Loading…

Flash Player 9 (or above) is needed to view presentations.
We have detected that you do not have it on your computer. To install it, go here.

Like this document? Why not share!

Ball drop

on

  • 1,444 views

 

Statistics

Views

Total Views
1,444
Views on SlideShare
1,443
Embed Views
1

Actions

Likes
0
Downloads
12
Comments
0

1 Embed 1

http://blogs.isb.ac.th 1

Accessibility

Categories

Upload Details

Uploaded via as Microsoft Word

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

Ball drop Ball drop Document Transcript

  • Non Bunpetch<br />Mr. Eles Physics HL1 P.8<br />11/1/10<br />BALL DROP EXPERIMENT<br />Analysis and Evaluation-<br />Figure 1 raw data taken of position as the ball is dropped to the ground with the motion detector on top. <br />In this case the motion detector is the 0 of the graph, when the ball is dropped it is moved away from the motion detector, and the point such as at 0.85 second is where the ball touches the ground. <br />Figure 2 Shows a manipulated position-time graph that is easier to understand, with the 0 at the ground<br />The graph was manipulated and adjusted so the 0 would be the ground. Interval from 0.85 s to 1.5 s shows a complete bounce, with 1.15 s is which where the ball is at top of a bounce, 1.4 s is a point when the ball is just about to touch the ground, 1.45 s is the instant the ball hit the ground, and 1.5 s is when the ball is just leaving the ground. <br />Figure 3 shows a relationship as Potential Energy to where the ball is located when dropped<br />Potential Energy in this experiment is the gravitational potential stored, it can also be represented as <br />PE = mgh [Eq. 1]<br />Where m is mass, g is gravity, and h is height. A complete bounce from the graph is from 0.85 s to 1.45 s. The point where the ball is at the top of a bounce is at 1.15 s , potential energy is maximum here due to the fact that height is greatest here. 1.4 s is when the ball is just starting to touch the ground, at 1.45 s is the instant where the ball touches the ground, here potential Energy is at it’s least, due to 0 height, and it is also where elastic energy is gained to bounce the ball back up but it does not bounce back as high due to energy lost as thermal. 1.5 is the instant the ball leaves the ground. Potential Energy is decreasing after each bounce due to it does not bounce back as high after each time it hits the ground. <br />Figure 4 shows kinetic energy in relationship to where the ball is.<br />Kinetic Energy is the energy of motion, and can be represented by <br />KE = 1/2mv2 [Eq. 2]<br />Where m stands for mass and v stands for velocity. A complete bounce in this graph is from the interval of 0.85s to 1.45s. The ball at highest point has 0 kinetic energy as the velocity of the ball becomes 0. And as the ball accelerates back toward the ground it gains more and more kinetic energy and reaches maximum at a point right before the ball hits the ground. At the instant the ball hit the ground kinetic energy becomes 0 as velocity of the ball is 0, as can be seen in the graph at point 0.85s. as the ball leaves the ground it has elastic energy stored to help it bounce back up, but not as high as some energy is lost as thermal. It decelerates as it goes up due to gravity acting on it and finally reaching 0 again. <br />Figure 5 shows the total energy of the system as the ball is dropped, the total energy is included of Potential and Kinetic Energy<br />Total Energy in this graph is the Potential and Kinetic Energy combined. A complete bounce is from the interval 0.85s to 1.45s. Throughout the bounce the ball have a constant total energy expect for when the ball hits the ground, then it becomes 0. It bounce have constant total energy because kinetic and potential added to be equal at every point in that bounce. At the point where the ball hits the ground, the energy doesn’t really become 0, it is conserved to elastic, but appeared as 0 because the motion detector does not measure elastic energy. The total energy after each bounce becomes less and less due to the lost in energy as thermal when the ball hits the ground. The percent energy lost can be calculated as <br />1st to 2nd bounce – |(3.8 – 2.2) / 3.8| x 100 = 42.1%<br />2nd to 3rd bounce - |(2.2 – 1.4) / 2.2| x 100 = 36.4%<br />It can be assumed that the percent energy lost becomes less and less after each bounce. <br />