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