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

[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 $m=\mathrm{const}.$