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How does sin(65)/sin(37) become sin(65)/tan(37)?

  1. Apr 1, 2012 #1
    1. The problem statement, all variables and given/known data
    This is collision in 2D problem. I already did most of the work except I'm stuck on the trigonometry and algebra for solving for the final velocity of object a.

    I will put the problem here just in case someone wants to show me a better way for getting to the answer.

    Collision between two pucks. Puck "a" has mass = 0.025kg Velocity of "a" = +5.5m/s

    along x axis "a" makes collision with puck "b" which has mass = 0.050kg and "b" starts at rest. Collision is not head on. So, after collision puck "a" flies apart from "b" at angle 65° and puck "b" flies off at angle 37°


    2. Relevant equations

    I am trying to solve for Vfa;

    Ma*Voa = Ma*Vfa(cos65) + [Ma*Vfa(sin65)/(sin37)](cos37)

    Ma = X
    Voa = Y
    Vfa = Z

    XY = XZ(cos65) + [ XZ(sin65)/(sin37)](cos37) solve for Z

    Hopefully that's better.

    The solutions in textbook managed to solve for Vfa or Z like this:

    Z = Y/[cos65 + (sin65/tan37)]

    My question how to get Z by itself and where did (sin65/tan37) come from?
     
  2. jcsd
  3. Apr 1, 2012 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Factor out the Z from the right hand side.
    What does tanθ equal? 1/tanθ?
     
  4. Apr 1, 2012 #3
    Okay What confused me was that I didn't know how to get rid of Z from inside the parenthesis.

    I'm not sure could I just plug in any number to θ and figure it out? I just get different numbers.
     
    Last edited by a moderator: Apr 2, 2012
  5. Apr 1, 2012 #4

    tanθ = sin/cos

    1/tanθ = cotθ = -sinθ/cosθ
     
  6. Apr 2, 2012 #5

    Doc Al

    User Avatar

    Staff: Mentor

    Right.

    So 1/tanθ = 1/(sin/cos) = cosθ/sinθ

     
  7. Apr 2, 2012 #6
    Thanks Doc Al. I always did have trouble with trigonometric identities. Fortunately this exact problem was on the exam, and I practically remembered every step with out much thought lol.
     
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