Trouble finding correct acceleration of a ball

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AI Thread Summary
The discussion revolves around calculating the acceleration of a pool ball after being struck by a cue. The user initially calculated the ball's acceleration as 31.16 m/s² but questioned the validity of this result due to the effects of friction and the ball's initial speed being greater than its final speed. Participants pointed out potential calculation errors and suggested that the assumption of the ball's initial speed being equal to the cue's final speed might be incorrect, given the cue's greater mass and elasticity. To accurately factor in elasticity, the user would need to know the relative masses and apply momentum and energy conservation principles. The conversation emphasizes the importance of considering these factors in collision dynamics.
Drake M
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Homework Statement


Well I need to find the acceleration of a pool ball after it has been hit by the cue. I measured the final velocity of the cue which was 2.9m/s and its acceleration was 41.42m/s*2. I know that the vf of the cue is the vi of the ball and because the ball has friction after it has had the force applied, the acceleration should be a negative value but it keeps coming positive and pretty large too.
OF CUE:
Vi=0m/s
t=.07s
d=.1016m

Used d=1/2(Vi+Vf)t to find Vf rearranged to vf=2d/t-Vi

OF BALL:
Vi=2.9m/s
t=.118s
d=.5588m

Used the ball values for equation below

Homework Equations


d=Vit+(1/2)at*2

The Attempt at a Solution


rearranging for a I get 2(d-(Vi t )) all divided by t*2=a

Substituting values in I get 2(.5588m-(2.9m/s x .118s) divided by .118s*2

That turns into 2(.5588-.3422)/.0139s

and once finished I got 31.16m/s*2 as my acceleration. But if that is the case then my Vf of the ball will be greater than the Vi and that wouldn't make sense considering its on a horizontal surface and friction is being applied. Did I make a mistake in the calculations or values. ANY help would be appreciated, thanks in advance.
 
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Check your multiplication of the ball's initial speed by travel time.
 
2.9 x .118 = .3422
 
Drake M said:
2.9 x .118 = .3422
Hmm.. my mistake.
Maybe you are wrong to assume the ball's initial speed is no more than the cue's final speed. The cue will be more massive than the ball (3 times seems typical), there is some elasticity in the tip, and quite a bit in the shaft of the cue. (See http://dbkcues.ru/articles-2/investigation-in-some-wave-properties-of-a-billiards-cue/?lang=en.)
If you were to go to the extreme and assume a fully elastic collision then I expect you would find the ball's initial speed could be substantially greater than the cue's maximum speed.
 
If I were to factor in elasticity, how would I figure out how it affects it. All I would have would be Vi(or Vf of the cue). Thanks for all the help by the way.
 
Drake M said:
If I were to factor in elasticity, how would I figure out how it affects it. All I would have would be Vi(or Vf of the cue). Thanks for all the help by the way.
You'd need to know the relative masses. With an estimate of the elasticity you could then use momentum and energy considerations to derive the two velocities after the stroke.
 
Ok, thanks
 
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