# Thread: Momentum of water from a hose

1. ## Momentum of water from a hose

(think this is the right section).

I'm stuck on a question in a physics paper:

Water of density $\displaystyle 1000kg\,m^{-3}$ flows out of a garden hose of cross-sectional area $\displaystyle 7.2\times 10^{-4} m^2$ at a rate of $\displaystyle 2.0 \times 10^{-4} m^3$ per second. How much momentum is carried by the water leaving the hose per second?

I think it may have something to with $\displaystyle F=ma$ and $\displaystyle F \Delta t = \Delta mv$. Maybe somehow using the rate of flow to work out acceleration and density to find mass but I'm clutching at straws there.

Thanks

edit, oh and probably $\displaystyle \rho = \frac{m}{V}$

2. I don't think you need Newton here. You just need your definition of momentum, as well as a little dimensional analysis to solve this problem. Think this way:

$\displaystyle \dfrac{p}{\text{sec}}=\dfrac{m}{\text{sec}}\,v=\le ft(\dfrac{\text{mass of water}}{\text{sec}}\right)(\text{velocity of water})=\left(\rho\,\dfrac{\text{volume of water}}{\text{sec}}\right)\left(\dfrac{\text{volum e of water}}{\text{sec}}\,\dfrac{1}{\text{area}}\right) .$

Do you follow?

3. I'm with you on:
$\displaystyle \frac{mv}{\Delta t} = \frac{\rho V}{\Delta t}v \because m = \rho V$

I'm not really sure what you're doing after that.

Are you saying this in the last bit? $\displaystyle v = \frac{V}{\Delta t} \frac{1}{A}$
Thank you

4. Oh wait, looking at the ratios of the units that does make sense, I just would of never have thought of it.
Think I get it now.
Thanks!

5. You're welcome. Have a good one!