Dynamics question- finding m in terms of rho, r and v

[Boogeyman]

Homework Statement

A toy rocket consists of a plastic bottle which is partially filled with water. the space above the water contains compressed air.

At one instant during the flight of the rocket, water is of density\rho is forced through the nozzle of radius r at speed v relative to the nozzle. Determine, in terms of \rho, r and v:

1. the mass of the water ejected per unit time from the nozzle

2. the rate of change of momentum of the water.

Homework Equations

F=ma
\rho=mass/volume
Area of nozzle through which water passes=\pir²

The Attempt at a Solution

This is what I did for part 1:
\rho=m/volume
therefore, m=\rho x volume

RE: area through which water exits:\pir²
therefore, m=(\pir²)(\rho) x length

I think the mass per unit time means divide by time?

m/t=[(\pir²)(\rho) x length]/t

I do not know how to do part 2..

 

[redargon]

Ok, well for number 1, i think you approached it a little too complicated. When the want mass per unit time, that is the mass flowrate. Volume per unit time is the volumetric flowrate. Mass flow rate is often used for gasses in rockets for example, because even if the density changes, the mass doesn't, whereas, the volume will change with changing density.

That aside,
m=rho*V
Q=Av (Volumetric flowrate is Area of nozzle times velocity of fluid, v, with a constant density)
mass flowrate, \dot{m}=rho*Q or rho*A*v
So mass flowrate, \dot{m}=\frac{dm}{dt}=rho*\pir²*v
where \frac{dm}{dt} is change in mass with time, or mass flowrate.

 

[redargon]

For part two, you need the relevant equations for momentum, what are they?

 

[Boogeyman]

Thanks, I got it.