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Conservation of Momentum Question

  1. May 25, 2016 #1
    1. The problem statement, all variables and given/known data
    The mass of a spaceship is 10012 kg. The spaceship is at rest. Then one part of the ship with a mass of 1000 kg is ejected and emerges with a speed of 112 m/s. What is the speed of the other part?

    2. Relevant equations


    3. The attempt at a solution
    I tried:
    pi=pf
    mivi=mfv2
    10012 (vi)= 1000 kg * 112 m/s
    solved for vi and got 11.19 m/s.
    Am I correct? or did i need to assume the other part of the spaceship is 12 kg?
     
  2. jcsd
  3. May 25, 2016 #2
    You have to apply conservation of momentum here, where pf is the sum of the momenta of the two pieces of the ship.
     
  4. May 25, 2016 #3
    so would it be :
    pi=pf
    mivi = m1v1 + m2v2
    (10012 kg)(0 m/s) =m1v1 + m2v2
    0= (1000 kg)(112 m/s)+ (12 kg) (v2)
    v2 = -9333.3 m/s
     
    Last edited: May 25, 2016
  5. May 25, 2016 #4
    That's more like it! Include some directions and you are golden. :)
     
  6. May 25, 2016 #5

    jbriggs444

    User Avatar
    Science Advisor

    Is the original post accurate? What do you get when you subtract one thousand from ten thousand?
     
  7. May 25, 2016 #6
    Good point.
     
  8. May 25, 2016 #7
    oh man... i totally messed up there, thank you so much for the correction! m2 will then = 9012
    (10012 kg)(0 m/s) =m1v1 + m2v2
    0= (1000 kg)(112 m/s)+ (9012 kg) (v2)
    v2 = -12.43 m/s
     
    Last edited: May 25, 2016
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