Calculations associated with a bouncing ball

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SUMMARY

The discussion focuses on calculating the energy dynamics of a bouncing ball experiment involving different masses, surfaces, and heights. Key equations include the potential energy (PE) retention formula, KE = mg(h0-h1), and the kinetic energy equation KE = ½mv². The calculations demonstrate that 68% of potential energy is retained after impact, with a subsequent kinetic energy of 0.56448 J and a velocity of 7.919 m/s at the second bounce. The discussion emphasizes the importance of accurate measurements of initial and final heights for precise calculations.

PREREQUISITES
  • Understanding of potential energy and kinetic energy concepts
  • Familiarity with basic physics equations related to motion
  • Ability to perform calculations involving mass, height, and gravitational force
  • Knowledge of LaTeX for formatting equations
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  • Research the principles of energy conservation in elastic collisions
  • Learn about the impact of surface materials on bounce efficiency
  • Explore advanced physics concepts such as energy loss and damping
  • Practice using LaTeX for scientific documentation and equation formatting
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Students conducting physics experiments, educators teaching energy dynamics, and anyone interested in the mechanics of motion and energy conservation in bouncing objects.

annamae
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Homework Statement


For our final project we have to design a lab, take the data, and write a lab report. My lab is bouncing balls of different masses, onto different surfaces, from different heights, and use balls of different materials in order to see what factor influences efficiency the most. I am now onto the calculation section and not sure of what equations would be relevant or if I'm even doing them right.


Homework Equations



PE1 x 100
PEinitial

KE = mg(h0-h1)

KE = ½mv2
mg(h0-h1) = ½mv2



The Attempt at a Solution


1. PE1 x 100
PEinitial

mgh1 x 100
mghinitial

(.18)(9.8)(.68) x 100
(.18)(9.8)(1)

68% - the amount of potential energy retained after impact

2. KE = mg(h0-h1)
KE = (.18)(9.8)(1-.68)
KE = .56448 J

a measurement of the joules that the ball has at the top of the second bounce

3. KE = ½mv2
mg(h0-h1) = ½mv2
(.18)(9.8)(1-.68) = ½(.18)v2
.56448 = .009v2
7.919 m/s = v

a measure of velocity when the ball is dropped
 
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Its a bit hard to follow your equations.

If you want to find the energy lost after one bounce it will be:

\Delta E = mg (h_i - h_f)

For the percent energy lost, it is

\frac{E_i - \Delta E}{E_i}

Is this what you were trying to do?

It is unclear what measurements you have made. I am assuming you measured initial height, and final height after one bounce. If so, these calculations would be the correct ones.

Try to learn to use LaTeX. It is quite easy. Here is a link that will help you learn it.

https://www.physicsforums.com/showthread.php?t=8997
 

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