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Effect of Gravity on impact

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Effect of Gravity on impact
Design Lab Report
Introduction:
This lab is looking at the conservation of energy in moving objects.
Research question: How do the height and kinetic energy of an object affect the diameter of its crater when it hits sand?
Theory:
The theory being looked at here is the transfer of gravitational potential energy (GPE) to kinetic energy (KE). At a point of drop of an object, the potential energy is equal to the kinetic energy of the object just before it impacts the floor:
GPEPosition of drop = KEBefore impact
The measure of the crater that the ball produces after impact should reflect on the change in kinetic energy and its transformation on all the small sand granules around it. If no energy is lost, all the KE of the ball should go into the sand granules and each granule should have a small amount of KE, that adds up to the same amount from the ball when all granules are grouped
Potentially the energy transformation should go:
GPEBall KEBall KESand granules.
Essentially an increase in GPEBall will also transfer into a larger KEBall, and thus larger KE per sand granule as well. The distance the granules move should increase as height does too.
Hypothesis:
My prediction is that as the height of where the ball is dropped increases, the diameter of the crater will also increase. However I do not believe that the kinetic energy of the drop will be directly proportional to the diameter of the crater and there will be a limit to its proportionality. The ball can only transfer so much KE to a small amount of small granules that it doesn’t seem like it can be proportional. Not only this but energy might be lost to thermal and sound energy, however that isn’t a main concern to affecting the proportionality.
Variables
Controlled Variable Independent Variable Dependent Variable
Variable The controlled variable in this case is the type of ball that is used. This includes the diameter, weight and density of the ball. The amount of sand is also something to be controlled and how flat it is The height of the drop will be my independent variable, as I will change and manipulate it to see whether there is any effect on the dependent variable. However ultimately the main independent variable is the Kinetic Energy, which in itself is controlled by the height of the drop. Since what I am changing is the height and KE of the drop, what will ultimately be affected is the diameter of the crater, or at least I believe that it is dependent on the KE of the ball before its impact
How to control/measure variable Because of this, the same ball is used throughout This was changed with the use of a steel rule, starting at 5 cm, going to 40 cm every 5 cm. It was also measured from its center of mass, as shown in fig 1 Since measuring the crater with a ruler could have potentially affected the recording, so a transparent sheet of square cm paper was used, as illustrated in fig 3 and fig 4. Initially, a picture is taken of each recording in fig 2.
8899678890Height of drop: from 5cm-40cm
Fig 1: Measuring of the independent variable
Measured from centre of mass
00Height of drop: from 5cm-40cm
Fig 1: Measuring of the independent variable
Measured from centre of mass

center-2904Square grid is placed after dropping the ball. It is used to measure the diameter of the crater
Picture is taken and diameter is measured later
Fig 2: A picture to measure dependent variable
Square grid is placed after dropping the ball. It is used to measure the diameter of the crater
Picture is taken and diameter is measured later
Fig 2: A picture to measure dependent variable

-341326151599Fig 3: The measuring of the dependent variable
Fig 3: The measuring of the dependent variable

3585739156541To scale
To scale

1175277181418
1206680466737Fig 4: Image of crater in 1 cm square paper
Fig 4: Image of crater in 1 cm square paper

Apparatus
Camera
Tray
Sand
One ball
Cups
Meter ruler
Steel ruler
Transparent grid paper
Bucket
Method
Using cups take sand out of the bucket and lay it onto the tray. When it fills the tray, drop the ball onto the sand at a height of 5 cm, measured by the steel rule (fig 1)
Drop the ball at that height.
Place the transparent grid paper onto the sand after the ball was dropped, carefully on the crater (fig 3). Make sure that the grid paper does not move the sand
Take a picture where the grid meets the crater, at bird’s eye view (fig 2).
Repeat the first five steps at 5 cm, four more times. This way the dependent variable, the height, is repeated five times before moving to a different measurement of the dependent variable
Repeat again at heights of 10, 15, 20, 25, 30, 35 and 40 cm. Five times at each height.
Print the pictures taken (fig 4).
Measure the diameter of the craters with the printed images, using a scale to translate into the cm squared paper.
Input all recorded data into a table of height of drop, uncertainty of height of drop, diameter D of crater and uncertainty of DCrater.
*All the values of the independent variable
Height of ball dropped, cm ± 0.001 m Diameter of crater, cm
Trial 1 Trial 2 Trial 3 Trial 4 Trial 5
0.050 2.71 2.64 2.83 2.73 2.63
0.100 3.00 3.24 3.33 3.50 3.08
0.150 3.64 3.78 3.64 3.67 3.69
0.200 3.83 4.60 4.08 4.00 3.81
0.250 4.67 4.42 4.09 4.53 4.68
0.300 4.63 4.52 4.48 4.63 4.29
0.350 4.82 5.18 5.23 4.94 5.15
0.400 5.25 5.30 5.18 4.99 5.32
Raw Data
The mass of the ball is exactly 21.1 grams, but is translated to 0.0211 kg for processing
Processing
Finding the average of the trials of DCrater: trial 1, trial 2…trial nn of trialsSince five trials were taken for each height, n of trials=5
For height of 0.050 m:
Average of trials=trial 1+trial 2+trial 3+trial 4+trial 55=2.71+2.64+2.83+2.73+2.635=2.71 cmThe same formula goes for results at height 0.100, 0.150, 0.200, 0.250, 0.300, 0.350 and at 0.400 cm.
To find the uncertainty of the diameter of the crater, the formula Maximum value-minimum value2 is used.
For the case of height 0.050 m, the maximum value for D is 2.83 and the minimum value is 2.64.
Uncertianty of Dheight 0.050 cm=2.83-2.642=±0.095 cmAs for the calculations of kinetic energy:
Kinetic energy at impact=Potential energy at beginning of fall12mass×velocity2=mass×gravitational field strength×heightFor the height of 0.050 m.
K.E=0.0211×9.8*×0.050KE=0.0103 J*We assume that gravitational field strength is exactly 9.8.
For the uncertainty of KEBall, we have to use percentages of two different uncertainties and add them for the final KE uncertainty, as both height and mass were multiplied in the processing of KE.
Percentage uncertianty of K.E=Percentage uncertianty of height+Percentage uncertianty of massFor the same height of 0.050 m
Percentage uncertianty of height=0.0010.050=± 0.020%Percentage uncertianty of mass=0.00010.0211=± 0.0047%Percentage uncertianty of K.E=0.020%+0.0047%=± 0.0247%Processed data table
KE of ball before impact, J Percentage uncertainty of KE of ball. Average of the diameter D of crater, cm Uncertainty of average diameter D, cm
0.0103 ±0.0247%2.71 ±0.095
0.0207 ±0.0147%3.23 ±0.250
0.0310 ±0.0117%3.68 ±0.070
0.0414 ±0.0097%4.06 ±0.395
0.0517 ±0.0087%4.49 ±0.295
0.0620 ±0.0080%4.51 ±0.170
0.0724 ±0.0076%5.06 ±0.205
0.0827 ±0.0072%5.21 ±0.165
-1532255152971500
To calculate the % Error of the gradient:
∆m=maximum slope-minimum slope2∆mm×100=% Error of gradientIf the maximum slope’s gradient is 38.00, minimum slope’s gradient is 31.00 and my line of best fit gradient is 34.05, then when implemented to the equations:
∆m=38-312=3.53.534.05×100=10.3% Error in gradientConclusion and Evaluation
Looking back at my hypothesis and theory at the start, there are some areas that I can agree with. Firstly, I was right about the proportionality between the KE of the ball and the diameter of the crater. However, I did not initially think that the relationship would be linear, or as linear as it is in the graph. I thought that there would not be much of a change in the diameter of the crater as the height of the drop, or KE increased, but it stayed constant throughout. However, looking at the % error of the gradient, random errors may have significantly affected the precision of the data that was collected. Even in the error bars it shows that problems with precision did occur. A % error of 10.3% is large and something to consider. However, in the third set of data we see none of the three slopes actually pass it or in its range of error bars. This could be considered as an outlier. The sixth data point in the graph may have the slopes path through it, but does look odd when looking at the general pattern of the data. Since the graph shows correlation and error bars are very big, random errors played more of a role in this experiment. This is potentially another outlier, which may have resulted in a systematic error for that set of data results. In one statement, the graph shows correlation between diameter of crater and KE of the ball, but as there is no line passing the origin the data cannot be directly proportional.
However, since there isn’t a directly proportional graph, systematic errors did occur as well. In this situation the difficulty of measuring the height of the drop was one that could have been a systematic error, as holding the ball in place and dropping it onto the sand was very difficult to do accurately. The controlled variable was managed easily, as I used a ruler to flatten the sand after every test, and the ball was kept constant throughout the experiment. Equipment wasn’t really much of a concern, however in some areas more equipment would have helped a lot for the accuracy of the data. For example a stand for the camera would have made it more accurate to use one scale for all pictures and translate it to the real value. The number of repetitions were good, at five for each height dropped. However, looking at the large error bars, the use of more would maybe make it more accurate. The range of values are very broad and convey a lot of problems with the precision of the results. Time however, could have caused a problem in the results as well. The experiment was done on two different days, and there may have been some changes with the equipment as well as my manipulation of the equipment, or method which might’ve been different one day or the other.

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