# Mass propelled backwards in 1 second?

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1. Apr 8, 2016

### ultralightbeam

1. The problem statement, all variables and given/known data
Cross sectional area: 4.5x10^-2 m
Speed: 3.5ms-1
Density of sea water: 1030 kg m-3

Calculate the mass of water propelled backwards in 1 s.
2. Relevant equations

3. The attempt at a solution
Could someone just guide me through it, maybe give me a starting point like which equation to use...Im unsure how to do this?

My thought was that its related to the power equation?

2. Apr 8, 2016

### Staff: Mentor

Cross sectional area of what? Speed of what?

You need to provide a clear description of the problem.

3. Apr 8, 2016

### ultralightbeam

Sorry, first post on here.

The motor on a small boat has a propeller attached that sends back a column of warer of cross-sectional area 4.5x10^-2 m^2 at a speed of 3.5ms. Density of seawater: 1030 kgm3
water is originally at rest

Last edited: Apr 8, 2016
4. Apr 8, 2016

### SophiaSimon

What is the context of the problem?

5. Apr 8, 2016

6. Apr 8, 2016

### SophiaSimon

:) Just slow.

First, how much distance does any cross-sectional area of water cover in 1s if it is traveling at $3.5\frac{m}{s}$?

7. Apr 8, 2016

### Staff: Mentor

That's gotta be a typo...

8. Apr 8, 2016

### ultralightbeam

Yes it is, sorry im pretty clumsy on the keyboard.

4.5x10^-2

9. Apr 8, 2016

### Staff: Mentor

Make yourself a drawing. Imagine a tube of water with your given cross sectional area moving at velocity v. Find the amount of water that moves past a given "starting line" in one second:

10. Apr 8, 2016

### ultralightbeam

So would that just be the cross-sectional area multiplied by 3.5 since the velocity is 3.5 metres per second? Giving a value of 0.1575m^3

11. Apr 8, 2016

### Staff: Mentor

Don't forget to include the time (1 s). It's important to make the units come out right. You should always check that the units come out right!

How are you going to convert that volume of water to a mass value?

12. Apr 8, 2016

### ultralightbeam

Would you divide it by the time so 1second? Since the question asks the mass in 1 second?

13. Apr 8, 2016

### Staff: Mentor

Nope. Distance is velocity x time. The length of the tube of water passing by is d = vt.

14. Apr 8, 2016

### SteamKing

Staff Emeritus
According to the OP, the density of seawater is 1030 kg / m3

You're not dividing by time here. The velocity of the flow is 3.5 m/s. The flow rate Q = A × v, and you know A and v.

15. Apr 8, 2016

### ultralightbeam

Not familiar with that equation:/

16. Apr 8, 2016

### SteamKing

Staff Emeritus
You should be. It's the standard equation of continuity. It's used all the time in pipe flow and such.

The mass flow rate is a modified form: $\dot{m} = ρ A v$

17. Apr 8, 2016

### ultralightbeam

The relevenat equation I know is density = mass /volume

I think writing that out gave me a brainwave..would you multiply the volume by the density, giving an answer of 162.2kg?

18. Apr 8, 2016

### SteamKing

Staff Emeritus
The quantity A ⋅ v is the volumetric flow rate, in cubic meters per second.

Multiplying the volumetric flow rate by the density of the fluid in kg / m3 gives the mass flow rate.
This is basic fluid mechanics stuff.

19. Apr 8, 2016

### SophiaSimon

Imagine there is a thin disk of water. The area of this disk is $A=4.5x10^{-2}m^2$. A force is going to be exerted on it, so that it moves in a straight line through the water around it. Because a net force was exerted on the disk of water, it changed its momentum and gained a velocity of $3.5\frac{m}{s}$.

As the disk of water moves, it's going to bump into all the water particles around it. You can find the distance the disk travels using the constant velocity equation. Let's call this distance $l$. You want to find how much mass the disk of water bumped into as it traveled the distance $l$. Let's take the disk to travel through a cylinder whose faces have an area of $A=4.5x10^{-2}m^2$ and whose length is $l$. What is the mass of water in this cylinder that the disk of water bumped into?

You aren't given the mass of water, but the mass density of water, $1030 \frac{kg}{m^{3}}$. This is how much water there is per unit volume. You can find the total mass of water in the volume of the cylinder by using the equation for mass density.