I Change in Momentum from mass contribution to the moving object

[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.