How Does Kinetic Energy Affect a Bouncing Ball's Height After Multiple Bounces?

Click For Summary

Homework Help Overview

The discussion revolves around the behavior of a bouncing ball and how kinetic energy influences its height after multiple bounces. The original poster presents a mathematical relationship involving a coefficient of restitution, denoted as 'e', and an initial height 'H'. Participants are tasked with demonstrating how the height after 'n' bounces can be expressed as H_n = e^(2n) * H.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between kinetic energy and height, with some considering the implications of energy loss during bounces. There is also a focus on the mathematical representation of the problem, including the correct interpretation of the coefficient 'e'.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some have provided insights into the energy transformations involved, while others have raised questions about the accuracy of the original statement regarding the fraction 'e' and its implications for the solution.

Contextual Notes

There is a noted ambiguity regarding the coefficient 'e' and whether it represents a loss or retention of energy, which may affect the mathematical formulation of the problem. Additionally, the original poster's request for help indicates a level of uncertainty about the underlying physics concepts.

Hangst
Messages
12
Reaction score
0

Homework Statement


A little ball bounces straight on a surface in a certain way so that at every bounce it loses a fraction of it's vertical speed (called e in this equation). Assume that the ball is released from a ''resting height' (called H in this equation).

Show that after the n'th bounce against the surface the ball has reached:

[itex]H_n=e^2n * H[/itex]

Homework Equations



I'm not sure.

The Attempt at a Solution



I'm thinking I need to maybe look at the kinectic energy but I'm sort of mystified. Any help is appreciated.
 
Physics news on Phys.org
Hangst said:

Homework Statement


A little ball bounces straight on a surface in a certain way so that at every bounce it loses a fraction of it's vertical speed (called e in this equation). Assume that the ball is released from a ''resting height' (called H in this equation).

Show that after the n'th bounce against the surface the ball has reached:

[itex]H_n=e^2n * H[/itex]

Homework Equations



I'm not sure.

The Attempt at a Solution



I'm thinking I need to maybe look at the kinectic energy but I'm sort of mystified. Any help is appreciated.

It is supposed to be e^2n not e^2 n
 
Consider the moment before impact with ground and the moment after impact.

Before impact, speed of the ball = v; Kinetic energy of the ball = 1/2 mv2
After impact, speed of the ball = e*v; Kinetic energy of the ball = 1.2 m (e*v)2

∴the total mechanical energy of the ball has been reduced to e2 of the original after one bounce, and similarly would be reduced to e2+2, after third bounce to e2+2+2, and so on.

∴after the nth bounce mechanical energy of the ball is E=e2n*E0, where E0 is the energy before the first bounce.

As all kinetic energy after impact is then converted into gravitational potential energy when the ball goes to the highest point, and because gravitational potential energy is proportional to the change of height...

Gravitational potential energy is reduced to e2n of the original.
∴the height is Hn =e2n∗H
 
Actually the OP said it loses a fraction 'e', but that must be a typo. To get the desired answer it must retain fraction e.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
13
Views
4K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 13 ·
Replies
13
Views
9K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K