Calculating Momentum of a Spacecraft with Thruster Firing?

  • Thread starter jamiescience
  • Start date
Just check your final answer to make sure the units are correct. In summary, a 1.0*10^4kg spacecraft traveling at a speed of 1200 m/s relative to earth has a thruster firing for 2.0 minutes with a continuous force of 25kN in the opposite direction of the spacecraft's motion. The initial and final momentum of the spacecraft can be calculated using the equation: Impulse = Change in Momentum. After converting the time to seconds, the equation becomes: (-25*10^3)(120) = (1.0*10^4)(mvi) - (1.0*10^4)(1200). Simplifying, the equation becomes (-3*10^
  • #1
jamiescience
8
0

Homework Statement



a 1.0*10^4kg spacecraft is traveling through space with a speed of 1200 m/s realtive to earth. a thruster fires for 2.0 min, exerting a continuous force of 25kN on the spacecraft in a direction opposite the spacecraft s motion . calculate the initial momentum and the fianl momentum kof the spacecraft .



Homework Equations





The Attempt at a Solution

 
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  • #2
You need to show work here in order to get help. Those are the rules. What have you tried so far? What are you thoughts on the problem?
 
  • #3
(-25*10^3)(2.0)=(1.0x10^4)(1200)+(1.0*10^4)(mvi)
 
  • #4
OK. Your big problem is with units. You have seconds on one side of the equation and minutes on the other side of the equation.

Also, remember the equation is:

Impulse =Change in Momentum.
[tex]J=\Delta p=p_f-p_i[/tex]

Check your signs on the RHS of your equation.
 
Last edited:
  • #5
ok so it would be 120s instead of 2.0min
 
  • #6
(-25*10^3)(120)= (1.0*10^4)(mvi)-(1.0x10^4)(1200)
 
  • #7
jamiescience said:
(-25*10^3)(120)= (1.0*10^4)(mvi)-(1.0x10^4)(1200)

That looks good to me.
 

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