1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Inelastic Collision Question

  1. Aug 19, 2009 #1
    Question:
    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 (Figure 7.7). 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.


    Equations:
    m1v1i + m2vi = m1v1f + m2v2f


    I honestly tried attempting this but I got a completely wrong answer, can someone at least help guide me in the right direction. It would be much appreciated, thanks
     
  2. jcsd
  3. Aug 19, 2009 #2
    Because they give you angles, its a hint that they require more than the standard conservation of momentum equation, and that x and y components are required
    what you should do is rewrite out that equation twice, first with an x subscript and than with a y subscript b/c momentum is conserved along both axes.
    Also, although your equation is correct, you can simplify things by rewriting it as:
    (m1+m2)vi=m1v1f+m2v2f as both masses have the same initial velocity since they are connected.
     
  4. Aug 20, 2009 #3
    so the equation would be (x1 + x2)yi = x1y1f + x2y2f ?
     
  5. Aug 20, 2009 #4
    its still: (m1+m2)vi=m1v1f+m2v2f

    but you have to rewrite it as:
    1)(m1+m2)vix=m1v1fx+m2v2fx
    2)(m1+m2)viy=m1v1fy+m2v2fy

    Basically, you have to find the final vertical and horizontal velocity components of each mass to find the original horizontal and vertical velocity components.
     
  6. Aug 20, 2009 #5
    ohh sorry i thought you said replace m and v with x and y. but that makes more sense. So im guessing i would then set up the 2 equations on 2 different triangles and find the hypotenuse of the triangles?
     
  7. Aug 20, 2009 #6
    You could create 2 triangles to find out each component for each mass after the separation, but you wouldn't want to find the hypotenuses of those triangles, because that would just give you the final velocity for each, which is what you started with!
    You want to find the initial components when they were stuck together, so that should tell you there will be one triangle at the end. The equation given will tell you how those 4 different components (2 for each mass) gets resolved into one system.
     
  8. Aug 20, 2009 #7
    and then once i find the 2 components i set up another triangle to find the final velocity?
     
  9. Aug 20, 2009 #8
    yup, although i guess in this case the "final" velocity is really the initial velocity :)
     
  10. Aug 20, 2009 #9
    sickkk thank you so much. just needed to be pointed in the right direction. I appreciate it!
     
  11. Aug 21, 2009 #10
    sorry i just want to make sure these equations are correct.
    I got
    1) vix=[m1v1fx+m2v2fx/(m1+m2)]cos5.1

    2) viy=[m1v1fy+m2v2fy/(m1+m2)]sin5.9
     
  12. Aug 21, 2009 #11
    Your equation is 99% correct, but you made one mistake with notation that might hurt you on an exam
    Remember:vx=vcos(theta)
    In your last equation, you have each final vx multiplied by cos(theta). This is kind of redundant, so you can get rid of the x subscript and have all the final velocities multiplied by cos, or keep the x subscripts and do the multiplying by cos earlier and then plug those values in.
    so yea, if you get rid of those x and y subscripts since you are have sin and cos in there you should be fine
     
  13. Aug 21, 2009 #12
    ohh okay, thank you very much. i appreciate the help
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Inelastic Collision Question
Loading...