# Quick question on force

1. Nov 28, 2004

### Flinthill84

Here is the problem:

Calculate the force exerted on a rocket, given that the propelling gases are expelled at a rate of 1500kg/s with a speed of 4*10^4 m/s (at the moment of take off)

I know that force is equal to the mass times the change in velocity over the change in time. Can I use a one second interval for the time and just make up a time like that?

2. Nov 28, 2004

### Sirus

Yes, you are on the right track. The gas exerts the same force on the rocket as the rocket exerts on the gas. Find the acceleration of the gas by setting time equal to one second, and displacement equal to $4\times10^{4}$ meters. Then use $F_{net}=ma$.

3. Nov 28, 2004

### Flinthill84

Solving for acceleration won't the acceleration be the same as the velocity in this equation?? I used the formula of Vf=Vo+at

4. Nov 28, 2004

### Andrew Mason

Use:

$$\vec F = \frac{\delta \vec p}{\delta t}$$

In this case:

$$\frac{\delta p}{\delta t} = v \frac{\delta m}{\delta t}$$

Where [itex]\frac{\delta m}{\delta t} = 1500 kg/sec[/tex]

AM

5. Nov 28, 2004

### Flinthill84

Thankyou very much!!!

6. Nov 28, 2004

### Sirus

Oh boy...forget this post. Not sure what I was thinking here...