On this challenge, we were tasked with trying to launch two carts with different spring constants at the same velocity. To do this we had to thoroughly understand the conceptual elements of energy conservation.
We began by determining the energy conservation equation.
Eel = Ek
We then manipulated the equation:
1/2kx^2 = 1/2mv^2
Through given information and initial calculations, we gathered the following information:
Red Spring Constant: 84Nm
Blue Spring Constant: 112Nm
Mass of cart: .542Kg
We knew that we had to get the same velocity for both carts so we decided to use .5m/s for our intended velocity. From there it was a plug and chug problem for both carts to find the x value.
Red Cart:
1/2kx^2 = 1/2mv^2
1/2(84)x^2 = 1/2(.542)(.5)^2
x^2 = .0016
x = 4cm
Blue Cart:
1/2kx^2 = 1/2mv^2
1/2(112)x^2 = 1/2(.542)(.5)^2
x^2 = .012
x = 3.5cm
From here, we tested our predictions on the track. The velocity of the red cart was .44m/s and the blue cart was .45m/s. These numbers left us with just a 2% margin of error, one of our lowest ones yet!
Tuesday, May 17, 2016
Friday, April 15, 2016
Rocket Projectile Practicum Post
Experiment:
Our task was to take a rocket and predict where it will land using our formulas from our unit on projectile motion.
Materials:
Rocket
Angles
Bike Pump
Steps:
1. We found the actual velocity of the rocket by testing it at a different angle and seeing how far it went. We worked backwards using trigonometric ratios.
2. We started with an angle of 25 degrees and found that the actual velocity of the rocket at the time of takeoff was 19.92m/s.
3. We then could predict the time using the viy (from the new angle (40 degrees)) and the equation
x = 1/2at^2+vit
We found the time as 3.114s
4. We next found the change in position using equation Vx = (change in) x/ t
Change in position: 39.87m
Conclusions:
The actual position of the rocket was 32.9m. We were somewhat off in our calculations, but I think it was because of the wind on the day that we first tested. If the wind speed had been constant, I believe that our prediction would have been correct.
Our task was to take a rocket and predict where it will land using our formulas from our unit on projectile motion.
Materials:
Rocket
Angles
Bike Pump
Steps:
1. We found the actual velocity of the rocket by testing it at a different angle and seeing how far it went. We worked backwards using trigonometric ratios.
2. We started with an angle of 25 degrees and found that the actual velocity of the rocket at the time of takeoff was 19.92m/s.
3. We then could predict the time using the viy (from the new angle (40 degrees)) and the equation
x = 1/2at^2+vit
We found the time as 3.114s
4. We next found the change in position using equation Vx = (change in) x/ t
Change in position: 39.87m
Conclusions:
The actual position of the rocket was 32.9m. We were somewhat off in our calculations, but I think it was because of the wind on the day that we first tested. If the wind speed had been constant, I believe that our prediction would have been correct.
 |
| This is a picture of the calculations I took |
Unknown Mass Cart Practicum
Procedure:
The task of our group was to try and find the mass of an object placed on a cart using our momentum equations. We were given a cart, a bag of salt (unknown mass), and capstone sensors.
Steps:
1. We knew that the mass of the empty cart was 0.5kg (Ma). We then measured the velocity of the cart so we found the Va.
2. From here we found the velocity of the cart with the bag of salt. (Vab)
3. Using the law of conservation of momentum.
mava + mbvb = (ma + mb)(Vab)
Evaluation and Manipulation of Data:
pbefore = pafter
mava + mbvb = (ma + mb)(Vab)
The task of our group was to try and find the mass of an object placed on a cart using our momentum equations. We were given a cart, a bag of salt (unknown mass), and capstone sensors.
Steps:
1. We knew that the mass of the empty cart was 0.5kg (Ma). We then measured the velocity of the cart so we found the Va.
2. From here we found the velocity of the cart with the bag of salt. (Vab)
3. Using the law of conservation of momentum.
mava + mbvb = (ma + mb)(Vab)
Evaluation and Manipulation of Data:
 |
| Diagram of our experiment. Drawing courtesy of Miranda Thompson (a group member) |
pbefore = pafter
mava + mbvb = (ma + mb)(Vab)
(0.5)(0.57) + (mb)(0) = (0.5 + mb)(0.24)
(0.28)+ 0 = .12 + .24mb
.28 - .12 = .24mb
.16 = .24mb
mb = .66kg
Conclusions:
The actual weight was .77kg, so we were a little off, but I think it's mostly due to lack of time to do a lot of trials for the velocity before and after. We took the average of 3 trial runs when calculating the velocity.
Tuesday, February 16, 2016
Properties of Projectiles
Overview:
We started a new unit with an introductory experiment. The experiment was simply to throw a ball forward and upward, and take a video. From there, we uploaded the video to a program called PascoCapstone. We tracked the ball's motion while in the air and then were able to find some key characteristics of projectile motion. The main thing that we discovered was that the motion in the horizontal direction is completely independent from the motion in the vertical direction.
Graphs/Raw Data:
Analysis of Data:
With the information that these graphs have given, we can manipulate the data to find certain unknowns.
Acceleration in y axis: 9.8m/s^2 -- We know this because on earth the force of gravity is always the same.
Acceleration in x axis: 0 m/s^2 -- We know this because there is no unbalanced force to accelerate the ball in the horizontal direction. It is at constant velocity.
Initial velocity in y: ~2.1m/s -- We know this because its the y intercept of the vertical plane
Initial velocity in x: 3.88m/s -- We know this because it is the y-intercept for the horizontal plane
Velocity at the top of the path on y axis: 0m/s^2 -- We know this because the ball is changing direction and stops momentarily.
Velocity at the top of the path on x axis: 3.8m/s -- constant velocity
Final Velocity in x: 2.2 m/s -- final point on v vs. t graph
Final Velocity in y: -0.5m/s -- final point on v vs. t graph
How High the ball got: 2.53m -- we can look on the x vs. t graph
How far the ball went: 3.05m -- I used Xfinal - Xinitial = Change in X
Time for the ball to reach the top of its path: 0.835s -- Used equation Vf=at+Vi
Total air time for ball: 1.57s -- x=1/2at^2+vit
Conclusions:
In this experiment, we learned that the vertical and horizontal motions are completely unrelated when dealing with projectiles. We also discovered that all horizontal motion was constant, so we could use formulas and equations from the BFPM unit. The vertical motion was all constant acceleration; we used the UFPM equations. The same principle applied for solving for displacement. The horizontal motion remained constant for the entire trip and the vertical motion stopped at the top of the path. Furthermore, we were able to use the graphs that we created to find many of the unknowns, which proved to be helpful.
We started a new unit with an introductory experiment. The experiment was simply to throw a ball forward and upward, and take a video. From there, we uploaded the video to a program called PascoCapstone. We tracked the ball's motion while in the air and then were able to find some key characteristics of projectile motion. The main thing that we discovered was that the motion in the horizontal direction is completely independent from the motion in the vertical direction.
Graphs/Raw Data:
 |
| This image shows the path of the ball that we threw. The red plus signs show the position of the ball. |
 |
| This graph shows the position of the ball vs the time. The red line is the horizontal displacement, and the green line is the vertical displacement. |
 |
| This graph shows the velocity of the ball vs. the time. The pink line shows the velocity of the Y-axis. The blue line shows the velocity of the X-axis. |
Analysis of Data:
With the information that these graphs have given, we can manipulate the data to find certain unknowns.
Acceleration in y axis: 9.8m/s^2 -- We know this because on earth the force of gravity is always the same.
Acceleration in x axis: 0 m/s^2 -- We know this because there is no unbalanced force to accelerate the ball in the horizontal direction. It is at constant velocity.
Initial velocity in y: ~2.1m/s -- We know this because its the y intercept of the vertical plane
Initial velocity in x: 3.88m/s -- We know this because it is the y-intercept for the horizontal plane
Velocity at the top of the path on y axis: 0m/s^2 -- We know this because the ball is changing direction and stops momentarily.
Velocity at the top of the path on x axis: 3.8m/s -- constant velocity
Final Velocity in x: 2.2 m/s -- final point on v vs. t graph
Final Velocity in y: -0.5m/s -- final point on v vs. t graph
How High the ball got: 2.53m -- we can look on the x vs. t graph
How far the ball went: 3.05m -- I used Xfinal - Xinitial = Change in X
Time for the ball to reach the top of its path: 0.835s -- Used equation Vf=at+Vi
Total air time for ball: 1.57s -- x=1/2at^2+vit
Conclusions:
In this experiment, we learned that the vertical and horizontal motions are completely unrelated when dealing with projectiles. We also discovered that all horizontal motion was constant, so we could use formulas and equations from the BFPM unit. The vertical motion was all constant acceleration; we used the UFPM equations. The same principle applied for solving for displacement. The horizontal motion remained constant for the entire trip and the vertical motion stopped at the top of the path. Furthermore, we were able to use the graphs that we created to find many of the unknowns, which proved to be helpful.
Friday, February 12, 2016
UFPM Practicum Lab
Goal:
The purpose of this lab was to test our knowledge of unbalanced forces and our applied use of kinematics. We demonstrated these skills with this lab. Our goal was to try and land a weight attached to a cart on another cart below going at constant velocity.
Procedure:
We first found the mass of the entire system. 0.693kg (We rounded to .64kg)
We then solved for acceleration. (The weight was .5N)
After, we found the distance from the starting position of the weight to where it would land on the cart. The distance was .74m.
Using the information that we now had, we were able to solve using a formula for time.
Next, using the formula V=(change in)x/(change in)t we found the velocity of the cart on the ground. We concluded that the cart went 1 meter in 3.6 seconds. We manipulated these numbers to find that the cart had a constant velocity of .28m/s. Using these equations we hypothesized that if we placed the constant velocity cart 0.38m away from the weight. We could release both at the same time and the weight would land on top of the cart.
After our initial test, we proved our hypothesis correct and were able to land the weight perfectly on top of the cart.
We filmed our test which can be seen HERE.
The purpose of this lab was to test our knowledge of unbalanced forces and our applied use of kinematics. We demonstrated these skills with this lab. Our goal was to try and land a weight attached to a cart on another cart below going at constant velocity.
Procedure:
We first found the mass of the entire system. 0.693kg (We rounded to .64kg)
We then solved for acceleration. (The weight was .5N)
a=Fnet/m
a=.5N/.64kg
a=0.78m/s^2
After, we found the distance from the starting position of the weight to where it would land on the cart. The distance was .74m.
Using the information that we now had, we were able to solve using a formula for time.
x=1/2at^2+Vit
(.74)=1/2(.78)(t)^2
1.89=t^2
t=1.37s
Next, using the formula V=(change in)x/(change in)t we found the velocity of the cart on the ground. We concluded that the cart went 1 meter in 3.6 seconds. We manipulated these numbers to find that the cart had a constant velocity of .28m/s. Using these equations we hypothesized that if we placed the constant velocity cart 0.38m away from the weight. We could release both at the same time and the weight would land on top of the cart.
After our initial test, we proved our hypothesis correct and were able to land the weight perfectly on top of the cart.
We filmed our test which can be seen HERE.
Wednesday, December 9, 2015
Fan Car Acceleration Experiment
Overview:
The purpose of this lab was to take two fan powered cars with different accelerations going in different directions, and try and predict where and when they would collide.
Materials:
The purpose of this lab was to take two fan powered cars with different accelerations going in different directions, and try and predict where and when they would collide.
Materials:
- Fan Powered Car
- Stop Watch
- Meter measuring stick
- chalk
Procedure:
We started the lab by finding the acceleration of our car. We did this by marking the displacement of the car at each second of motion. We then shared our data with another group so we could calculate the point of intersection. Once we compared data, we manipulated the accelerations to find where and when the two cars would collide.
Raw Data:
Time Their Data Our Data
Manipulated Data:
 |
| Here is the position vs time graph for the two cars |
Summary:
We collaborated with Miranda and Morgan's group to find the position and time of intersection. They were our partner group, so we were able to use the same test and the same numbers as them. We ended up being within the allowed 5% error margin at the end of the experiment.
Prediction:
Time: ~2.5 seconds
Position: ~61 cm
Actual:
Time: 2.9 seconds
Postion: 63 cm
Sunday, December 6, 2015
Forces: Challenge Lab
Introduction:
In this lab, we tried to find the weight of an object based on the tension of the wire that was holding it up.
Data:
We used trigonometric ratios to find the unknown tensions.
Cos58 = fty1 / 1.6
-> fty1 = 0.85N
Cos22 = fty2 / 3.5
-> fty2 = 3.25N
With these tensions, I concluded that the weight of the object must have been 4N. I was correct.
In this lab, we tried to find the weight of an object based on the tension of the wire that was holding it up.
Data:
We used trigonometric ratios to find the unknown tensions.
 |
| FBD for the experiment |
Cos58 = fty1 / 1.6
-> fty1 = 0.85N
Cos22 = fty2 / 3.5
-> fty2 = 3.25N
With these tensions, I concluded that the weight of the object must have been 4N. I was correct.
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