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2D Momentum Question on pool balls

  1. Oct 10, 2014 #1
    1. The problem statement, all variables and given/known data
    Standard pool balls have a mass of 0.17kg. Before a collision, ball A is at rest and ball B is traveling at 4.5 m/s [N]. After the collision, ball B is travelling at 1.6 m/s [N20E]. What is the velocity of ball A?


    2. Relevant equations
    pinitial = pfinal
    m1v1 + m2v2 = m1v1 + m2v2

    3. The attempt at a solution
    I believe that the initial momentum would be 0.17 * 4.5 m/s = 0.765kg m/s [N].
    The momentum for B is 1.6 * 0.17 = 0.272
    The initial momentum has to equal the final momentum, so:

    0.765kg/m/s = 0.272 + A

    I drew a triangle for B's angle.

    For X component: 0.272 cos 20 deg = 0.255
    For Y component: 0.272 sin 20 deg = 0.09

    The X component will just be 0.255 [W] because the X component is 0.
    The Y component will have to be the difference between 0.765 and 0.09302, because you need the north to have the same momentum. The answer to that is 0.7168.

    Then I plug them all in and use Pythagorean theorem to find the momentum, which was 0.76 kg m/s.
    I divide that by the mass of 0.17kg and got a velocity of 4.2 m/s.

    But unfortunately, the answer is 3 m/s.

    How?
     
  2. jcsd
  3. Oct 10, 2014 #2

    RUber

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    The North components do not have to have the same momentum. The total momentum in the system cannot change.
     
  4. Oct 10, 2014 #3

    Andrew Mason

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    This last statement is not correct. The north component of B's momentum (i.e. mv(cos(20)) + the north component of A's momentum has to equal B's initial momentum. I think you are also assuming that this is an elastic collision. You cannot assume that.
    AM
     
  5. Oct 11, 2014 #4

    haruspex

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    You may have cos and sin switched over. N20E is 20 degrees E of N, so if you are taking the positive X axis as E then the X component will involve sin of 20 degrees.
     
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