. Proving this quantity is invariant under the Lorentz Transformation is a good exercise for the student.

 

[DaTario]

I wish I could. Here in this forum, there are people way more capable of explaining the theoretical background and providing renowned sources. In fact, I feel you have a deeper background than myself already from all the books you mentioned.
I'm just a ME but I happen to have battled against this particular concept in the past because I couldn't make sense of it during my time in university. If you allow $\frac{dm}{dt}v$ to contribute to the force then you run into situations that are not possible as the one I described with the rocket expelling fuel symmetrically.
This of course applies to classical mechanics. How things work when Quantum Physics and Relativity are involved is beyond me so I'm gonna sit this one out and read your conversation that looks interesting.

I have never discussed this topic in this depth. But regarding the symmetric expelling, I think may be we are neglecting something important as, for instance, the vectorial nature of this equation. The spatial distribution of the mass may have some consequences when taking into account its time derivative (as a kind of continuity equation, which will relate the time derivative of the mass with some mass current,...)

 

[DaTario]

We can rearrange the equation to get:
$$m^2c^4 = E^2 - p^2c^2$$The invariance of $m$ then depends on the invariance of the RHS, which we recognise as the magnitude squared of the four-vector $(E, \vec p c)$ in Minkowski space. Proving this quantity is invariant under the Lorentz Transformation is a good exercise for the student.

Ok, but it seems that we can use this argument also in the Newtonian equation: $$F = \frac{dm}{dt}v + m\frac {dv}{dt} $$ so that it becomes $$m\frac {dv}{dt} = F - \frac{dm}{dt}v.$$

 

[PeroK]

Ok, but it seems that we can use this argument also in the Newtonian equation: $$F = \frac{dm}{dt}v + m\frac {dv}{dt} $$ so that it becomes $$m\frac {dv}{dt} = F - \frac{dm}{dt}v.$$

Sorry, that's doesn't work at all. Only $v$ is frame dependent. Or, alternatively, you have to make $F$ frame dependent.

 

[DaTario]

why the term $\frac{dm}{dt}$ does not depend on the reference frame, since it is somehow associated with a continuity equation that brings with it mathematical elements such as divergence of a stream of matter?

I am deliberately taking risks in this argument

 

[PeroK]

why the term $\frac{dm}{dt}$ does not depend on the reference frame, since it is somehow associated with a continuity equation that brings with it mathematical elements such as divergence of a stream of matter?

$m$ and $t$ are frame invariant, hence so is the time derivative of $m$.

I am deliberately taking risks in this argument

That's not how I would describe it!

In any case, this is my last post on the subject.

 

[DaTario]

As my last post too (I think I should not be trying here to invent new physics...) consider the figure below. Although the rate of change of mass seems to be the same in both frames of reference (typical S and S' comparison), the dynamical effects (taking those collision aspects we mentioned in the begining) seems to introduce an unbalance and therefore a proper force in the system. The square is the big mass and the circles represent small masses that are leaving the system.
My point here is that the rate of variation of the mass with respect to time, mainly in Newtonian physics, may be more complex than we think and may involve spatial aspects somewhat related to a continuity equation and collision processes.

[Updated figure attached by a Mentor (vector u in lower left now shows as a vector)]

 

[kuruman]

I will try one more time to say the same thing that has been said before hoping that this time it will stick. Your original question was "How to use Newton's second law to treat differently the two cases below in which a system has its mass increased?" If I look at the "two cases below", I see no system that has its mass increased. Specifically in case 2 all I see is a drop with zero horizontal velocity and a cart with some water in it moving with horizontal velocity $V$. This is the "before" picture. In the "after" picture you have the drop added t the water in the cart and the ensemble moving with common velocity $V_{\text{after}}.$

Now, there could be 3 choices of system here:

1. The drop that has mass $m_{\text{drop}}$
2. The cart with the water before the drop falls into it that has mass $m_{\text{cart}}$
3. The cart + the drop that has mass $m_{\text{cart}}+m_{\text{drop}}.$

Before you apply Newton's second law, you have to declare clearly and unambiguously your choice of system. If I look at the choices above, I see no system that has its mass changed. There is momentum transfer between the drop and the cart. In order to apply Newton's 2nd law in this case, you must have a consistent choice of system throughout the momentum transfer process.

To put it bluntly, if you say that you choose system 2 (cart) before the drop falls into it and system 3 (cart + drop) after the drop falls into it, you cannot expect to use Newton's 2nd law to say anything meaningful. Changing systems midway through the calculation is analogous to changing the origin of coordinates midway through a derivation of equations of motion.

Here is how one would apply Newton's laws to case 2.

The system is drop + cart
The external horizontal force acting on the system is $~F_{\text{net,x}}=0.$ The momentum of the system is $$P_{\text{sys,x}}=m_{\text{drop}}*v_{\text{drop}}+m_{\text{cart}}*v_{\text{cart}}.$$Applying $~\dfrac{dP_{\text{sys,x}}}{dt}=F_{\text{net,x}}~$, $$0=\Delta P_{\text{sys,x}}=m_{\text{drop}}*\Delta v_{\text{drop}}+m_{\text{cart}}*\Delta v_{\text{cart}}.$$ The last expression gives the final velocity of the system $$V_{\text{drop+cart}}=\frac{m_{\text{cart}}V}{m_{\text{cart}}+m_{\text{drop}}}.$$The system is one of cart or drop
As noted above there is no external horizontal force acting on the system of both the drop and the cart. This means that $$0=\frac{dP_{\text{(cart+drop)}}}{dt}=\frac{dP_{\text{cart}}}{dt}+\frac{dP_{\text{drop}}}{dt}\implies m_{\text{cart}}\frac{dv_{\text{cart}}}{dt}=-m_{\text{drop}}\frac{dv_{\text{drop}}}{dt}$$ This says
(a) if the drop is the system, the net horizontal force acting on it is $F_{\text{on drop}}=m_{\text{cart}}\dfrac{dv_{\text{cart}}}{dt}.$

(b) if the cart is the system, the net horizontal force acting on it is $F_{\text{on cart}}=-m_{\text{drop}}\dfrac{dv_{\text{drop}}}{dt}.$

Note that the net horizontal external forces acting on each system are Newton'w 3rd law counterparts.

 

[DaTario]

I will try one more time to say the same thing that has been said before hoping that this time it will stick. Your original question was "How to use Newton's second law to treat differently the two cases below in which a system has its mass increased?" If I look at the "two cases below", I see no system that has its mass increased. Specifically in case 2 all I see is a drop with zero horizontal velocity and a cart with some water in it moving with horizontal velocity $V$. This is the "before" picture. In the "after" picture you have the drop added t the water in the cart and the ensemble moving with common velocity $V_{\text{after}}.$

Now, there could be 3 choices of system here:

1. The drop that has mass $m_{\text{drop}}$
2. The cart with the water before the drop falls into it that has mass $m_{\text{cart}}$
3. The cart + the drop that has mass $m_{\text{cart}}+m_{\text{drop}}.$

Before you apply Newton's second law, you have to declare clearly and unambiguously your choice of system. If I look at the choices above, I see no system that has its mass changed. There is momentum transfer between the drop and the cart. In order to apply Newton's 2nd law in this case, you must have a consistent choice of system throughout the momentum transfer process.

To put it bluntly, if you say that you choose system 2 (cart) before the drop falls into it and system 3 (cart + drop) after the drop falls into it, you cannot expect to use Newton's 2nd law to say anything meaningful. Changing systems midway through the calculation is analogous to changing the origin of coordinates midway through a derivation of equations of motion.

Here is how one would apply Newton's laws to case 2.

The system is drop + cart
The external horizontal force acting on the system is $~F_{\text{net,x}}=0.$ The momentum of the system is $$P_{\text{sys,x}}=m_{\text{drop}}*v_{\text{drop}}+m_{\text{cart}}*v_{\text{cart}}.$$Applying $~\dfrac{dP_{\text{sys,x}}}{dt}=F_{\text{net,x}}~$, $$0=\Delta P_{\text{sys,x}}=m_{\text{drop}}*\Delta v_{\text{drop}}+m_{\text{cart}}*\Delta v_{\text{cart}}.$$ The last expression gives the final velocity of the system $$V_{\text{drop+cart}}=\frac{m_{\text{cart}}V}{m_{\text{cart}}+m_{\text{drop}}}.$$The system is one of cart or drop
As noted above there is no external horizontal force acting on the system of both the drop and the cart. This means that $$0=\frac{dP_{\text{(cart+drop)}}}{dt}=\frac{dP_{\text{cart}}}{dt}+\frac{dP_{\text{drop}}}{dt}\implies m_{\text{cart}}\frac{dv_{\text{cart}}}{dt}=-m_{\text{drop}}\frac{dv_{\text{drop}}}{dt}$$ This says
(a) if the drop is the system, the net horizontal force acting on it is $F_{\text{on drop}}=m_{\text{cart}}\dfrac{dv_{\text{cart}}}{dt}.$

(b) if the cart is the system, the net horizontal force acting on it is $F_{\text{on cart}}=-m_{\text{drop}}\dfrac{dv_{\text{drop}}}{dt}.$

Note that the net horizontal external forces acting on each system are Newton'w 3rd law counterparts.

Thank you, kuruman. Ok, but you recognize this process as accretion (a system gaining mass) according to the reference given by PeroK, don't you?

 

[kuruman]

Ok, but you recognize this process as accretion (a system gaining mass) according to the reference given by PeroK, don't you?

What's in a name? I'd rather ignore mass accretion because it doesn't add anything new in my opinion. The quoted Wikipedia article derives an "accretion" expression using the same reasoning I used in post #38 under the heading "The system is drop + cart". Say the system consists of all the water that could conceivably drop in the cart plus the empty cart. Since the drops are added one at a time it makes no difference to the form of the derived equation whether 10 drops or 20 drops have already fallen in.

The Wikipedia article says, "In mechanics, a variable-mass system is a collection of matter whose mass varies with time. It can be confusing to try to apply Newton's second law of motion directly to such a system." I don't see what is so confusing about applying Newton's second law if one starts with a closed system on which no external force is acting. There is no doubt in anyones's mind that the total momentum change of such a system is zero. That's the starting point. Now, if one wishes to divide this isolated system into two subsystems, 1 and 2, and wants to figure out its acceleration, one can easily apply (without the alleged confusion) Newton's second law.

First one writes $~0= \dfrac{d(P_1+P_2)}{dt}=\dfrac{dP_1}{dt}+\dfrac{dP_2}{dt}\implies \dfrac{dP_1}{dt}=-\dfrac{dP_2}{dt}.$

Next one declares "Subsystem 1 is the cart with whatever water has collected in it and subsystem 2 is the next drop that is about to fall into it. We ignore all the other drops because they do not interact with each other or with either subsystem." Then one writes
$\dfrac{dP_1}{dt}=m_{\text{cart}}\dfrac{dv_{\text{cart}}}{dt}~$ and $~F_{\text{net on 1}}=-\dfrac{dP_2}{dt}.$

Note specifically that there is no $~v_{\text{cart}}\dfrac{dm_{\text{cart}}}{dt}~$ term because the mass of the cart is not changing while the drop is in the air. Thinking of the cart as accreting mass might be a temptation to add such a term.

To finish the derivation, we write $~m_{\text{cart}}d v_{\text{cart}}=-m_{\text{drop}}d v_{\text{drop}}$

We note that $d v_{\text{drop}}=v_{\text{cart}}$, redefine $v_{\text{cart}}\equiv V$, $m_{\text{cart}}\equiv m$ which makes $m_{\text{drop}}=d m$ and write
$md V=-Vd m.$ Divide by $dt$ and you get the "accretion" equation in the Wikipedia article with $\mathbf{u}=0$, which is justified here because the drops have zero horizontal velocity.

I am satisfied with this method which I find straightforward and not at all confusing. Why bother with accretion when, by choosing an appropriate system of unchanging mass, one can simply consider momentum transfer from one part of the system to another?

 

[DaTario]

I am satisfied with this method which I find straightforward and not at all confusing. Why bother with accretion when, by choosing an appropriate system of unchanging mass, one can simply consider momentum transfer from one part of the system to another?

Ok, I think the message is clear. But what about those classical exercises where a system has a mass given by $m(t) = 5 + 2t$ and is being subjected to a force of $20N$? The exercise then asks for the acceleration when $t = 6 s$. Is it the correct way to solve the differential equation: $$ \frac{dv}{dt} + \frac{2v}{5 + 2t} - \frac{20}{5 + 2t} = 0 $$ ?

Which case in the OP is this exercise most similar to?

 

[kuruman]

Ok, I think the message is clear. But what about those classical exercises where a system has a mass given by $m(t) = 5 + 2t$ and is being subjected to a force of $20N$? The exercise then asks for the acceleration when $t = 6 s$. Is it the correct way to solve the differential equation: $$ \frac{dv}{dt} + \frac{2v}{5 + 2t} - \frac{20}{5 + 2t} = 0 $$ ?

Which case in the OP is this exercise most similar to?

What about those exercises? To apply Newton's second law to a growing mass subjected to a constant force, one must know exactly how the mass is added. It can't just grow. You need to describe the physical situation. I strongly suspect that this is a contrived mathematical exercise masquerading as physics because there are no units in the expression for the time-dependent mass. In any case, assuming that the mass is in kg when $t$ is in seconds and that Newton's second law applies in this case, the question can be answered simply by saying, "At $t = 6~$s the mass is $m=5+12=17~$kg. By Newton's second law, the acceleration is $a=F/m=20/17=1.2~$m/s²."

This is a plug and chug exercise not similar to anything in the OP. If you are itching to solve a differential equation related to it, solve $$(5+2t)\frac{dv}{dt}=20$$ to find $v(t)$ subject to the initial condition of your choice.

 

[DaTario]

Thank you for your attention and for your time, kuruman. I'm sorry for the lack of units. In fact it was an exercise elaborated yesterday, in a short space of time and in the heat of the discussion. Your assumption that these are values in the international system aligns with how I would complete the statement of the question. As for the initial velocity, we could assume that at $t =0$, $v = 0$.

IMHO the above exercise has physical sense if it is completed with a description similar to case 1 of OP (in which the incoming mass arrives with the same instantaneous velocity of the body). It is interesting to see that in your solution the term $\frac{dm}{dt} v$ present in $F = \frac{dp}{dt}$ was completely ignored.

It seems that you align with the same understanding as PeroK and Juanda, namely that there is no rate of change of mass with time in Newtonian mechanics. I must say that seeing this behavior of yours is very instructive for me. I saw very few times in my physics course systems in classical mechanics where it would be reasonable to assume that there is mass variation (for example, the problem of having a bag of powdered salt with a hole through which the powder flows, while acts on the bag a constant force, or also problems with rockets).

 

[kuruman]

I too have said all I have to say and remove myself from this thread.

 

[DaTario]

I too have said all I have to say and remove myself from this thread.

No problem. Thanks, anyway.

 

[DaTario]

I have just found a paper (Revista Brasileira de Ensino de Física -- Brazilian Journal of Physics Teaching) in which the author discusses this subject in some detail. May be of use for those interested in this topic.

Newton’s Second Law of a particle with variable mass
https://www.scielo.br/j/rbef/a/GfKTvx3fH5X6SwpgzdWd6tn/?lang=en&format=pdf

One interesting conclusion of this paper is "that it is not possible to mimic the dynamics of a particle with variable intrinsic mass by studying the temporal evolution of composite bodies whose mass varies over time."

 

[DaTario]

Just to make this discussion more complete, I will put down here the analysis of the exercise that I proposed above solved in the two ways that we consider possible, so that we can examine the difference between the two solutions.

A system, initially at rest, has a mass given by $m(t)=5+2t$, expressed in kilograms, and is being subjected to a force of $20 N$. The exercise then asks for the velocity and the acceleration as a function of $t$.

Solution 1 (red graph)
$$ \frac{v(t)}{dt} = \frac{20}{5 + 2t}$$
with $v(0) = 0$.
$$ \Rightarrow v(t) = -10 ( \log(5) - \log(5 + 2t) $$
$$\Rightarrow a(t) = \frac{20}{5+2t} $$

Solution 2 (green graph)
$$ \frac{v(t)}{dt} = \frac{20}{5 + 2t} - \frac{2v}{5 + 2t}$$
with $v(0) = 0$.
$$ \Rightarrow v(t) = \frac{20t}{5+2t} $$
$$\Rightarrow a(t) = -\frac{40 t}{(5+2t)^2} + \frac{20}{5+2t} $$

This yields the following results.

Now it remains to be seen which of these two results is more in line with what is obtained in the laboratory in a setting similar to that shown in case 1 of OP (see below).

 

[A.T.]

A system, initially at rest, has a mass given by $m(t)=5+2t$, expressed in kilograms, and is being subjected to a force of $20 N$. The exercise then asks for the velocity and the acceleration as a function of $t$.

The problem statement seems incomplete without specifying the initial velocity of the added mass.

 

[ergospherical]

If $F= (d/dt)(mv) = m \dot{v} + v\dot{m}$ were true for $\dot{m} \neq 0$, then rockets ejecting mass at a fixed rate $\dot{m}$ would have the same acceleration regardless of how fast the mass shoots out the back (i.e. the speed of ejection does not even appear in this equation). You see why that’s a bit silly?

 

[DaTario]

The problem statement seems incomplete without specifying the initial velocity of the added mass.

I understand your preoccupation. But I have informed that the experimental setup is similar to what is shown in the last figure of the post. Therefore the initial velocity of the added mass is always the same as that of the main system.