Cart Challenge - Ally and Nikki

Finding the Unknown Mass

Ally and I solved to find the mass of a variable on top of a cart. The variable we found was the weight of the soup can.

First, we weighed the mass of the cart:  .49 kg. We put the center of the cart at 110 inches and the markers on the side at 53 inches and 173 inches.


After taking 5 trials of the empty cart, the average velocity taken was .5 m/s. We then put the soup on the cart and measured the velocity. After taking 5 more separate trials, we found the average velocity with the soup can to be .25 m/s.  

Then our group needed to solve for the mass of the cart with the soup can.  We used the formula mava + mbvb = (ma + mb) (Vab).

mava + mbvb = (ma + mb) (Vab)
.49 (.5) +0 = (.49 + mb) .25
.245 = .1225 + .25mb
.1225 = .25mb
.49 kg = mb

To check our percent error, our group measured the cart with the soup can on it: 478 grams and converted it to kg: 0.478 kg

Therefore our percent error= 2.51%

To calculate the percent error:
=(.478)(.49)/ .478*100
= 2.51%

Video Blog Analysis

  Nikki Montlick
Video Blog Analysis

The video to the right is the video my group used to collect the majority of our data for this lab. The yellow line against the blue wall represents a meter within the context of the video. My group uploaded our video to capstone and identified the points of motion the ball in the air. The proceeding image is the points of motion for the ball along the arc. The balls movement is forward and up.

Next, using capstone, my group created position versus time graphs and velocity versus time graphs and position versus time graphs in both the x and y for the motion of the ball.

3. Analysis:

Fg is the only force in the vertical direction. The greatest possible acceleration is -10m/s^2. When I highlighted my points on my graph my acceleration was within a a tenth of 10m/s^2. The FBD is the same as a. because the same object is being thrown. The Fnet=0 because there is no force in the horizontal. Therefore the acceleration is equal to 0m/s^2. I know this because it is the same as the velocity- since there is no acceleration, the velocity would remain constant. The velocity of the object will remain constant throughout its path because there is no force acting in the horizontal direction. 

c
   c. The initial velocity in the x direction is 0.6m/s. I know because I looked on the vty graph and on the equation of the line, b is always the initial velocity. I can identify the initial velocity in the x and y direction by looking at the equation on the graph. The equation is y=mx+b, and I know that b is equal to the initial velocity, then the initial velocity based off my equation of the graph is 0.6m/s. The initial velocity is .3m/s. I know this because on the velocity versus time graph in the y direction, the b in the equation of the line is .3m/s.The velocity at the top of its x path is .6m/s. 

f.      The velocity at the top of its path in the y direction is 0m/s. I can distinguish this by looking at my v versus t graph in the y direction. By looking to find a straight diagonal on the graph, I can see that the line cuts through the y value of 0. This represents the point where the ball hits the vertex at 0m/s.

  

The final velocity in the x direction is 2.94 m/s. This can be found be looking at the equation on the graph and identifying the number plugged in for b in y=mx+b. The final velocity in the y direction is -2.06   m/s. The ball traveled 1.13m total in the y direction. The ball traveled .71 meters.  How much time did it take for the ball to get to the top of the path? It took the ball 4.04 seconds to get to the top of the path.The ball was in the air for 4.348 seconds.

4. Conclusions

This project introduced a new perspective into thinking about force for our class. I split forces into vertical and horizontal directions. The vertical direction in this project incorporated the CAPM and UFPM model. The vertical direction used the following equations: (Keep in mind that any of these equations could solve for any of the variables used in them. Algebraically you could solve for whatever variable you didn't know by plugging in the known factors.)

Vf=at+Vi                 (equation for final velocity)

x=1/2(a)(t^2)+Vit    (equation for displacement)

Vf^2=Vi^2+2a        (equation for final velocity)

In the horizontal direction, there is no acceleration. Since there is constant velocity and no acceleration the following equation can be easily used. The horizontal direction applied the BFPM and CVPM models.

The vertical forces could be found using the following equations:

V= change in X

     change in t

UFPM: Nikki and Elise

Materials:

-cart

-buggy

-pulley

-Atwood machine

-500 kg weight

First we took the mass in grams of all variables in the experiment, and converted them to Newtons.

In this experiment, the goal was to predict the point that the buggy and the pulley would intersect. To make this prediction, we had to find the following variables:

•   net force

•   velocity of the buggy

mass and net force on the Atwood Machine

In this experiment, we did not take the acceleration of the buggy became the force was balanced since it moved at constant velocity.

Item	Mass (kg)
Buggy	.477
Cart	.9892
Pulley	.0509
Total system	1.04

First, we plugged in Fg, which was -10.04N, and could conclude that the diagram was balanced in the vertical direction, so therefore Fn was also 10.4N.  We knew that it was frictionless along the surface, so we concluded friction was 0N. Next, to solve for Ft, we used the equation: Fnet= -Ff+ Ft (which is equal to ma).

Using the data we found for the mass of the pulley, we found the force of this mass due to gravity.

.0509 kg

Fg2= 0.509N

After I found the Fnet in the horizontal direction, I plugged it and the mass into A=Fnet.

              Mass

Then I used the equation for displacement to solve for time of the hanging mass to drop.

ΔX= 1/2at^2+Vit

.86 = 1/2at^2 + Vit

(2) .86 = .5 t^2

1.8547 = t

Prediction:

1.85s =t

I also timed how long it would go for the buggy to go one meter. Since it is moving at constant speed, acceleration is equal to 0. So the equation becomes:

ΔX= Vit

Vi= ΔX  = 1 m= 0.25 m/s

       T         4

Now that we found the velocity, we can use the time it took for the hanging mass to drop and solve for the displacement of the buggy.

Δx= Vit

(.25)(1.85)

ΔX= .4625m

=46.25cm back from where the pulley will land.

Our pulley landed on the cart, so the experiment had 0% error.

Fan Cart Blog

Group A: Nikki and Elise
Group D: Hudson and Maren

Purpose: The purpose of the lab was to find the point of intersection between two cars from different groups. The cars accelerated from rest from different starting points, and using our data we predicted the point where they would collide.

Materials: Chalk, 2 Fan carts, timer, meter stick

Lab Description: Our group lost our work but however, received our original information from Group D (who had the opposing cart whose collision we predicted). Using this data we calculated our prediction for collision point.

Group D- Initial velocity was Vi:-.18

Starting point: 1m away from motion sensor

Group D then plugged these values in the equation:

∆x=1/2a(∆t)^2+(-.18)(t)

The found the value 4.3 seconds.

Our group found our initial velocity and plugged it into the same equation.

This equation produced: ∆x=1/2(.1203)^2+(.18)(t)

After testing our estimates to find the point of intersection, we found these values.

We predicted:

Cart C: 97cm

Cart D: 1.072m

Percent error: 0%

The results showed an intersection at exactly our predicted distance, therefore our group had zero percent error.

BFPM Practicum

a.

b.

c. First, our group measured the angles of the wire and drew a picture of the hanging object. Next, we drew free body diagrams to understand the forces involved. In our diagrams, we wrote the forces, then added the angle measurements, and then calculated the force of each Ft. We used Sin to solve for Ft on each wire and added the amounts together to get a total weight of 2.51N.

Texting While Driving Corrections

Texting While Driving Data:

Prediction: We predict the car will travel 0.02815 in 2.38 seconds at a constant velocity of 45 m/h.

distance = 45 miles = 45 miles = x miles 
time           1 hour        60 min      1 min

60x = 45       x= .75 miles per 7 min
60      60

.75 miles =  75 miles
1 min             60 sec

After setting up a proportion and canceling the units until it was miles/ seconds, we cross multiplied the equation to find the variable x, which represents miles:

75 miles  =  x miles        x= 0.02875
60 sec          2.3 sec

Therefore, in 2.3 secounds the car would travel about .03 or .02875 seconds at a constant velocity of 45 m/h.

This motion map describes the cars motion because it shows that it is moving at a constant velocity. The velocity vs. time graph shows how the car moves at a constant velocity, from an origin of 45 m/h. The position vs. time graph shows how the cars position changes, but the distance per second is consistent throughout the movement of the car.

Nikki's Blog Post Unit 1

Unit One Summary:

Constant Velocity Particle Model

What did I learn from this unit?
This unit taught how to identify relationships on a graph, read a graph, and make predictions based on the data on a graph. I also learned how to identify relationships in an experiment, and be able to manipulate data using my independent variable and excel. I also learned how to create and read motion maps, create a graph with excel, position verses time graphs, and identify the factors in a graphs. These factors include velocity, time, position, distance verses displacement, and variables. In this blog, I will convey the skills we have learned and examples of how I used them.

Constant Velocity Particle Model Example Questions and Analysis:

If the curve is straight it indicates constant velocity and a linear relationship. In the picture to the left, I solved to find the velocity of the line by using the equation rise. 

                                   run

Then, to find the mathematical equation to describe the object's motion I used:
X= Vt + Xo

I plugged in the variables from the graph that I knew and solved for X. 

How to Make a Line with Excel:

1.     Enter data in columns.

2.     Highlight data

3.     (In insert menu) insert chart

4.     Choose XY on the marked scatter.

5.     Add trend line and equation on the graph.

Motion Map (Diagrammatic):

The motion map represents the velocity, position, and acceleration of an object at equally spaced times. The purpose of the motion map is to show you these factors at various time reading in a visual representation other than a graph.

Here is an example of a motion map. This map displays an object moving at a constant velocity. 

Velocity:

Velocity is the rate of change of an object.

V = slope     D= Velocity

T                  T

Position vs. Time Graphs:

The position verses time graph helps distinguish the displacement and total distance of an object. It also reveals key information about the velocity of the object. For example, a steep slope means a faster velocity. A straight (linear) slope means a constant velocity. A curved line means the velocity changes over time. Here is an example of a constant slope:

Equation of a graph:

X= Vt + X⁰

X⁰= position

V= velocity

T= time

The equation X= Vt + X⁰ describes the motion of an object.This equation can also be used to predict the position of an object at a certain point.

Path length:

The path length in the total distance something traveled.

Independent variable:
The variable over which the experimenter has complete control (x axis).

Dependent variable:
The variable that responds to change in the independent variable (y axis).

Displacement:

The movement of something from its original position; difference in position from origin and final point.

Speed:

Speed is how fast the object is going. A faster speed can be identified by a steeper slope on a graph.

Describe the Motion of the Object:

The motion of the graph can be read on a motion map, position verses time graph, or the constant velocity particle model. The image below shows an example of how all the same information can be displayed in the different models.

Shapes on a Graph: By being able to identify these shapes and trends on a graph it improved our ability to interpret graphical relationships and express it in written form. 

Connections:

This unit can be applicable in everyday life because it is important to understand graphs and charts. The ability to understand and make a graph is crucial in business. It is a required skill for making predictions and analyzing data. Also, these skills are applicable in our every day life. Everywhere we go, we are in a specific position. Although the starting point could be anywhere, it can still be displayed on a graph. We consider distance vs. time every day when calculating how long it will take to arrive at a specific point of location. People often try to predict these times themselves, but it could easily be predicted using the CVPM. For me personally, I could use the CVPM to measure how fast I run when exercising. I can use the CVPM to see if I am staying at a constant speed, of if my speed decreases sporadically.