# Investigate impulse and rebound height

Extracts from this document...

Introduction

Name: Jonathan Tam Class : 12A

Physics

Experiment Report – Planning (b), Data Analysis

## Investigate impulse and rebound height

Planning (a): To investigate the relationship between impulse and rebounce height.

Planning (b): In order to measure rebounce height, we first have to select an apparatus which has a considerable rebounce. This also means something which undergoes a highly elastic collision.

Tennis ball seems to be a good object to use since it rebounces at a considerable and measurable height.

Measuring the height is a problem. Since the tennis ball reaches its highest rebounce height and stays there for not more that a second, it is very difficult, if not impossible, to measure this height at this split second.

From literature formula, we have the followings:

s = ut + 1/2at2

where s= displacement, u = initial velocity, a = acceleration, t = time

Middle

1

0.56

0.495

0.31

0.2

0.145

2

0.8

0.41

0.235

0.16

0.11

3

0.83

0.35

0.265

0.24

0.175

4

0.82

0.445

0.25

0.17

0.125

5

0.85

0.365

0.285

0.16

0.11

6

0.795

0.385

0.265

0.155

0.105

7

0.825

0.425

0.22

0.155

0.11

s (m)

1

1.53664

1.200623

0.47089

0.196

0.103023

2

3.136

0.82369

0.270603

0.12544

0.05929

3

3.37561

0.60025

0.344103

0.28224

0.150063

4

3.29476

0.970323

0.30625

0.14161

0.076563

5

3.54025

0.652803

0.398003

0.12544

0.05929

6

3.096923

0.726303

0.344103

0.117723

0.054023

7

3.335063

0.885063

0.23716

0.117723

0.05929

Due to the fact that the time is the time it takes to complete the whole displacement, which is double of the height. So we divide time by two to get the time it takes to complete the displacement from the highest point to the ground.

From the graph, we can see there is a relationship between the different rebounces. This is due to the fact that as the ball first hit the gorund, much energy is lost to overcome the significant air resistance and some kinetic energy is changed into heat energy and transferred into the ground. So the rebound heights are not as high as the first one, and they have a decreasing trend.

The rate of decreasing rebound height slows down as the number of rebounces increases.

Conclusion

t2 (s2)

1

0.3136

0.245025

0.0961

0.04

0.021025

2

0.64

0.1681

0.055225

0.0256

0.0121

3

0.6889

0.1225

0.070225

0.0576

0.030625

4

0.6724

0.198025

0.0625

0.0289

0.015625

5

0.7225

0.133225

0.081225

0.0256

0.0121

6

0.632025

0.148225

0.070225

0.024025

0.011025

7

0.680625

0.180625

0.0484

0.024025

0.0121

s (m) | |||||

1 | 1.53664 | 1.200623 | 0.47089 | 0.196 | 0.103023 |

2 | 3.136 | 0.82369 | 0.270603 | 0.12544 | 0.05929 |

3 | 3.37561 | 0.60025 | 0.344103 | 0.28224 | 0.150063 |

4 | 3.29476 | 0.970323 | 0.30625 | 0.14161 | 0.076563 |

5 | 3.54025 | 0.652803 | 0.398003 | 0.12544 | 0.05929 |

6 | 3.096923 | 0.726303 | 0.344103 | 0.117723 | 0.054023 |

7 | 3.335063 | 0.885063 | 0.23716 | 0.117723 | 0.05929 |

We can also see that the graph appears to be a curve. In order to prove this is a graph with the correct formula, we can deduce the formula by:

s = 1/2at2

log s = log(1/2at2)

log s = log(1/2) + log(at2)

log s = log(1/2) + log(a) + log(t2)

log s = log(1/2) + log(9.8) + 2log t

log s = 2log t + log(4.9)

#### As from above, if we plot log s against log t, we should be able to get a straight line graph with slope 2 and a y-intercept of log 4.9 = 1.7

From the graph, there is a straight trendline. So it has been proved that the results are valid to prove that displacement increases proportionally to (time)2 .

Conclusion:

Displacement increases proportionally to (time)2 .

This student written piece of work is one of many that can be found in our International Baccalaureate Physics section.

## Found what you're looking for?

- Start learning 29% faster today
- 150,000+ documents available
- Just £6.99 a month