Solving Pool Ball Problem: Speed & Force After Impact

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In the pool ball problem, two balls (6 and 7) strike the 8 ball simultaneously at the same speed and force. Upon impact, the 8 ball is expected to travel at a 45-degree angle, with its speed calculated as the initial speed multiplied by the square root of 2. The 6 and 7 balls will stop moving if they hit the 8 ball directly in the center. The concept of force is clarified, indicating that a ball does not travel with force, but rather with velocity. The discussion emphasizes the physics of momentum transfer and vector addition in this scenario.
pallidin
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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. Let's 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.
 
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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...

-Jason
 
Originally posted by pallidin
at what force does the 8 ball travel?
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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