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Change in Momentum

  1. Oct 12, 2009 #1
    1. The problem statement, all variables and given/known data

    A billiard ball of mass m = 0.15 kg strikes the cushion of a billiard table at θ1 = 48° and a speed v1 = 21 m/s. It bounces off at an angle of θ2 = 670 and a velocity of v2 = 16 m/s. What is the magnitude of its change in momentum (in kg·m/s)?

    http://schubert.tmcc.edu/res/msu/mmp/kap6/picts/pool.gif


    2. Relevant equations



    3. The attempt at a solution
    Not sure how to solve this one. Any help will be appreciated.

    Sincerely,
    CaptFormal
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Oct 12, 2009 #2

    rl.bhat

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    Homework Helper

    Take the vertical and horizontal components of v1 and v2.
    Find the difference in vertical and horizontal components. Take care of signs.Then take the resultant of these components.
     
  4. Oct 12, 2009 #3
    Ok, so here is what I got so far.

    Vx = 21cos(48) - 16cos(67) = 7.8

    Vy = 21sin(48) - 16sin(67) = 0.8779

    Now I am not sure what to do. I tried the following:

    (7.8^2 + 0.8779^2)^(1/2)

    and then took that answer and multiplied it by the mass but it was incorrect. What am I missing?
     
  5. Oct 13, 2009 #4

    rl.bhat

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    Vx = 21cos(48) - 16cos(67) = 7.8
    This is wrong. vx components are in the opposite direction. So the change in vx is
    Vx = - 21cos(48) - 16cos(67) = ?
     
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