- Level: International Baccalaureate
- Subject: Physics
- Word count: 3256
Investigate the Size of Craters in Sand Due to Dropped Object.
Extracts from this document...
Introduction
To Investigate the Size of Craters in Sand Due to Dropped Object
________________________________________________________________________
Name : Mohd Hafiz B.Mohd
Partner : Mohd Fhati B.Azait
Subject : Physics
Teacher : Mrs. Marzini
Class : A03A
Title:
To Investigate the Size of Craters in Sand Due to Dropped Object
Abstract:
The aim of this experiment is to investigate the size of craters in sand due to dropped object. Our research question is; what is the correlation between the depth of craters in sand due to slotted mass and the height of dropping slotted mass? The size of craters here is represent by the volume of craters itself. We predicted that at the greater height, as the result the depth of the craters will also greater. In other words, when the height of dropping slotted mass is increased in corollary the depth of the craters will increase too. We also predicted that the volume of craters will also increase as the depth of craters increase.
Based on the theory, when the potential energy is increased, at the same time kinetic energy will increase too in virtue of its motion. If energy is conserved which indicates that PE is equal to KE and no energy loses to the surroundings. The velocity of the slotted mass dropped at each height and it impulse can be obtained.
Middle
Radius, r = 1.9 ± 0.05 cm
∏ = constant value, 22/7
Height of Dropped, cm | Depth, cm (d) | Calculation for the Volume of Craters, cm ³ |
2.00 | 0.30 | = ∏ r ²d = (22/7) x (1.9) ² x 0.3 = 3.40 cm ³ |
4.00 | 0.50 | = ∏ r ²d = (22/7) x (1.9) ² x 0.5 = 5.67 cm ³ |
6.00 | 0.60 | = ∏ r ²d = (22/7) x (1.9) ² x 0.6 = 6.81 cm ³ |
8.00 | 0.70 | = ∏ r ²d = (22/7) x (1.9) ² x 0.7 = 7.94 cm ³ |
10.00 | 0.90 | = ∏ r ²d = (22/7) x (1.9) ² x 0.9 = 10.21 cm ³ |
12.00 | 1.10 | = ∏ r ²d = (22/7) x (1.9) ² x 1.1 = 12.48 cm ³ |
- Third Reading Calculation
Diameter of slotted mass = 3.8 ± 0.05 cm
Radius, r = 1.9 ± 0.05 cm
∏ = constant value, 22/7
Height of Dropped, cm | Depth, cm (d) | Calculation for the Volume of Craters, cm ³ |
2.00 | 0.50 | = ∏ r ²d = (22/7) x (1.9) ² x 0.5 = 5.67 cm ³ |
4.00 | 0.60 | = ∏ r ²d = (22/7) x (1.9) ² x 0.6 = 6.81 cm ³ |
6.00 | 0.70 | = ∏ r ²d = (22/7) x (1.9) ² x 0.7 = 7.94 cm ³ |
8.00 | 0.90 | = ∏ r ²d = (22/7) x (1.9) ² x 0.9 = 10.21 cm ³ |
10.00 | 1.00 | = ∏ r ²d = (22/7) x (1.9) ² x 1.0 = 11.34 cm ³ |
12.00 | 1.10 | = ∏ r ²d = (22/7) x (1.9) ² x 1.1 = 12.48 cm ³ |
- Fourth Reading Calculation
Diameter of slotted mass = 3.8 ± 0.05 cm
Radius, r = 1.9 ± 0.05 cm
∏ = constant value, 22/7
Height of Dropped, cm | Depth, cm (d) | Calculation for the Volume of Craters, cm ³ |
2.00 | 0.20 | = ∏ r ²d = (22/7) x (1.9) ² x 0.2 = 2.27 cm ³ |
4.00 | 0.40 | = ∏ r ²d = (22/7) x (1.9) ² x 0.4 = 4.54 cm ³ |
6.00 | 0.60 | = ∏ r ²d = (22/7) x (1.9) ² x 0.6 = 6.81 cm ³ |
8.00 | 0.80 | = ∏ r ²d = (22/7) x (1.9) ² x 0.8 = 9.07 cm ³ |
10.00 | 0.90 | = ∏ r ²d = (22/7) x (1.9) ² x 0.9 = 10.21 cm ³ |
12.00 | 1.10 | = ∏ r ²d = (22/7) x (1.9) ² x 1.1 = 12.48 cm ³ |
Calculation of Uncertainties for the Volume of the Craters:
In order to calculate the uncertainties for the volume of the craters, we can use the formula below:
- First Reading Calculation
Height of Dropped, cm | Depth, cm (d) | Calculation of Uncertainties for the Volume of Craters, cm ³ |
Uncertainties = ± 0.05 | Uncertainties = ± 0.05 | |
2.00 | 0.10 | ± 0.60 |
4.00 | 0.20 | ± 0.63 |
6.00 | 0.40 | ± 0.67 |
8.00 | 0.50 | ± 0.72 |
10.00 | 0.70 | ± 0.78 |
12.00 | 0.80 | ± 0.81 |
- Second Reading Calculation
Height of Dropped, cm | Depth, cm (d) | Calculation of Uncertainties for the Volume of Craters, cm ³ |
Uncertainties = ± 0.05 | Uncertainties = ± 0.05 | |
2.00 | 0.30 | ± 0.66 |
4.00 | 0.50 | ± 0.72 |
6.00 | 0.60 | ± 0.75 |
8.00 | 0.70 | ± 0.78 |
10.00 | 0.90 | ± 0.84 |
12.00 | 1.10 | ± 0.90 |
- Third Reading Calculation
Height of Dropped, cm | Depth, cm (d) | Calculation of Uncertainties for the Volume of Craters, cm ³ |
Uncertainties = ± 0.05 | Uncertainties = ± 0.05 | |
2.00 | 0.50 | ± 0.72 |
4.00 | 0.60 | ± 0.75 |
6.00 | 0.70 | ± 0.78 |
8.00 | 0.90 | ± 0.84 |
10.00 | 1.00 | ± 0.87 |
12.00 | 1.10 | ± 0.90 |
Conclusion
that we used is in the unsatisfied condition. Actually, many laboratory apparatus is already outdated. For example, the scale is not accurate or precise than it supposed to be.
...read more.
There are several ways to overcome these problems from occurred again. We can prevent it by making some improvements:
- In order to get an accurate result we should use the laboratory apparatus which in the best condition.
- The experiment should be repeated for many times. This is because we can take the mean of all the reading in order to get accurate result.
Conclusion:
- Hypothesis is valid and accepted.
- At the end of the experiment, we realized that the correlation between the depth of craters and the height of dropping slotted mass is the height of dropped slotted mass is proportional to the depth of craters. Or in other words, we can say that the depth of craters is also proportional to the volume of craters. Once the depth of craters increased, as the result the volume of craters will increase too.
Contribution:
- I contributed some notion about how we will conduct this experiment in the early discussion.
- Besides that, I also surf the internet in order to get some information about this experiment.
- I tried to find the method that we were going to use in order to carry out this experiment from the reference books in library.
________________________________________________________________________
- -
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
