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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