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Conservation of momentum of a rocket

  1. Mar 18, 2005 #1
    The last stage of a rocket is traveling at a speed of 8000 m/s. This last stage is made up of two parts that are clamped together, namely, a rocket case with a mass of 330.00 kg and a payload capsule with a mass of 155.00 kg. When the clamp is released, a compressed spring causes the two parts to separate with a relative speed of 910 m/s. What are the speeds of the two parts after they have separated? Assume that all velocities are along the same line. What is the speed of the payload? What is the velocity of the rocket case?

    i got 8291m/s and 7381m/s but the computer says its wrong
    :mad:
     
  2. jcsd
  3. Mar 18, 2005 #2

    BobG

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    Yes, the smaller payload should have a larger change in velocity than the heavier rocket if momentum is to be conserved.

    It would be nice if you had shown what you did to get your answers. From your numbers, it looks like you made a silly mistake at the end and swapped which objects you should have added/subtracted velocities to.
     
  4. Mar 18, 2005 #3
    when I solve the problem, I get an answer of the samller parts speed being 6284m/s and the bigger parts being 8806m/s.
    This is my set up:

    [tex] (m_1 + m_2)V_i = m_{1}(V_i - (910m/s - x)) + m_{2}(V_i + x) [/tex]

    Regards,

    Nenad
     
  5. Mar 20, 2005 #4
    ok the computer had your 8806m/s as the speed for the smaller one and the bigger one was 7906 (remember the difference in speed is 910).
     
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