Ch2_Doomank

=Honors Physics Rocks Wiki =

=Class Notes 9/7/11=

Lab Partner: Jake G.

Purpose: How fast does a CMV go? What does a position-time graph tell us? How precisely are you required to measure? Hypothesis: The CMV (yellow) is about 1.4 m/s while the CMV (blue) is about 1.2 m/s. A position-time graph tells us how far something has traveled in a certain amount of time. You should measure as precisely as possible.

Data Table for Time and Position of a Yellow CMV 
 * Time (s) || Position (cm) ||
 * 0.0 || 0 ||
 * 0.1 || 1.85 ||
 * 0.2 || 4.77 ||
 * 0.3 || 7.65 ||
 * 0.4 || 10.56 ||
 * 0.5 || 13.57 ||
 * 0.6 || 16.57 ||
 * 0.7 || 19.5 ||
 * 0.8 || 22.51 ||
 * 0.9 || 25.09 ||
 * 1.0 || 28.12 ||


 * Conclusion: ||  ||   ||   ||   ||   ||   ||   ||   ||   ||   ||   ||   ||   ||   ||   ||
 * |||||||||||||||||||||||||| Our results showed our CMV moving at 28.12 cm per second. Our hypothesis was fairly accurate, as we guessed it would move at 30.48 cm (1 foot) per second. ||  ||   ||
 * |||||||||||||||||||||||||||||| Errors may have come from false measurements. These may have come from the limits of the accuracy of our ruler, as well as how well we placed it at the start and determined with the naked eye ||
 * |||||||||||||||| where it ended. Also, any problem with the spark timer or lack of accuracy will give us inaccurate results. ||  ||   ||   ||   ||   ||   ||   ||
 * |||||||||||||||||||||||||||||| Next time, we could run the experiment multiple times with different spark timers to ensure a more accurate reading. Also, we could use a more accurate measuring stick to get a more accurate ||
 * || result. ||  ||   ||   ||   ||   ||   ||   ||   ||   ||   ||   ||   ||   ||   ||
 * || result. ||  ||   ||   ||   ||   ||   ||   ||   ||   ||   ||   ||   ||   ||   ||

1. Why is the slope of the position-time graph equivalent to average velocity? ||  ||   ||   ||   ||   ||   ||   ||   ||   ||   ||   ||   ||   ||
 * Questions:
 * a. The x-axis displays time, while the y-axis displays distance. Rate (the velocity) is distance over time, which exactly what the graphed line shows, therefore, the slope, which is x/y is now distance/time, the equation to find slope. ||  ||   ||
 * 2. Why is it average velocity and not instantaneous velocity? What assumptions are we making? ||  ||   ||   ||   ||   ||   ||   ||   ||   ||   ||   ||   ||
 * a. The average takes into account all the points, and with more data, the margin for error decreases. ||  ||   ||   ||   ||   ||   ||   ||   ||   ||   ||   ||   ||
 * 3. Why was it okay to set the y-intercept equal to zero? ||  ||   ||   ||   ||   ||   ||   ||   ||   ||   ||   ||   ||   ||   ||   ||
 * a. Because in this case, at zero seconds, the CMV has moved zero centimeters, which is the ideal starting point when trying to find out average velocity. ||  ||   ||   ||   ||   ||   ||   ||   ||
 * 4. What is the meaning of the R2 value? ||  ||   ||   ||   ||   ||   ||   ||   ||   ||   ||   ||   ||   ||   ||   ||   ||
 * a. It is a measurement of how accurate the line of best fit is. The closer to 1.0 the value is, the more accurate the line of best fit is. ||  ||   ||   ||   ||   ||   ||   ||   ||   ||   ||
 * 5. If you were to add the graph of another CMV that moved more slowly on the same axes as your current graph, how would you expect it to lie relative to yours? ||  ||   ||   ||   ||   ||   ||   ||
 * a. It would be more of a gradual slope, therefore being lower on the graph than our original line. Since slope and average velocity are the same for this graph, it would make sense that a slower CMV would have a graph that is less steep than our original line. ||

Homework 9/8/11
1. I already understood the idea of scalars and vectors. From our classroom discussion, I already knew that scalars are only described by a numeric value. Vectors, on the other hand, are described by both a numeric value //and// a direction. 2. I was not quite clear on how exactly to describe displacement, but the way the reading explained it really helped clear things up. I kept getting distance and displacement confused, but the example problem about the teacher walking 12 meters (the //distance//) but then explaining that her //displacement// is 0 meters, as she is not "out of place," meaning that there was not any displacement as a result of her motion. Giving an example to showed clearly what distance and displacement were, and demonstrating the clear difference between the two helped me understand what exactly displacement is. 3. There is nothing that I am still unclear about. 4. We did not discuss the definitions of mechanics and kinematics during class, and those two terms were explained and defined in the reading. Mechanics is the study of motion of objects, whereas kinematics is "the science of describing the motion of objects"

Class Notes 9/9/11
-Average Speed: take total distance and total time and that's your average speed -Constant Speed: going the same speed all the time -Instantaneous Speed: the speed you're going now at this very moment -At Rest v=0 a=0 *A= acceleration V=velocity -Constant --> --> --> --> --> (a=0) (arrows all the same length, represent velocity) -Increasing (speeding up) ---> -> ---> (a>) (acceleration points in same direction as velocity) -Decreasing (slowing down) -> -> ---> (a<-) (acceleration points opposite direction as velocity)
 * //__Types of Motion__ __Motion Diagram__ //**
 * motion diagrams can be rotated to fit situation (ex. if something is falling down, arrows can point down)
 * if arrows are pointing down, the velocity is negative
 * if arrows are pointing up, the velocity is negative
 * arrows going right, velocity is positive
 * arrows going left, velocity is negative
 * HOWEVER, even if acceleration and velocity are negative, you can still be increasing speed

-Signs are arbitrary: someone made it up and then we agreed with it; it's not right or wrong, it just is -mostly useful for labs
 * //__Ticker Tape Diagram__//**
 * //decreasing speed// means dots get closer together
 * //increasing speed// means dots get further apart
 * //constant speed// means are evenly space

Describe motion with words: [|Physics Classroom]

Lab 9/12/11
Partner: Jake G.


 * Objectives:**

Hypothesis: Static equilibrium is when something is stationary and has a steady velocity (of 0) and dynamic equilibrium is when something is moving and has a steady velocity. "At rest" is a straight line on each. On a position-time graph, the straight line will be at the numerical value representing the distance between the object/person and the detector. On a velocity-time graph, as well as a position-time graph, the straight line will be at the origin. Position-Time, Velocity-Time, and Acceleration-Time Graphs of a Person Walking This graph reflects the 5 times someone walked in front of the sensor. Run 1 represents a person at rest. Run 3 represents someone walking fast. Run 4 represents someone walking slowly. Run 6 represents someone walking away from the sensor. Run 7 represents someone walking towards the sensor. __//**Discussion Questions**//__ > Conclusion: Our hypothesis was fairly accurate. In our graphs, though, some errors came from the person walking losing balance or tripping accidentally. These caused spikes in our graphs that should not have been present. Next time, we could make sure the area is clear without any objects to trip over, and we could really focus on walking at a steady speed, instead of just walking.
 * What is the difference between static and dynamic equilibrium?
 * How is “at rest” represented on a position vs. time graph? On a velocity vs. time graph? On an acceleration vs. time graph?
 * How is constant speed represented on a position vs. time graph? On a velocity vs. time graph? On an acceleration vs. time graph?
 * How are changes in direction represented on a position vs. time graph? On a velocity vs. time graph? On an acceleration vs. time graph?
 * 1) How can you tell that there is no motion on a…
 * 2) position vs. time graph: the line is straight, with no slope
 * 3) velocity vs. time graph: there is no slope to the line
 * 4) acceleration vs. time graph: there is no slope to the line
 * 5) How can you tell that your motion is steady on a…
 * 6) position vs. time graph: the slope is steady with little to no major bumps
 * 7) velocity vs. time graph: the line is steady, not increasing or decreasing in slope
 * 8) acceleration vs. time graph: the line is steady, not increasing or decreasing in slope
 * 9) How can you tell that your motion is fast vs. slow on a…
 * 10) position vs. time graph: the graph for a fast graph is shorter than the line for the slow graph
 * 11) velocity vs. time graph: the line is straight, but for a fast graph, the line is shorter than the line for a slow graph
 * 12) acceleration vs. time graph: the line for the slow graph is longer than the graph for the fast graph
 * 13) How can you tell that you changed direction on a…
 * 14) position vs. time graph: if you were going towards it, the graph would slant down, if you were going away then the graph would slant up
 * 15) velocity vs. time graph:
 * 16) acceleration vs. time graph: acceleration will be negative
 * 17) What are the advantages of representing motion using a…
 * 18) position vs. time graph
 * 19) velocity vs. time graph
 * 20) acceleration vs. time graph
 * 21) What are the disadvantages of representing motion using a…
 * 22) position vs. time graph: if you hesitate or misstep, the graph will be thrown off with a sudden spike
 * 23) velocity vs. time graph: there is room for human error; the subject could trip of lose balance which will adversely effect the graph showing spikes or dips, or not giving an accurate representation of time
 * 24) acceleration vs. time graph: someone could stumble and increase speed, or they could hesitate and decrease speed accidentally, which could lead to a wrong representation
 * 25) Define the following:
 * 26) No motion: an object is at rest, not moving forward or backwards; staying in place
 * 27) Constant speed: an object is traveling at the same speed, not increasing (accelerating) or decreasing (decelerating); a steady speed.

Position-Time Graph -Slope tells you velocity -negative slope means you're walking towards something-the steeper slope is faster

Velocity-Time Graph -tells you how fast you're going -At rest- line at 0 -Constant speed: if you're moving at 2 m/s the whole time, it should be a straight (horizontal) line at 2 -Positive values are moving away, negative values are towards (negatives only apply to direction, not actually numerical value)
 * slope=acceleration

Lab 9/13/11
Partner: Jake G.

//Hypothesis//: For a position-time graph of increasing speed, the line will be a curve; as an object accelerates, the position will get further and further from the last point on the graph. From the graph, you can find the slope (velocity), y-intercept.
 * Objectives:**
 * What does a position-time graph for increasing speeds look like?
 * What information can be found from the graph?

a) Interpret the equation of the line (slope, y-intercept) and the R2 value. b) Find the instantaneous speed at halfway point and at the end. (You may find this easier to do on a printed copy of the graph. Just remember to take a snapshot of it and upload to wiki when you are done.) c) Find the average speed for the entire trip.
 * Analysis:**
 * See Graph for this information.
 * Halfway Point: 12.2 cm/s
 * Endpoint: 28.1 cm/s
 * distance/time= 13.96/1= 13.96cm/s

//Procedure//: Step 1: set up track on top of one textbook in order to create a slanted slope for the cart to travel down. Step 2: attach ticker-tape to cart Step 3: Thread ticker-tape through the ticker-tape machine Step 4: turn machine on and allow cart to proceed down the track, catching at the end so that I does not fall off the table Step 5: label points on ticker-tape by .10 seconds up until 1 second Step 6: measure the distance from the first dot to the second, then the first to the third, etc. and include those in the data table Step 7: Using the data collected in the data table, make a position-time graph of the results. **Discussion Questions:** What would your graph look like if the incline had been steeper? What would your graph look like if the cart had been decreasing up the incline? Compare the instantaneous speed at the halfway point with the average speed of the entire trip. Explain why the instantaneous speed is the slope of the tangent line. In other words, why does this make sense Draw a v-t graph of the motion of the cart. Be as quantitative as possible.
 * The Curve would probably have been steeper
 * <span style="font-family: Arial,Helvetica,sans-serif;">It would start moving up, but curve downward
 * The instantaneous speed at the halfway point was actually a bit less than the average speed for the entire trip. While the instantaneous speed at the halfway point was 12.2 cm/s, the average speed was 13.96 cm/s. However, this makes sense, as the cart increased in speed throughout the experiment, which means that it did get faster after the halfway point, and therefore, the average speed should be expected to be higher than the average speed for a vehicle that is increasing in speed.
 * This makes sense because slope=velocity and if the line is straight then the velocity is the same throughout the entire line.

Homework 9/14/11
1.I already knew that acceleration is the rate at which something changes velocity. 2. This reading clarified the fact that acceleration actually has nothing to do with going fast; it has to do with changing how fast an object is moving. 3. There isn't anything that I am still unsure about after doing this reading. 4. It was helpful to see the animation that depicted the three cars each moving at different speeds, and being able to look at each to discern whether they were accelerating or moving at a constant speed. The visual helped to solidify my understanding of the meaning of acceleration.

Homework 9/15/11
Lesson 1. I already knew the concept of ticker tape diagrams since we used them in our lab earlier this week. 2. Though I already understood vector diagrams pretty well, it was good to have a review and see it explained in words. 3. There is not really much that I am confused about still. 4. I already knew what a vector-diagram was, but I think it was useful to see multiple examples of them, each applying to a different circumstance.

Lesson 3. 1. I already knew what a position-time graph was and how to make one. 2. It was helpful to see a constant-motion position-time graph right next to an acceleration position-time graph so that I could compare the different shapes and be able to fully visualize the difference between the two. It was also helpful to see the "negative velocity slow to fast" which curved down, and the "leftward velocity, fast to slow" which sloped downwards in a curve. 3. I understood everything well; there is nothing that I am still confused about. 4. I already knew what a position-time graph was, but it was helpful to see examples of different graphs from different circumstances, like a graph of leftward velocity, fast to slow.

<span style="color: #a600ff; font-family: 'Comic Sans MS',cursive;">Lab: A Crash Course in Velocity (Part II) CMV
<span style="color: #a600ff; font-family: 'Comic Sans MS',cursive;">Lab Partners: Jake, Lerna,

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 110%;">**Two CMV's Used**: Yellow Slow and Yellow Fast


 * Objectives ** : Both algebraically and graphically, solve the following 2 problems. Then set up each situation and run trials to confirm your calculations.

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 110%;">**Calculations from Lab** <span style="font-family: Arial,Helvetica,sans-serif; font-size: 110%;">

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 110%;">**Data Table:** Catch-Up || Percent Error Intersection || Percent Difference Catch-Up || Percent Difference Intersection || % error: __ l theoretical - experimental l __ x 100 theoretical
 * || Intersection || Catch-Up || Percent Error
 * Experimental || 190.5 cm || 186.8 cm ||  ||   ||   ||   ||
 * Test 1 || 210 cm || 150 || 19.70 || 10.24 || 2.44 || -2.69 ||
 * Test 2 || 200 cm || 152 || 18.63 || 5.00 || 1.14 || 2.20 ||
 * Test 3 || 205 cm || 160 || 14.35 || 7.61 || 4.07 || 0.24 ||
 * Test 4 || 203 cm || 153 || 18.09 || 6.56 || 0.50 || 0.73 ||
 * Avg. Experimental || 204.5 || 153.75 ||  ||   ||   ||   ||

Example: __ l 190.5 - 200 l __ x 100= 18.63 190.5

% difference __ l avg experimental - individual experimental l __ x 100 avg. experimental

Example: __ l 204.5 - 203 l __ x 100=0.50 204.5
 * Discussion**: The percent error for the catch-up was much greater than that of the intersection. However, my lab group and I knew that this ouwld happen. Looking at our results, we noticed that the experimental proved to be much less than the experimental that we had calculated. After checking the math, though, we noticed that we had not made any errors in calculations. The percent difference for both was relatively the same, except for the answer of 4.07 in the percent difference for catch-up. In contrast to the catch-up, the intersection calculations, barring the results for the experimental value 210, were fairly accurate, compared to those for catch-up.

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 110%;">**Discussion questions**
 * 1) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 110%;">Where would the cars meet if their speeds were exactly equal?
 * 2) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 110%;">If they were facing one another, they would meet in the middle, but if one started in front of the other, they would never meet up.
 * <span style="font-family: Arial,Helvetica,sans-serif; font-size: 110%;"> 2. Position-Time Graph of Meeting Point vs. Catch-Up Point of CMV's **

For the catch-up case, a v-t graph cannot show when the CMVs are in the same place at the same time because their initial, or starting, positions are separated by one meter.
 * 3.Velocity-Time Graph of Catch-Up vs. Meeting Point of CMV's**
 * Videos of Experiments- Meeting-Point and Catch-Up Scenarios**

media type="file" key="Lerna Movie on 2011-09-20 at 13.47.mov" width="300" height="300" media type="file" key="Movie on 2011-09-20 at 13.37.mov" width="300" height="300"


 * Conclusion**: We had a relatively easy time figuring out the equations based on the drawings provided for each scenario. However, in our experiments, the CMV's (yellow fast and yellow slow) did not meet up exactly where we thought they would. Our calculations showed that they should catch up at about 186.8 cm; however, the closest to this number we got in our experiments was a catch-up point of 160 cm, with the average catch-up point being 153.75 cm. We believe this could be due to people letting go of the CMV's at different times, which would provide an inaccurate result. It could also have to do with the curved path that one CMV traveled. We calculated the catch-up point as if the two CMV's were each traveling in straight lines, but one of ours did not, which was not accounted for in the theoretical calculations. For the intersection, we found the theoretical to be 190.5 cm, however, one result was 210 cm. This could be due to either the fact that one CMV traveled a curved path or it could be due to human error, misreading the meterstick or not not looking soon enough to catch the exact moment of catch-up. For part a, the meeting-point scenario, they met up earlier than expected, however, they consistently met up at the same relative area, which would suggest that our calculations were incorrect; however, upon re-doing our calculations, we found that we had not made any mistakes. We were a bit off in part b, the catching up part, as well, however this was much less drastic of a difference. We believe that the differences may be caused by the fast CMV, which tended to go off in a curve, rather than a straight line, which could have taken more time, which allowed the slower CMV to get further, thus making the meeting point slightly off. I would try to use a rail to make both CMV's go in a straight line to prevent error that came from the divergence of one of the CMV's. This way, there would be a clear, measurable intersection and catch up point, instead of observing the "ballpark" intersection and catch up points.

Our acceleration was 12 m/s^2 which is much to high to be possible for the highest possible acceleration is 9.8 m/s^2. The reason why our acceleration is so off is because the timing of the drop and impact of the egg wasn't properly done. If we were to do it again we would count down from three to zero and then let go. We would also have more than two people time the experiment giving us a higher range to average from. If we had more runs it would have been possible to get an even more accurate average time, which would result in a more correct acceleration. Next time I would try and make a cone shaped one that all the impact was taken by the tip of the cone. The egg would sit about five inches from the tip so there would be no impact on the egg. Also possibly packing the egg with paper shreds or string

Pictured above is the drawing of the final prototype. Our design started with full-length straws held together with tiny rubber bands, and we made the straws into a cylinder. Then, inside that, we placed two layers of shorter straws, also held together with tiny rubber bands. Inside the two layers of rubber bands, we placed the egg. The egg was wrapped in newspaper, and we placed straws inside the newspaper as we wrapped the egg, in an attempt to create a barrier. Then we used larger rubber bands to secure the egg (which was now inside the newspaper and two smaller cylinders of straws) at the top of the large cylinder. This held it suspended, which when tested, prevented the egg from taking the shock of the impact. Into the bottom of the large cylinder, we threaded additional small straws in order to increase the surface area of the bottom, hoping that it would make for a more stable landing, also absorbing more of the shock, instead of having the shock go to the egg.

Class Notes 10/3/11
v=0, acceleration decreasing Example Problems
 * Freefall: an object moving under the influence of gravity //only//
 * when an object is going up during freefall, the velocity is up and the acceleration is down, when it is going down during freefall, velocity and acceleration are down
 * **ignore air resistance**
 * cannot use v= d/t <--- that is constant speed and we freefall is NOT constant speed

Interpreting Position-Time Graphs Classwork
A)

D) E) F) G)

Homework 10/3/11
__1-D Kinematics Lesson 5__
 * Section A**[[image:http://www.physicsclassroom.com/Class/1DKin/U1L5a1.gif width="94" height="272" align="right"]]
 * A free falling object is an object that is falling under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall . There are two important motion characteristics that are true of free-falling objects:
 * Free-falling objects do not encounter air resistance.
 * All free-falling objects (on Earth) accelerate downwards at a rate of 9.8 m/s/s (often approximated as 10 m/s/s for //back-of-the-// calculations)
 * Because free-falling objects are accelerating downwards at a rate of 9.8 m/s/s, a [|ticker tape trace] or dot diagram of its motion would depict an acceleration. The dot diagram at the right depicts the acceleration of a free-falling object. The position of the object at regular time intervals - say, every 0.1 second - is shown. The fact that the distance that the object travels every interval of time is increasing is a sure sign that the ball is speeding up as it falls downward.
 * Section B**
 * The acceleration of gravity - the acceleration for any object moving under the sole influence of gravity; physicists have a special symbol to denote it - the symbol g . The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s/s.
 * Even on the surface of the Earth, there are local variations in the value of the acceleration of gravity (**g**). These variations are due to latitude, altitude and the local geological structure of the region.
 * rate at which an object changes its velocity. It is the ratio of velocity change to time between any two points in an object's path. To accelerate at 9.8 m/s/s means to change the velocity by 9.8 m/s each second.
 * If the velocity and time for a free-falling object being dropped from a position of rest were tabulated, then one would note the following pattern.
 * ~ Time (s) ||~ Velocity (m/s) ||
 * 0 || 0 ||
 * 1 || - 9.8 ||
 * 2 || - 19.6 ||
 * 3 || - 29.4 ||
 * 4 || - 39.2 ||
 * 5 || - 49.0 ||

vf = g * t At t = 6 s > vf = (9.8 m/s2) * (6 s) = 58.8 m/s > At t = 8 s > vf = (9.8 m/s2) * (8 s) = 78.4 m/s > The distance that a free-falling object has fallen from a position of rest is also dependent upon the time of fall. This distance can be computed by use of a formula; the distance fallen after a time of t seconds is given by the formula. d = 0.5 * g * t2 where g is the acceleration of gravity (9.8 m/s/s on Earth). Example calculations for the distance fallen by a free-falling object after one and two seconds are shown below. At t = 1 s > d = (0.5) * (9.8 m/s2) * (1 s)2 = 4.9 m > At t = 2 s > d = (0.5) * (9.8 m/s2) * (2 s)2 = 19.6 m > At t = 5 s > d = (0.5) * (9.8 m/s2) * (5 s)2 = 123 m > (rounded from 122.5 m)
 * Section C**
 * position versus time graph for a free-falling object is shown below.
 * A curved line on a position versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9.8 m/s/s), it would be expected that its position-time graph would be curved.
 * The object starts with a small velocity (slow) and finishes with a large velocity (fast). Since the slope=velocity of the object, small initial slope indicates a small initial velocity and the large final slope indicates a large final velocity. Negative slope of the line indicates a negative (i.e., downward) velocity.
 * A velocity versus time graph for a free-falling object is shown below.
 * The line on the graph is a straight, diagonal line; diagonal line on a velocity versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9,8 m/s/s, downward), it would be expected that its velocity-time graph would be diagonal.The object starts with a zero velocity and finishes with a large, negative velocity; the object is moving in the negative direction and speeding up(negative acceleration)
 * slope=acceleration of the object, constant, negative slope indicates a constant, negative acceleration
 * This analysis of the slope on the graph is consistent with the motion of a free-falling object - an object moving with a constant acceleration of 9.8 m/s/s in the downward direction.
 * Section D**
 * If dropped from a position of rest, the object will be traveling 9.8 m/s (approximately 10 m/s) at the end of the first second, 19.6 m/s (approximately 20 m/s) at the end of the second second, 29.4 m/s (approximately 30 m/s) at the end of the third second, etc. The velocity of a free-falling object that has been dropped from a position of rest is dependent upon the time that it has fallen. The formula for determining the velocity of a falling object after a time of t seconds is
 * ( g is the acceleration of gravity)
 * The value for g on Earth is 9.8 m/s/s.
 * The above equation can be used to calculate the velocity of the object after any given amount of time when dropped from rest. Example calculations for the velocity of a free-falling object after six and eight seconds are shown below
 * Example Calculations: **
 * Example Calculations: **
 * Section E**
 * The questions are often asked "doesn't a more massive object accelerate at a greater rate than a less massive object?" "Wouldn't an elephant free-fall faster than a mouse?"
 * This question is probably based in part upon personal observations made of falling objects in the physical world. (nearly everyone has observed the difference in the rate of fall of a single piece of paper (or similar object) and a textbook. The two objects clearly travel to the ground at different rates - with the more massive book falling faster.)
 * The answer to the question (doesn't a more massive object accelerate at a greater rate than a less massive object?) is absolutely not!
 * More massive objects will only fall faster if there is an appreciable amount of air resistance present.
 * Increasing force tends to increase acceleration while increasing mass tends to decrease acceleration-->the greater force on more massive objects is offset by the inverse influence of greater mass-->All objects free fall at the same rate of acceleration, regardless of their mass.

=<span style="color: #a600ff; font-family: 'Comic Sans MS',cursive; font-size: 130%;">Falling Object Lab 10/4/11 =
 * Objective:**What is acceleration due to gravity? Predict what v-t graph will look like for this. How are you going to get acceleration due to gravity off of the v-t graph?


 * Hypothesis**: Acceleration due to gravity is 9.8 m/s/s.The v-t graph would be below the axis, and have a negative slope, considering that the object is being dropped. The slope of the v-t graph is equal to acceleration so once we find the slope, we will know the acceleration.

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">**Procedure**: Drop 100 gram mass. Use a spark timer, plugged in, and hold it over the edge of where you are dropping the mass. Take a piece of masking tape, attach ticker tape to mass, letting it drop to the ground. Try not to wrinkle the tape or coil it up. Make sure it's all laying out and no one is standing on the end. Then lay the ticker tape out and get a measuring tape to get the data with which we will make the v-t graph and do the lab.

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">Example Calculations: <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">Graphs: <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">Velocity Time Graph for freefall lab: <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">The slope of this v-t graph is the acceleration, which is going upwards in a positive linear fashion, meaning that it is accelerating constantly. The m value, or slope, represents acceleration of the object. The R2 value should ideally be 1, but our value is close enough to one that it is accurate enough, meaning that our data matched the line of best fit very well. The first dot, the starting point is not exactly at zero because the first dot on the ticker-tape was not made exactly when we dropped the mass.

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">Position Time Graph for Freefall Lab:

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">This position-time graph shows the acceleration from zero of the mass that we dropped. The m value, the slope, represents the velocity, which increased, or accelerated, as the object dropped further. The curve of the line matches that of an expected X-T graph where the object is accelerating away from the origin. Again, for the R<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%; vertical-align: super;">2 value, ours was not exactly 1, which is ideal, but it was close enough to 1 for it to be pretty accurate, meaning that the line of best fit worked very well with our data.

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">Analysis: <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">Discussion Questions:
 * 1) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">The shape of the V-T graph is similar to what was expected, which means that my hypothesis is fairly accurate. My prediction and the real graph have a linear line (the linear line is acceleration; the increase in speed). We were wrong in our hypothesis, though; we thought the slope would be negative (a downward line), but in reality, the lab proved that the slope was actually positive (an upward line).
 * 2) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">The prediction was on-track with what our position-time graph showed. In terms of the prediction, I thought that the graph would have a curved, positive slope that traveled away from the origin. This was what the actual graph displayed, meaning that my prediction was correct.
 * 3) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">The percent difference between the results for the average experimental, or results for the class, and our individual experimental results is 4.53%. The value that we got, 4.53% was higher than the class average.
 * 4) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">Our mass, which weighed 100 grams, showed a constant acceleration. We were able to get this observation by looking at the velocity-time graph, which had a linear line. Since the slope of a V-T graph is the acceleration, and ours accelerates over equal increments, we were able to find that the acceleration is therefore constant.
 * 5) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">The acceleration due to gravity could be greater than the normal value, 9.8 m/s<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%; vertical-align: super;">2, if the data was gathered incorrectly. It could also be greater than normal if someone misread the data, such as beginning and end points. The acceleration due to gravity could turn out to be lower than normal if the tape attached to the (100g) object was being dragged or if there was too much friction; this tape also must go through the spark timer, which creates additional friction, which in turn slows down the speed at which the object falls. If the spark timer was held differently, either vertically or horizontally, or if the tape encountered any “road blocks” like the clock, a foot, hand, or notebook, this could throw off the time significantly.

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">Conclusion: <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">The first part of our hypothesis was not too bad; the results came out as 842.38cm/s<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%; vertical-align: super;">2 or 8.42cm/s<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%; vertical-align: super;">2. The typical accepted value of acceleration due to gravity is 980cm/s<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%; vertical-align: super;">2 or 9.80m/s<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%; vertical-align: super;">2. In general, our results were fairly close to the known value. If we compared our results to the known value, it would result in a percent error of 14.08%. This is not absolutely terrible, but it is not perfect, either. <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;"> It turns out that the second portion of our hypothesis was also right; our actual graph displayed a linear line speeding up, accelerating, away from the origin; we predicted the same result, since we knew that the velocity was considered positive. <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">The third part of our hypothesis was correct as well. We determined g (acceleration due to gravity) based on the slope of the VT graph because we knew that the slope of a V-T graph is equivalent to acceleration.This is also demonstrated in our acceleration of gravity result: 842.38cm/s<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%; vertical-align: super;">2 or 8.42cm/s<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%; vertical-align: super;">2. <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">In terms of errors, one possible source could have happened when the tape passed through the spark timer, creating friction, which in turn slows down the tape passing through, which would then effect the acceleration of the object that was dropped. To fix this, we could apply a lubricant of sorts that would decrease friction, or we could feed the tape directly through the timer, instead of letting it drag along the floor then through the timer. Another possible source of error in our experiment could come from measuring the dots on the ticker-tape. With the dots being so small, it would be easy to accidentally skip one, or the person measuring could misread the numbers. To fix this, we could go back and re-measure the dots to ensure that the values we wrote down were correct.

Class Example Problems 10/7/11
<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;"> <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%;">