How Much Energy Does a Squash Ball Lose After Bouncing?

AI Thread Summary
To determine the energy loss of a squash ball after bouncing, the gravitational potential energy (PE) at the initial height of 3.0 meters must be compared to the PE at the height of 0.076 meters. The relevant equations are Ek = 1/2mv² and Eg = mgh, where 'g' is 9.80 m/s². The calculation involves finding the two mgh values and determining the percentage of energy lost during the bounce. The discussion emphasizes understanding the heights involved and applying the correct formulas to solve the problem. This approach will yield the percentage of gravitational potential energy lost after the bounce.
Araiz
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Homework Statement


This is a simple question but I can't seem to solve it( I don't know if this is advanced physics or not first time on this website, it said Work and that is part of this energy loss problem) here is the problem: What percentage of its gravitational potential energy does a squash ball lose if it falls from 3.0 m and returns to a height of .076 after bouncing once?

Homework Equations


(100-x)(Ek1 + Eg1)/100 = Ek2 + Eg2
Ek = 1/2mv²
Eg = mgh

[/B]

The Attempt at a Solution


Honestly, my Teacher did not give any example on a question with no "x" variable like this one, so literally I had no clue on how to attempt this question.
 
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Araiz said:

Homework Statement


This is a simple question but I can't seem to solve it( I don't know if this is advanced physics or not first time on this website, it said Work and that is part of this energy loss problem) here is the problem: What percentage of its gravitational potential energy does a squash ball lose if it falls from 3.0 m and returns to a height of .076 after bouncing once?

Homework Equations


(100-x)(Ek1 + Eg1)/100 = Ek2 + Eg2
Ek = 1/2mv²
Eg = mgh

[/B]

The Attempt at a Solution


Honestly, my Teacher did not give any example on a question with no "x" variable like this one, so literally I had no clue on how to attempt this question.
Welcome to the PF.

You would just compare the E=mgh values of PE for the start and end positions of the ball. But what does "to a height of 0.076" mean? 0.076 meters, or 0.076*3.0 meters?
 
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berkeman said:
Welcome to the PF.

You would just compare the E=mgh values of PE for the start and end positions of the ball. But what does "to a height of 0.076" mean? 0.076 meters, or 0.076*3.0 meters?
Thanks !

Its H1= 3.0 m and H2= .076, I know that for sure because, the ball is first at 3.0 M then it bounces and reaches a height of .076m. that's all i got from the question, oh and g= 9.80.
 
Araiz said:
Thanks !

Its H1= 3.0 m and H2= .076, I know that for sure because, the ball is first at 3.0 M then it bounces and reaches a height of .076m. that's all i got from the question, oh and g= 9.80.
So all you have to do is calculate the two mgh values, and figure out what percentage of the original PE was lost in the bounce. :smile:
 
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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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