Why is $F = ma$ at an instant, in variable mass systems? [duplicate]

Question

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)?

Answer 1

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.