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Momentum in Billiards


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Use this link to view a YouTube video from the correct time (2m3s).

Copyright Notice: I do not own any portion of this video. All credit goes to the posters of it on Youtube.com and to the owners of each clip.

Here's a question... how can one pool ball knock six others into the pocket? The answer: conservation of momentum. I like to think of the stagnant six pool balls as one single object. If the cue has mass 'M,' the object including the corner balls has mass 6M. Using conservation of momentum... [ATTACH]50[/ATTACH]

and that after this (nearly) elastic collision, the velocity of the cue ball is 0...

MV + 6M(0) = M(0) + 6MV'

Therefore, V', or the velocity of the six balls, is V/6.

The smaller velocity of each of the six balls explains why one ball can move six. While each ball after the collision is moving slower, the sum of all of their momenta equals that of the cue ball before the collision.

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"momenta" - love it!

Solid post.

Just to add on, inorder to sink all 6 balls, you would need to put a greater force on the cue ball. Greater force means greater impulse since it is equal to F\Delta t . With a greater impulse, the velocity of the cue ball would be greater since F\Delta t = m\Delta V . With a greater velocity, V/6 would be a fast enough speed for the balls to make it to the pockets before the deceleration due to friction brought them to a halt.

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Just because you analyzed billiards physics in some way you get five stars. I would say that the orientation of the six billiard balls if they were grouped would usually create different resulting velocities from the initial momentum of the cue ball.

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Guest Rellseli66

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is this really true though, because I've played pool, and the cue ball does not always come to a stop right after it hits the other ball, and as a result that would change your equation and adjust your speed.

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(sorry for the late reply) I think that the angle is what determines whether the cue ball is stagnant after the collision. In this case, the cue does stop moving and thus I would think that my equation works out, but it certainly is not all-inclusive of all billiards.

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