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Oblique collision involving two spheres

  1. Jul 16, 2013 #1
    1. The problem statement, all variables and given/known data
    A smooth sphere of mass 4 kg collides obliquely with another smooth sphere of mass m which is at rest.After impact the two spheres move at right angles to each other.If the coefficient of restitution was (4/7), calculate the value of m.


    2. Relevant equations

    m1u1 + m2u2 = m1v1 + m2v2

    (v1 - v2)/(u1 - u2) = -e


    3. The attempt at a solution

    i is the horizontal unit vector,where i is along the line of the centres(of the spheres) at impact
    j is vertical unit vector

    since i is along there centres j is unchanged before and after the impact

    conservation of momentum along the i axis

    4u1 + m(0) = 4v1 + mv2 ................ mv2 = 4u1 - 4v1


    (v1 -v2)/(u1) = -4/7

    I'm not sure what to do from here, I'd assume its something to do with them being at right angles after the impact?
     
  2. jcsd
  3. Jul 16, 2013 #2

    tiny-tim

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    Hi Woolyabyss! :smile:
    No, the collision is oblique, which means that the initial velocity is not parallel to the line between the centres on contact. :wink:
     
  4. Jul 16, 2013 #3
    For the previous oblique collision questions I've done in my book.It says the i axis is between the centres at impact.
     
  5. Jul 16, 2013 #4
    Since the m-sphere was at rest initially and the exchange of momentum happens only along i, what is the direction of its velocity after impact? What is the direction of the other sphere then? What can be said about its velocity in the i-direction?
     
  6. Jul 16, 2013 #5

    tiny-tim

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    the i axis is wherever you choose it to be!

    in this case the initially moving ball was not moving parallel to the i axis (the line between the centres at impact)
     
  7. Jul 16, 2013 #6
    Would that mean the m sphere moves on the i axis and the 4kg sphere move on the j axis( meaning the 4kg sphere would have 0 momentum in the i direction)? EDIT (this is after the collision)
     
  8. Jul 16, 2013 #7
    Ya I just worked it out there.

    conservation of momentum along the i axis.

    m(0) + 4u = mv + 4(0) ...................... v = 4u/m

    (0 - v)/(u-0) = - 4/7 ............... v =4u/7

    4u/m = 4u/7

    m = 7

    this is the answer at the back of my book thanks for the help.
     
  9. Jul 17, 2013 #8

    tiny-tim

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    hold on … that's not oblique!

    start again, with the initial velocity at angle θ to the i axis :wink:
     
  10. Jul 17, 2013 #9

    haruspex

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    Woolyabyss never actually defined u. You seem to be assuming it's defined as the speed of the first mass, but that would make several errors in Woolyabyss' equations. With a suitable different definition it all works.
     
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