Experiment on Motion in One and Two Dimensions
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23/09/19 Sciences Reference this
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Report for Experiment 3
Motion in One and Two Dimensions
Abstract
This experiment is designed to replicate Galileo experiment. The key equipment’s used in this experiment were an air table to provide a frictionless environment and two pucks. While the carbon paper and spark timer were used in determining the distance traveled of the puck by providing dots at every time interval. By tilting the air table, motion in both dimensions could be stimulated. The investigation support Galileo’s discovery that if an object being dropped vertically and another was fired horizontally from the same height will reach the ground at the same time. Thus, the acceleration in the y direction of both objects should be equal. The acceleration of two different scenarios in this experiment fall within range of error, which is approximately 0.56 m/s². While the acceleration in the x direction should remain 0. By measuring the height and length of table, the angle in which the table is tilted was determined. This helped in calculating the average gravitational acceleration which is equal to 9.04±0.277 m/s². The value did not fall within range of error with the theoretical value 9.80 m/s². This could be due to systematic errors.
Introduction
In this experiment, we will study one and twodimensional motion of a puck on an inclined plane. This can be done by using an air table with pucks along with a spark timer to determine the movement of the pucks in a nearly frictionless surface. The relationship between position, velocity, and acceleration is determined and analyzed in both dimensions. The goal of this experiment is to create a better understanding of the component of motion as it will replicate a famous experiment performed by Galileo that demonstrated that a ball fired horizontally at the same time as a ball is dropped from same exact height will reach the ground at the same time. This experiment determined that the downward motion under the influence of gravity had to be independent of the horizontal motion. In first two investigations, the given air table will be tilted by a wooden block allowing the puck to slide across the table. The spark timer set at 100ms will be used to determine the movement of the puck by leaving marks of the pucks position on a carbon paper in a 100ms interval.
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Find out moreThe first investigation will mimic a onedimensional motion of an object being dropped from a specific height by simply letting the puck slide down the air table. The change in the distance in the y component can be calculated to determine the velocity of the puck at a given period. A ruler will be used to measure the change in position between mark 1 and 3, 2 and 4, 3 and 5 until it reached the 8^{th} mark. Then, a plot was made to show the relationship of velocity and time in a onedimensional motion. The derivative of the equation of the line will be equivalent to the acceleration of the puck.
The second investigation will mimic an object being fired horizontally from the exact same height of the object from the first investigation. This is done by simply pushing the puck from one side of the table to the other side, creating a parabolic mark on the carbon paper. The change in distance on the x and y direction can be calculated which is then used to find the velocity of the puck in both dimensions. A ruler will be used to measure the change in position between mark 1 and 3, 2 and 4, 3 and 5 until it reached the 8^{th} mark. Two plots are then made to show the relationship of velocity in the y and x direction and time in a twodimensional motion.
Investigation 1
Setup and Procedure
The first investigation will carry out a onedimensional motion of an object being dropped vertically from a specific height. The equipment used throughout the entire experiment are, pucks, spark timer, a carbon paper, a block, and a ruler. The air table will be tilted by placing a block on one of the legs to make an inclined plane. A carbon paper is then placed on the table which will be used to show the motion of the puck. An air hose is attached to the air valve. This will allow the puck to move in a nearly frictionless surface. The spark time is set to the frequency of 10Hz or 100ms. The puck is then placed on the upper left corner of the table. My partner and I worked collectively, where one is responsible for dropping the puck, and the other is to press and release the spark timer. The spark timer will be pushed at the same time the puck is released and the spark timer is released right before the puck hits the edge of the table. The motion of the puck is shown on the carbon paper, where the marks were to be numbered from the second to the end. This will help determine the distance in the y direction between mark 1 to 3, 2 to 4, 3 to 5 and so on.
The data provided are in meters, thus we need to convert the time from milliseconds to seconds to fit with the measurements.
Data and Analysis
Dot Interval 
Time (s) 
Δt (s) 
δΔt (s) 
Δy(m) 
δΔy(m) 
δΔy/Δy (m) 
v (m/s) 
δv/v 
δv 
1 to 3 
0.2 
0.1 
0.0002 
0.044 
0.0005 
0.01136364 
0.220 
0.0113636 
0.0025 
2 to 4 
0.3 
0.1 
0.0002 
0.055 
0.0005 
0.00909091 
0.275 
0.0090909 
0.0025 
3 to 5 
0.4 
0.1 
0.0002 
0.067 
0.0005 
0.00746269 
0.335 
0.0074627 
0.0025 
4 to 6 
0.5 
0.1 
0.0002 
0.080 
0.0005 
0.00625000 
0.400 
0.0062500 
0.0025 
5 to 7 
0.6 
0.1 
0.0002 
0.089 
0.0005 
0.00561798 
0.445 
0.0056180 
0.0025 
6 to 8 
0.7 
0.1 
0.0002 
0.100 
0.0005 
0.00500000 
0.500 
0.0050000 
0.0025 
Table 1.1 – Measurement of distance traveled and average velocity in each dot interval, considering any necessary errors.
The above table shows the results of the onedimensional motion of a puck. The distance measure between each time interval is shown by Δy, and the error is ±0.0005 meters. The relative error is given by dividing the uncertainty of the distance over the distance, which is given by the equation: δΔy/Δy. The velocity was then calculated by taking the distance and dividing it by time.
$\textcolor[rgb]{}{v}\textcolor[rgb]{}{=}\frac{\textcolor[rgb]{}{\mathit{distance}}}{\textcolor[rgb]{}{\mathit{time}}}$
But since the distance Δy is between dots 1 and 3, 2 and 4 etc, the time is therefore equal to 2 times the divisions as shown in the equation below.
$\textcolor[rgb]{}{v}\textcolor[rgb]{}{=}\textcolor[rgb]{}{}\frac{\textcolor[rgb]{}{\mathrm{\Delta y}}}{\textcolor[rgb]{}{2}\textcolor[rgb]{}{\mathrm{\Delta t}}}$
The propagated error of the velocity was then calculated by using the formula:
$\frac{\textcolor[rgb]{}{\mathrm{\delta v}}}{\textcolor[rgb]{}{\mathrm{v}}}\textcolor[rgb]{}{=}\sqrt{{\left(\frac{\textcolor[rgb]{}{\mathrm{\delta \Delta}}\textcolor[rgb]{}{y}}{\textcolor[rgb]{}{\mathrm{\Delta}}\textcolor[rgb]{}{y}}\right)}^{\textcolor[rgb]{}{2}}\textcolor[rgb]{}{+}{\left(\frac{\textcolor[rgb]{}{\mathrm{\delta \Delta t}}}{\textcolor[rgb]{}{\mathrm{\Delta t}}}\right)}^{\textcolor[rgb]{}{2}}}$
The propagated error of the velocity δv/v, and distance δΔy/Δy are equal as shown in Table 1.1. The uncertainty of the velocity is then given by multiplying the velocity by the propagated error, giving the value of ±0.0025 m/s
Figure 1.1 – Plot showing the relationship between velocity and time, including a trendline, equation of the line, and error bars.
The derivative of the equation of the line will result in the acceleration of the puck, which is 0.5643 m/s² . Then the IPL best fit line calculator was used to determine the slope and error. The slope of the line was 0.5642x uncertainty of ±0.00598223m/s². Meaning that the slopes from equation of the line and IPL best fit line calculator are both within the range of error.
Investigation 2
Setup and Procedure
The second investigation analyzes a motion in two dimensional. This will stimulate an object being fired horizontally, causing the object to experience velocity in both the x and y direction. The general setup of investigation 2 is the same to that of the first investigation. However, in this investigation, the puck will be pushed from one side to the other and if done correctly, the puck will move parabolically from one side of the table to the other. The labeled x and y axis carbon paper will show marks in a parabolic pattern. The data was then used to determine the velocities of the puck in the x and y direction as shown in Table 1.3.
Data and Analysis
Dot Interval 
Time (s) 
Δt (s) 
δΔt (s) 
Δy(m) 
δΔy(m) 
δΔy/Δy 
Δx(m) 
δΔx(m) 
δΔx/Δx 
1 to 3 
0.2 
0.1 
0.0002 
0.023 
0.0005 
0.021739 
0.095 
0.0005 
5.26E03 
2 to 4 
0.3 
0.1 
0.0002 
0.038 
0.0005 
0.013158 
0.099 
0.0005 
5.05E03 
3 to 5 
0.4 
0.1 
0.0002 
0.049 
0.0005 
0.010204 
0.092 
0.0005 
5.43E03 
4 to 6 
0.5 
0.1 
0.0002 
0.060 
0.0005 
0.008333 
0.091 
0.0005 
5.49E03 
5 to 7 
0.6 
0.1 
0.0002 
0.070 
0.0005 
0.007143 
0.088 
0.0005 
5.68E03 
6 to 8 
0.7 
0.1 
0.0002 
0.081 
0.0005 
0.006173 
0.079 
0.0005 
6.33E03 
Table 1.2 Measurement of distance traveled in both dimensions in each dot interval, considering any necessary errors.
The above table shows the results of the twodimensional motion of a puck. The distance in the y direction is Δy and the distance in the x direction is Δx, which were measured directly from the carbon paper. Where their uncertainty in both dimensions is equal to ±0.0005 m. the relative error can be derived by using the formulas
$\frac{\textcolor[rgb]{}{\mathit{\delta \Delta y}}}{\textcolor[rgb]{}{\mathit{\Delta y}}}$and
$\frac{\textcolor[rgb]{}{\mathit{\delta \Delta x}}}{\textcolor[rgb]{}{\mathit{\Delta x}}}$. Their values are shown in Table 1.2
Figure 1.2 – Plot of the relationship of Δx and t, including a trendline, equation of the line, and error bars at each point.
The above plot clearly shows that distance in the x direction increases for each 0.1 seconds time interval. In a frictionless environment, the change in the distance in the x direction should be constant. However, the change in the x direction in both the graph and the table shows a slight variation. This could be due to friction of the puck and carbon paper or just merely air friction. This error could have not been avoided, as it is caused by the environment.
Figure 1.3 – Plot of the relationship of Δy and t, including a trendline, equation of the line, and error bars at each point.
The above plot clearly shows that distance in the y direction decreases for each 0.1 seconds time interval. From Figure 1.3, the trend of the marks is parabolic unlike the distance in the x direction (linear). The puck has a negative acceleration due to the acceleration of the puck in the downwards direction. While, the distance of x vs time is linear which means that it has a constant and positive velocity and therefore it should experience zero acceleration in the horizontal direction.
Dot Interval 
Time (s) 
Δt (s) 
Vy (m/s) 
δVy/Vy 
δVy 
Vx (m/s) 
δVx/Vx 
δVx 
1 to 3 
0.2 
0.1 
0.115 
0.005267 
0.00061 
0.475 
0.021740 
0.010327 
2 to 4 
0.3 
0.1 
0.19 
0.005054 
0.00096 
0.495 
0.013159 
0.006514 
3 to 5 
0.4 
0.1 
0.245 
0.005438 
0.00133 
0.460 
0.010206 
0.004695 
4 to 6 
0.5 
0.1 
0.3 
0.005498 
0.00165 
0.455 
0.008336 
0.003793 
5 to 7 
0.6 
0.1 
0.35 
0.005685 
0.00199 
0.440 
0.007146 
0.003144 
6 to 8 
0.7 
0.1 
0.405 
0.006332 
0.00256 
0.395 
0.006176 
0.002440 
Table 1.3 Calculation of vertical and horizontal velocities and their respective relative errors
The above table shows the results of the vertical and horizontal velocity. By using the measured distance in the x and y direction of the puck at every time interval, the vertical and horizonal could be determined by using the formula:
$\textcolor[rgb]{}{v}\textcolor[rgb]{}{=}\textcolor[rgb]{}{}\frac{\textcolor[rgb]{}{\mathrm{\Delta y}}}{\textcolor[rgb]{}{2}\textcolor[rgb]{}{\mathrm{\Delta t}}}$
The relative error for both the velocity in the y and x direction can be calculated by using the formula:
$\frac{\textcolor[rgb]{}{\mathrm{\delta v}}}{\textcolor[rgb]{}{\mathrm{v}}}\textcolor[rgb]{}{=}\sqrt{{\left(\frac{\textcolor[rgb]{}{\mathrm{\delta \Delta}}\textcolor[rgb]{}{y}}{\textcolor[rgb]{}{\mathrm{\Delta}}\textcolor[rgb]{}{y}}\right)}^{\textcolor[rgb]{}{2}}\textcolor[rgb]{}{+}{\left(\frac{\textcolor[rgb]{}{\mathrm{\delta \Delta t}}}{\textcolor[rgb]{}{\mathrm{\Delta t}}}\right)}^{\textcolor[rgb]{}{2}}}$
The propagated error is given as δVy/Vy and δVx/Vx, where the uncertainty can be calculated by multiply the velocity (in the y direction and x direction) by the specific propagated error to each velocity. It can also be noted that if the environment was completely frictionless, the velocity in the x direction should be equal in every time interval. But as shown in the Table 1.3, the velocity in the x direction slightly differ which could be due to air resistance of friction of the carbon paper on the puck.
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View our servicesIn addition, the velocity in the y direction should be increasing after each time interval because the gravitational acceleration should speed up the velocity by 9.8 m/s every second. However, since the puck is moving in a downward direction, the velocity in the y direction should get more negative after each time interval. This is clearly shown in the Table 1.3.
Figure 1.4 – Plot of the relationship of velocity in the x direction and t, including a trendline, equation of the line, and error bars at each point.
The graph demonstrated the relationship between the velocity in the x direction and time. As stated above, the graph is almost showing a pattern that is parallel to the x axis. However, due to friction, the result showed otherwise. The acceleration of the graph is 0.1629 m/s². The IPL best fit line calculator was then used to determine the uncertainty of the acceleration which is ± 0.0114945 m/s².
In a nonfrictionless surface, the acceleration should be equal to 0. This can be taken to be systematic error of friction in testing.
Figure 1.5 – Plot of the relationship of velocity in the y direction and t, including a trendline, equation of the line, and error bars at each point.
The graph demonstrated the relationship between the velocity in the y direction and time. The plot is showing a downward linear pattern. The velocity in the y direction should be increasing after each time interval because the gravitational acceleration should speed up the velocity by 9.8 m/s every second. Since the puck is moving in downward direction, the velocity in the y direction should get more negative in this case.
The y intercept of the line should equal the initial velocity of the puck in the y direction.
The acceleration in the y direction is 0.5671 m/s². The IPL best line calculator was then used to determine the uncertainty of the acceleration which is ± 0.00327275 m/s². This can be due to a systematic error and/or random error in measuring the distance of the dots on the carbon paper.
As a check, we compared the acceleration of Vy and the acceleration measured in investigation 1. The acceleration of both investigations should be almost equal or fall within the range of error, since they are experiencing the same gravitational acceleration value of 9.8 m/s². The acceleration from the y direction in investigation 2 is equal to 0.5671 m/s² ±0.00327275, while the accusation for investigation 1 is 0.5643 m/s² ±0.00598223. The negative sign shows the direction of the puck; thus it can be dismissed in this case. The acceleration of both investigations is almost entirely equal.
Investigation 3
Setup and Procedure
No necessary setup was required for this investigation, as only calculation was performed. The goal of this investigation to calculate the acceleration due to gravity by using our results from investigation 1 and 2. This can be done by measuring the length and height of the incline.
Data and Analysis
The goal of this investigation is to calculate the acceleration of free fall both investigations and their uncertainty. Then both gravitational accelerations are used to obtain their average and their respected errors.
The formula to derive g using the angle of the table is:
$\left[\textcolor[rgb]{}{1}\right]\textcolor[rgb]{}{\mathit{g}}\textcolor[rgb]{}{=}\frac{\textcolor[rgb]{}{\mathit{ay}}}{\textcolor[rgb]{}{\mathit{sin\theta}}}$
Sinθ can be obtain by diving the height by the length of the table (also known as the hypotonus). The height and length obtained in this investigation is 0.04 and 0.645 m respectively with an uncertainty of ± 0.0005 m. Thus, sin will equal:
$\mathrm{Sin\theta}=\mathrm{}\frac{0.04}{0.645}=$
0.062016
With this information, g of both investigations could be calculation using the formula [1]. Using the y acceleration of Investigation 1 at 0.56m^{2}/s,
$g1=\frac{0.56}{0.062016}=9.04{m}^{2}/s$.
Using the y acceleration of Investigation 2 at 0.56m^{2}/s,
$g2=\frac{0.56}{0.062016}=9.04{m}^{2}/s$.
If done correctly, the gravitational acceleration of both should be equal, which was obtained in this case. However, the gravitational acceleration is supposed to be 9.80
${m}^{2}/s$, which was not obtained. This could be due to a systematic error that as a group could not figure out. It could also be since one of the legs of the air table given to us was broken, making the height of the table uncertain.
Their uncertainties can be found using the rules of error propagation in Appendix A by first finding error for the
$\mathit{sin}\left(\theta \right)\left[2\right],$and then for the
$g$.
$\left[2\right]\mathit{}\frac{\mathit{\delta sin\theta}}{\mathit{sin\theta}}=\sqrt{{\left(\frac{\mathit{\delta h}}{h}\right)}^{2}+{\left(\frac{\mathit{\delta l}}{l}\right)}^{2}}=0.012524$
The errors of which are derived using the propagation formula:
$\frac{\textcolor[rgb]{}{\mathrm{\delta g}}}{\textcolor[rgb]{}{\mathrm{g}}}\textcolor[rgb]{}{=}\sqrt{{\left(\frac{\textcolor[rgb]{}{\mathrm{\delta a}}}{\textcolor[rgb]{}{\mathrm{a}}}\right)}^{\textcolor[rgb]{}{2}}\textcolor[rgb]{}{+}{\left(\frac{\textcolor[rgb]{}{\mathrm{\delta sin\theta}}}{\textcolor[rgb]{}{\mathrm{sin\theta}}}\right)}^{\textcolor[rgb]{}{2}}}$Uncertainty of
${g}_{1}$is:
$\frac{\delta {g}_{1}}{{g}_{1}}=\sqrt{{\left(\frac{\delta {a}_{1}}{{a}_{1}}\right)}^{2}+{\left(\frac{\mathit{\delta sin\theta}}{\mathit{sin\theta}}\right)}^{2}}=0.016408$
Uncertainty of
${g}_{2}$is:
$\frac{\delta {g}_{2}}{{g}_{2}}=\sqrt{{\left(\frac{\delta {a}_{2}}{{a}_{2}}\right)}^{2}+{\left(\frac{\mathit{\delta sin\theta}}{\mathit{sin\theta}}\right)}^{2}}=0.059054$
The resulting error for g_{1} is ±0.016408m^{2}/s, and the error for g_{2} is ±0.059054m^{2}/s. Finally, we need to calculate the averages of the accelerations and their uncertainties. We can do this by using a normal average calculation so that
${\textcolor[rgb]{}{g}}_{\textcolor[rgb]{}{\mathit{av}}}\textcolor[rgb]{}{=}\frac{{{\textcolor[rgb]{}{g}}_{\textcolor[rgb]{}{1}}\textcolor[rgb]{}{+}\textcolor[rgb]{}{g}}_{\textcolor[rgb]{}{2}}}{\textcolor[rgb]{}{2}}$, or
${\textcolor[rgb]{}{g}}_{\textcolor[rgb]{}{\mathit{av}}}\textcolor[rgb]{}{=}\frac{\textcolor[rgb]{}{9}\textcolor[rgb]{}{.}\textcolor[rgb]{}{04}\textcolor[rgb]{}{+}\textcolor[rgb]{}{9}\textcolor[rgb]{}{.}\textcolor[rgb]{}{04}}{\textcolor[rgb]{}{2}}\textcolor[rgb]{}{=}\textcolor[rgb]{}{9}\textcolor[rgb]{}{.}\textcolor[rgb]{}{04}{\textcolor[rgb]{}{m}}^{\textcolor[rgb]{}{2}}\textcolor[rgb]{}{/}\textcolor[rgb]{}{s}$. Since both accelerations are equal, their average acceleration will not change.
Uncertainty of
${g}_{\mathit{av}}$is:
$\delta {g}_{\mathit{av}}=\frac{\sqrt{{\left(\delta {g}_{1}\right)}^{2}+{\left(\delta {g}_{2}\right)}^{2}}}{2}=0.277\frac{m}{{s}^{2}}.$
The final acceleration is equal to 9.04±0.277 m/s². This value does not agree with the theoretical value of gravity 9.81 m/s². This is mainly caused by a systematic error when measuring the length and height of the table. Also noting that the air table had a deformed leg, which could result In the change in height. However, the calculated uncertainty is equal to ±0.277 which means the experiment was done well but not perfect. One of the reasons for the error is frictional force of the puck with the carbon paper, or even just air friction. And the other would be a systematic error as stated above.
Conclusion
Even though the experiment was performed well, there will still be uncertainties and errors. This experiment replicated Galileo experiment by using a puck and an air table. A carbon paper and a spark timer were used to mark the movement of the puck every 100ms. A replica of the onedimensional motion was performed in the first investigation. Values of velocities were calculated by measuring the change in distance of the puck at every time interval. This provided the data to calculate the acceleration of the puck by taking the derivative of the equation of the line. The calculated acceleration for the first investigation is 0.5643 m/s² with an uncertainty of ±0.00598223m/s²
In the second investigation, a replica of Galileo second dimensional motion was performed. The velocities in both the x and y direction were calculated by measuring the change in distance at each time interval (in the x and y direction). The relationship between velocity and time for each direction were graphed separately. This provided the data to calculate the acceleration of the puck by taking the derivative of the equation of the line. The calculated acceleration for the velocity in the x direction 0.1629 m/s² ±0.0114945 The calculated acceleration for the velocity in the y direction 0.5671 m/s² ± 0.00327275 The acceleration values of investigation 1 and investigation 2 in the y direction were compared to show if the experiment was done correctly. Both accelerations were within range of error.
For the third investigation, the gravitational acceleration for each investigation and its average (in the y direction) was calculated by measuring the height and length of the table. By doing these measurement, the angle of the incline was determined which is necessary in solving for the gravitational acceleration. The calculated average acceleration was
${g}_{\mathit{av}}=9.04\frac{m}{{s}^{2}}\textcolor[rgb]{}{\pm}\textcolor[rgb]{}{0}\textcolor[rgb]{}{.}\textcolor[rgb]{}{277}\textcolor[rgb]{}{\mathrm{}}$. The actual value did not fall within error of the theoretical value of
$9.81\frac{m}{{s}^{2}}.$This is mainly due to systematic error when measuring the distance of the puck at each time interval. This could also be due to a slight friction between the puck and the carbon paper. Another issue is that the air table given had a unstable leg, which could affect the height of table and affect our results. To get a more accurate gravitational acceleration, the experiment needs to be done several times to get consistent measurement and to be given a stable air table.
Questions

Find the total time to reach the bottom of the air table for Investigation 1 and 2. Are there times all the same, as shown in the picture of the cannon? Why or why not?

Using the kinematics equations, x = v_{0x}t +
$\frac{1}{2}$
at^{2}, and y = v_{0y}t +
$\frac{1}{2}$at^{2}. We can calculate the total time to reach the bottom for both investigations. The only type of acceleration acting on the puck in the y direction of both investigations is the gravitational acceleration.
 The total time to reach the bottom for investigation 1 is:
 The total time to reach the bottom for investigation 2 is:
 The slight difference in time is likely due to random error.

Using the kinematics equations, x = v_{0x}t +
$\frac{1}{2}$

If the acceleration of a puck falling down on an incline of angle
$\theta $
(where
$\theta $<
${45}^{0}$) is
${\alpha}_{\theta}$, what is
${\alpha}_{2\theta}$?

Using formula for acceleration of a puck falling down an incline of angle for
${a}_{\theta},$
${a}_{\theta}=g*\mathit{sin}\left(\theta \right)$
, and for
${a}_{2\theta}=g*\mathrm{sin}\left(2\theta \right)=g*2\mathit{sin\theta cos\theta}={2a}_{\theta}*\mathrm{cos}\theta .$

Using formula for acceleration of a puck falling down an incline of angle for
${a}_{\theta},$
${a}_{\theta}=g*\mathit{sin}\left(\theta \right)$

You determined the acceleration from the slope of you plot v vs. t. What is the meaning of the intercept of your line with the v axis? Do you expect this value to be close to the origin?
 The initial velocity will be equal to the intercept of y axis. In the first investigation I would expect the initial velocity to equal zero. While in the second investigation, I would expect the initial velocity to equal the value of the push of the puck.

If you performed Investigation 1 and 2 again with an incline of
$\theta =0$
, what would the trajectories look like in each case?
 If the angle of incline equals zero, then there would be no motion or trajectories in investigation one as the puck will not move. While in investigation 2, the puck will experience a constant velocity, as there is no gravitational acceleration.

If there were insufficient air flow through the air table, how would the slope of your graphs be affected?
 Insufficient air flow will cause the puck to experience a frictional force. This would mean that the slope of the y velocity versus time graphs would be less steep and the x velocity versus time graphs would be more steep. Thus, slowing down the puck.
Honors Questions

How would your results of Investigation 1 and 2 change if the masses of the pucks were doubled?
 If the mass of the puck were double would not affect the result of acceleration.
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