Why not use the product rule to expand Newton's 2nd law?

[mridul]

Its as simple as this .actually mass of the body does not change with time so it is a constant .Now the constant comes out of the factor that's why.sorry for bad eng

 

[Stephen Tashi]

This note is given in the current Wikipedia article on "Newton's laws of motion":

Halliday; Resnick. Physics. 1. p. 199. ISBN 0-471-03710-9. It is important to note that we cannot derive a general expression for Newton's second law for variable mass systems by treating the mass in F = dP/dt = d(Mv) as a variable. [...] We can use F = dP/dt to analyze variable mass systems only if we apply it to an entire system of constant mass having parts among which there is an interchange of mass.

However, don't we put a similar restriction on applying any equation of classical physics?

Technically a "mass" is not "an object". "Mass" is a property of an object (or system) and, as far as I can see, in classsical mechanics, "object" and "system" have no rigorous definition. They are left as terms of common speech.

Our habit is to use the phrase "a mass" to designate an object. If that is to be an unambiguous designation, when "a mass" breaks into 5 pieces, all those 5 pieces must be considered part of the designated object and designated mass.

I we allow ourselves to define "a system" in a way not related to mass, we could make definitions like "System B: consists of all things having a mass that are in my bedroom". By that definition, a person who walked out of the bedroom would no longer be part of System B. So there would be no conservation of mass in System B. Likewise there need be no conservation of momentum in System B.

So the restriction that Halliday and Resnick state: "only if we apply it to an entire system of constant mass" must also be enforced on other equations of mechanics that apply to "a system".

It's interesting to ask whether "conservation of mass" (in classical physics) is merely a definition! - i.e. mass is conserved "in a system" because we require that "a system" be specified in such a way that it always includes a constant amount of mass".

 

[conscience]

@haruspex , do you mind giving your views on this topic .

 

[haruspex]

@haruspex , do you mind giving your views on this topic .

If you insist.
$\vec F=\frac{\vec{dp}}{dt}$ is correct, where $\vec p$ is the momentum of a system of particles and $\vec F$ is the net external force acting on the system.
$\frac{\vec{dp}}{dt}=m\frac{\vec {dv}}{dt}+\vec v\frac{dm}{dt}$ is correct, but normally we would take the mass of a particle to be invariant. A nonzero dm/dt term means mass is (particles are) being added to or removed from the system represented by $\vec p$.
If you consider a rocket plus its fuel (used or unused) as the system, there is no external force and no net change in momentum.
If you consider the rocket as one system and all its fuel as another, neither is changing mass, so for each there is only the m dv/dt term, and this corresponds to the force each exerts on the other.
If you consider the rocket plus unused fuel as one system and spent fuel as the other, $\vec F=\frac{\vec{dp}}{dt}$ no longer applies, because we are no longer considering a consistent system of particles. We can illustrate this with a simple experiment. I throw a ball to which there is a particle of dirt attached. The particle falls off in flight. If we define a system as "the ball plus anything currently attached to it", the momentum suddenly dropped, yet there is no evident force.

A possible approach is to extend the concept of force to include the rate of momentum change resulting from addition or removal of mass.

 

[vanhees71]

I usually don't like forces, because they are a murky concept to me. The true laws of nature are all describable with the action principle, but here, I'd say it's pretty simple to see at the rocket example that Newton's law holds true at least in some cases of systems with changing mass. The force on the rocket is the recoil of the streaming gas on the rocket, and of course the product rule is correct. It's a mathematical theorem and has nothing to do with physics except that you can use it in physic-related calculations ;-). Also in your ball example you can use this concept: When the dust particle falls of it gains momentum relative to the ball and thus the ball also gets a recoil, which is a force acting on it (in addition to other forces obviously involved here). Of course, for the entire closed system of ball+dust particle momentum is conserved, and you can derive the equation of motion of the ball and the dust particle also by using momentum conservation, as in the example with the rocket.

 

[haruspex]

When the dust particle falls of it gains momentum relative to the ball and thus the ball also gets a recoil,

You appear to have misunderstood my example. Perhaps I should have written "becomes detached from". The dust particle, as such, does not gain momentum from the ball. It keeps exactly the momentum it had, as does the ball. They are simply no longer joined. The problem comes from defining the system as "ball plus whatever is attached". When the particle becomes detached, this system loses momentum, yet no forces were involved.
I believe the same issue applies to the rocket if you try to define the system as rocket plus unused fuel.

 

[PeroK]

An extreme example of what @haruspex is talking about:

You are driving along in "your car", which has a trailer attached, which is considered part of "your car". A passenger is on his/her mobile phone and tells you that the trailer has just been sold on ebay, so it's no longer yours. "Your car", therefore, no longer includes the trailer and has lost mass. Analyse the change in momentum and forces acting on "your car" during the ebay transaction.

 

[vanhees71]

You appear to have misunderstood my example. Perhaps I should have written "becomes detached from". The dust particle, as such, does not gain momentum from the ball. It keeps exactly the momentum it had, as does the ball. They are simply no longer joined. The problem comes from defining the system as "ball plus whatever is attached". When the particle becomes detached, this system loses momentum, yet no forces were involved.
I believe the same issue applies to the rocket if you try to define the system as rocket plus unused fuel.

Well, that doesn't make sense. Either the particle changes it's momentum relative to the ball or it still is comoving with the ball. You cannot loose momentum without a force, and total momentum is conserved. For the rocket, I've shown both approaches in this thread above. For this purpose you need to consider only the center of mass of the fuel and of the rocket as an effective description of the closed system, and total momentum is conserved, which leads to the equation of motion ("rocket equation") for the rocket.

 

[jbriggs444]

Well, that doesn't make sense. Either the particle changes it's momentum relative to the ball or it still is comoving with the ball. You cannot loose momentum without a force, and total momentum is conserved. For the rocket, I've shown both approaches in this thread above. For this purpose you need to consider only the center of mass of the fuel and of the rocket as an effective description of the closed system, and total momentum is conserved, which leads to the equation of motion ("rocket equation") for the rocket.

One can gain or lose momentum from a system by re-drawing the boundaries around the system. Of course, that means that the system is not closed. That is the point.

 

[haruspex]

Well, that doesn't make sense. Either the particle changes it's momentum relative to the ball or it still is comoving with the ball. You cannot loose momentum without a force, and total momentum is conserved. For the rocket, I've shown both approaches in this thread above. For this purpose you need to consider only the center of mass of the fuel and of the rocket as an effective description of the closed system, and total momentum is conserved, which leads to the equation of motion ("rocket equation") for the rocket.

That's what this thread is all about, that if you apply the standard equations
1. F=ma
2. F=dp/dt
3. dp/dt=m dv/dt+v dm/dt
to the 'system' consisting of rocket plus unburnt fuel, something goes wrong.
Some authorities argue that (3) is false because, for Newton, mass does not change. Another way of saying that is that (3) should only be applied to materially closed systems. I am arguing that (3) is always valid, but if the dm/dt term is nonzero then the system is not materially closed, which makes (1) invalid, because loss or gain of mass does provide a way to change momentum without a force.

No-one disputes that there are ways of viewing the system which work with all those equations; the question is what goes wrong with rocket+unburnt fuel view.

Note my references to "materially closed". Technically, a closed system would have no external forces either, but (1) can handle those. Hence the last sentence of my post #34, suggesting that an extension of the notion of external force would allow (1) to extend to systems not materially closed.

 

[vanhees71]

Have a look at

Sommerfeld, Lectures on theoretical physics, Vol. 1

There's a short chapter on the question discussed here. Of course 2 is correct by definition of what a force is (combination of Newton's Leges I+II), and 3 is a mathematical identity. 1. is only true for $m=\mathrm{const}.$