# Homework Help: Need help with Momentum Principle problem!

1. Jan 27, 2012

### fillipeano

1. The problem statement, all variables and given/known data
Suppose you are navigating a spacecraft far from other objects. The mass of the spacecraft is 3.0 x 10^4 kg (about 30 tons). The rocket engines are shut off, and you're coasting along with a constant velocity of ‹ 0, 22, 0 › km/s. As you pass the location ‹ 7, 7, 0 › km you briefly fire side thruster rockets, so that your spacecraft experiences a net force of ‹ 8.0 x 10^5, 0, 0 › N for 22.5 s. The ejected gases have a mass that is small compared to the mass of the spacecraft. You then continue coasting with the rocket engines turned off. Where are you an hour later? (Think about what approximations or simplifying assumptions you made in your analysis. Also think about the choice of system: what are the surroundings that exert external forces on your system?)

2. Relevant equations
Δp = ƩFΔt
rFinal = rInitial + (pFinal/mass)Δt

3. The attempt at a solution
The hint I was given said this: Apply one step of the Momentum Principle, then one step of the position update equation using the new velocity just after the short burn of the thruster rockets. At this low speed the momentum is approximately m.

pFinal = <18.0 x 10^6, 6.6 x 10^5, 0>
so my final position equation looks like this:
<7,7,0> + <2160000,79200,0>

2. Jan 27, 2012

### Staff: Mentor

Recheck the calculation of your pFinal, paying particular attention to the powers of ten. Could be that you've been bitten by a unit conversion somewhere along the line.

3. Jan 27, 2012

### fillipeano

I checked, my answer is still wrong ):

4. Jan 27, 2012

### Staff: Mentor

Well, let's go through your calculation step by step. Suppose you start by presenting your calculation for the change in momentum, Δp.

5. Jan 27, 2012

### fillipeano

Okay, for delta p I did force x time, I get 18000000

6. Jan 27, 2012

### Staff: Mentor

Okay. That's 1.8 x 107 and it looks good. You should specify that it is a change in the x-component of the momentum (Momentum is a vector quantity). The units, of course, will be kg*m/s.

You can calculate the new velocity as $v_1 = v_o + \frac{\Delta p}{M}$
where v1, vo and Δp are all vectors. What do you get for the components of v1?

7. Jan 27, 2012

### fillipeano

Converting 22km/s into m/s I get 22000 m/s. Delta p/m = 600.

So, <0,22000,0> m/s + 600 kg m/s = <0,22600,0> kg m/s

Last edited: Jan 27, 2012
8. Jan 27, 2012

### Staff: Mentor

The force applied has only an x-component, so the Δp should have only an x-component, and the thus the velocity change should also be to the x-component. You've added the velocity change to the y-component.

9. Jan 27, 2012

### fillipeano

<600,22000,0> for velocity.

10. Jan 27, 2012

### Staff: Mentor

That's better . Now, how about the position after 1 hour? Keep in mind that the initial position is specified in km (and likely the final answer is supposed to be in km units also).

11. Jan 27, 2012

### fillipeano

Do I use rInitial + (pFinal/mass)*delta t?

12. Jan 27, 2012

### Staff: Mentor

rf = ri + v*t

Just like in 1D kinematics, only in this case the variables (except for time) are vectors.

You calculated v above. t and ri are given in the problem statement.

Last edited: Jan 27, 2012
13. Jan 27, 2012

### fillipeano

I'll convert my initial position to meters since the answer requires it in meters
rfinal = <7000,7000,0> + <600,22000,0>*3600s = <2167000, 29000, 0>

14. Jan 27, 2012

### Staff: Mentor

Your y-component is not correct; you failed to multiply the y-velocity component by the time.

15. Jan 27, 2012

### fillipeano

rfinal = <2167000,79207000,0>

16. Jan 27, 2012

### Staff: Mentor

Okay, that looks good. Be sure to include the units on any results that you submit!

17. Jan 27, 2012

### fillipeano

Okay, thank you so much for your help! My problem was that I was not using rFinal = rInitial + v*t correctly!