1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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...
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted