1. The problem statement, all variables and given/known data A long wagon with mass m0 and inital velocity v0 on a frictionless floor. On time t=0 we open a faucet which starts dripping water in a given pace Q=Δm\Δt=α. 1. What's the speed of the wagon on time t. 2. Answer 1. if on t=0 the wagon is full of water and it's mass is M0 2. Relevant equations F=dp\dt 3. The attempt at a solution It was actually an example shown in a recorded lecture course I'm taking. It's a part of a set of examples about changing mass systems. The answer to 1. was: F=(dm\dt)v=(-αv) (because α is given and the professor said that the minus sign is because of the diretion of the force acting on the wagon). F=ma=m(dv\dt) (newton's second law) ⇒ -αv=m(dv\dt) m = m0 + αt (wagon's initial mass + change) ⇒dv\v=(-αdt)\(m0+αt) integrate both sides between v0 and v and 0 to t and get: v=(m0v0)\(m0+αt) I kind of understand what's going on here, but not really. Why F=(dm\dt)v instead of deriving both v and m using the chain rule [(dm\dt)v + (dv\dt)m)]? both are changing, no? the wagon gains mass but also loses velocity to the water drops it accelerates. The answer to 2. was: The wagon's mass is now constant (M0), but there's still a change in the momentum in order to accelerate the new water drops which fall down on it (the excess water fall of the wagon): F=(dm\dt)v=(-αv) F=ma=M0(dv\dt) -αv=M0(dv\dt) integrate both sides ang get: V=V0e-αt\M0 I don't understand why the following equation: F=(dm\dt)v=(-αv) is valid in this situtation. dm\dt = 0 cause the mass of the wagon is constant, no? the only thing that changes is the velocity of the water drops which I'm not sure how to calculate (the wagon accelerates the water drops but how can I know the magnitude? well, it doesn't matter cause it seems it's not the right answer anyway). The main thing I want to understand is the logic behind all of this. I guess I don't have a good enough grasp of this concept. Thanks and let me know if what I wrote down here is clear enough!