Inelastic Collision and Finding Initial Velocity

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Collisions and Determining the Original Speed

Homework Statement



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.

Homework Equations



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

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 =)
 
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Answers and Replies

  • #2
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Momentum is conserved along the direction perpendicular to the initial motion.
Also since the andle of divergence is given you could add them vectorially...
 
  • #3
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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" [Broken]
 
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