1. Not finding help here? Sign up for a free 30min 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!

Cue ball collision physics

  1. Nov 21, 2004 #1
    63. A pool shark is forced to do a tricky shot in order to win a game. He needs to sink the target ball in the corner pocket at an angle of 30 degrees away from the collision location.

    The Cue ball needs to bounce off of the other ball at an angle of 315 degrees with a velocity of of 0.75 m/s. If he gives a cue ball a velocity of 1.00 m/s @ ) degrees what is the velocity of the target ball after the collision?

    --I'm just having a hard time visualizing this one, and if someone could help start me off that would be swell.
  2. jcsd
  3. Nov 21, 2004 #2
    thinking of 315 degrees as -45 degrees might help some. It's just a simple matter of resolving the momentums in the x and y directions. Let's take the path the cueball takes to the site of collision to be our 0 degrees reference point. Before the collision, it's moving at 1m/s at 0 degrees from the reference angle. After the collision, the cueball is moving at -45 degrees at a velocity of 0.75 m/s. You should have all the information you need to determine the velocity of the other ball given its direction is 30 degrees from the reference angle. Tell me if you'd like to expound or put into equations what I just said.
  4. Nov 22, 2004 #3
    Okay so what I've done so far...

    I did the M-V-P for the cue ball before collision

    M: x
    V: 1 m/s
    P: 1x

    Then I did the after collision mvp for both cue ball and target ball.

    Cue: M:x V:0.75 P: 0.75x
    Target: M: x V:? P:?

    What should I do next?
    I'm stuck again, but I'll continue to work on it. Hopefully you can help me out soon. =p
    Last edited: Nov 22, 2004
  5. Nov 22, 2004 #4
    Anyone? =(
  6. Nov 22, 2004 #5
    (Didn't see the private message, so absent-minded I am) You have to work out the momentums in the x and y directions separately. That's the only way you can resolve this question. Tell me if you need help setting them up but I don't want to tell you more than you'd like to know :)
  7. Nov 22, 2004 #6
    I do I'm completely stumped on this question. If you could maybe write out what your telling me in formula form, that might even help more. Thanks.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Cue ball collision physics
  1. Cue Ball (Replies: 1)