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Conservation of momentum - Collision of pool balls

  1. Apr 8, 2005 #1

    I am having trouble recalling my concepts in momentum. I just remember its conserved :rolleyes:

    Plus, I am horrible with directions, so need help understanding this problem, as well as the underlying concepts.

    Two pool balls are rolling along a pool table. The orange pool ball has a velocity of 15 cm/s west and the green ball a velocity of 15 cm/s north 30° east. What are the velocities of the balls after the collision?

    What I am doing is as follows:

    15*m + 15*m = m(15 + 15)
    which doesn't take me anywhere

    Do I need to resolve into x and y components?

    So any help on this question would be greatly appreciated. I will try and return the help by helping others. (i.e. pay it forward!)
    Last edited: Apr 8, 2005
  2. jcsd
  3. Apr 8, 2005 #2
    Yeah, you need to resolve the velocities into components.
  4. Apr 8, 2005 #3
    Okay, so if I resolve them, what exactly does North 30 degrees east mean? Is it north or east or north east? If its north east, whats the 30 degrees about?

    How do I draw the diagram? The way I am drawing it right now, they never appear to collide?
  5. Apr 8, 2005 #4
    The oragne ball will be a vector in the negative x direction magnitude 15cm. The green ball has 15cm north+30 degrees east. I would interpret this to be 30 degrees east of north, which would be 60 degrees from the x axis. At the point of collision, the two vectors will be nose-to-nose, so you can start from that point and draw the vectors from there. You can break down the green ball's vectors by making a triangle and using sin and cosin to find each component.

    Then conserve momentum in each direction.
  6. Apr 8, 2005 #5
    remember the i component is (magnitude)cos(angle from + x axis) and the j component is (magnitude)sin(angle from + x axis) then all you have to do is add them, and convert them back into angles and magnitudes (tan^-1 j/i is angle from + x axis, magnitude is sqrt (i^2 + j^2))

    Or you can just draw a vector diagram and work it out with sin and cos rules.
  7. Apr 9, 2005 #6
    So then would this be correct?

    Momentum is conserved

    x direction:
    [tex]mv_{1} + mv_{2} = 2mv_{x}[/tex]
    [tex]10 + 10cos(60) = 2v_{x}[/tex]
    [tex]15 = 2v_{x}[/tex]
    [tex]7.5 = v_{x}[/tex]

    y direction:
    It remains the same [tex]10sin(60)[/tex] for [tex]m_{2}[/tex]

    I just want to know if I did this correctly :) After this, I would simply have to find the direction and composite velocity for the green ball.
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