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Physics problem 2D collisions

  1. Apr 5, 2013 #1
    1. The problem statement, all variables and given/known data
    In a game of billiards the 0.165 kg cue ball is hit toward the 0.155 kg eight ball which is stationary. The cue ball travels at a speed of 6.2 m/s and after impact rolls away at an angle of 40.0 degrees counterclockwise from its initial direction with a velocity of 3.7 m/s. Find the velocity of the eight ball after the collision.


    2. Relevant equations
    m1v1 + m2v2 = m1vf1 + m2vf2


    3. The attempt at a solution
    I cut the coordinate system up into the x y and z axis. In the y axis, there is no movement, as the ball stays on the table. So, we have x and z to work with. I drew a Cartesian graph with the axis 'X' and 'Y' where 'Y' represents 'Z'. I will also interpret the cue ball being shot up the Y axis into the eight ball, instead of being shot horizontally.

    So I cut the motion up like this, in the y axis vectors:
    0.165kg * 6.2 m/s + 0 = 0.155kgVf + 0.165kg*3.7m/s*sin40
    Vfy = 4.4 m/s [up]

    x vectors:
    0 = 0.155kgVf + 0.165kg*3.7m/s * cos40
    Vfyx = -3.0m/s

    Now, I add the vectors using Pythagorean theorem and my result is 5.3 m/s [W 55 degrees N] or 5.3 m/s [35 degrees clockwise].

    However, the book gives me the answer of 4.4 m/s [35.2 degrees clockwise].

    This is the part that confused me, why is the book not taking the X forces into account? Its just using the answer I got in the Y coordinate system. This is confusing, because my teacher did the same thing in school and answered with "there is no movement in the Y axis".

    The problem is that there are three dimensions: X,Y,Z and Y is set to 0, but we still have movement in the Z axis (which I represented as Y in this answer).

    So I have no idea why its done like this, and am very confused. Did the book make a mistake, or am I wrong?
     
  2. jcsd
  3. Apr 5, 2013 #2
    Pay attention... when you compute what you call ##V_f##, you have to consider that also the second ball goes away with an angle... therefore the correct equations should be

    Along y:
    $$ m_c V_{0c}=m_c V_{fc}\cos(40)+m_8 V_f\cos\theta $$
    Along x:
    $$ 0=m_c V_{fc}\sin(40)-m_8 V_f\sin\theta $$

    Now solve for $V_f$ and if you want also for ##\theta##. In this way indeed $V_f$ comes 4.4 =)
     
    Last edited: Apr 5, 2013
  4. Apr 5, 2013 #3
    mmmh... sorry I noticed something just now... your way is also correct because, even if you would have better to call in different ways ##V_f## in the equations in the two directions, it is the same thing (you are leaving implicit the ##\sin\theta## and ##\cos\theta## I am writing explicitely).

    The problem is in the definitions... along your Y you have to put the ##\cos## (measuring the angles leaving from the Y axis) while along X you put ##\sin##. In this way it works...
     
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