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Relative velocity and conservation of linear momentum

  1. Sep 11, 2014 #1
    1. The problem statement, all variables and given/known data
    The last stage of a rocket, which is travelling at a speed of 7600m/s, consists of two parts that are clamped together: a rocket case with a mass of 290.0 kg and a payload capsule with a mass of 150.0kg. When the clamp is released, a compressed spring causes the two parts to separate with a relative speed of 910.0m/s. What are the speeds of
    (a) the rocket case and
    (b) the payload after they have separated? Assume that all velocities are along the same line.


    2. Relevant equations

    m1V1i + m2V2i = m1V1f + m2V2f

    Vac = Vpc + Vap
    Subscripts being ac=case velocity according to A, pc=relative velocity between case and payload, ap=velocity of payload according to A

    3. The attempt at a solution
    Howdy folks, I hope I haven't botched the equation writing part, being a newcomer here.

    I took the perspective of an observer, A, and used the Galilean transformation for velocity which resulted in

    Vac = 910 + Vap

    Then with the conservation of momentum:

    150Vap + 290Vac = 440(7600), 440kg being the total mass.

    Solving for the case's velocity gave an answer of 7910m/s with the correct answer being 8200m/s. I noticed that dividing the relative speed of 910m/s in the same ratio as the masses gives the correct answer. Two attempts at the problem without success so far.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 11, 2014 #2

    jbriggs444

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    Science Advisor

    Those simultaneous equations look correct to me. Can you show your work solving them?
     
  4. Sep 11, 2014 #3

    DEvens

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    Education Advisor
    Gold Member

    An important method for dealing with this sort of problem is to treat the interaction in the centre of momentum (COM) frame. Then when you have that figured out, transform the whole thing back to the lab frame.

    In the lab frame here, the whole system starts as one "lump" moving at 7600 m/s. The COM frame starts with the whole system as a lump at rest. You should be able to very easily figure out what happens to this system if it starts as one lump at rest and splits as described. Then just add the 7600 m/s back on to get what happens in the lab frame.
     
  5. Sep 11, 2014 #4
    150VAP + 290VAC = 440(7600)

    VAP = (440(7600) - 290VAC)/150

    Substituting yields VAC = 910 + (440(7600) - 290VAC)/150

    150VAC = 910(150) + 440(7600) - 290VAC

    VAC = (910(150) + 440(7600))/440

    VAC = 7910m/s
     
  6. Sep 11, 2014 #5

    Doc Al

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    Staff: Mentor

    Try assuming that the payload goes forward and the capsule goes backwards (relatively).
     
  7. Sep 11, 2014 #6
    With this in mind and the fact that the total momentum after the spring action must be zero, I get

    290VC = 150VP, ignoring signs for now. This then means that the velocity the spring imparts to each must be 910m/s divided into an opposite ratio, so to speak. Obviously one or the other must then be a negative quantity and adding the 7600m/s yields the correct answer. Am I correct in thinking about it this way?
     
  8. Sep 11, 2014 #7

    Doc Al

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    Staff: Mentor

    That's fine and equivalent to my comment earlier.

    Your analysis was fine as well. But you assumed that the capsule went forward and payload backward. Reverse that assumption and you'll get the answers you need. (Don't give up on your original analysis!)
     
  9. Sep 11, 2014 #8
    I see what you mean and the correct answers have indeed appeared. I actually thought the entire calculation was a complete balls-up but a single sign change makes all the difference. Many thanks.
     
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