# [stumped]Momentum: Force exerted on a rocket

1. Oct 18, 2009

### Senjai

1. The problem statement, all variables and given/known data

Calculate the force exerted on a rocket, given that the propelling gases are expelled at a rate of 1000 kg/s with a speed of 60,000 m/s (at takeoff)

2. Relevant equations
$$p=mv$$
$$F\Delta{t} = \Delta{p}$$

3. The attempt at a solution

This is probably the first question im posting where i don't reall know how to start. Im given a ratio of 1000 kg/s which represents no unit that i know of. Momentum is kg x m/s. I dont know what the initial velocity is given for. Presuming for the force at the exact time of takeoff..

I'd appreciate a couple hints..

Thanks,
Senjai

2. Oct 18, 2009

### Pengwuino

In regards to your first equation, can you convince yourself that the change in momentum is equal to the change in mass if the velocity this mass is being expelled at is constant?That is, $$\Delta p = \Delta mv$$?

Now if you can convince yourself that is true, if you divide both sides by $$\Delta t$$, what result do you have?

3. Oct 18, 2009

### Senjai

I can't, i dont have a Delta t, yes the velocity the mass is being expelled from the rocket is constant, but wouldn't mass be inversely proportional to velocity in this case? so momentum wouldn't change? Uh.. i think i just said it myself.. are you treating 1000 kg as the mass?

4. Oct 19, 2009

### Senjai

*bump*

5. Oct 19, 2009

### cepheid

Staff Emeritus
1000 kg/s is, as the unit suggests, the rate at which mass is being expelled (with time). hence, kilograms PER second. In other words, it is the mass flow rate.

Δm/Δt = 1000 kg/s = const. ​

Which means that EACH second, a mass of Δm = 1000 kg is expelled at speed v = 60 000 m/s, thus carrying away with it an amount of momentum equal to Δm*v.

Conservation of momentum therefore dictates that EACH second, the rocket gains the same amount of momentum Δm*v in the opposite direction.

So now you know the change in momentum over a time interval of Δt = 1s. This means that you know the rate of change of momentum (= force).

6. Oct 19, 2009