How Does Hitting a Cue Ball Above the Equator Affect Its Motion?

AI Thread Summary
Hitting a cue ball above the equator introduces torque, affecting its spin and motion. The average force applied by the cue stick is 6000 N over a contact time of 20 microseconds. To determine the ball's speed after impact, one must consider both linear acceleration and the torque resulting from the off-center hit. The equations of motion, including F=Ma and the displacement formula, are relevant for calculating the ball's speed. Understanding the impact of the cue's contact point is crucial for accurate results.
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Homework Statement


A cue ball is hit 4.5 cm above the horizontal surface uk = 0.35 it is on.
The ball is 6 cm in diameter and has a mass of 250 g. It is hit
by a cue stick with an average force of
6000 N during a contact time of 20 usec.

(a) What is the ball’s speed after it is hit?

Homework Equations



F=Ma
x=x0+v0xt+1/2at^2

The Attempt at a Solution


I can't figure out the effect being hit above the equator has and how to incorporate it?
Thanks
 
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It sounds like they want you to take into account the torque on the cue ball due to the fact that it was hit off center. I guess you assume that the cue does not slip and that it hits parallel to the horizontal surface, but it doesn't actually say that, so I'm just guessing.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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