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Finding speed of objects after elastic collison

  1. Nov 18, 2012 #1
    1. The problem statement, all variables and given/known data

    7I8OI.png

    2. Relevant equations

    1/2m1v12 + 1/2m2v22 = 1/2m1v1f + 1/2m2v2f
    m1v1 + m2v2 = m1v1f + m2v2f
    v1 - v2 = -(v1f - v2f)
    3. The attempt at a solution
    So I solved the momentum of conservation for the final velocity of object 1. I then plug that equation into the third equation listed above, as v1f. But the thing is the equation i get for v1f has v2f in it and that is what I am solving for and this is where I am stuck.
     
  2. jcsd
  3. Nov 19, 2012 #2
    I don't think any of the velocity equals 0 at any point. Initially they are moving in the same direction. Then since it says its perfectly elastic, one object will bounce back and have a velocity in the opposite direction. I think the 60g tennis ball will bounce back in the negative direction but I still can't figure out how to find the velocities after the collision :(
     
  4. Nov 19, 2012 #3

    ehild

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    Show your whole work please.

    ehild
     
  5. Nov 19, 2012 #4
    First use conservation of momentum
    ( m1v12 + m2v22 - m2v2f2 ) / m1 = v1f2

    Then plug this v1f into this helper equation, the third equation listed in first post.
    v2f = v1 - v2 + [( m1v12 + m2v22 - m2v2f2 ) / m1 ]1/2


    And I'm stuck because I can't get both v2f to one side so i can solve for it. then once i get v2f i plug it back into the first equation, [( m1v12 + m2v22 - m2v2f2 ) / m1 ], and solve for v1f
     
  6. Nov 19, 2012 #5

    ehild

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    Use the last two equations. Plug in the numbers first. Express v1f from the third equation.

    ehild
     
  7. Nov 19, 2012 #6

    ehild

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    Use the last two equations. Plug in the data first, express vif form the last equation, and substitute into the second one.

    ehild
     
  8. Nov 19, 2012 #7
    v1f = -v1 + v2 - v2f

    Plugged that into,
    (m1v1 + m2v2 - m1v1f) / m2 = v2f

    v2f = -.368m/s
     
  9. Nov 19, 2012 #8

    ehild

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    Check it.
     
  10. Nov 19, 2012 #9
    v1f = -v1 + v2 - v2f is really v1f = -v1 + v2 + v2f. ..

    (m1v1 + m2v2 - m1v1f) / m2 = v2f turns into (m1v1 + m2v2 + m1v1f) / m2 = v2f

    so v2f = .307m/s
     
  11. Nov 19, 2012 #10

    ehild

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    Check.

    ehild
     
  12. Nov 19, 2012 #11

    ehild

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    Plug in the numbers for v1 and v2: v1f = v2f - 1.35.
    Plugging in the data into the equation m1v1+m2v2=m1v1f+m2v2f and also substituting v2f - 1.35 for v1f: 0.2535 = 0.06(v2f - 1.35)+0.09v2f.
    Expand and collect the terms with v2f, solve.

    ehild
     
  13. Nov 20, 2012 #12
    Its 3.75 or 1.35?
     
    Last edited: Nov 20, 2012
  14. Nov 20, 2012 #13

    ehild

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    What is 3.75 or 1.35?

    ehild
     
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