I can therefore choose to calculate the proportional change at either the top or the bottom of
every bounce, I have chosen to take the reading from the top of every bounce because it is
more likely to be accurate. The reason for which is shown below in the graph
This graph in theory should always come to rest at a base point after every bounce at o
metres i.e. the x-axis. However as can be seen from the graph this is not apparent in my
graph. The blue line roughly represents the base point for the graph (where the ball actually
impacted) this is not at 0 because when taking the footage the glass tile was actually in the
picture and therefore took up some of the frame which the computer then counted as height.
Other programs allow you to account for this, If I wanted to solve this The tile would be
placed just out of frame.
The reason that the ball does not consistently impact with the blue line is because of the
sample rate, obviously the ball does impact with the glass tile but it is moving so fast that
within 1/25 of a second (sample rate) the chances are that the ball will have impacted and
reversed. So the chances of capturing the ball in the split second when it is close to or
touching the base are remote. As can be seen in two places on the graph the ball was
captured close to the base line by chance.
Whereas at the top of each bounce the ball slowly loses momentum turning its kinetic energy
into potential energy until it reverses direction and begins to drop. The ball takes a longer
time to achieve this as can be shown by the larger amount of plotted points at the top of
each bounce when compared to the bottom of each bounce. This greater amount of
recording at the top of each bounce enables a more accurate reading of what is perceived
by this data to be the top of each bounce.
The proportion of lost energy is given by h /h and so on. The general rule for this is shown
below:
This has cancelled down to just H because M and G are just constants in this case which
means that and when divided by each other than simply cancel out to form 1.
I have now taken my raw data and made a table of all of the peaks, shown below:
PEAK NUMBER
Y COORDINATE / METRES
PEAK
0.573
PEAK
0.488
PEAK
0.413
PEAK
0.356
PEAK
0.306
PEAK
0.242
I must now work out p / p to find out what proportion of the original height the ball will
reach.
PEAK NUMBER
Y /
METRES
% OF ORIGNAL
BOUNCE HEIGHT
PEAK
0.573
-----------------
PEAK
0.488
85.1657941
PEAK
0.413
84.6311475
PEAK
0.356
86.1985472
PEAK
0.306
85.9550562
PEAK
0.242
79.0849673
This table highlights that the % of the previous bounce height reached by the ball is on
average round about 85.49% (2d.p) when excluding the peak six result. This result is over
6% of the mean of the other numbers which would indicate that it is an anomalous result, If it
were to be included in the mean it would lower the result to 84.21% (2d.p). If I had more
time it would be nice to get some more results from the same experiment to find out if this
result really was wrong, however the graph below also indicates towards a false result.
These results when plotted on a graph with a line of best fit give the show a linear
relationship. Which would indicate that this is a constant figure, with more results I would
have obviously got closer to this result as I could average the results to lessen the impact of
anomalous results.
To indicate how close this result is to the others I am going to work out the average distance
away from the average and compare the result i.e. the standard deviation.
x
x
PEAK
85.1658
7253.21
PEAK
84.6311
7162.42
PEAK
86.1985
7430.18
PEAK
85.9551
7388.28
PEAK
79.085
6254.44
TOTAL
421.036
35488.53
From this it can be worked out that
N = 5 x = 7090.85 and x = 35488.53
Now I can work out the standard deviation which is:
The standard deviation is 2.62 (2d.p). Despite the fact that there will be some small
rounding errors the result still shows that the anomalous figure of 79.085 is roughly twice as
far off the average result as most other figures. This may because of a program error,
because the camera has not caught the apex of the curve in the footage causing me to plot a
lower point and as a result get a lower returning height.
This result may also be due to an error with the programming, If I for example placed the
point to low when plotting the points in the program. I worked out how far in cm I would
have had to have misplaced the point from the average to cause such a degree of error the
answer was 1.56826 (working shown below)
This is a plausible answer to the problem. It may also be explained by interlacing. This
occurs because of the way the camera works. It takes two images and combines them for
every frame for example if a video camera has 600 video lines every other line would take
the image (300 lines) this would be processed and then the other 300 would take an image
this is to help the camera process the information. However if an object is moving at great
speed then the camera frame can have two balls which are split into lines. You must choose
just one ball and plot it. But just after impact it can be hard to tell which ball is the correct
one, as one can be going into the bounce while the other exits and it is hard to distinguish.
This may also have caused the error.
When the ball hits the floor and stops, that energy has to go somewhere. The energy goes
into deforming the ball -- from its original round shape to a squashed shape. When the ball
deforms, its molecules are stretched apart in some places and squeezed together in others.
As they are pushed about, the molecules in the ball collide with and rub across each other
causing a build up of heat, and although immeasurable in this experiment it can be used by
dropping Lead shot in a cardboard tube. Some energy is also lost through sound.
CONCLUSION
From this experiment I can conclude that yes the same proportion of energy is lost per
bounce irrelevant of height. I can say that this is a constant of around 85.5% for this
material.
I can say this because I have carefully analysed the data and its limitations for example, I
did not have enough readings at the base of the bounce to properly analyse the velocity to
find out the kinetic energy. So instead I worked the result out using the 1potential energy at
the apex of the curve. I also worked out which and why certain bits of data proved to be
anomalous.
With more results I could have got a more exact figure for this material but I feel that The
results which I do have indicate towards the correct figure and are on the whole consistent
despite the limitations of the video footage.
After looking on the internet I have found the results for glass on glass and the return height
for this is 96%, with steel on glass this result is 90%. This result is interesting because it
shows how much more energy is lost from a rubber ball that glass or steel.
BIBLIOGRAPHY
source: ADVANCING PHYSICS
Author: Edited by Jon Ogborn and Mary Whitehouse
Published: 2nd Edition 2001
Publisher: Institute Of Physics Publishing’s
Source: www.exploratorium.edu/baseball/bouncing_balls.html
Author: Paul Doherty
Published: 1997
Source: www.phys.hawaii.edu/~teb/java/ntnujava/bouncingBall/bouncingBall.html
Author: Gfu-Kwun Hwang
Published 1997
APPENDIX
RAW DATA DIRECTLY PRINTED FROM MULTIMEDIA MOTION
