What Happens to Energy When a Ball Bounces off a Wall?

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SUMMARY

The discussion centers on the principles of momentum and energy conservation when a ball bounces off a wall. The key equation referenced is the conservation of energy, specifically the relationship between initial and final energy states, denoted as Ein = Eout. The change in momentum is calculated using the formula m(v), where 'm' is mass and 'v' is velocity. Additionally, the conversation touches on energy transfer in the context of archery, emphasizing the transition of potential energy (Ep) to kinetic energy (Ek) when the arrow is released.

PREREQUISITES
  • Understanding of momentum and its calculation
  • Familiarity with the conservation of energy principle
  • Knowledge of potential energy (Ep) and kinetic energy (Ek)
  • Basic physics equations related to work and power (W - P)
NEXT STEPS
  • Study the concept of momentum in elastic and inelastic collisions
  • Explore detailed examples of energy conservation in mechanical systems
  • Learn about the equations governing potential and kinetic energy transformations
  • Investigate the physics of archery, focusing on energy transfer during arrow release
USEFUL FOR

Students studying physics, educators teaching mechanics, and anyone interested in understanding energy dynamics in collisions and projectile motion.

yamugushi
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Change in momentum of a ball

Please note, instead of making a new thread I just edited my new question into this thread

Homework Statement


A ball is rolled into a wall and bounces back at the same speed, the change in momentum is?

Homework Equations


Ein = Eout

The Attempt at a Solution


I thought it was m(v) but I got it wrong (doing test corrections ATM)
 
Last edited:
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Conservation of energy.

Where does that 100J go when the bow is pulled back and held in place? After the archer releases, where is the energy at the instant when the arrow leaves the bow, and how does that help you estimate its velocity at that instant?
 
JaWiB said:
Conservation of energy.

Where does that 100J go when the bow is pulled back and held in place? After the archer releases, where is the energy at the instant when the arrow leaves the bow, and how does that help you estimate its velocity at that instant?
Oh, I get it, set Ep
to Ek
I was looking at W - P equations
 

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