- Make sure that I am isolated from other people and their experiments so I can be safe.
Key Factors:
Here are some key factors that will affect the experiments. I need to carefully monitor these factors because they may affect whether or not the test is fair and accurate.
Mass/Weight
The mass of paper helicopter will change because I will be adding paper clips, increasing the mass and weight. (These are different: mass is a measurement, weight is a force caused by gravity) The equation I will need to work out weight in relation to mass is:
F = ma
F (weight) = mass (kg) x acceleration (earth)
Surface area
The surface area of the helicopter could affect on the air resistance, which would in turn affect terminal velocity. If the helicopter had a larger surface area, there would be a larger surface for the air molecules to hit against it. This would mean that terminal velocity would be slower/faster depending on the surface area. To prevent this happening and causing an unfair test, I am going to use the same helicopter for each experiment I perform.
Height
The height from which I drop the paper helicopter is important. I must allow enough vertical distance for the helicopter to reach terminal velocity (when the helicopter starts to spin) before I start timing. If I did not the test would be ruined. I will use the same height for timing in each experiment – this will make it easier and a fair test.
Human error
It is going to be very hard to be accurate with the timing of terminal velocity of the paper helicopter. My reactions are quick, but my results may be affected by the fact that I will be doing lots of experiments after another and my reaction time will decrease as the time goes on. To prevent this I will try and do my experiments with short breaks.
Air currents
Air currents or drafts would affect the way in which my helicopter fell, ruining my results. The air molecules would bombard my helicopter as it fell, causing the forces to be upset. This will increase the amount of time spent in the air and producing an unfair test. To prevent this happening I will work away from any doors or windows and make sure that doors and windows are closed before performing my experiments.
Method
Apparatus list:
Paper helicopter
Stool
Meter stick
Paper clips
Electronic scale
Stopwatch
My partner will stand on the stool and drop the helicopter. When the helicopter starts to spin (means it has reached terminal velocity) I will start to time. I will stop time when the helicopter has hit the floor. I shall the record the time in my tables and then repeat using a number of different masses.
Range
To decide on an appropriate range to use, I made some preliminary results. These will help me decide on the following factors:
- Height dropped from.
- Height to time from
- How many paperclips to use
- Whether my helicopter is suitable for the final experiments
Here are my preliminary results –
Helicopter weight – 2.4g (the fact that I am using 2.4g instead of an easier number like 2.5g is relevant because it means that my results will be more accurate)
Paper clip weight – varies so keep a cumulative weight
I decided that 7 different masses would be enough to see whether my prediction was correct and accurately plot a graph. I tested the extremes to save time.
These results were taken from a drop at the ceiling (2.72 metres) and time started at 1.5m.
These results showed that my height of 1.5m allowed for all my helicopters to reach terminal velocity before they reached that point. They also showed that the test wouldn’t be too long or too short for me to time accurately.
I decided to use a range of 0-6 based on my preliminary results.
I decided to use a height of 1.5m based on my preliminary results.
I decided to use a drop height of 2.72m (ceiling) based on my preliminary results.
I decided that my helicopter is suitable for my final experiments.
Repeats
The aim of repeating an experiment is to find out whether what you have recorded is accurate and true. I am going to have five repeats. I think five is the correct number to use because there are many things that could affect the drop of the helicopter and taking five results will mean that I can accurately identify any anomalous results that I might have.
Obtaining
I worked out terminal velocity using this formula:
s = d/t
s – speed
d – distance
t – time
Example:
Time 2, Paperclips 4
s = 1.5/0.75
s = 2.00 m/s
Analysing
GRAPH 1 – The average terminal velocity
From my results, I can see that as I increase the weight, terminal velocity also increases. This is a continuous pattern throughout my results.
From graph number one and two, which both have upward, sloping curves, I can see that as the weight increases in kilograms, terminal velocity increases in metres per second. The increase between each weight is not the same, but is very similar especially in the lower weights
I was correct in my prediction. I said that:
I predict that if I increase the weight, then terminal velocity will be faster.
This has proved true because terminal velocity occurs when acceleration and air resistance (drag) are equal. Air resistance is caused by air particles hitting an object repeatedly in the opposite direction to which it is moving. The greater the speed of the object’s velocity, the harder the air molecules hit the object causing a greater air resistance, which means that when terminal velocity is reached it is at greater magnitude. A greater weight on the paper clip caused it to fall faster, which in turn increased terminal velocity.
From my graphs and results, I can see that my prediction was correct. My line of best fit on both graphs shows that terminal velocity increases as the weight increases. This is why my line curves upwards on both of my graphs.
I said in my prediction that:
‘I think that if the weight (variable) is doubled, the terminal velocity will also double.’
‘I think this because if the amount of paperclips (weight) was doubled, the air resistance, must double for terminal velocity to take place. Therefore, the particles that hit the helicopter must also double. The way this happens is if the speed doubles because of the weight doubling.’
I am now going to see if I was correct in my numerical prediction.
From graph number one, I am using the points (1.00 , 0.14) and (2.00 , 0.29)
These points are also represented by the equations:
-
= 0.14 cm3/s = (1.00 , 0.14)
733
-
= 0.29 cm3/s = (2.00 , 0.29)
342
I predicted:
That if the concentration of acid (variable) is doubled, the rate of reaction will also double.
When I used the concentration 1M, the rate of reaction was 0.14 cm3/s. When I doubled the concentration to 2M, the rate of reaction was 0.29 cm3/s. From these results I can see that my numerical prediction was nearly correct for these two points.
From graph number two, I also going to use the points (1.00 , 0.23) and (2.00 , 0.48).
The reason for using the same points on both graphs is to see if my prediction is true for these points. By comparing them I can do this.
These points are also represented by the equations:
-
= 0.23 cm3/s = (1.00 , 0.23)
213
50 = 0.48 cm3/s = (2.00 , 0.48)
104
I predicted:
That if the concentration of acid (variable) is doubled, the rate of reaction will also double.
When I used the concentration 1M, the rate of reaction was 0.23 cm3/s. When I doubled the concentration to 2M, the rate of reaction was 0.48 cm3/s. From these results I can see that my numerical prediction was nearly correct for these two points.
In both sets of points from the different graphs, I was nearly correct with my numerical prediction. In graph one I was out by 0.01 and 0.02 in graph number 2. These differences may be due to human error. It was very hard to take the precise time and know exactly when a certain amount of gas was produced. This is most likely to have caused the slight errors that wronged my prediction.
However, my prediction was not correct for other points I my graph. For different points, my prediction was not correct. This could have been due to inaccuracy on my part or something was wrong with the experiments. A change in temperature could have had an effect on this.
The reason for my prediction and my results is because if the amount of acid molecules were doubled, the number of collisions per second would also double. This, in turn, would double the rate of reaction.
This also shows a reasonably high degree of accuracy in my experiments.
