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Homework Help: Conservation of momentum question - elastic collisions

  1. Feb 16, 2015 #1
    1. The problem statement, all variables and given/known data
    A 2.0 kg ball moving with a speed of 3.0 m/s hits, elastically, an identical stationary ball as shown. If the first ball moves away with angle 30° to the original path, determine

    a. the speed of the first ball after the collision.

    b. the speed and direction of the second ball after the collision.

    2. Relevant equations
    p = mv

    Conservation of momentum
    maVa + mbVb = ma'Va' + mb'Vb'

    Conservation of KE

    1/2maVa^2 + 1/2mbVb^2 = 1/2maVa'^2 + 1/2mbVb'^2

    3. The attempt at a solution
    I have looked at a few solutions on here and elsewhere, but the concept is just not quite clicking yet... I'm not really sure how to mess around with the equations to get what I need. I understand that you have to break it down into components, so:
    ma = mb

    p(x) = mVa = mVa' + mVb'
    mVa = m(Va' + mVb')
    Va = Va' + Vb' (and the horizontal components of the velocities are *cosθ)
    Va = Va' *cosθ + Vb' *cosθ (the image included with the question shows ball B going below the x axis at an angle = θ, so I am wondering if the second θ is negative...)
    I am not too sure what to do with this information ... maybe solve for Va' or Vb' and leave it there for now? So it would be:

    Va' = (Va - Vb' *cosθ)/cosθ
    Vb' = (Va - Va' *cosθ)/cosθ

    p(y) = mVa + mVb = mVa' + mVb'
    0 = Va' + Vb'
    0 = Va' *sinθ + Vb' *sinθ (again, not sure if the second θ is negative, since it is below the x axis)
    Va' *sinθ = -Vb' *sinθ

    Also, from the conservation of energy:

    0.5m(Va^2) + 0 = 0.5m(Va'^2 + Vb'^2)
    Va^2 = Va'^2 + Vb'^2

    I'm stuck after this. I don't know what I'm looking for after this point... Any help is appreciated! Thanks!
  2. jcsd
  3. Feb 16, 2015 #2
    How can you express the resultant velocities of the two masses, after the collision, in terms of their vertical and horizontal components?
  4. Feb 16, 2015 #3
    *the resultant speeds of the two masses.
  5. Feb 16, 2015 #4
    Do you mean like:

    Va = sqrt(Va'(x)^2 + Va'(y)^2)

  6. Feb 16, 2015 #5
    Yes! Now where can you substitute that information?
  7. Feb 19, 2015 #6
    Using the components above?


    Va = sqrt[ (Va - Vb' *cosθ/cosθ)^2 + (-Vb' *sinθ/sinθ)^2 ]

    My problem here is simplifying this.. I'm not really familiar with trigonometric functions, although I was tipped off that I need to somehow incorporate the function of (sinθ)^2 + (cosθ)^2 = 1.

    I was able to get [simplified Va'(x) and Va'(y)]:

    Va = sqrt[ { (Va/cosθ)-Vb' }^2 + (-Vb')^2]

    But I'm stuck now :S Is anyone able to give some hints on how to simplify this further?
  8. Apr 7, 2015 #7
    In the work that has been done here, θ has been used for both the angles involved however, that would denote they are equal correct? when they may not be. Shouldn't we be putting 30 in for the known angle and leaving the unknown angle as θ since we are not sure if they are the same?
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