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Elastic collision between pool balls

  1. Jan 24, 2017 #1
    1. The problem statement, all variables and given/known data
    A super cue ball is made of the same material with the target ball (radius r) but slightly larger: rc = (3)^(1/3)*r The cue ball collides with the target ball on a frictionless table, as shown below, with initial speed of v0. The collision is not head-on, as shown below.

    a) What are the velocities of the two balls after the collision? You need to determine both the speeds and directions!

    b) In the CM frame, what is the target ball’s direction, relative to the x-axis?

    2. Relevant equations
    ρ = m/V
    ρ = 4/3 π r3

    3. The attempt at a solution

    so for the masses,

    m1 = ρV1 , m2 = ρV2

    taking the ratio of those guys gives m1 = 3 m2

    if i use conservation of linear momentum and kinetic energy conservation, i end up with two sets of equations but one is a vector equation and the other is not, so i dont know how to even start this problem
     

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    Last edited: Jan 24, 2017
  2. jcsd
  3. Jan 24, 2017 #2

    ehild

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    There should be a figure to the problem. Attach it please!
    The vector equation can be written as two scalar equations for the components of the linear momentum. So you have three equations.
     
  4. Jan 24, 2017 #3
    Oh right! its there now.
     
  5. Jan 24, 2017 #4

    ehild

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    OK. So the velocity of the target ball encloses the angle pi/3 with the original velocity of the cue ball after the collision. The direction of velocity of the cue ball is unknown after collision. And energy is conserved. Write the equations for the components of the momenta, and also for the conservation of energy.
     
  6. Jan 24, 2017 #5
    Ok so this is what I've done so far

    m1<u1x,u1y> = m1<v1x, v1y> + m2<v2x, v2y>

    where u1 is the velocity of mass 1 before the collision, v1 is the velocity of mass 1 afterwards, and similarly for mass 2.

    I end up with

    (√3)/6 v2 + v1cos(φ) = v0 where v0 = u1x these are the magnitudes of the velocities

    v2 = -(6/√3)v1sin(φ)

    and then the energy conservation expression

    3v02 = 3v12 + v22

    does this look ok so far?
     
  7. Jan 24, 2017 #6

    ehild

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    Check the equations, there are mistakes in both.
    The energy equation is OK.
     
  8. Jan 24, 2017 #7
    m1<u1x, 0> = m1<v1x, v1y> + m2<v2x, v2y>

    <m1u1x, 0> = <m1v1x, m1v1y> + <m2v2x, m2v2y>

    <m1u1x, 0> = < m1v1x + m2v2x, m1v1y + m2v2y>

    therefore,

    m1v1x + m2v2x = m1v0

    m1v1y = -m2v2y

    3v1cos(φ) + ½v2 = 3v0

    v2 = -(6/√3)v1sin(φ) still got the same thing for this one
     
  9. Jan 24, 2017 #8

    ehild

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    It is OK. But do not forget, that you get negative angle φ. The cue ball moves under the x axis after collision.
    Isolate cos(φ) and sin(φ), square and add the equations, so as φ eliminates.
     
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