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Inelastic Collisions in 2-D

  1. Nov 20, 2007 #1
    1. The problem statement, all variables and given/known data
    A car with mass of 950 kg and speed of 16 m/s approches an intersection in the positive x direction. A 1300-kg minivan traveling at 21m/s is heading towards the same intersection in the positive y direction. The car and minivan collide and stick together. Fine the speed and direction of the wrecked vehicles just after the collision, assuming external forces can be ignored.


    2. Relevant equations
    p = mv
    [tex]\sum[/tex] p before = [tex]\sum[/tex] after


    3. The attempt at a solution

    Well this question was grabbed from my physics textbook as an example question.

    Here are the steps they preformed:

    1. Set the initial x component of momentum equal to the final x component of momentum
    m[tex]_{1}[/tex]v[tex]_{1}[/tex] = (m[tex]_{1}[/tex] + m[tex]_{2}[/tex])v[tex]_{f}[/tex]cos[tex]\vartheta[/tex]

    2. Do the same of the y component of momentum.
    m[tex]_{2}[/tex]v[tex]_{2}[/tex] = (m[tex]_{2}[/tex] + m[tex]_{2}[/tex])v[tex]_{f}[/tex]sin[tex]\vartheta[/tex]

    I understand everything up to this part, but I got confused when they divide the y momentum equation by the x momentum equation. They did it to eliminate the final velocity, giving an equation solving for [tex]\vartheta[/tex] alone.

    My question is: Why is it that you can divide to eliminate variables??

    Oh, and those are suppose to be subscripts, not superscripts.....
     
    Last edited: Nov 20, 2007
  2. jcsd
  3. Nov 20, 2007 #2
    [tex]\sin^2\theta+\cos^2\theta=1[/tex]

    Just put [tex]\sin\theta[/tex] and [tex]\cos\theta[/tex]

    And unknown left is v(f)
     
  4. Nov 20, 2007 #3

    nrqed

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    You can always divide two equations. If A = B and C=D, then the ratio A/C must equal B/D as long as C and D are not zero. This is a common trick.
     
  5. Nov 22, 2007 #4
    ahh okay thanks.

    i never did it this way before :shy:
     
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