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Direct collisions on an inclined plane

  1. Jun 29, 2013 #1
    1. The problem statement, all variables and given/known data

    A smooth sphere falls vertically and strikes a fixed smooth plane inclined at an angle of ∅ to the horizontal.If the coefficient of restitution is (2/3) and the sphere rebounds horizontally,
    Its speed before impact is u and after impact v
    calculate the fraction of kinetic energy lost during impact.


    2. Relevant equations



    3. The attempt at a solution

    Diagram of the question is attached to this post.

    i = horizontal component vector and j = vertical component vector

    Taking the inclined plane as the x axis
    v, vcos∅i + vsin∅j ..... alternate angle to inclined angle

    u, vcos(90 - ∅ )i - v sin(90 - ∅)j = usin∅i - ucos∅j


    the I component remains the same

    vcos∅ = usin∅

    v = utan∅

    vsin∅/(ucos∅) = 2/3 (v/u)tan∅ = 2/3 (utan∅/u)tan∅ =2/3

    (tan∅)^2 = 2/3 tan∅ = (√6)/3


    v = u(√6)/3

    .5mu^2 - .5mv^2 = energy loss

    .5mu^2 -.5m(6/9)u^2 = .5mu^2 - m(6/18)u^2 = .5mu^2 - (1/3)mu^2 = (1/6)mu^2

    fraction of kinetic energy lost is 1/6


    my book says the answer is 1/3 but I got 1/6. Any help would be appreciated.
     

    Attached Files:

    Last edited: Jun 29, 2013
  2. jcsd
  3. Jun 29, 2013 #2

    Andrew Mason

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    Correct so far

    It is a bit convoluted there. I can't follow what you have done. Just use:

    vsin∅/(ucos∅) = 2/3

    so: tan∅ = 2/3(u/v)

    Since you also have tan∅ = v/u

    ∴ v^2/u^2 = 2/3

    AM
     
  4. Jun 29, 2013 #3
    I used the equation

    v^2/u^2 = 2/3

    v^2 = (2/3)u^2

    so .5mu^2 - .5m(2/3)u^2 = 1/6mu^2

    fraction of kinetic energy lost during impact = 1/6

    I got the same answer do you think my book could be wrong?
     
  5. Jun 30, 2013 #4

    haruspex

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    What fraction is 1/6mu^2 of .5mu^2?
     
  6. Jun 30, 2013 #5
    1/3!!! thanks
     
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