Confusion about Newton's third law The question:

A toy rocket consists of a container of water and compressed air.Water is pushed vertically downwards through a nozzle by the compressed air. The rocket moves vertically upwards. The nozzle has a circular cross-section of radius 7.5mm. The density of the water is 1000kgm–3. Assume that the water leaving the nozzle has the shape of a cylinder of radius 7.5mm and has a constant speed of 13ms–1 relative to the rocket. Given t the mass of water leaving the nozzle in the first 0.20s after the rocket launch
is 0.46kg, find the force exerted on the this mass of water by the rocket.

My Problem:
My confusion is that will the force calculated using the formula "change in momentum over time" only be the force exerted by the rocket because the weight of the water is being balanced by some other force, as suggested by the constant speed statement or is there another reason? I thought this formula gives the resultant force, so if my logic is incorrect wouldn't the weight have to be subtracted from the answer to get the action reaction force?
 A: In this simple model for a rocket of mass M you have a constant mass flow $\dot{M}$ (i.e. the mass of the water that leaves the rocket per second) away from the rocket which you can calculate with the data given in the exercise. The matter that leaves the rocket is assumed to have a velocity $v$ relative to the rocket. This leads to a resultant force
$F = \dot{M} v$
I think that the concerns about the mass of the rocket that you are referring to in the end of your question is only relevant when you are interested in the resulting accelaration. For this you will have to integrate:
$M(t) = M(0) + \dot{M} t $
and then calculate
$a(t) = F(t)/M(t) = \frac{\dot{M}}{  (M(0) + \dot{M} t )} v = \frac{v}{ M(0)/\dot{M} + t}$
A: "I thought this formula gives the resultant force" You are correct. The question is how much of the momentum gain can be attributed to the water's weight and (trickier) the weight of how much water?
Consider this...  Suppose the top of the rocket were cut off leaving the water exposed to the atmosphere (rather than to compressed air). Water will still flow out of the nozzle, but at a speed $v_0$ (say). So if the nozzle area is $A$ and the density of water is $\rho$, the mass flow rate out of the nozzle will be $\rho A v_0$, and the rate of gain of momentum of the water will be
$$\frac{dp}{dt}=\rho A v_0^2=2\rho gh\ A$$
The last step follows from Torricelli's theorem: assuming that the magnitude of the rocket's acceleration is much less than $g$, then if the depth of water in the rocket is $h$,
$$v_0^2=2gh.$$
Let $h$ be 0.10 m – ample for a toy rocket, I'd have thought. Then $v_0=1.40 \ \text{m s}^{-1}$
and the rate of gain of momentum due to this essentially gravitational effect is
$$\left(\frac{dp}{dt}\right)_{gr}=2\rho gh\ A=2\times 1000\ \text{kg m}^{-3} \times 9.81\ \text{N kg}^{-1} \times 0.10\ \text m \times \pi (0.0075\ \text m)^2=0.35\ \text N $$
But we are told that the water actually emerges at $13 \ \text{m s}^{-1}$ and at a rate of 0.46 kg in 0.20 s. So the actual rate of gain of momentum of the water is
$$\frac{dp}{dt}=\frac{0.46\ \text{kg}}{0.20\ \text s} \times 13 \ \text{m s}^{-1}=30\ \text N$$
I would therefore attribute almost all the 30N to the force from the compressed air and only about 1 % of it to weight, if the depth of water in the rocket were 0.10 m, and the rocket's acceleration were much less than $g$.
A: You need to be careful when conserving momentum in a system with nonzero dm/dt. The naïve interpretation of
$F = dP/dt$
might be: "$P = mv$ therefore $dP/dt = vdm/dt + mdv/dt$"
But this is not a reasonable physical description of a system with nonzero $dm/dt$. If we conserve mass, it is clear that for the system to be losing mass, the mass that is leaving the system must be going somewhere else, and therefore it can't possibly be traveling at velocity v. So multiplying $dm/dt$ by $v$ doesn't describe any physically meaningful quantity.
Couldn't find a proof so here's mine:
If all the mass leaving the system at a given time is traveling at the same velocity $v_2$, we have
Let $m_u+m_v=m$
Let $m_u$ be mass leaving the system, which is traveling at $v_2$
Let $m_v$ be mass staying with the system, which is traveling at $v$
Note: $dm_v/dt + dm_u/dt = 0$
$F = dP/dt = \frac {d}{dt} (vm_v + v_2m_u) = [m_v\frac {d}{dt}v + m_u\frac {d}{dt}v_2] + [v\frac {d}{dt}m_v + v_2 \frac {d}{dt}m_u]$
$ = [m_v\frac {d}{dt}v + m_u\frac {d}{dt}v_2] + [(v_2 - v)\frac {d}{dt}m_u]$
For a rocket with an exhaust at constant velocity $u$ relative to the rocket, this can be greatly simplified using
Let $v_2-v = u$
then $\frac {d}{dt}v = \frac {d}{dt}v_2$
$F = [m\frac {d}{dt}v] + [u\frac {d}{dt}m_u]$
So, for a rocket in free space ($F_{external} = 0$)
$m\frac {d}{dt}v = -u\frac {d}{dt}m_u$

A note on signs for sanity checking:
$\frac {d}{dt}m_u$ is positive because mass is leaving the system
$u$ is the direction the exhaust is going relative to the rocket, so negative for a positive-direction acceleration
$m$ is positive, obviously
