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Inelastic collision. Two angle variables.

  1. Sep 12, 2011 #1
    http://session.masteringphysics.com/problemAsset/1007938/22/6318.jpg
    Two cars, both of mass m, collide and stick together. Prior to the collision, one car had been traveling north at speed 2v, while the second was traveling at speed v at an angle phi south of east (as indicated in the figure). After the collision, the two-car system travels at speed v_final at an angle theta east of north.

    Now I solved Vfinal into its i and j components

    i: vicos[itex]\phi[/itex]=2vfcos[itex]\theta[/itex]

    j: 2vi-sin[itex]\phi[/itex]vi=2(vfsin[itex]\theta[/itex])

    The problem is if I solve for vfinal, I get the answer in terms of Vinitial, phi, and theta. The problem only asks for Vfinal in Vinitial and theta. How do I get rid of phi to only have theta and Vi?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 12, 2011 #2
    I think you changed the equations.It should be
    i: vicosϕ=2vfsinθ
    j: 2vi-sinϕvi=2(vfcosθ)
    And my sugestion to get rid of phi is to isolate cosϕ on eq.i and use the fundamental relation cos²ϕ+sin²ϕ=1 to find sinϕ, and then substitute sinϕ on eq.j.
     
  4. Sep 13, 2011 #3
    do you mean isolate cos(phi) and square it, and then isolate sin(phi) and then square it and then sub it into cos^2phi + sin^2phi =1? I did that but then I got an equation Vf and Vf^2 . The fraction looks unsolvable for Vf. Cannot factor or anything.
     
  5. Sep 13, 2011 #4
    if i am not wrong

    you can rearrange it to

    2vi - 2vfcos(t) = vsin(phi)

    2vfsin(t) = vcos(phi)

    square both sides

    your right hand side will be vi2 , since c and s cancel

    your left hand side will be

    4vi2 + 4vf2 - 8vivfcos(t)

    this is quadratic, so use -b+- squareroot b^2 - 4ac over 2a.
     
  6. Sep 13, 2011 #5
    I don't see how the sin and cos cancel when you square it? There must be a step I'm missing because if you square the first equation, on the right side you get Vi2sin[itex]\phi[/itex]. On the left side you would get a long quadratic with variables Vi, Vf, and theta.
     
  7. Sep 14, 2011 #6
    the aim is to get rid of the phi term right?

    so that means you have to express phi in terms of everything else

    so squaring both sides of both equations give you

    4vi2 - 8vivfcos(t) + 4vf2cos2(t) = vi2sin2(phi)

    4vf2sin2(t) = vi2cos2(phi)

    so now look at the RHS of both equations, you add them so that the cos and sin (phi) "cancel off" since cos2x + sin2x = 1, so you effectively left with vi2 , i.e, you have factored out vi2

    so similarly, for lhs, you combine the sin and cos (t) to get

    4vi2 + 4vf2 - 8vivfcos(t) = vi2

    so you are now effectively solving for vf in a quadratic equation, without the phi terms
     
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