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Pool Ball Problem

  1. Dec 14, 2003 #1
    This is probably simple, so please excuse my ignorance, but it's bugging me.

    I place the 8 ball in the center of a pool table. Then the 7 ball is place in the front of the table, say, 1 foot from the 8 ball for a direct line shot. I also place the 6 ball on the side, again 1 foot away for a direct side shot on the 8 ball.
    Two players hit the 6 and 7 ball towards the 8 at exactly the same time and with exactly the same speed and force. Lets say, 10 mph with 1 lb of force.
    Those two balls impact the 8 ball, and the 8 ball flys-off at a 45 degree angle(I think). Fine.
    But at what speed and force does the 8 ball travel? And what happens to the 6 and 7 ball after this impact?

    Thank you for your time.
  2. jcsd
  3. Dec 14, 2003 #2
    Supposing you hit the two ball's exactly in the center, they would both stop dead in their tracks when they hit the 8 ball. As far as what velocity the 8 ball would travel at, you can look at it its Velocity as a magnitude, being the sum of two vectors (one ball being horizontal, and one being vertical). Hope this is correct...

  4. Dec 14, 2003 #3
    This question makes no sense. A ball doesn't travel with force.

    I agree with Jason's answer. The 8 should go off at a 45 degree angle to the initial velocity of either ball with speed v*sqrt2 where v is the initial speed of the 6 ball.
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