Trouble finding correct acceleration of a ball

  • Thread starter Thread starter Drake M
  • Start date Start date
  • Tags Tags
    Acceleration Ball
Click For Summary

Homework Help Overview

The original poster attempts to determine the acceleration of a pool ball after being struck by a cue. They provide measurements for the cue's final velocity and acceleration, as well as the initial conditions for both the cue and the ball. The poster expresses confusion regarding the positive acceleration value obtained, which seems inconsistent with the expected effects of friction on the ball's motion.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculations made by the original poster, particularly focusing on the multiplication of the ball's initial speed and travel time. There is also a consideration of the mass difference between the cue and the ball, and how elasticity might affect the ball's initial speed.

Discussion Status

Some participants have pointed out potential errors in the original poster's calculations and assumptions. There is an ongoing exploration of how to incorporate elasticity into the analysis, with suggestions to consider relative masses and momentum principles. The discussion is active, with participants seeking to clarify and refine their understanding of the problem.

Contextual Notes

The original poster is working under the constraints of a homework assignment, which may limit the information available regarding elasticity and mass ratios. There is an implicit assumption that friction will affect the ball's acceleration, which is being questioned in the context of the calculations presented.

Drake M
Messages
19
Reaction score
0

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.
 
Physics news on Phys.org
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
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
20
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 14 ·
Replies
14
Views
6K