03-May-2017: Ballistic Pendulum

Lab: Ballistic Pendulum

03-May-2017
Idea: Demonstrate an inelastic collision and determine the launch velocity of a ball fired from a spring loaded launcher.

Summary:

1) Having a ballistic pendulum, as shown below, we first make sure that:

1. The device is level. We make sure that the ball does not roll around when placed flat.
2. The block connected to the strings is also directly facing the gun and that when fired, the ball will enter the block.

2) After measuring the length of the string attached to the block, mass of the ball, mass of the block, we proceed to fire the gun five times and record the angle that the block-ball mass goes to, as indicated by the metal arm, and then I took the average of these angles. This will allow me to calculate the launch velocity.

3) Then, we proceeded to launch the ball horizontally, and see how far the ball would land, which would allow us to again calculate the same launch velocity, using a black carbon paper. This is so that we could verify our calculations done in step 2, and perform propagated uncertainty calculations.

Analysis:

Part 1:

Our goal in the first part of the lab was to use conservation of momentum in order to find the firing speed of the gun. The ball undergoes an inelastic collision with the block once fired, and the two swing to a certain height with a certain angle, theta. In order to calculate Vo, I had to use the equations for conservation of momentum and energy, and furthermore, know exactly how high the block-ball swings up to.

To do that, I simply said that the starting level of the gun-block was zero, and because the block was connected to a string and rose to a certain angle theta, that would allow me to find the Y component of the string when tilted at that particular final angle. It turned out to be L-Lcos(theta).

Therefore, I was able to calculate the launch velocity using my momentum and energy equations by setting up two different equations, and solving for Vo.

Part 2:

Next, the task was to verify this calculated result by putting the block to the side, and launching the gun off a tabletop. By finding the horizontal distance that the ball travels, we could use kinematics in order to determine the launch speed of the ball based on the trajectory measurements.

To do this, we recorded the distance from the gun to the table edge, and then the edge to where the ball lands, as the X distance the ball travels. We do so similarly for the Y distance, we measure the height of the table, and then the height from the tabletop to the gun. These allow us to do our kinematic equations.

With such measurements, we also went ahead and calculated our propagated uncertainty for our result, based off the measurements we recorded with our meter sticks, and the mass scale. Doing so yielded a value of Vo very close to those of part 1, which was to be expected.

Measured Data:

Data Table of Measured Data

Calculated Data:

Data Table of Calculated Data for Part 1 and 2

Calculations of Vo for Part 1

Calculations of Vo for Part 2

Propagated Uncertainty Calculations for Part 2

Conclusion:

Comparing the calculations for velocity from part 1 and part 2, I notice that the two are remarkably similar, especially once I take my propagated uncertainty into account. Doing percent error calculations gives me a difference of little more than 1%, meaning that whether I find the speed using kinematics or momentum/energy, because both methods are equally valid, I should end up with final values that are very similar--if my measurements are accurate.

In terms of uncertainty,

• In measuring the height and length the ball travels in part 2, there were so many measurements to take (distance between the table and gun, distance between gun and black carbon paper, height of table, etc) that my measurements, if they are, are most likely off due to human error. I would not be surprised if I mistakenly missed or added a centimeter here or there, which would contribute to inaccurate lengths for X and Y, and subsequently an inaccurate value for V in part 2.
• The collision with the ball and the block is not perfect, the position of the gun required some meddling so that it would at least go into the block, which of course would lead to an inaccurate angle, on top of the fact that the angle indicator itself has an uncertainty of +- 0.5 degrees. An inaccurate angle leads to a value of height too tall or too short, subsequently messing up our solving for the initial speed with the conservation of momentum and conservation of energy.

24-April-2017: Collisions in Two DImensions

Lab 15: Magnetic Potential Energy Lab

24-April-2017
Idea: Create two-dimensional collisions and verify whether momentum and energy are conserved.

Summary:

1) Using a leveled glass table, I set up my phone camera that will capture the collision of

1. Steel ball with steel ball, one is initially stationary
2. Steel ball with smaller aluminum ball, also initially stationary, at 240 fps.

2) Thus the setup looked like such:

3) I record two videos for the two cases, and export the videos to Logger Pro, where I begin to plot the position of the two balls. Then, I scale the image based on the width of the glass table, and position an xy-axis over the video in order to create my graphs, like the following.

4) Therefore this gives me four sets of data for each of the two test cases: positionof x before and after the collision for both balls, position of y before and after the collision for both balls. Because I am trying to verify whether momentum & energy are conserved, what is important here is only the components of velocity in the x-y direction before & after the collisions: these will be used to calculate the momentum and energy, and as such the slopes of the position-time graphs gives me these velocities.

5) Lastly, I produce graphs of the position x-y center mass for both systems throughout the collisions, and also the velocity x-y center mass for both systems.

Data Analysis/Discussion

1)  In order to demonstrate that momentum is conserved, I obtain the slopes of the position-time graphs, which would allow me to find my values of velocity for the balls. After collecting this data, I calculated for momentum by writing out momentum (M=mv) before and after the collision, and the results I got did indeed verify this, getting results for Momentum initial versus Momentum final that were relatively close to each other:

Steel vs Steel ball

Mx: 0 = -.0508 N*S

My: .6548 = .5367 N*S

Steel vs Aluminum Ball:

Mx: .00556 = .0059 N*S

My: -.0163 = -.01459 N*S

To verify this further, I then performed percent error calculations, which confirmed this (12-27%), exception being for the momentum in the x direction for the balls of equal mass--since the way I positioned my axis meant that the initial x-component of velocity was zero, I was left trying to compare 0 to 0.0508, which was impossible to do in the percent error calculations, so I decided to leave that one to a percent error of 51%, definitely a source of uncertainty. But overall, my calculated data pointed towards demonstrating conservation of momentum for both cases as a result.

In proving conservation of energy I had a similar process of writing out the energy of the system before and after the collision with the changing kinetic energies, but unlike momentum, energy is a scalar--thus I used the following equation to obtain the actual speed of the two masses, NOT velocities in the x and y directions.

Sample calculation of how I calculated these velocities for my energy equations.

In doing so, and calculating for the energy of the system, I observed that notable amounts of energy were lost:

1. I calculated the final and initial energies to not be close enough to one another for them to be considered equal. When the masses were equal, I got a percent error of approximately 53% for that first case. The initial energy, .42J, was much greater than the final energy, .20J, showing that a significant amount of energy was lost during the collision. 
2. This was less of a case on the second trial where the final energy, .00709J, was much more closer to the initial energy, .0095J with a percent error of 25%, showing that the loss of energy was not as significant in this case. However, it is nevertheless a relatively high degree of error.

Therefore, I could see that the total kinetic energy before the collision was not the same as total kinetic energy after the collision, and could therefore conclude that unlike momentum, in this collision, energy unfortunately was not conserved.

I obtained numerous graphs that show:

• Initial and final position of x for each of the balls for both cases
• Initial and final position of y for each of the balls for both cases
• Initial and final position of x AND y for each of the balls for both cases

2) Afterwards, I had to create graphs of the x&y centers of mass vs time for both position and velocity of the balls. Reason being, I concluded, that these graphs would show that the center mass of position and velocity would demonstrate the conservation of momentum and energy further; that the system behaves as if all of its mass were at the center mass.

To demonstrate this, I used the following equations to solve for the various center masses:

I entered these equations for xcm and ycm for both position and velocity in new calculated columns, and the resulting graphs demonstrate the original assumption: the center mass moves linearly, and its velocity of this point appears to be constant for BOTH cases.

Overall, I obtained numerous graphs that show:

• Xcm + Ycm for case 1
• Vxcm and Vycm for case 1
• Xcm + Ycm for case 2
• Vxcm and Vycm for case 2

Measured Data/Table:

Measured data of:
steel ball, aluminum ball, x-y components of velocity for case Equal Masses (table 1) and Different Masses (table 2)

Calculated Data:

Demonstrating the Conservation of Momentum in the X and Y directions for both cases.

Calculation of the Conservation of Momentum in the X and Y directions for both cases

Percent error of Conservation of Momentum

Calculating the Conservation of Energy for both cases
showing that energy is not conserved except for case 2 , with a percent error of 25%.

Sample of how I calculated center mass, in this case, for position in the Xcm. The same was done for Ycm, Vxcm, Vycm, and so forth.

Graphs:

Steel balls of equal Mass:

X vs Time for steel ball with initial velocity

Y vs Time for steel ball with initial velocity

X vs Time for steel ball with zero initial velocity

Y vs Time for steel ball with zero initial velocity

All of the above graphs overlayed into one.

Different Masses of steel and aluminum ball:

X vs Time for steel ball AND  aluminum ball

Yvs Time for steel ball AND  aluminum ball

X-Y Center Mass & V Center Mass for Equal Steel Masses

Xcm vs t and Ycm vs t overlayed

Vx-cm vs t & Vy-cm vs t overlayed

X-Y Center Mass & V Center Mass for Steel and Aluminum Balls

Xcm vs t and Ycm vs t overlayed

Vx-cm vs t & Vy-cm vs t overlayed

Conclusion:

Based upon my calculated and measured data, I can reasonably conclude that momentum was conserved due to how numerically close their initial and final values were, even if the percent error on one of them was 51% because of zero. Besides that, the percent error was very reasonable:

Percent Error for momentum

 However, the same could not be said for energy. Because of just how large the numerical difference, not just the percent error, of the initial and final energies of the cases were from each other. For us, energy cannot, at least for the two balls of equal mass, be said to have been conserved. There was a clear and significant loss of energy:

Percent Error for energy: Far right column

Uncertainty/Error

• We can determine a level of uncertainty in the measurement of the ball mass. The scales had an uncertainty of .01 grams, which would affect the second case calculations for momentum and energy more so than the first, as the variable mass simply cancels out when the two balls are of equal mass.
• Next, human inaccuracies (compounded by video quality) in positioning the balls in the position vs time graph would lead to an inaccurate slope; velocity, and therefore introduce a degree of uncertainty in our calculations for momentum AND energy.
• Most importantly, we also see that real-world collisions see losses of energy from a variety of sources: friction, sound, vibration, etc, all of which clearly affected our data--most significantly, our calculation of energy of the equal masses.

19-April-2017: Impulse-Momentum

Lab 14: Impulse-Momentum

19-April-2017
Idea: Verify the impulse-momentum theorem by observing collisions in which we observe a cart to simulate an inelastic or elastic collision.

Summary:

EXPT 1:

1. We fasten a cart-track system as shown below. There is a rubber stopper on the cart connected to our force sensor, and once given a pull via a wire, meant to hit and bounce off of an obstacle at the end of the track. In this lab, another cart with a plunger.
   
2. Then, we have a motion sensor on the opposite end of the track, and in Logger Pro make sure that the position towards the motion sensor is the positive direction.
3. After calibrating and zeroing our force sensor, we give the cart a gentle pull towards the stopper multiple times until we get a good set of graphs. These graphs give us Force vs Time and Velocity vs Time, which we need in order to find Momentum, and consequently demonstrate the impulse-momentum theorem.

EXPT 2:

1. With the same setup, we add 500 grams to the cart and repeat the experiment, in order to compare whether the impulse and change in momentum were still equal to each other with a larger cart.
2. Thus we obtain a similar graph to EXPT 1 and compare their results.

EXPT 3:

1. Leaving the extra mass on the cart, we replace the cart at the end of the track with a vertical piece of wood with a lump of clay attached onto it. This experiment is meant to simulate an inelastic collision, wherein the cart stops immediately after collision.
2. I again obtain graphs of Force and Velocity, and calculate the impulse-momentum theorem.

Analysis:

EXPT 1:

Question 1: The net force exerted on the cart just before it starts to collide must be the force exerted by the spring on the extended cart's plunger, which increases as it compresses.

Question 2: The magnitude of the force is at its maximum when the rubber stopper's maximum compression of the plunger 

Question 3: After the collision, it is again the opposing force exerted by the plunger on the rubber stopper.

Question 4: The collision takes approximately a tenth of a second.

Question 5: *Calculations explained below*

Question 6: *Conclusion of EXPT1 explained below*

1) Our Force vs Time graph allows us to find the impulse. We can also calculate the momentum using the data recorded by the motion sensor, using p=mv. This allows me to demonstrate the impulse-momentum theorem by finding the change in momentum, and showing that it is in fact equal to the momentum applied to it.

2) Upon first inspection, on impact, the velocity suddenly increases to zero until it begins travelling in a constant positive direction towards the motion sensor.

3) Our graph gives an Impulse of .5290 N·S. Following my calculation of the change in momentum, I get a value of .560 N·S, relatively accurate to that of the graph, with a percent error of approximately 5%, very accurate. Therefore, I could reasonably determine that this adequately demonstrates the impulse-momentum theorem.

EXPT 2:

Question 7: *Conclusion of EXPT2 explained below*

1) Again on impact, the velocity suddenly increases to zero until it begins travelling in a constant positive direction, towards the motion sensor.

2) In order to demonstrate that regardless of the mass, the change in momentum is equal to the impulse, we again find the impulse and change in momentum of our new graph. Doing so produces an impulse of .841 N·S, and a change in momentum of .892 N·S. This again is accurate to a percent error of approximately 5%, so this indeed demonstrates that regardless of the mass, the impulse-momentum theorem holds.

EXPT 3:

Question 8: *Conclusion of EXPT3 explained below*

1) This time, the velocity does not continue to increase after 0. Because the cart sticks to the clay, the whole system stops. However, I reason that because momentum is still conserved, the impulse still equals the change in momentum.

2) To demonstrate this, I find the impulse and calculate the change in momentum for the last time, and I get values of 0.4314 N·S and 0.455 N·S, respectively. These values are very close to one another once again within 5%, and I can thus reason that even in the inelastic collision the calculated momentum equals the measured impulse applied to it.

Calculate Data

Results for change in momentum for experiments 1, 2, and 3.

Graphs:

EXPT 1
Impulse: .5290 N·S∆Momentum: .560 N·S

EXPT 2Impulse: .841 N·S∆Momentum: .892 N·S

EXPT 3Impulse: 0.4314 N·S∆Momentum: 0.455 N·S

Conclusion:

As demonstrated in experiments 1, 2, and 3, inelastic or elastic, we demonstrated that the impulse is in fact equal to the change in momentum, as for all three cases our calculations showed just how close the two values were. Therefore, we can reasonably conclude that this indeed demonstrates the impulse-momentum theorem. Between the three experiments, my calculations for percent error yielded 5.5%, 5.7%, and 5.2% respectively, showing that indeed this is a very reasonable conclusion.

• As for sources of error: the scale was inaccurate to .1 grams, and although it isn't very significant,  it is important because it does affect our change of momentum calculations to a certain extent.
• Our force sensor is approximately inaccurate to about .001 Newtons, as even in a zeroed state it would randomly read values like .0021 N. This would clearly lead to an inaccurate Force x Time graph, and ultimately an inaccurate value of Impulse. In addition, we had to make sure that the force never exceeded 10N, because apparently the sensors would read inaccurate data.
• In addition, similar inaccuracies to the motion sensor would create an erred velocity x time graph, inevitably leading to a value of a change in momentum with an uncertainty.
• Loss of energy. Clearly, these were not perfect elastic or inelastic collisions, as energy is inevitably lost due to friction, and for all three experiments this means that there is an error in both Impulse and Momentum, Friction would affect the cart's change in velocity.

17-April-2017: Magnetic Potential Energy

Lab 13: Magnetic Potential Energy Lab

17-April-2017
Idea: Model the conservation of energy for a cart-track system with a magnetic potential energy, with an equation we do not yet have. 

Summary:

1. After leveling an air track like the one below, where a cart on an air track opposes a magnet attached to the other end, I tilt the track at various angles (recorded with a smartphone), and I obtain a Force vs Position graph by plotting these test cases. In Logger Pro, I use a power curve to fit this graph, and record the given uncertainties, and reasonably ascertain that this graph gives me the force exerted by the magnet.
   
2. Based on the values given by the curve I determine an approximate function for Force, which refers to the magnet. Its integral would give me the PE of the magnet.
3. Given this function, I now attach an aluminum reflector to the cart, level the track again, weigh my cart, and run the motion detector whilst fixed near to the magnet.
4. From the distance that the motion detector records, I would subtract the distance between the sensor and the magnet in order to obtain the distance between the cart and the track. This is important in order to calculate the Potential Energy. PE = F*x
5. I create a new calculated column given this distance. I then turn on the air track, and give the cart a gentle push.
6. I am then able to obtain a single graph showing KE, PE, and total energy (PE + KE) of the system as functions of time.

Analysis/Explanations:

1) From the first part of the lab wherein we recorded the force at various angles of the incline air track, we obtained a force vs position power fit graph that gave us a value of f(r) as a function of y: 4.281*10^-5 * r^-2.282.

• By graphing Force vs Position, we could obtain it as a function of position along the track. This, we could reasonably assume, was the only force present in the track-system, assuming that friction was zero and loss of energy in general was zero. Therefore, we could further infer that this force must reference the force exerted against the cart by the magnet. This is why it was important that we record record the force using our force sensor.

2) By taking the integral of this function of force, we could obtain the PE of this magnetic force over the distance along the track. Doing so gave me a value for PE of 3.34*10^-1.282 * r^-1.282. From our base assumption, that in this case we could find the PE of the magnet by measuring force and taking its integral, we should see that its change in energy should be equal and opposite to KE. Furthermore, we expect the sums of PE and KE to result in a straight line, because as the KE of the cart decreases, we expect the PE of the magnet to increase. 

3) Therefore, we graph KE, PE, and their sum in order to visually demonstrate this idea.

After graphing the following: KE, PE, and KE+PE vs time, we get a visual representation of the two and if our assumptions were true, then we expect KE to be the opposite of PE, and their sums should result in a straight line.

4) Judging from the graphs that we got, we see that this is accurate: the PE of the magnet peaks just as the cart's KE is at its minimum, and their sum, while not exactly a straight line, demonstrates our base assumption as aforementioned.

Measured Data:

Distance between the magnets at various angles, Theta

Calculated Data:

Calculating the distance(s) between the cart and the magnet for each case.

Integration of function of Force, resulting in a PE of 3.34*10^-5 * r^-1.282

Graphs:

Graph of Force vs Position. The power fit curve as aforementioned gives a function of force as 4.281*10^-5 * r^-2.282.

Graph of KE, PE, and KE+PE.

Conclusion:

We were able to model conservation of energy for a cart-track system and extrapolate from this the equation of PE for our magnet, something we did not know beforehand, because our graphs showed that the change in energy of KE and PE were largely equal and opposite. Therefore, their sum resulted in a graph where we could see this occur. While the graph of their sums isn't a perfect line by any means, it adequately demonstrates this idea with the fact that the graph of KE reaches its absolute minimum just as PE reaches its maximum.

• With sources of uncertainty, there were the usual culprits like the scale, which has an uncertainty to about 0.1 grams. It would damage the graph of KE, as it is the product of a-half, mass, and velocity squared. However, I'd imagine that it didn't hurt the accuracy of my graph very much because this lab was worked entirely in meters, kilograms, etc, units larger upwards to a magnitude of a thousand.
• The other comes from the 'protractor' apps on our phones, as the accuracy of its readings depend on:
• The hardware. For a lot of smartphones the G sensor isn't meant to be fine tuned for accurate scientific measurements.
• The App itself. Various apps had wildly varying levels of accuracy. With the app we used, we determined it had an uncertainty of .1 degrees. Overall, while this would not directly damage the accuracy of our calculations later on, it does create an inaccurate representation of the system.
• The Force sensors and motion tracker. The former was inaccurate to at least.001 N and as a result it would lead to my integration being inaccurate, which means an inaccurate value of PE. The latter, I would not be aware of any errors, but if there were, it would of course lead to an inaccurate KE graph.
• Friction between the cart and the track. Although with the track running, we can reasonably assume friction to be zero, we realize that it isn't a perfect system and there must be, to a certain extent, a loss of energy to such forces. This would affect our calculations for the PE due to force readings that rely on how much the cart accelerates on its way across the track.