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Inelastic Collision and Finding Initial Velocity

  1. Jul 15, 2008 #1
    Collisions and Determining the Original Speed

    1. The problem statement, all variables and given/known data

    Two spacecrafts from different nations have linked in space and are coasting with their engines off, heading directly toward Mars. The spacecrafts are thrust apart by the use of large springs. Spacecraft 1, of mass 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, and spacecraft 2, of mass 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 spacecrafts when they were linked together.

    2. Relevant equations

    mv1 + mv2 = mv1' + mv2'
    p = mv

    3. The attempt at a solution

    Since there are angles, I tried looking for the components. So, the components of the 3.4 x 10^3, are x= 3.4x10^3cos5.9 and y=3.4x10^3sin5.9. The components of the 3.5 x 10^3 km/h are x= 3.5x10^3cos5.1 and y=3.5x10^3sin5.1. I'm having a difficult time understanding what to do from here. So we're given the final velocities of both, and their angles. How do I incorporate the components? Where do I sub them in? So, both must have the same initial velocity, so you can simplify the equation slightly:

    since v1=v2
    v(m1 + m2) = mv1' + mv2'

    If anyone could give me a hint, that'd be GREAT =)
     
    Last edited: Jul 15, 2008
  2. jcsd
  3. Jul 16, 2008 #2
    Momentum is conserved along the direction perpendicular to the initial motion.
    Also since the andle of divergence is given you could add them vectorially...
     
  4. Jul 18, 2008 #3
    Suppose their common speed before separation is v,
    Spacecraft 1's speed after separation is v1
    Spacecraft 2's speed after separation is v2

    Note that in this case, the momentum is conservative along both the x and y directions.
    So
    Along X axis: (m1 + m2)v = m1v1cos(5.10) + m2v2cos(5.90)
    Along Y axis: 0 = m1v1sin(5.10) - m2v2sin(5.90)

    http://www.idealmath.com
     
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