Investigating the Speed at Which a Ball Bounces Off a Surface

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Investigating the Speed at Which a Ball Bounces Off a Surface

Plan

Introduction

I intend on investigating the speed at which a ball rebounds of a

given surface. I will try and find a relationship between the speed it

hits the surface and the speed it comes off the surface.

Background Information

======================

The principle of conservation of energy states "Energy cannot be

created or destroyed, only converted from one form to another."1 This

is the reason that the ball does not rebound off the block at the same

speed that it hits it at. As some of the energy has been converted

into a different form when it hits the block.

Prediction

==========

[IMAGE]I predict that as the velocity of the ball approaching the

surface increases then the speed at which it rebounds off it also

increases proportionally. So if I double the speed of Va (velocity of

approach) then the speed at which it rebounds is also doubled.

e.g. Va Vr. This means that the graph of my results should produce a

straight-line graph as they are proportional (as shown in the

diagram). This also means that the wooden block takes the same

proportion of energy every time the ball hits the block

The reason why I think that the rebound speed will be slower than the

approaching speed is that some of the kinetic energy from the moving

ball is converted to the form of heat and sound when it hits the

block. The kinetic energy of the ball has not been destroyed only

changed its form, as quoted in my background information.

If an object of mass 'm' moves at speed 'v'. Then we can say it has a

kinetic energy of ½ x m x v²

This means that the formula for the kinetic energy of the approaching

ball is

K = ½ x m x Va² and so the formula for the rebounding ball must be K =

½ x m x Vr²

As I predict that they are both proportional to each other then I can

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