# Momentum with changing mass

1. Sep 15, 2008

### PhyStan7

1. The problem statement, all variables and given/known data
Hi, our teacher set this in class. I think its above what we have to know for the syllabus and have tryed to work it out but i dont think im doing it right. I just want a few pointers, not the answer!

ok...

A rocket with mass 5000kg of which 4000kg is propellant is launched vertically. The fuel is consumed at a rate of 50kg per second. What is the least velocity of the exhaust gases if it is to release the rocket immediately after firing.

2. Relevant equations

F=ma
Suvats
Momentum=mv
F=Change in mv/t

3. The attempt at a solution

Ok i experimented a bit with it but had too many unknowns to really make much of it. I saw you could model the rocket and gas given off as 2 different particles acting in opposite directions. The momentum of the rocket equaling the momentum of the gas given off. This would mean the velocity of the gas cloud would be much greater than the velocity of the rocket at the 1 second mark.
I multiplied the mass of the rocket (and propellant) after 1 second by g to get 48510. I figured this is the weight of the rocket which acts downwards. The force provided by the gas cloud pushing the rocket upwards must exceed the weight of the rocket downwards for the rocket to move up. The mass of gas released is 50kg so i divided the minimum force needed by the amount of gas to get the acceleration which was 970.2.

Im pretty sure the working ive done is wrong as i haven't used the formula for momentum. I tryed expressing the velocity of the gas with the velocity of the rocket in the momentum formula and played around with substituting in stuff but havnt got anywhere.

I dont think the working can be too complicated, i just think there is something im missing. Any hints or pointers would be appreciated to get me on the right track. Thanks

2. Sep 15, 2008

### Topher925

Is that word for word the problem statement? If I understand it correctly you are looking for the minimum velocity of the propellant gas in order for the rocket to lift off.

If that is the case, you only need f = ma. However you may prefer it in the form:

F = dp/dt = m2v2 - m1v1

3. Sep 15, 2008

### LowlyPion

I think you are right. I misread the actual question. That is a much simpler calculation for lift off since you are not worried about mass loss.

4. Sep 16, 2008

### PhyStan7

Ok ive had another look at it, thanks for the help. Is this right? At t=1 the rocket will have mass 5000-50kg, 4950kg. g downwards is 9.81 so the total force downwards is 48559.5

By modeling the gas cloud as a particle, it will have to exert an upward force of over 48599.5 for the rocket to move vertically.

F= change in mv/t and as F must equal 48599.5 (and t=1) F/m=v. Tge mass of the gas cloud is 50kg so the velocity of the cloud = 971.2 m/s.

It seems logical to me but i dont think the answer seems right. Any pointers would be appreciated, again i want to try and find the answer myself. Thanks

5. Sep 16, 2008

### LowlyPion

Let U = the combustion velocity of the gas

$$F = M \frac{dv}{dt} = \frac{dm}{dt}*U = M*g$$

At lift off then you must have sufficient velocity of the combusted gas to achieve a minimum of M*g

Actually it is a little more complicated because M is not exactly M, because as you noted it is M - dm/dt. So I suppose it's a matter of whether lift off is at T=0 or T=1.

6. Sep 16, 2008

### Topher925

Your going to need a little calculus if you want to go this route. Your solution for the velocity of the gas will then not be constant but a function of time. For the question, are you sure it is not asking for the velocity at a time instant of 0? The other route seems a bit to complicated for the principles you are learning.

If you want to go the complicated route, then to get started

dmrocket/dt = - dmgas/dt

Thrust = $$\int$$$$\int$$ $$\ddot{m}$$$$\dot{V}$$ dm dv

*sorry if its hard to read, I suck at using latex

7. Sep 16, 2008

### PhyStan7

Yeh im not sure about the time. I didnt think it was t=0 as i thought no gas would have been released. But im sure which ever it is as long as ive done the right method my teacher wont mind. Thanks for the help guys!

8. Sep 16, 2008

### LowlyPion

$$Thrust = \int \int \ddot{m} \dot{V} dm dv$$

Better?