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While studying systems with varying mass, I have come across examples using thrust, and summing all the forces on the system including thrust, as being equal to $F_{net} = M\bar{a}$.

(My system's definition in an example: If the situation deals with a rocket, my system is just the shell of the rocket + unburnt fuel, and not the ejected mass.)

Newton's Second Law of motion states that the net force acting on any system is equal to the rate of change of linear momentum, i.e., $F_{net} = \frac{d\bar{P}}{dt}$. Since $\bar{P} = m\bar{v}$, by the product rule:

$$F_{net} = \frac{d(M\bar{v})}{dt}= M\frac{d\bar{v}}{dt} + \frac{dM}{dt}\bar{v}.$$

My two questions are given below:

  1. What is the physical meaning of the second term ($\frac{dM}{dt}\bar{v}$)?

  2. Why do we ignore it (with the given definition of my system)?

Qmechanic
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  • We don't ignore the second term. We just make sure we are applying $d(mv)/dt$ correctly. There are a few ways to show that your suggestion leads to motion not observed in nature, i.e. nonsense. Consider a rocket at constant speed shedding mass in two streams diametrically opposed, and perpendicular to the velocity. There is no net force on the rocket, so the velocity will remain constant. But your equation says that the rocket will slow down. – garyp Jul 24 '23 at 17:20
  • Exactly! That's why I was confused. But then, I thought, since P is only the momentum of the rocket as a function of time, dP/dt should give us net force force acting on the rocket only, right? – Science done right Jul 24 '23 at 23:46
  • The question and most of the answers wrongly assume that ${\bf F}_{\rm ext} = d{\bf v}/dt$ is valid for variable mass systems. – Farcher Jul 25 '23 at 08:18

1 Answers1

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The physical meaning of the second term, $\frac{dM}{dt}\bar{v}$, is as follows:

$\frac{dM}{dt}$ is the mass flow rate of the rocket exhaust leaving the rocket engine. In SI units, that would be $\frac{kg}{sec}$. Obviously, the remaining part of that expression is the velocity at which that exhaust is leaving the rocket engine. When these two terms are multiplied together, the resulting units are $\frac{kg m}{s^2}$, which is the thrust contribution from the second term in Newtons.

David White
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