For anyone who has difficulty writing up their practicals, this article helps to explain the data collection and processing component of writing an IB Physics Practical.
Basically, the practical involved calculations of momentum using a machine that allowed collisions with minimal friction. The aim was to verify the law of conservation of momentum.
Below is an example of one such practical:
Data collection and Processing:
Qualitative Observations:
It was noticed that the metal of the ramp heated up slightly after each recording and that unless the ramp was adjusted, the glider would move due the metal changing shape. By adjusting the knob at the bottom of the ramp, the height of the ramp can be shifted so that the ramp is roughly parallel to the ground, but since the track has bent, the glider is likely to be influenced by the effects of gravity on the sides of the ramp. This would have added to the uncertainty of the experiment, since the effects of gravity would now be taking effect.
Furthermore, the VELA records the acceleration based on infra-red light readings during two separate times. The glider consists of two pieces of metal that block out the infra-red light and the time that it is blocked out is used by the machine to calculate the acceleration. However, the two pieces of metal reflect light and may have reflected some of the natural light in the room, onto the reader. This may also have added to the uncertainty in the experiment.
When the ramp was set such that it was horizontal and the glider was placed in the centre, the glider oscillated back and forth. The air rising from the holes in the ramp would brush up against the edge of the glider and push it back. Then the air behind the glider would also brush against the edge of the glider and push it forward.
After the collision of the two gliders, it was noticed that the gliders' final speeds (separating after the collision) was similar to their initial speeds (before the collision). This suggests that the value of the final momentum should be similar to the value of the initial momentum.
After the string between the gliders was cut, the magnetic force of repulsion between the magnets attached to them caused them to travel in separate directions. It was noticed that the heavier glider had a slower speed than the lighter glider. This is necessary for their momentum to be equal (and opposite since their directions are opposite), since p = mv. If m is greater, v will need to be smaller for p to be equal for both gliders.
Raw Data:
Note: The units will be converted into the correct SI units in the processed data.
There were 2 gliders, glider A and glider B. Glider A had a mass of 261g. Glider B had a mass of 309g. Let the uncertainty be half the last digit since the scales used can only measure accurate to 1 gram.
Glider A will always be on the left initially and Glider B will always be on the right initially. Let the direction of the velocity to the right be positive.
The uncertainty of the mass of each glider is +/- 0.5 grams since the scales can only measure accurately to that degree. The uncertainty of the initial and final velocities would in theory be +/- 0.05 mms-1. This will be recorded into the raw data.
However, there are more uncertainties involved in this measurement. It is not possible for the slope of the ramp to be perfectly horizontal, therefore gravity will take effect. Furthermore, the air rising from the holes would also alter the measurement slightly. The light reflecting off the metal plates may also interfere with the result. Also, the light recorder calculates the velocity based on the length of the metal plates. The length of these plates is 100mm +/- 0.5mm. Thus there is much more uncertainty than +/-0.05mms-1.
The uncertainty cannot be quantified, but it would mean that the data can no longer be measured to 4 significant digits. Thus a rough estimate of the uncertainty will be 5mms-1. This is half the last digit of a measurement to 3 significant figures.
The correct number of significant digits will be used in the processed data. For now, it will be left as 4 significant digits since it is the raw data.
Collision Experiment
Glider
Mass+/-0.05(g)
u +/-5(mms-1)
v +/-5(mms-1)
A
261
+1493
-836.0
B
309
-1186
+780.0
Magnetic Repulsion Experiment
Glider
Mass+/-0.05(g)
u+/-5(mms-1)
v+/-5(mms-1)
A
261
0.000
-280.0
B
309
0.000
+230.0
Processed Data: The data will now be converted into SI units as well as their uncertainties.
The mass in the raw data will need to be converted into kg. Thus the uncertainty will become+/-0.0005kg.
The velocity in the raw data will be converted into ms-1. This will only be to 3 significant digits, as explained above and the uncertainty will be 0.005ms-1.
Collision Experiment
Glider
Mass+/-0.0005(kg)
u+/-0.005(ms-1)
v+/-0.005(ms-1)
A
0.261
+1.49
-0.836
B
0.309
-1.19
+0.780
Magnetic Repulsion Experiment
Glider
Mass+/-0.0005(kg)
u+/-0.005(ms-1)
v+/-0.005(ms-1)
A
0.261
0.000
-0.280
B
0.309
0.000
+0.230
Calculating Momentum:
Let the mass of glider A be mA and its initial velocity be uA and its final velocity be vA Let the mass of glider B be mB and its initial velocity be uB and its final velocity be vB
Let the initial momentum of glider A be piA and the final momentum be pfA. Let the initial momentum of glider B be piB and the final momentum be pfB.
Collision Experiment:
piA= mAuA = 0.261x 1.49 = 0.389 Ns
The uncertainty can be calculated by adding the relative uncertainties of the mass and the initial velocity:
Relative error of mass = 0.0005/0.261 = 0.002 Relative error of initial velocity = 0.005/1.49 = 0.003 Sum of relative errors = 0.005 or 0.5% Absolute error = 0.005 x 0.389 = 0.002
Thus piA = 0.39 +/- 0.002 Ns
Using the same method, the other values of momentum, as well as their absolute error can be calculated:
pfA= mAvA = (0.261+/-0.0005) x (-0.836+/-0.005) = - 0.22+/- 0.002 Ns
piB= mBuB = (0.309+/-0.0005) x (-1.19+/-0.005) = - 0.36 +/- 0.002 Ns
pfB= mBvB = (0.309+/-0.0005) x (0.780+/-0.005) = 0.24 +/- 0.002 Ns
Magnetic Repulsion Experiment:
Since uA and uB are both equal to 0, the initial momentum of glider A and glider B are both equal to 0. This is because pi = mu, and m x 0 = 0.
pfA= mAvA = (0.261+/-0.0005) x (-0.280+/-0.005) =- 0.07 +/- 0.001 Ns
pfB= mBvB = (0.309+/-0.0005) x (0.230+/-0.005) = 0.07 +/- 0.002 Ns
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