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Momentum in Two Dimensions

  1. Jan 28, 2010 #1
    1. The problem statement, all variables and given/known data

    Two spacecraft from different nations have linked in space and are coasting with their engines off, heading directly toward Mars. The spacecrafts are thrust apart by large springs. Spacecraft 1, with a mass of 1.9 x 10^4 kg, then has a velocity of 3.5 x 10^3 km/h at 5.1 degrees to its original direction. Spacecraft 2, whose mass is 1.7 x 10^4 kg, has a velocity of 3.4 x 10^3 km/h at 5.9 degrees to its original direction. Determine the original speed of the two craft when they were linked together.

    m1 = 1.9 x 10^4 kg
    v1 = 972 m/s

    m2 = 1.7 x 10^4 kg
    v2 = 944 m/s

    2. The attempt at a solution

    Pf1 = (1.9 x 10^4 kg)(972 m/s) = 1.8 x 10^7 Ns
    Pf2 = (1.7 x 10^4 kg)(944 m/s) = 1.6 x 10^7 Ns

    The initial momentum will be the parallel components of the final momentum, so

    Pti = (1.8 x 10^7 Ns)(cos5.1) + (1.6 x 10^7 Ns)(cos5.9) = 3.4 x 10^7 Ns
    v = (3.4 x 10^7 Ns)/(3.6 x 10^4 kg) = 944 m/s

    I'm not sure of it at all, help appreciated.
     
  2. jcsd
  3. Jan 29, 2010 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    There is no need to convert to m/s. You have to use more significant figures than 2. This is because the angles are small and the difference between the actual speed and the component in the original direction is lost if you stick to 2 significant figures. The problem is poorly set, in that respect.

    Pf1 = (1.9 x 10^4 kg)(3.5 x 10^3 km/hr) = 6.65 x 10^7 kg km/hr
    Pf2 = (1.7 x 10^4 kg)(3.4 x 10^3 km/hr) = 5.78 x 10^7 kg km/hr

    Pti = (6.65 x 10^7)(cos5.1) + (5.78 x 10^7)(cos5.9) = 12.4 x 10^7 kg km/hr
    v = (12.4 x 10^7 kg km/hr)/(3.6 x 10^4 kg) = 3.44 x 10^4 km/hr

    AM
     
  4. Jan 18, 2011 #3
    Your velocity seems to be one x10 too big.

    v = (12.4 x 10^7 kg km/hr)/(3.6 x 10^4 kg) = 3.44 x 10^3 km/hr

    Noted for reference.
     
  5. Jan 18, 2011 #4

    gneill

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    Staff: Mentor

    I note that the sum of the momenta in the y-direction is not zero if you interpret the given data in a straightforward way. This has implications for the initial speed, for the initial direction of motion, and for the angles given. Again, the problem seems poorly posed.
     
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