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Collisions in Two Dimensions (Perfectly Elastic)

  1. Nov 12, 2008 #1
    1. The problem statement, all variables and given/known data
    Assume an elastic collision (ignoring friction
    and rotational motion).
    A queue ball initially moving at 4.3 m/s
    strikes a stationary eight ball of the same size
    and mass. After the collision, the queue ball’s
    final speed is 2.3 m/s .

    Find the queue ball’s angle with respect
    to its original line of motion.

    v1i = 4.3
    v1f = 2.3

    2. Relevant equations
    p1i + p2i = p1f + p2f
    KEi = KEf

    3. The attempt at a solution

    sqrt(4.3^2 - 2.3^2) = v2f = 3.6 m/s

    4.3 = 2.3*cos θ + 3.6*cos φ (for x direction)

    0 = 2.3*sin θ - 3.6*sin φ (for y direction)

    2.3*sin θ = 3.6*sin φ

    Not quite sure how to proceed from here.
     
  2. jcsd
  3. Nov 12, 2008 #2
    [tex]m_1u_1+m_2u_2=m_1_v_1+m_2v_2\rightarrow m(u_1+0)=m(v_1+v_2)[/tex].
     
  4. Nov 12, 2008 #3
    I'm not sure how that helps . The masses are equal so they drop from the equations and the velocities are known.
     
  5. Nov 12, 2008 #4
    4.3 = 2.3 + v...
     
  6. Nov 12, 2008 #5
    Still don't know what you're getting at :).

    But I managed to solve it using pythagoras: (4.3-2.3 cos θ)^2 + (2.3 sin θ)^2 = 3.6^2

    With that I get around 57 degrees which is correct.

    Just out of curiosity, would there have been a simpler way of solving this? My solution required using power-reducing formulas for sin/cos...which seems too complicated considering the level of the course.
     
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