Change in Momentum from mass contribution to the moving object

[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.

 

[DaTario]

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?

Yes, I see. There seems to be something missing. I guess I agree with most of the comments here. I would like to understand in what specific category of problems the equation $$F = \frac d {dt}(mv) = \frac{dm}{dt}v + m\frac {dv}{dt}$$ is used in a direct and well defined way.

 

[A.T.]

Therefore the initial velocity of the added mass is always the same as that of the main system.

Then it's solution 1.

 

[DaTario]

Then it's solution 1.

OK. That is, we are saying that the force has absolutely no participation in the mass gain process and therefore the term containing the time derivative of the mass must not be in Newton's second law, is that it?

* I note that the implementation of this experiment in a laboratory seems to be very artificial, requiring a monitoring system so that a robot can follow the main mass with the same velocity and deliver the water with no relative motion.

** But one can also try a self-consistent methodology, so we created the robot with a timetable $\vec r(t)$ being consistent with one of the two shown in the graphs above (red or green). That one that follows the car most faithfully will be the most correct.

 

[PeroK]

Yes, I see. There seems to be something missing. I guess I agree with most of the comments here. I would like to understand in what specific category of problems the equation $$F = \frac d {dt}(mv) = \frac{dm}{dt}v + m\frac {dv}{dt}$$ is used in a direct and well defined way.

Which part of the Wikipedia page don't you understand?

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

 

[DaTario]

Which part of the Wikipedia page don't you understand?

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

Some pieces are not quite fitting in this puzzle.
On the one hand we have the very popular definition of the resultant force concept and Newton's second law, arranging the resultant force as a time derivative of linear momentum. On this same side we see a term containing the derivative of the mass multiplied by the instantaneous velocity of the studied system.

On the other hand, you and others claim that this equation is wrong, and there is also this Wikipedia page (I don't know who wrote it and who endorsed it) that puts a relative velocity that is equivalent to adding a $-u \cdot dm$ term to original equation.

Okay, I understand that relative velocity comes up when we deal with the same phenomenon by looking at it as a collision (or explosion, or fission) problem.

Perhaps what resolves this understanding problem is to clarify which physics problem is solved by adopting the equation
$$F_{ext} = \frac{dm}{dt}v + m\frac {dv}{dt}$$.
A kind of jeopardy (which question has this answer...).

 

[PeroK]

A kind of jeopardy (which question has this answer...).

The equation is wrong. A simple analysis of momentum entering and leaving the system shows you that. Any able student should be able to figure out for themselves the equation is wrong - despite its "popularity".

 

[DaTario]

The equation is wrong. A simple analysis of momentum entering and leaving the system shows you that. Any able student should be able to figure out for themselves the equation is wrong - despite its "popularity".

I am convinced that the equation is wrong in some situations we discussed here. Do you know which physics problem is solved by adopting the equation
$$F_{ext} = \frac{dm}{dt}v + m\frac {dv}{dt}$$?

 

[DaTario]

Do you agree with A.T. in #52 on the problem I posed in #47?

 

[PeroK]

I am convinced that the equation is wrong in some situations we discussed here. Do you know which physics problem is solved by adopting the equation
$$F_{ext} = \frac{dm}{dt}v + m\frac {dv}{dt}$$?

That equation is frame-dependant and, therefore, not part of standard Newtonian physics, with the usual definition of force.

 

[A.T.]

OK. That is, we are saying that the force has absolutely no participation in the mass gain process..

In this particular case, with no relative velocity, no force is needed to accelerate the added mass to the velocity of the cart on addition.

 

[DaTario]

In this particular case, with no relative velocity, no force is needed to accelerate the added mass to the velocity of the cart on addition.

until the moment the element of mass enters the cart. After the addition this element of mass $dm$ will produce its effect on the way the whole system reponds to the constant force.

 

[jedishrfu]

It seems that now is a good time to close this thread. The OPs question has been answered and not much else can be added.