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Conservation of Kinetic Energy Proof

  1. Oct 22, 2009 #1
    [EDIT]Title is incorrect, it is a proof using Conservation of Energy/Momentum


    1. The problem statement, all variables and given/known data
    You take a pool shot for the win. The perfectly [EDIT]elastic[/EDIT] collision is such that after the collision the Que ball has a non-zero velocity along angle [tex]\theta[/tex], and the 8 ball has a non-zero velocity along angle [tex]\beta[/tex]. Prove that [tex]\theta[/tex]+[tex]\beta[/tex]=90[tex]\circ[/tex]. The two pool balls have exactly the same mass.


    2. Relevant equations
    No equations are given, but I am fairly certain only [tex]\Delta[/tex]K=[tex]\Delta[/tex]P=0 is needed. My teacher vaguely hinted that those were the equations/principals he recommend we utilize.


    3. The attempt at a solution
    I came up with 2 equations:
    1) [1/2]M*VIQ2=[1/2]M*VFQ2*cos([tex]\theta)[/tex]+[1/2]M*VF82*cos([tex]\beta[/tex])
    2) M*VIQ=M*VIQ*cos([tex]\theta)[/tex]+M*VF8cos([tex]\beta[/tex])
    I solved for VFQ and plugged that into the 1st equation. After simplifying I got;
    2[(VF8cos([tex]\beta[/tex]))(VFQcos([tex]\theta[/tex]))]=0
    I figured out that this means one of those variables must be 0.I know that VFQ is not, and VF8 is not, therefore, one of the angles must be equal to 90[tex]\circ[/tex] for that equation to equal 0.
    I believe my initial equations are wrong, though i could have easily made an algebraic mistake. Any pointers would be very nice, Thanks!
     
    Last edited: Oct 22, 2009
  2. jcsd
  3. Oct 22, 2009 #2

    cepheid

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    Hi Davey Boy, welcome to PF.

    Are you sure that the problem statement shouldn't say "the perfectly elastic collision?" If not, then why are you using conservation of kinetic energy?

    Your first equation, (the energy equation), is wrong. Why do you have those cosine factors in there?
     
  4. Oct 22, 2009 #3
    It should be elastic rather than inelastic, my mistake. I'll edit that part. I have the cosine factors to account for the angle at which the pool balls are traveling after the collision. I know those are needed for the momentum equation because my teacher did a brief example for 2-D momentum problems, but are they not for kinetic energy?
     
  5. Oct 22, 2009 #4

    cepheid

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    Energy doesn't have a direction. The kinetic energy of a ball ( its energy due to motion) depends upon its *speed*, i.e. the magnitude of its velocity. So KE = (1/2)mv2, period. The energies of the two balls after the collision must add up to the energy of the que ball before.
     
  6. Oct 22, 2009 #5
    Awesome, that's what I had suspected. Thanks for the help, I'll try that out and get back to you if it works or not.
     
  7. Oct 22, 2009 #6
    Thanks to the hint of energy not being a vector, thus not requiring a direction, I was able to find equations for VFQ and VF8. I have no idea what to do now though. The equations I got are:
    VFQ=[tex]V_{IQ}(cos\beta-1)/(cos\beta-cos\theta)[/tex]
    and
    VF8=[tex]V_{IQ}(1-cos\theta)/(cos\beta-cos\theta)[/tex]
     
  8. Oct 22, 2009 #7

    cepheid

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    Hey, well I noticed that your momentum conservation equation was only for conservation of momentum in one direction. There's also conservation of momentum in the other, perpendicular direction (which gives you another equation to play with). That equation will relate VF8 to VFQ.
     
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