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Angle of Queue Ball after elastic collision

  1. Oct 24, 2007 #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 2.2 m/s strikes a stationary eight ball of the same size and mass. After the collision, the queue ball's final speed is 0.61 m/s.

    Find the queue ball's angle [tex]\theta[/tex] with respect to its original line of motion. Answer in units of degrees.


    2. Relevant equations

    [tex]\frac{1}{2}mv_1_i^2+\frac{1}{2}mv_2_i^2=\frac{1}{2}mv_1_f^2+\frac{1}{2}mv_2_f^2[/tex]


    3. The attempt at a solution

    [tex]\frac{1}{2}mv_1_i^2+\frac{1}{2}mv_2_i^2=\frac{1}{2}mv_1_f^2+\frac{1}{2}mv_2_f^2[/tex]
    [tex]v_1_i^2+v_2_i^2=v_1_f^2+v_2_f^2[/tex]
    [tex]v_2_f=2.114 m/s[/tex]

    From here I am unsure of how to come to the angle [tex]\theta[/tex] the question is asking for.

    I thought it might be:

    [tex]tan\theta=\frac{v_2_f}{v_1_i}[/tex]

    This is incorrect, though.

    Any guidance would be appreciated!
     
    Last edited: Oct 24, 2007
  2. jcsd
  3. Oct 24, 2007 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You can't determine the angle just by using the speeds. You have to consider that momentum is also conserved, which is a vector quantity. Choose x and y axes and split into components (each of which is conserved).
     
  4. Oct 24, 2007 #3
    You have written equations for kinetic energy conservation. Write an equation for conservation of momentum also. Then solve.
     
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