Water Rocket - Rate at which water exits

AI Thread Summary
The discussion revolves around calculating the rate at which water must exit a water rocket to achieve a thrust of 300 N. The key equations used include momentum and force relationships, specifically F = Δp/Δt and the conservation of momentum. The participant struggled with incorporating the necessary variables without a given delta time, which complicated their calculations. Guidance provided helped clarify the relationship between the speed of the water and the mass flow rate, leading to the correct answer of 10.7 kg/s. Ultimately, the assistance enabled a better understanding of the problem-solving process.
NMS09
Messages
2
Reaction score
0

Homework Statement



We're currently studying Momentum & Collisions and Work and Energy

Use algebra – no Calculus

I’d appreciate help in setting it up. I seem to be missing one equation since I keep coming up with two unknown variables.

A water rocket (2-liter half filled with water and pressurized with compressed air) can develop a thrust of 300 N.

Question: At what rate would water have to come out of the rocket to develop that thrust?

The jet of water emerging from the bottle has a diameter of 2.2 cm. (.022 m.).
The water density is 1,000 kg/cu. m.

The answer provided is 10.7 kg/s.


Homework Equations



Equations we’ve been using recently:

Momentum: p = mv

Change of momentum: F = delta p / delta time

Force: F = ma

Force & Impulse: F = mvf - mvi / delta time

Conservation of Momentum: m1v1,i + m2v2,i = m1v1,f + m1v1,f + m2v2,f Note: Those are subscripts after v1 , ve, etc. I need to learn how to use the formulas on this site.



The Attempt at a Solution



I calculated:

The area of the exit hole = 1.52 x 10^-3 m^2
Water in the rocket = 1 kg

I then tried the convservaton of momentum equation but could not get the 300 N into a form to use on one side of the equation. Everything I tried seem to need a "delta t" value, which I don't know and was not given.

The free body diagram has the forward thrust of the rocket created by the rearward explusion of the presurized water. The 300 N must overcome Earth's gravity (-9.81 m / sec^2).

Sorry for not being more organized, but I've spent 3 or 4 hours going around on this one.
 
Physics news on Phys.org
The force depends on the speed of the water being ejected.

Assume that the water comes out at a speed v relative to the container. The rocket thrust is, as you have stated:

F = \Delta p/\Delta t = v\Delta m/\Delta t

where \Delta m is the mass of a volume of water ejected in time \Delta t.

Since the water is being ejected at the speed v, \Delta m is \rho \dV = \rho A\Delta s = \rho Av\Delta t. So the rate of mass expulsion is \Delta m/\Delta t = \rho Av

Work out the expression for force and then find the speed that is needed to reach 300 N. From that you can calculate the rate at which mass has to be ejected.

AM
 
Andrew --

Thank you for your time and that direction. It helped a lot and I was able to work through the problem to get the correct answer.

The steps made sense and helped me understand what was going on.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top