Mr. Incredible
The system was composed of two tension ropes, a figurine, a weight and two stands.
The two ropes were attached to the figurine at angles of 45 and 35 degrees. The figurine was at the bottom with a weight of .2Kg attached.
Using trigonometry I found the horizontal and vertical force of the two diagonal tensions. By creating an equation with the two opposing fores of the vertical and horizontal tensions I was able to calculate the tension of Mr incredible and the force of the weight. I subtracted the force of the weight from the equation, and divided by 9.8 to get the weight of Mr. Incredible, which was 195 grams 0r .195 Kg.
Modeling the Force of Gravity
The purpose of this lab was to determine the mathematical and graphical relationship between weight and force of an object.
To find an answer to the hypothesis objects were put on a spring scale in order to find the newtons of said objects. An object with a known force was put on the scale, in order to have a control during the experiment, and to see if the scale and the known force differed.
During the experiment I used 10 instead of 9.8 newtons in my equation to determine the newtons of specific objects. This really affected my data’s preciseness.
Using the data from this lab I concluded that the force of gravity multiplied by the weight of an object equals the Newtons of that object.
For example if something weighed 7kg in order to find the Newton’s you would use the equation (9.8m/s)(7kg) = -68.6N
This video explains how to use a spring scale.
Changing Motion Activity
Which cart/ container combo does your group think is heavier? How did you decide ?
Cars and trucks designed to tow trailers usually have larger engines and more powerful brakes than those without a tow package. For Example, the new ford F-250 super duty tow package boasts a”6.7 Turbo Diesel V-8 with integrated Exhaust Braking”
Use the results of this experiment explain why this is significant.
Is the toy really “thrown forward” when the cart hits the barrier? From your observation, what really happens to the toy when the crash occurs.
As of September 19, 1997 Maine law requires drivers to ensure that they and their passengers are buckled-up. Use your observation to write a brief scientific explanation of why it is important to wear a seat belt when riding in an automobile.
Describe as accurately as you can what you observe happens to the toy when the cart starts to move. Does it really Fall Backwards when the cart is first pulled?
RSU5 policy JICC states that students should immediately take their seats and remain seated while the bus is in motion. Use your observations from this experiment to explain why bus drivers wait for passengers to be seated before getting under way.
Object at rest stays at rest.
This is a picture of a pendulum. With no outside forces acting upon it, it does not move.
Object at rest stays at rest.
This is the same pendulum. The outside force (the wood) pulls the balls away from their original position and when the board releases the balls they swing. This is an example of an outside force acting upon an object at rest.
Object in motion states in motion.
If a baseball is thrown in outerspace it will not stop until an outside force acts upon it, such as a meteor. There is no outside force, such as gravity or friction, acting on the ball. Therefore it will not stop moving until it comes in contact with another object.
Object in motion states in motion.
This image shows how the ball (the outside force) moves the mans head. This is an example of how an object stays in motion until acted upon by an outside force.
Effect of mass on inertia
This is an excerpt from a mythbusters experiment. A vehicle going 60 mph shot a ball out of the back going 60 mph. The ball stays in the same exact position horizontally, but falls vertically due to it’s mass and the effect of gravity.
The Road Runner Problem
The task of this lab was to predict when a ball rolling down the ramp would hit a car moving along the bottom of the ramp.
In order to successfully time the collision of the ball to hit the car I had to measure the velocity and of the ball and the car. First, the velocity of the ball had to be calculated so that when it was released it would hit at the exact time that the car was passing the bottom of the ramp. Secondly, the amount of time it would take for the car to reach the bottom of the ramp would need to correspond with when the ball was released from the top of the ramp.
Calculations:
Our first attempt was a fail. The ball missed the mark by about four inches. What lead to us succeeding was realizing that we had to release the ball when the tape on the ground was lined up where we wanted it to hit on the car. As you can see in the video above, the ball was released when the tape was lined up with the middle of the car. Therefore, the ball hit the middle of the car. This shows that I am proficient in calculating constant velocity because me. and my group, were able to calculate the exact time it would take the ball to roll down the ramp, and the time that would be needed for the car to reach the end of the ramp the same time that the ball did. This is the same reason that I am proficient in calculating and representing constant acceleration.
Acceleration Lab
Samantha Jordan
(Nate Smail and Yacob Olins)
Mr. Smail
AP Physics
Acceleration
To establish a graphical and mathematical relationships between position and time for an object moving down an incline.
In the experiment a car was set on a ramp. The ramp was situated so that one of its ends was more elevated than the opposite end. This experiment was run in three different ways. The first time, a stopwatch was used. Everytime the car passed by a line, marked every 10 centimeters, on the ramp the stopwatch was hit and the time was recorded. So, everytime the car went down the ramp 10 centimeters was added onto the distance until it reached the end of the ramp. This made the experiment especially hard because the car had to be set down the ramp multiple times to get an average of time that it took for the car to go a certain distance. The second time this experiment was done, a video was taken of the car going down the ramp, the time was noted every time the car passed by one of the markers. This was easier because it was not necessary to set the car down the ramp multiple times. It was difficult to pinpoint the exact time that the car passed by a line because the frames of the video were so spread apart. The last time the experiment was done a motion detector was used. The motion detector measured the position of the car on the ramp, related to the time it took for the car to get to that position.
This data table represents the data that was collected in the first experiment. A stopwatch was used to determine the amount of time that the car took to go each distance.
During the experiment it was observed that as the cart moved down the ramp it began to pick up speed. When the information of position vs. time was graphed, a top opening parabola was derived. When the information of velocity vs. time was graphed a linear set of data points, increasing diagonally, was formed. For the velocity vs. time graph, the line remained flat, just above zero.
Because the experiment was run multiple times, using different methods, fairly good data was collected. If just one method was used, such as using the stopwatch, the graphs would’ve been more inaccurate. The equations that were derived for the three mathematical models of x vs. t, v vs. t, and a vs. t, were almost identical to the actual mathematical equations. Acceleration explains the rate at which velocity is changing. Velocity is the speed of something moving in a certain direction, therefore acceleration down a ramp stays constant. When observing the class’s models it was established that all of the graphs were relatively similar. Most of the graphs had points that were very random.
Independent vs. Dependent
Samantha Jordan
(Lindsay Cartmel)
Mr. Smail
AP physics
September 21, 2015
Circles
The purpose of this lab is to experimentally determine specifical graphical and mathematical relationships between independents and dependent variables. Circle example (diameter Circumference Area) For this lab a small white stick (about an inch long) was used to measure the circumference, diameter, and radius of different circular objects in order to create equations for circumference and area of circles. This lab posed different problems with measuring due to slipping and misreading of the stick against the circular objects
| Object | Diameter | Circumference | Area |
| Coin | .8sL | 2.8sL | 8sL² |
| Wooden Circle | 2.4sL | 7.3sL | 5sL² |
| Metal Weight | .5sL | 2sL | .47sL² |
| Can | 1.6sL | 4.6sL | 1.9sL² |
| Field Hockey Ball | 2sL | 6sl | 3.9sL² |
| Frisbee | 8sL | 24.8sL | 44sL² |
| Yellow Lid | 3sL | 10.3sL | 7.5sL² |
Linearized Data
Raw Data
| Derived Equation | Correct Equation | |
| Circumference Formula | C=(3.084sL/sL)D+0sL | C=πD |
| Area Formula | A=(0.6738)D ² | A=πr² |
Slope is the incline of a line. In the equation y=mx+b the slope is written as the variable m. The equation derived from the relationship between circumference and diameter using the white stick [C=(3.084)D+0sL] is very similar to the equation used to find the circumference of any circle (C=πD). Compared to other groups during this lab the equation formed from our experiment was average. There were some groups closer to the original equation than ours, but most were further away (instead of π they found theirs to be either 2.8,2.9, or 3.3). The equation varied due to the slope of the line, which was suppose to be pi.
The equation formed by the relationship between area and diameter using the white stick [A=(0.6738)D ²] is similar to the actual equation [A= πr² or A=(π/4)D²]. Compared to other groups the equation that was created using the data from the white stick was about the same. In this equation the variable that changed between equations was suppose to be pi/4.
Problems that were created during the lab came from the measuring circles wrong or from misjudgment of a measurement. This caused issues in the graphs that were created and the derived equations. To work through these problems we remeasured the circles multiple times then averaged the measurements.
| Area of Soccer Ball | A=(πr²)
A=π(4.3Eraser)² A=71.5Eraser |
| Circumference of Soccer Ball | C=πD
C=π(8.6Eraser) C=27.017Eraser |
Physics 