Kevin Tran
Lab partners: Kevin Nguyen, Jose Rodriguez
29 April, 2017

Collisions in two dimensions

Purpose: To determine if momentum and energy are conserved by looking at a two dimensional collision.

Introduction: We performed two different experiments: collisions in two dimensional with two ball of the same weight and different weights. We performed these experiments on a leveled glass table. We set one ball in the middle of the leveled glass table at rest and we collided it with another ball with some initial velocity.

Below is the set up we used in order to perform this experiment. We have a level glassed table with an apparatus to record the collision. We used our phones as the recorder. We made our phone capture 60 frames per second to get a more accurate data.

Below is a picture of the results of the collisions between two balls.

This is the process after we video capture the collisions of two balls in 60 frames per second.

First, we started off with the collisions of two balls with a similar mass. To be exact, we used two marbles with a mass of 0.004 kg

Below is the results of both marbles in a two dimensional graph. In order to get this graph, students need to

1. Options-> Movie Options

2. Override frame rate to 60 to 120fps(frames per second))

3. Advance the movie 2 or 4 frames after adding a new point.

This graph is a position/time graph. In the y-axis, we have two x's and two y's since we have the positioning of both marbles.

For the video, we used two different colors for each ball in order to determine its position in a certain time frame, which will result in the creation of our position/time graph.

As a result from this graph, we got the following velocities from the two balls from the slopes.

Intial:

Vy= 418.5 cm/s

Final:

Vx= -138 cm/s

Vy= 284 cm/s

Vx2= 136.2 cm/s

VY2= 108.5 cm/s

With the velocities, we can determine if momentum and energy are conserved by using the formula Pintial= Pfinal. 

As a result, both the initial and the final momentum are relatively the same. So, momentum and energy are conserved.

Next, we performed the 2nd experiment when two balls of different masses result in a collision. To be exact, we will be using a marble of 4g and a metal ball of 30g for this case. The marble will be placed at rest in the middle of the leveled glass table while we apply some velocity to the metal ball in the direction of the marble ball. 

The graph is a position/time and it is set up similar to the 1st experiment. We have time(x-axis) and the positions of both balls in the x and y direction in the y-axis.

As a result from this graph, we got the following velocities from the two balls from the slopes.

Intial: 

Vy= 288.6 cm/s

Final:

Vx= -35.54 cm/s
Vy= 305.7 cm/s
Vx2= 140.1 cm/s
Vy2= 402.9 cm/s

With the velocities, we can determine if momentum and energy are conserved by using the formula Pintial= Pfinal. 

As a result, the initial and final momentum in the x direction are not the same; therefore momentum and energy is not conserved, but the initial and final momentum in the y direction are relatively the same.

Conclusion: We were able to determine if momentum and energy is conserved by performing this lab. We realized that momentum and energy is conserved when we collided two balls with a similar mass in comparison to the collision with two different masses.

Kevin Tran
Lab Partners: Kevin Nguyen, Jose Rodriguez
22 April, 2017

Impulse- Momentum 

Purpose: To test the impulse- momentum theory with inelastic and elastic collision activities.

Introduction: We performed 3 EXPT: Observing Collision Forces that change with time, A larger momentum change, and impulse momentum theorem in an inelastic collision.

First, we performed EXPT 1: Observing collision forces that change with time.

First we set up a track with one end having the motion detector and the other end with a cart with its stringy bit extended. The cart is clamped to a rod clamped to the table. We will be setting our main cart around the middle of the track where the motion detector can accurately collect data. First we calibrated the motion sensor, which is attached to the main cart. We mounted the force probe onto the cart with a rubber stopper to replace the hook mounted on the protruding part of the force sensor. We collided the cart with the plunger few times and observed what happens to the spring plunger.

First, I like to say that the mass of the cart + motion probe= 0.692 kg

During the collision, the force is not constant since the impulse-moment theorem states that the amount of momentum change for the moving cart is equal to the amount of the net impulse acting on the act, which is J=delta P

Now, we pushed our cart and collided with the clamped cart with a stringy bit extended. As a result with have an elastic collision between a cart and the clamped cart with a stringy bit extended.

Above, we have two graphs: a velocity/time graph and a Force/time graph of the cart before and after the elastic collision. 

In the Force/time graph (2nd graph), we highlighted a portion of the graph from one time to another during the time frame of the collision, and we took the integration of it. Since we know that Impulse= Force*time, we can take the integration to find our impulse. As a result, we got an impulse of 0.7347 N*s. Now we want to test if the momentum is similar to the result of impulse. In order to find out, we have to use the formula of momentum:

delta p= mvf - mvi

From our graph, we have an initial velocity of -0.542m/s before the collision and a final velocity of 0.503m/s after the collision.

Since our mass of the cart is 0.692kg, we can solve for momentum.

m(vf-vi) = 0.692kg(0.503m/s-(-0.542m/s)) = 0.7251 kgm/s.

If we compare 0.7347 N*s to 0.7251 kgm/s, the results are relatively the same, which proves the impulse-momentum theorem for this case.

EXPT 2: A Larger Momentum Change.

For this experiment, we performed the same exact task as EXPT 1, but the only difference is increasing the mass of the cart.

We added 500 grams onto our cart and got a total mass of 1.192kg

In performed the same ritual as EXPT 1. We highlighted a portion of the Force/time graph from one time to another during the time frame of the collision, and we took the integration of it . As a result, we got an impulse of 0.9055 N*s. We will be using the same formula to test that J=delta p

Delta p =mvf - mvi

From our graph, we have an initial velocity of -0.448m/s before the collision and a final velocity of 0.323m/s after the collision.

Since now our mass of the cart is 1.192kg, we can solve for momentum.

m(vf-vi) = 1.192kg(0.323m/s-(-0.448m/s)) = 0.89996kgm/s

If we compare 0.9055 N*s to 0.89996 kgm/s, the results are relatively the same, which also proves the impulse-momentum theorem for this case.

EXPT 3: Impulse-Momentum Theorem in an Inelastic Collision.

First, I like to say that we will be using the same mass of the cart from EXPT 2 of 1.192 kg. Now we have a case when we have an inelastic collision. In other words, after the collisions, they will stick to each other instead of bouncing off each other.

Below is the picture of our set up. We will be replacing the rubber stopper with a nail. Also, we removed the dynamics cart from its clamp and replaced it with a wooden wall and attached a blob of clay to the wall at the height of nail. Now we have created an elastic collision set up. 

Next, we made sure we zeroed our force probe and then collide the cart with the clay.

Above is our velocity/ time graph and Force/time graph. We realized during the graph we have an initial velocity, but our final velocity becomes zero since the collision was inelastic. Similarly to the other two EXPT, we highlighted a portion of the Force/ time graph and took the integration to find impulse. As a result, our impulse is 0.5045 N*s. Again, we proved that J=delta P in this case.

Delta P= mvf -mvi

From our graph, we have an initial velocity of -0.408m/s before the collision and a final velocity of 0m/s after the collision.

Since the mass of the cart is 1.192kg,

m(vf-vi) = 1.192kg(0m/s-(-0.408m/s))= 0.486kgm/s

If we compare 0.5045 N*s to 0.486 kgm/s, they are relatively the same, which proves the impulse-momentum theorem.

Conclusion:

After performing three experiments, we were able to prove the impulse-momentum theorem that J=delta P. During an elastic collision, we know that there will be an initial an a final velocity since both carts will bounce off of each other. While in an inelastic collision, we know that there will be an initial velocity but no final velocity since the cart and the clay will stick to each other; therefore there will be no final velocity.

Kevin Tran
Lab Partners: Kevin Nguyen, Jose Rodriguez
22 April, 2017

Magnetic Potential Energy Lab

Purpose: This experiment was set up to verify the conservation of energy applied to an apparatus. In order to verify the conservation of energy that is applied to the apparatus, we needed to create an equation for magnetic potential energy.

In the picture below:

r= distance between the magnet in the end of the trash and the magnet on the glider.

h= height of air track from the ground

mass of cart: 0.344 kg

First, we set up our apparatus similarly to the model above. We made our air track as level as possible to make sure we get an angle of 0.0 degrees. We tilted the air track at 5 different angles to find the relationship between magnetic Force (F) and separation distance (r). We plotted a graph of magnetic force (y-axis)/ separation distance (x-axis). Since we assume that the relationship takes the form of a power law: F= Ar^n, we created a power line fit of the graph. We created a magnetic force/ separation distance because integrating the magnetic force in respect to separation distance will give us the magnetic potential energy. By finding the magnetic potential energy, we were able to verify the conservation of energy applied to the apparatus. 

-Above is the magnetic force/ separation distance graph.

-The uncertainty of the "A" parameter is plus or minus 4.62*10^-6

-The uncertainty of the "B" parameter is plus or minus 0.08493

After finding the equation from the box, we integrated the equation of magnetic force in order to find the magnetic potential energy equation. The magnetic potential energy equation that we got was  U(r) = (8.76756*10^-6)r^-1.566.

After finding the equation for magnetic potential energy, we were able to verify the conservation of energy. We attached an aluminum reflector on top of the air track cart in order to record the speed of the cart accurately with a motion detector. We placed the cart close to the magnet and ran the motion detector. By doing so, we are able to determine the relationship between the distance the motion sensor reads and the separation distance of the magnets.

We created a new column that allowed us to get the separation between the magnets from the position measured by the motion detector. We made our motion detector capture 30 samples per second. We also created additional columns in order to find the kinetic energy, the magnetic potential energy, and the total energy. We set the cart at the far end of the air track and gave it a small push. After, we made a single graph that displays kinetic energy, magnetic potential energy, and total energy in the same time frame.

Data

    angles                            "r"                        Force

1.8 degrees                    32 mm                  0.106 N

3.0 degrees                    25 mm                  0.176 N

3.6 degrees                    23 mm                  0.212 N

5.3 degrees                    20 mm                  0.312 N

8.2 degrees                    17 mm                  0.481 N

Calculated Data

Below is a graph that represents kinetic energy, potential magnetic energy, and total energy.

Kinetic Energy= purple

Potential magnetic energy= red

Total energy= blue

We created this graph in order to help us verify the conservation of energy by determining if the total energy is a linear line. If the total energy is a linear line, conservation of energy does apply to this apparatus.

Below is a picture of calculating magnetic potential energy.

Conclusion:

By using the graph that contains kinetic energy, magnetic potential, and total energy, we found out that the total energy line was not linear; therefore the conservation of energy does not apply to the apparatus. But when we we adjusted the separation distance from .257 to .254 meters, the total energy line became straight. The uncertainty in the separation distance affects our results tremendously. 

Kevin Tran
Lab partners: Dylan, Chris
15 April, 2017

Work-Kinetic Energy Theorem Activity

Purpose: To understand how the Work-Kinetic Energy Theorem works.

Introduction: We performed four EXPTs to see how the theorem works: Work done by constant force, work done by a nonconstant spring force, kinetic energy and the work-kinetic energy, and lastly the work-KE theorem.

First EXPT: Work done by a Constant Force

1st. we set up a track, cart, motion detector, force probe, pulley, cart stop and hanging mass as shown in the picture
2nd. We zero the force sense. Then we verify that the force sensor reads 4.9N.
3rd. We added 500 grams onto our cart and entered this value as the mass. Our total mass of the cart is 1190g.
Cart mass: 545 g
force probe: 145 g
added mass: 500g
                    =1190g

Next, we hanged 50 grams onto the hanging mass, which will allow the cart to move. While it is moving, we collected data. We have created a Position/time graph, Velocity/time graph, and Force(N)/position graph for the cart moving at a constant speed. As the cart reaches the cart stop, the position will remain where it is at that point and the velocity will drop. On the third graph, the red line represents our Force/position analysis and the purple represents the kinetic energy.

In order to find kinetic energy, we went under the data menu, choose New Calculated column. Next we gave it a name: Kinetic Energy(KE) and units: J. The equation with be 0.5*mass*"Velocity"^2. After, we should get the results for kinetic energy(purple line).

Next, in the Force(N)/position graph, we highlighted a section of our graph from our leftmost position going to some distance to the right. We went to the Analyze menu and chose Integral to allow us to get a result(black letters)




























We have tried two experiments: highlighting a smaller area of this graph and highlighting a larger area of this graph to see the difference.

For the smaller area, our integral value is 0.06N*M and for the kinetic energy of the cart is 0.07J. In conclusion, we can draw from our results that both values are relatively the same.











For the larger area, our integral value is 0.166N*M and for the kinetic energy of the cart is 0.156J. In conclusion, we can also draw from our results that both values are relatively the same.

The idea here is that work done on the cart by the tension force in a string should equal the kinetic energy gained by the cart at any point during the acceleration of the cart.

2nd EXPT: Work Done by a Nonconstant Spring Force

Now we measure the work done when we stretched a spring through a measured distance.

The model looks like the figure below.








We stretched the spring 0.6 meters from the starting point. Below is our Force(N)/Position(m) graph. From this graph, it gives us a slope of 3.460 N/m, which tells us the spring constant of our spring.


Next, I will display the Force vs distance graphs of both constant and nonconstant force.


First, we have the Force vs Distance graph of the constant force. By performing the integration routine, we are able to find the area under the graph of 0.5146 N*m

By knowing the spring constant of 3.46N/m, we are able to find the work by using the equation: 0.5kx^2, which will result in 0.662J. In comparison with the area 0.5146N*M and work 0.662J, they are relatively the same.



Next, we have the Force vs Distance graph of the nonconstant force. By performing the integration routine, we are able to find the area under the graph of -0.4021N*m . The reason for the results being negative because we verified that the motion detector is set to "Reverse Direction", so that toward the detector is the positive direction.

By knowing the work of the string 0.662J, we are able to compare that result with the integration result. In comparison with the area 0.4021N*m and 0.662J, they are relatively the same.



























EXPT 3: Kinetic Energy And The Work-Kinetic Energy Principle

We used the same information from EXPT 2 in order to perform this. This time we will be finding the kinetic energy of the cart.

First to note that the cart's mass is 545 kg.

Under Data-> New Calculated Column, we entered a formula that would allow us to calculate the kinetic energy of the cart at any point.

Again, we stretched the spring for 0.6 m, then we released to collect data.


As a result, the kinetic energy reads 0.639 J and the integration(area under the curve) is -0.6583N*m. We are trying to prove that the kinetic energy equals to the area under the Force vs position graph. As a result, they are relative the same. Also, in comparison to the work done on the cart by the spring and the change in kinetic energy is relatively the same as well.

Since we know the formula Work=Fd, we can assume that the net work=net force*d. Since we know that net force=ma, we can replace "net force" with ma and the equation will be W=mad. I am assuming that the acceleration will be constant so I will be using a kinetic equation "Vf^2-Vo^2=2ad". In the equation, we can isolate the "ad", and the equation will be ad="(Vf^2-Vo^2)/2". Then we plug the ad in the equation "W=mad", Once plug it in we will have~

W=0.5m(Vf^2-Vo^2), which is kinetic energy.

EXPT 4: Work-KE theorem

We watched a movie filed entitled "Work KE theorem cart and machine for Phys 1.mp4. In the video, the professor pulls back on a large rubber band by using a machine. The force being exerted on the rubber band is recorded. Below is the Force/ Position graph for the machine stretching the rubber band in the video.

Below is the force vs position graph.


Below is the work done at each point and total work.

In conclusion, we were able to understand how the work-kinetic energy theorem works. We learned that values for the integral and the kinetic energy from EXPT 1 should have the same result. Same for EXPT 2 that the area under the graph should be equal to the work done by the applied force. Also for EXPT 3, we realized that kinetic energy should equal to the area under the Force vs Position graph. This lab gave us a better understanding of how the work-energy principle relates to kinetic energy.

Kevin Tran
Lab partners: Kevin Nguyen, Jose Rodriguez
8 April, 2017

Work and Power

       Purpose: Students will understand how work and power works.

       Introduction: Me and my two lab partners have performed the three experiments required to gather data. The activity was held outside of the science building. First, one person from our group will be pulling down on a rope that goes over a pulley to a backpack containing a known mass. We chose the 5kg backpack in this case. We timed our member for lifting the backpack from the ground to the bottom of the backpack being level with the top of the balcony. Next, another member walked up the stairs and get timed. (stairs on the left of the picture below) Finally, another member runs up the stairs while being timed. (stairs on the left of the picture below).

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First, our members measured the height of the stairs. In order to find height, we took the measurement of one stairs and multiplied it by the total number of stairs, which in this case is 26 steps. As a result, the height is 4.39m 

1st experiment: Pulling on the rope.

-As a result, one of the members pulled on a rope of a 5kg back in 4.4 seconds
Below is the result of work and power while performing this experiment:





















Since we know the height is 4.39m, we were able to find the average velocity of 1m/s with kinematics (v=d/t or v=4.39m/4.4sec)

Our teacher gave us the angle of the angles of 40 degrees. With the given angle, we can find h(hypotenuse). As a result, h= 6.8m and the x distance is 5.23m

2nd experiment: Walking up the stairs.
-As a result, one of the members walked up the stairs in 12.58 seconds.
Below is the result of work and power while performing this experiment:



Since we have the distance h=6.8m, were able to find the average velocity of 0.54m/s with kinematics (v=d/t or v=6.8m/12.58 sec)

3rd experiment: Running up the stairs.
-As a result, one of the members ran up the stairs in 5.8 seconds.
Below is the result of work and power while performing this experiment:


Since we have the distance h=6.8m, were able to find the average velocity of 1.16m/s with kinematics (v=d/t or v=6.74m/5.8sec)

Conclusion:
a) Using the results from walking: KE=.5mv^2 ---> .5(87.63kg)(0.54m/s)^2 = 12.8J
b) -Since we have a result of 300 watts when one of our members walked, it would take about 3.66 flights of stairs(95 steps) to equal the power output of a microwave oven.
     -Since we have a result of 540 watts when of our members walked, it would take about 2 flight of stairs(32 steps) to equal the power output of a microwave oven.
c) 

d) (1) Since Power= Work/time, then (12.5*10^6)/600 sec = 20,833.33 watts.
     (2) 20,833.33 watts/ 100 watts = 208 people.
     (3) Since Power = Work/time, then 12.5*10^6)/t = 100 watts. So it would take me about 125,000 seconds(2,083.33 minutes) to ride on a bicycle-powered generator in order to heat water for my 10 minute shower.

By performing the 3 activities: pulling on the rope, walking up the stairs, and running up the stairs, we realize that the work and power for each activity is different according to our calculations. The harder the task, there will be a higher result in work and power. Example, if we were to replace our 5kg backpack with a 10kg backpack, our work and power will be bigger due to knowing that we will have a bigger force applied. Another example is if we were to walk or run up the stairs faster, the value for power will increase due to less time.

Kevin Tran
Lab Partners: Kevin Nguyen, Jose Rodriguez
8 April, 2017

Centripetal force with a motor

     Purpose: Students will come up with a solution to find the relationship between θ and omega.

    Theory: We will be using an apparatus to find the relationship between  θ and omega. (picture shown below). As the motor spins at a higher angular speed (w), the mass revolves around the central shaft at a larger radius and the angle θ increases.

Below is the apparatus that we used to determine our values.

Below is the apparatus that is set up in order to find our individual values.

W= omega

R= radius 

H= height of the entire apparatus

h= height of the mini apparatus

L= length of the string

H-h= the height of the string when angled

Lsin θ= distance from the edge of the pole to the mini apparatus

θ= angle when angular speed is changing.

After measuring:

W: (listed below)

R= 0.8m

H= 1.8m

h= (listed below)

L= 1.61m

H-h= (listed below)

θ= (listed below)

1st, our professor activated the apparatus and performed 5 trials. Each trial represents a different omega.

Trials:

10 rotations x 2pi/time= rad/sec

1: 1.9 rad/sec (10 rotations in 33 seconds)

2: 2.3 rad/sec (10 rotations in 27 seconds)

3: 2.62 rad/sec (10 rotations in 24 seconds)

4: 3.14 rad/sec (10 rotations in 20 seconds)

5: 4.19 rad/sec (10 rotations in 15 seconds)

Next: We adjusted the height of mini apparatus to find the height of where the end of the string hits the paper.

As a result for a each angular speed:

Trials: (height)

1. 0.36m

2. 0.652m

3. 0.87m

4. 1.134m

5. 1.36m

Results of H-h:

1. 1.44m

2. 1.148m

3. 0.93m

4. 0.666m

5. 0.44m

Below is the equation created to find the individual angles of the 5 trials and the results:

Next, we created a formula to assume the angular velocity of each of the 5 trials

Below is the results of the 5 trials.

We compared the calculated omega with the omega given to us. While we have calculated omega(x-axis) and recorded omega(y-axis). As a result, the slope gives us a number of 1.136, which is really close to 1. This proves that our calculated result is close to the recorded. 

Conclusion: By comparing our calculated results with the recorded, the angular speeds of the 5 trials are almost similar but very close. It is true that the higher the angular speed, the bigger the angle θ will be.

Kevin Tran
Lab partners: Kevin Nguyen, Jose Rodriguez
1 April, 2017

Centripetal Acceleration vs. angular frequency

     Purpose: To determine the relationship between centripetal force and angular speed.

      Introduction: Our teacher used a rotational system to determine the centripetal force. The centripetal force is a force that allows a body to follow a curved path. We used the formula for centripetal to determine the centripetal force. With the rotational system, the professor will vary few things: the radius of the string, the mass of the object, and omega by changing the amount of power in the power supply.

Below is the force for centripetal force where:

F= Centripetal force

m= mass of object

r= radius

w= omega

In order to test this formula, we need to create several graphs.

Below is the rotational system we used to determine centripetal force.

1) First, we performed 5 trials when radius is varied, but everything else is constant. (data collected below).

With the collected data, we are able to construct a radius(x)/ force(y) graph that will give us mw^2 (slope), and we got it by linear fitting the graph. Our slope happens to be 6.930

2) Next, we performed 3 trials when mass is varied, but everything else is constant. (data collected below).

With the collected data, we are able to construct a mass(x)/ force(y) graph that will give us rw^2 (slope), and we got it by linear fitting the graph. Our slope happens to be 8.869.

3) Next, we performed 3 trials when omega is varied, but everything else is constant. (data provided below). All we were given is the change in the power supply, so we have to find omega.

Since 10 rotations = 10 revolutions, in order to find omega, we have to convert the 10 revolutions into rad/sec. (work shown above for each trial).

From the collected data, we are able to construct a omega^2 (x)/ Force(y) graph in order to find mr(slope), and find it by linear fitting the graph. As a result, the slope is 0.1078

4) Next, we made a mw^2(x)/ Force(y) graph. We used the same results from the previous data when omega is varied to make this graph. By creating a linear fit, we are able to find slope (r). Our slope (r) is 0.3667.

In this graph: 1) Mass is supposed to vary.

                        2) Omega wasn't supposed to vary

                        3) Radius was for sure constant

5) Next, we made a rw^2(x)/ Force(y) graph. We used the same results from the previous data when omega is varied to make this graph. By creating a linear fit, we are able to find slope (m). Our slope (m) is 0.2292

In this graph: 1) Radius is supposed to vary.

                        2) Omega wasn't supposed to vary.

                        3) Mass was for sure constant.

Conclusion: Overall, we were able to test for the formula F(Centripetal force)=mrw^2. What we noticed is when radius is increased, the average force for 10 rotations increases as well. Also, when mass is increased, the centripetal force increases, and the case is opposite when the mass is decreased. When our teacher increased the power supply, our omega increases as well. A problem that could of occurred during the lab is if we only collected data for 5 rotations instead of our usual 10 rotations. Also, another error that could happen is if we were to calculate everything in centimeters and grams, but our usual units is meters and kilograms. We were able to learn a lot about centripetal force, and we can use this knowledge to understand how everything around us works.