Purpose
The purpose of this lab was to evaluate the increase in velocity with time during a free fall. Also, to determine the position from the start of the fall.
Hypothesis
It is hypothesized that the object that is dropped will endure gradual change acceleration in acceleration. The acceleration for the object in the velocity-time graph will be gravity (9.81 m/s^2). And the velocity-time graph will be a straight line.
Materials
– C Clamp
– Metre-stick
– Mass (50g)
– Cushion
– Recording timer tape
– Graph paper
– Recording timer
– Power supply
– Long retort stand
– Masking tape
Procedure
I used the table on the given lab sheet to design a table for my results. Me and my lab partners clamped a recording timer in a vertical position above the floor as shown in figure 1. We used masking tape to attach a 50g mass to the end of the 121cm recording tape. We threaded the recording tape through the timer and held the upper end vertical to minimize the friction between the timer and the tape.
We started the timer and then released the 50g mass. Afterwards, individually we analyzed the tape which had been marked into 6 dot intervals. Every 6th dot is 0.1s. I measured and recorded the displacements corresponding to the 6 dot intervals in table 1. I calculated and recorded average velocity for each 6 dot interval. I plotted a graph of velocity against time using the half time intervals.
I calculated the slope of the velocity-time graph, which gave me acceleration in cm/s^2. I measured and recorded the position from the start of the tape corresponding to each half time interval and then plotted a velocity-position graph using the velocities I recorded. I then answered all the given questions.
Observations
|
Time t (s) |
0.1s |
0.2s |
0.3s |
0.4s |
0.5s |
0.6s |
|
Displacement (cm[down]) |
2.5cm |
9.2cm |
15.6cm |
26.5cm |
31.3cm |
36.2cm |
|
Average velocity (cm/s[down]) |
25cm/s |
92cm/s |
156cm/s |
265cm/s |
313cm/s |
362cm/s |
|
Time t (s) |
0.05s |
0.15s |
0.25s |
0.35s |
0.45s |
0.55s |
|
Position from the start (cm[down from the start]) |
1.25cm |
5.85cm |
13.65cm |
26.9cm |
42.55cm |
60.65cm |
Calculations
Average velocity was found by dividing change in time by displacement. The equation was as follows.
Vave = Δd/ Δt
One example calculation would be at 0.2 seconds, when displacement is 9.2cm. To find average velocity I did the following:
Vave = Δd/ Δt
= 9.2cm/0.1s
= 92cm/s
Therefore at 0.2s, the average velocity was 92cm/s.
Acceleration was found by finding the slope of the velocity time graph.
m = (y2-y1)/(x2-x1), In this case
a = (v2-v1)/(t2-t1)
= (362cm-25cm)/ (0.55s-0.05s)
= 337cm/0.50s
= 674 cm/s^2 (*note: To convert into meters, I divided (674 cm/s^2) by 100)
= 6.74 m/s^2 [down]
Therefore the acceleration throughout was 6.74 m/s^2 [down] this may be caused by experimental errors and calculation errors.
Analysis
1. The velocity-time graph is in a straight line, meaning it was going at a constant acceleration. It was heading in a positive direction (refer to graph 1).
2. The relationship between the change in velocity and elapsed time was that the velocity increases constantly as the time increases therefore they were proportionate.
3. The slope of the velocity time graph was accelerating 6.74 m/s^2 [down].
4. The velocity position graph (refer to graph 2) was curved because changing velocity means change in slope.
5. The change in velocity of an object in free fall was directly proportional to the displacement. It is directly proportional to the time. This was because as time increases, velocity changed at a constant rate. Due to a change in velocity, there was a change in displacement. Given the formula vf = vi +at and d = (vf +vi/2) . We know that velocity changes with t = time and displacement changes with v = velocity.
Errors
After performing the experiment, several errors were noted. One error might be in the calculations in Table 1. The anticipated error was that the numbers would have been rounded and not completely accurate.
Another systematic error is that the ruler is not 100% accurate when measuring; it may be off by .5mm. A random error that occurred was that one group member released the object, not in synchronization with the ticker timer which may have hindered the process of dots being recorded on the tape.
Conclusion
I concluded that the velocity-time graph did go in a straight line. My hypothesis was only partially correct. Instead of the acceleration being 9.81 m/s^2 [down] as I hypothesized, it was actually 6.73 m/s^2 [down] when calculated due to calculation and experimental errors.
