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Conservation of Momentum: Spacecrafts

  1. Mar 19, 2006 #1
    Hi again,

    Could someone take a look at my solution please? I have answered the following question but I'm not sure if I address the problem correctly.

    [​IMG]

    Answer:

    Page 1

    Page 2

    Thanks,

    James
     
  2. jcsd
  3. Mar 19, 2006 #2

    Fermat

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    The majority of it is correct, but falls down at the conservation of momentum part.
    Momentum is a vector. Your statement,
    P1 + P2 = P'1 + P'2
    would have been correct if you had been talking about vectors, but you were talking about absolute values.
    P1 and P2 are in the original direction of motion, whereas P'1 and P'2 are at an angle to this direction.
    The correct eqn would have been,
    P1 + P2 = P'1x + P'2x
    Apart from this, your method is OK, but you should have worked to a greater accuracy.
    I have P'2 = 1.605556x10^7 Ns
    and P'2x = 1.597051x10^7 Ns
    I ended up with V = 3,437 km/hr or 954.71 m/s
     
  4. Mar 19, 2006 #3
    Ahh ok Fermat, I understand. Thank you!
     
  5. Mar 19, 2006 #4
    Remember.

    Momentum is not a scalar quantity as energy is. Don't get the two mixed up... Momentum needs to be broken down into X and Y components.

    Most books don't state this especially, but the correct way to describe the conservation of momentum is...

    The momentum in the X-direction initially is the same as the momentum in the X-direction finally. The same goes for Y-direction. Remember to split into components WHENEVER you can.
     
  6. Mar 20, 2006 #5
    Actually Fermat, I do have something to ask. A clarification, you can say.

    Why would P1 + P2 = P'1x + P'2x ?

    Is it because originally, P1 and P2 have an angle of zero degree, hence its x component = to the vector's magnitude?
     
  7. Mar 20, 2006 #6

    exactly. because the ships both travel along the x-axis only, their momentum has only a x-component. since momentum is conserved in the x and y axes, the momentum in the x-axis of the ships before separation is all the momentum the ships carry so the sum of the x-components of the momenta of the ships after separation would be the same as the initial momenta of the ships.
     
  8. Mar 20, 2006 #7

    Fermat

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    Looks like I got beat to it :(
    As a little extra, getting into the conservation of momentum bit, The y-components of momentum should be equal and opposite.
    So, P'1y = P'2y
     
  9. Mar 20, 2006 #8
    Thanks everyone for your help! Much appreciated :=)
     
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