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Question: Conservation of Momentum

  1. Sep 12, 2009 #1
    I'm not very confident with my final answer for this question, wondering where I might have gone off track...

    2 spacecraft are thrust apart, what was the original speed of the 2 craft when they were linked together?

    Mass spacecraft 1 = 1.9 x 10^4kg
    Velocity spacecraft 1 after separation is 3.5 x 10^3 km/h @ 5.1º [NE]

    Mass spacecraft 2 = 1.7 x 10^4kg
    Velocity spacecraft 2 after separation is 3.4 x 10^3 km/h @ 5.9º [SE]

    For spacecraft 1
    p = m*v
    p = 1.8 x 10^7 kg.m/s

    For spacecraft 2
    p = m*v
    p = 1.6 x 10^7 kg.m/s

    The first vector is 1.8 x 10^7 kg.m/s @ 5.1 [NE]
    The second vector is 1.6 x 10^7 kg.m/s @ 5.9 [SE]
    The angle between these 2 vectors is 169º.

    Using law of cosines...
    c^2 = (1.8 x 10^7)^2+(1.6 x 10^7)^2 -[2(1.8 x 10^7)(1.6 x 10^7)cos(169)]
    c^2 = 3.24 x 10^14 + 2.56 x 10^14 -[2(2.88 x 10^14)cos(169)]
    c^2 = 5.8 x 10^14 + 5.7 x 10^14
    c^2 = 11.5 x 10^14

    c = 3.4 x 10^7kg.m/s

    Final momentum = Initial momentum
    p = m.v
    and combined mass before separation is 36000kg
    3.4 x 10^7kg.m/s = 36000kg.v

    v = 944m/s
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 15, 2009 #2

    Redbelly98

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    Welcome to Physics Forums :smile:

    I really don't follow your derivation, in particular:

    • Are the angles 5.1° and 5.9° taken from the x axis or the y axis?
    • I don't see the significance of c here. I get that it's supposed to be the difference between the momenta of the two spacecraft, but that isn't useful in any way I'm aware of.

    That being said -- the usual way to solve these problems is to set up two equations, expressing the conservation of momentum in the x and y directions. This will give you the x and y components of the initial velocity.
     
  4. Sep 15, 2009 #3
    Hi, thanks for responding!

    My thinking was...first I'd calculate the total final momentum of each craft after separation.

    Then create a diagram of the vector addition and use Law of Cosines where the resultant would be the "c" variable.

    The arrow representing P after separation of the first craft would be 5.1° below the x-axis.
    The arrow representing P after separation of the second craft would be 5.9° above the x-axis.
    The angle between the 2 vectors is 169°

    Since initial momentum = final momentum, the resultant here will equal total initial momentum which I arrived at 3.4 x 10^7kg.m/s

    And two masses combined before separation is 36000kg

    3.4 x 10^7kg.m/s = 36000kg.v

    v = 944m/s
     
  5. Sep 15, 2009 #4

    Redbelly98

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    Ah, now I understand. Your answer looks pretty good, though there may be some roundoff error. You might try expressing your intermediate calculations in 3 significant figures, then round off to the correct sig figs at the very end.
     
  6. Sep 15, 2009 #5
    OK, Thanks so much!!
     
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