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Cue Ball

  1. Nov 13, 2009 #1
    1. The problem statement, all variables and given/known data

    Figure 9-48 gives an overhead view of the path taken by a 0.160 kg cue ball as it bounces from a rail of a pool table. The ball's initial speed is 1.47 m/s, and the angle θ1 is 62.9°. The bounce reverses the y component of the ball's velocity but does not alter the x component. What are (a) angle θ2 and (b) the magnitude of the change in the ball's linear momentum? (The fact that the ball rolls is irrelevant to the problem.)

    http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c09/fig09_47.gif

    2. Relevant equations

    p = mv where p and v are vectors.

    3. The attempt at a solution

    before the moment of impact, I found the Vx and Vy of the ball traveling through the pythagorean theorem and I know that the velocity along the x axis is constant. I'm just having a hard time trying to comprehend this problem.

    I tried this equation:

    J = Pf - Pi = M (Vf-Vi) where J, P and V are vectors

    Jx = 0 because the Vx is equal on before and after the collision.

    But still found myself stuck.
     
  2. jcsd
  3. Nov 13, 2009 #2

    Delphi51

    User Avatar
    Homework Helper

    Having found Vx and Vy initially, all you have to do is put a minus sign on the Vy and you have the components after the collision. It should be easy to use a tangent calc to get the angle. The change in velocity is the new Vy minus the old Vy. That will not be zero because of the sign change. The change in momentum is the mass times the change in velocity.
     
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