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Finding angle of billiard ball after collision

  1. Feb 25, 2013 #1
    1. The problem statement, all variables and given/known data

    A billiard ball of mass 0.162 kg has a speed of 1.86 m/s and collides with the side of the billiard table at an angle of 51.8°. For this collision, the coefficient of restitution is 0.841. What is the angle relative to the side (in degrees) at which the ball moves away from the collision?

    m = 0.162 kg
    vi = 1.86 m/s
    θ = 51.8°
    CR = 0.841


    2. Relevant equations

    CR = v/u
    mvxi + mvyi = mv'xf + mv'yf

    3. The attempt at a solution

    CR = v/u
    0.841 = v / 1.86 m/s
    v = 1.56 m/s

    mvxi = mvxf
    mvcosθ = mvcosθ'
    1.86 * cos 51.8° = vxf
    vxf = 1.15 m/s

    cosθ' = vxf/v
    cosθ' = 1.15/1.56
    θ' = cos-1(1.15/1.56) = 42.5°
     
    Last edited: Feb 25, 2013
  2. jcsd
  3. Feb 25, 2013 #2

    rude man

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    I have a problem with these two. E.g. the second says θ = α and I don't believe the first one either!

    Is momentum really conserved in the billiard ball? What about momentum transferred to the Earth via the pool table etc.?

    M.E.'s out there, help?

    1.86 * cos 51.8° = vxf
    vxf = 1.15 m/s

    cosα = vxf/v
    cosα = 1.15/1.56
    α = cos-1(1.15/1.56) = 42.5°[/QUOTE]
     
  4. Feb 25, 2013 #3
    Hmm not sure about the momentum transfer to the Earth etc. I checked a couple sites where they said momentum is conserved in the x direction in this situation.
     
  5. Feb 25, 2013 #4

    rude man

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    Yeah, that makes sense. Momentum transfer to earth should be only in the direction perpendicular to the table's side. Good point. Did you pick x as going along the side of the collision?
     
  6. Feb 25, 2013 #5
    Yes, I think so. I chose the x-axis to be the side of the table. On a x-y coordinate, the ball travels from the 3rd quadrant, hits at the origin, and deflects towards the 4th quadrant. And, I chose the angles to be along the x axis. Don't know if that's the right way.
     
    Last edited: Feb 25, 2013
  7. Feb 25, 2013 #6

    Andrew Mason

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    I think you are supposed to assume that the speed is reduced only in the direction perpendicular to the cushion.

    AM
     
  8. Feb 26, 2013 #7
    Yeah, that's what I was assuming but i found out my mistake. I noticed that CR should've been CR = [itex]\vec{v}[/itex]sinθ' / [itex]\vec{u}[/itex]sinθ

    Thanks rudeman and Andrew for the suggestions!
     
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