Ch2_KwarkR

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Chapter 2 -->Lesson 1 Reading a,b,c,d: 9/7/11 1. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully.I learned about vectors (just a plain number) and scalars (the number with a direction). Also, I read about 4 definitions we had gone over in class: Distance (how far something went), Displacement (how far away from its original position something is), speed (how fast something is going), and velocity (rate of change of position, with a direction). I also reviewed the formula

2. What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding. I sort of confused velocity and speed. We did not really go over that velocity had to do with displacement (or how far it moved from its original placement). I thought that it was simply speed with a direction; however, it turns out that no matter how much distance you cover, if you always return to the same spot there is a velocity of 0 because you have not moved anywhere in the long run. On the other hand, speed takes into account ALL the times you moved, the distance moved, and the amount of time it took to get there and back. For example, a turtle that moves north 1 foot in 1 second and south 1 foot in 1 second would have a speed of 1ft/sec, but it would have a velocity of 0. I also learned another formula that is very similar to the speed formula; the Velocity formula.

3. What (specifically) did you read that you still don’t understand? Please word these in the form of a question. I understood everything fine. 4. What (specifically) did you read that was not gone over during class today? Everything was gone over in class.

-->Lesson 1 Reading E (9/13/11)
1. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully. That acceleration was change in velocity over time and that it was a vector quality. I also read about the difference between constant velocity and a changing one; the latter is acceleration, the former is just velocity. The last thing was that positive acceleration means its speeding up, negative acceleration means its slowing down. 2. What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding. I didn't really have any misconceptions. 3.What (specifically) did you read that you still don’t understand? Please word these in the form of a question. I understood all of it. 4. What (specifically) did you read that was not gone over during class today? I learned the formula for average acceleration (change in velocity/time).

-->Class Notes: Constant Speed 9/8/11 __At rest__: V=0__Average Speed__: all the speeds in a certain time period averaged together. __Constant Speed__: unchanging speed. __Instantaneous Speed__: the speed at a certain time __Increasing speed__ __Decreasing speed__ There are only 4 types of motion: at rest, constant, increasing, and decreasing speed.

-->Lesson 2 Reading a,b,c: (9/8/11)
1. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully. I read about ticker tape and vector diagrams, both of which are fairly simple concepts. Ticker tape diagrams uses a device that leaves a trail of dots on a piece of special tape. It usually marks it at intervals of 10 dots per second. This allows us to determine whether the object was gaining speed, decreasing speed, at rest, or at constant speed from the distance between the dots (or lack thereof). Vector diagrams are merely arrows showing direction that also includes acceleration. To show that an object is increasing speed, you would just make consecutively longer arrows and write an //**a**// with an arrow pointing in the direction that it is accelerating; for decreasing speed, you would make consecutively shorter arrows and write an **//a with an arrow pointing in the opposite way. //** 2. What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding. There was nothing really confusing. 3. What (specifically) did you read that you still don’t understand? Please word these in the form of a question. I understand everything clearly. 4. What (specifically) did you read that was not gone over during class today? Everything was gone over in class.

-->Lesson 3 Reading a,b,c (9/14/11)
1. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully. On a position time graph, I read that the slope was indicative of velocity, and that a curved line indicated an acceleration. Look at notebook for illustrations and examples of these. In addition, I read about examples of where there would be changes in the velocity, which would lead to changes in the slope. I also read about how to calculate slope (rise/run). 2. What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding. This was relatively easy stuff, but it helped clarify a bit with the graphs; I would have trouble sometimes reading what the graphs meant ut the pictures of the curved lines really helped me to figure out what meant what. 3. What (specifically) did you read that you still don’t understand? Please word these in the form of a question. I understand everything. 4. What (specifically) did you read that was not gone over during class today? We went over everything.

-->Lesson 4 Reading a,b,c (9/14/11)
1. What (specifically) did you read that you already understood well from our class discussion? Describe at least 2 items fully. I read about the slope of a velocity/time graph. A constant velocity graph would be a horizontal line, while a graph of a changing velocity would have a slope of some sort. Also, lines in the positive part of the graph mean that the object is heading in the forward or right direction, while lines in the negative would mean that the object is heading in the backward or left direction; changes from positive to negative indicate changes in direction. Speeding up means that it goes from a lower to higher number, like 3 to 9 or -3 to -9 (disregard the ± sign for the magnitude of speed), while slowing down means the opposite. Basically, if it has a positive slope in the positive area or negative slope in the negative area, it is speeding up, and if it has a negative slope in the positive area or positive slope in the positive area, then it is slowing down. The reading also mentions acceleration, which only applies to lines with a slope. That means that the velocity is speeding up or slowing down every unit of time (so 10m/s/s would mean a slope of 10). This has an example of trends of graphs: [] They're also in my notes. 2. What (specifically) did you read that you were a little confused/unclear/shaky about from class, but the reading helped to clarify? Describe the misconception you were having as well as your new understanding. Some of the graph reading can be a little confusing at first, but once you think about it it makes sense. 3. What (specifically) did you read that you still don’t understand? Please word these in the form of a question. I understand it. 4. What (specifically) did you read that was not gone over during class today? Nothing.

-->Motion-detector graphs with cart 9/14/11
Camera at Bottom: Camera At Top:

Classwork (9/15/11)

-->Position Time graph wksht:
1) Qualitative description: AB: at rest BC: Constant towards (negative) CD: at rest DE: constant negative EF: at rest FG: constant positive GH: constant positive 2)Change in position AB: 0 BC:-10 CD: 0 DE: -16 EF: 0 FG: 16 GH: 14
 * THIS IS NOT ON THE RUBRIC*

3)Velocity AB: 0 BC: -10/2= -5m/s CD: 0 DE: -32 m/s EF: 0 FG: 16 GH: 7

4)Average speed for the entire graph: Total distance/ total time = avg. speed 56m/11 sec = 5.1 m/s 5)Average velocity for the entire graph: Displacement/time = avg. velocity. 4m/11s = .44 m/s 6)Acceleration: 0 for each segment b/c each segment is constant.

7)Draw a v-t graph for this x-t graph.

-->Interpreting Position-Time Graphs Wksht (Quantitative Graph Interpretations)
Homework: 9/19 D: Answers and Work :

E: Answers and Work:

-->Lesson 5 (a,b,c,d,e)
a)I ntroduction to Freefall

A free falling object is an object that is falling under the sole influence of gravity. two important motion characteristics 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

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 position of the object at regular time intervals - say, every 0.1 second - is shown. 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.

b) ** The Acceleration of Gravity ** 9.8 m/s is such an important value that it is given a special name. It is known as the ** acceleration of gravity ** - the acceleration for any object moving under the sole influence of gravity. The symbol for acceleration of gravity is “** g **”. There are slight variations in this numerical value (to the second decimal place) that are rely upon on altitude. Acceleration is the rate at which an object changes its velocity, or 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.  c) ** Representing Free Fall by Graphs ** One way of describing the motion of objects is through the use of position vs. time and velocity vs. time graphs. A position versus time graph for a free-falling object is shown below. Observe that the line on the graph curves. A curved line on a position versus time graph shows accelerated motion. Since a free-falling object is undergoing acceleration, its position-time graph should be curved. A further look at the position-time graph reveals that the object starts with a small velocity (slow) and finishes with a large velocity (fast). The small initial slope indicates a small initial velocity and the large final slope indicates a large final velocity. The negative slope of the line indicates a negative/downward velocity.
 * g = 9.8 m/s/s, downward ( ~ 10 m/s/s, downward) **

Velocity vs. Time graph

The graph is a straight, diagonal line. A diagonal line on a velocity versus time graph signifies acceleration. Since a free-falling object is accelerating its velocity-time graph should 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. This is known as negative acceleration. Since slope of a velocity-time graph means acceleration, the constant negative slope indicates a constant, negative acceleration. d) ** How Fast? and How Far? ** The velocity of a free-falling object that has been dropped from 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 **  V f  = g * t **  where ** g ** is the acceleration of gravity. The value for g on Earth is 9.8 m/s/s. This equation can be used to calculate the velocity of the object after any given amount of time when dropped from rest.  **﻿ Example Calculations: ** At t = 6 s vf = (9.8 m/s2) * (6 s) = 58.8 m/sAt t = 8 s vf = (9.8 m/s2) * (8 s) = 78.4 m/s  The distance 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 * t^2 **** Example Calculations: ** At t = 1 s d = (0.5) * (9.8 m/s2) * (1 s)2 = 4.9 mAt t = 2 s d = (0.5) * (9.8 m/s2) * (2 s)2 = 19.6 m   The diagram below shows the results of several distance calculations for a free-falling object dropped from a position of rest.

e) ** The Big Misconception ** 9.8 m/s is the same for all free-falling objects regardless of how long they have been falling, or whether they were initially dropped from rest or thrown up into the air. Yet many wonder “doesn’t a bigger object fall faster than a smaller one?” This question is a reasonable inquiry; nearly everyone has observed the difference in the rate of fall of a single piece of paper and a textbook. The textbook clearly falls faster. The answer is no! That is, no if we are considering Free-fall; objects do not encounter air resistance. More massive objects will only fall faster if there is an appreciable amount of air resistance present. The actual explanation involves the concepts of force and mass. The acceleration of an object is directly proportional to force and inversely proportional to mass. Increasing force increases acceleration while increasing mass decreases acceleration. Thus, 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.

-->Class Notes: Freefall
Examples:

-->Speed of a CMV (constant motion vehicle)
9/9/11 Lab Partner: Gabby Leibowitz

Objectives: 1) How precisely can you measure distances with a meterstick? 2) How fast does your CMV move? 3) What information can you get from a position-time graph?

Materials: spark timer + spark tape, meterstick, masking tape, CMV

Hypotheses: 1) Distances can be measured accurately to the nearest mm with our metersticks. 2) The CMV probably moves around 30 cm per second (I estimated around a foot). 3) We can figure out its distance at a certain time, its average speed/velocity, and whether an object accelerated, slowed down, or its speed remained constant.

Data Table: Position vs. Time 1. Why is the slope of the position-time graph equivalent to average velocity? Slope=change in y/change in x. Since y=distance and x=time in this graph, the equation would mean: Slope=change in position/change in time. Oh, look, the formula for average velocity is the exact same: Avg. velocity=change in position/change in time. Since they both are equal to the same thing, they can be set equal to each other (substitution property). slope=avg. velocity 2. Why is it average velocity and not instantaneous velocity? What assumptions are we making? It is average velocity because we put together an average of 10 points. Instantaneous velocity would be the velocity at just one point. We are assuming that the CMV is moving at a constant speed and not changing speeds erratically. 3. Why was it okay to set the y-intercept equal to zero? The time we started was at 0 sec, and at that time the CMV's distance was 0 (it hadn't moved anywhere). 4. What is the meaning of the R2 value? It shows how accurate the trend line was to our points on the graph. The higher the R2 value is, the more accurate the slope is in accordance with our given points. 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? It would go at a flatter slope (which means a lower velocity), meaning that it would be under our line. Because it would be slower than my CMV, at any given time the distance it traveled should be lower than the distance my CMV would have traveled at that exact time.

Conclusion: The average speed of my CMV was 31.084 cm/sec. I calculated this by finding out the slope of the trend line. Since the R2 value was .9996 (or 99.96%), I figured this was really accurate. My first hypothesis was true; the metric stick only measured to the millimeter, so we had to guess the measurement in between each mm tick mark. My second hypothesis was pretty accurate for an estimate; 30 cm/sec as compared to 31.084 cm/ sec. My third hypothesis was also true; from the graph, we can pick any point of time and figure out its position at that time. We found its average velocity by finding the slope of the trend line. In addition, since it was a flat line, we could tell it was moving at a constant speed; if something sped up, the position-time graph would have shown an upwards curve, and vice versa for something slowing down. Some sources of error may have been the equipment; the spark timer, at first, had to warm up, and during that period it would have given inaccurate data. To fix this problem, we could just measure the last 10 dots on the spark tape. In addition, the surface on which the CMV was moving could have caused inaccuracies in data. Any incline or decline of the surface would cause the CMV to go slower or faster, and any bumps or miniscule particles on the surface would certainly affect its speed. In order to prevent this, we can measure out the slope of the surface and make sure it is flat, and make sure to clean off the surface scrupulously. The third source of error could have come from the measurement of distance on the spark tape. The ruler or spark tape may have moved just a little bit while measuring, or was not perfectly aligned with the tape. To minimize these issues, we could use an adhesive like tape to hold the ruler and spark tape in place, and use a transparent ruler to better judge alignment. In addition, we measured each space dot by dot, then added them together as we went along. This was not a great idea, though it ended up fine, because every time we moved from dot to dot to measure the distance we could have made slight errors in the placement of the ruler (either by not putting it exactly on the 0cm mark or by doing the things mentioned before). A better way would have been to leave the meter stick on the tape and measure the whole thing without moving the ruler from dot to dot.

9/12/11 At rest: Constant Slow: Fast:

1. a)The line is flat, meaning that at any time the position is still the same. b)The line is flat at 0 m/s, meaning that there is no motion. c)The line is flat at 0, meaning that there was no acceleration at all.

2. a)The line is very steady, with no outliers. b)The points all remain around one single velocity, meaning that the velocity remains unchanged. On my graph, the points all stay near the 1 m/s mark, meaning that I was moving at 1 m/s steadily. So basically, constant value and no slope. c)The points should be steady, with no slope and a constant value; however, humans do not walk steadily, so on our graph there are many hills and valleys (which show the steps we took).

3. a)For the fast motion graph, the slope is steeper than the slow motion graph. b)The velocity should remain a constant, since we are not accelerating (but we're human so its not perfect). c)The velocity of the fast graph has a constant value, but its value is higher than the velocity of the slow graph.

4. a)Whatever the slope was before, it will be the opposite or the negative. b)Whatever the velocity constant was before, the values become the opposite (so from positive velocity at 1 m/s to negative velocity at -1 m/s). c)It should stay the same, since we didn't really speed up or slow down.

5. a)It shows how far or close the object has traveled, as well as whether you are going away or towards the sensor. You can also calculate total distance as well as displacement. You can also calculate average speed via the slope. b)It is also useful for noting constant and instantaneous speed, and also whether you are going away or coming towards the sensor (if positive, away, if negative, towards). You can also find acceleration (the slope) and displacement. c)You can easily tell whether the object is moving at a constant speed, accelerating, or slowing down.

6. a)You cannot tell the instantaneous speed. b)You cannot tell the position at a certain time without doing significant amounts of calculation (which, even then, might not be accurate). c)You cannot tell either the position or velocity.

7. a)No motion: When the object no movement, change of position, or acceleration. It is at rest. b)Constant speed: When the object is moving at one speed. It does not speed up or slow down, and moves steadily.

-->Lab: Acceleration on an Incline (9/17/11)
Lab Partner: Gabby Leibowitz

Objectives: 1. What does a position-time graph for increasing speeds look like? Hypothesis 1: They would look like a curved slope upwards (a parabola opening upwards) because for every increment of time, the object would be moving farther per unit of time than the point before it. 2. What information can be found from the graph? Hypothesis 2: We can find acceleration, instantaneous speed, and its position at a certain time.

Spark tape, spark timer, track, dynamics cart, ruler/meterstick/measuring tape
 * Available Materials:**

Procedure: 1.Get the materials 2)Prop the metal ramp/track on the textbook. 3)Put the spark tape through the spark timer, then taped the spark tape to the dynamics cart. 4)Next, place the spark timer at the top of the track, turn it on, and let the cart roll down the track. 5)Repeat step 3. 6)Place the spark timer at the bottom of the track, turn it on, and push the cart up the track. 7)Clean up 8)We measured the space between each dot on the tape and recorded it on Excel.

Data Table: Time vs. Position

Graph:

Analysis: a) Interpret the equation of the line (slope, y-intercept) and the R2 value. The slope of the line is a polynomial (y=Ax^2 +Bx, which can then be turned into y=1/2at^2 + bt and the y-intercept is at 0 because it started at 0 m/s at 0 sec. The R^2 value shows the accuracy of the points to the slope of best fit. For the linear lines, the r^2 values was not very good, but for the polynomial graphs, the r^2 values were very good (99.99%). 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). Time (sec) Uphill halfway instantaneous speed: 40 cm/s end instantaneous speed: 7.38 cm/s Downhill halfway instantaneous speed: 38.35 cm/s end instantaneous speed: 73.3 cm/s c) Find the average speed for the entire trip. The average speed for the uphill graph is 43.4 cm/s. The average speed for the downhill graph is 39.1 cm/s.

1. What would your graph look like if the incline had been steeper? On the uphill graph, the cart would have slowed down more quickly, so the curved slope would have been flatter and lower than the current one. On the downhill graph, the cart would have sped up faster, so the curved slope would have been steeper and higher than the current one. 2. What would your graph look like if the cart had been decreasing up the incline? If the cart had been decreasing speed up the incline, the graph would start off with a high velocity, but as gravity and friction worked against it it would slow down, gaining less distance per unit of time (meaning a flatter, curved slope). This would mean that on a position-time graph, it would look like an upside down parabola; the steep beginning would become flatter and flatter until it reached a certain position where its velocity would be 0 m/s, then it would change direction and go back down the incline, gaining speed. 3. Compare the instantaneous speed at the halfway point with the average speed of the entire trip. The instantaneous speed of both uphill and downhill graphs match very closely with the average speeds. For the uphill graph the average is 43.4 cm/s while the instantaneous speed is 40 cm/s. For the downhill graph the average is 38.35 cm/s compared to the instantaneous speed of 39.1 cm/s. 4. Explain why the instantaneous speed is the slope of the tangent line. In other words, why does this make sense? The slope of a graph relates to the speed of an object, but only the average speed (or the collective amount of distances over a certain time). But a curved graph like a parabola does not have a specific slope, as the slope changes at every point. So by picking a point and drawing a line tangent to that one point, we can determine the slope for that one point. 5. Draw a v-t graph of the motion of the cart. Conclusion: My hypothesis about the position-time graph for increasing speed was correct. The graph did curve upwards in a parabolic shape. My second hypothesis was also correct; we can find instantaneous speed by drawing a line tangent to the graph at the specific point, we can find position at a certain time simply by looking at the graph, and we can find average acceleration by doing (change in velocity)/(change in time). Some errors that we could have made would have been with the handling of the cart, and with the measurement of the dots. When I pushed the cart up the incline, the dots on the tape could have been all jumbled up because maybe I had moved it back a bit before I pushed. Also, at the top of the incline when the cart was almost stopped, we may not have stopped the cart in time before it stopped and started going back down the incline. However, these errors would only affect the ends of the tape, not the middle. Thus, we can avoid this problem by measuring only the middle of the ticker tape. Now, because we are human, we could have made mistakes with the measurements of the dots (such as not putting the tape exactly parallel to the measuring tool), leading to incorrect data and weird graphs. We can try to minimize this by double checking and making sure that the ruler and the tape are exactly lined up.

-->CMV Crash Lab (9/23/11)
Lab Partners: Jon Itskovitch, Lauren Kostman, Gabby Leibowitz

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

1. Find another group with a different CMV speed. Find the position where both CMV’s will meet if they start //at least// 600 cm apart, move towards each other, and start simultaneously. Procedure: media type="file" key="Movie on 2011-09-23 at 11.21.mov" width="300" height="300" Calculations: Theoretical Result: They will collide at 204.84 cm. Trial Results: Collision course towards each other:
 * Trial1 || 200cm ||
 * Trial 2 || 204cm ||
 * Trial 3 || 208cm ||
 * Trial 4 || 210cm ||
 * Trial 5 || 205cm ||
 * Average: || 205.4cm ||

2. Find the position where the faster CMV will catch up with the slower CMV if they start //at least// 1 m apart, move in the same direction, and start simultaneously. Procedure: media type="file" key="Movie on 2011-09-23 at 11.31.mov" width="300" height="300"

Calculations: Theoretical Result: The faster cmv will catch up to the slower cmv at 207.55 cm. Trial Results: Catch Up
 * Trial 1 || 198cm ||
 * Trial 2 || 210cm ||
 * Trial 3 || 205cm ||
 * Trial 4 || 211cm ||
 * Average: || 206cm ||

Discussion Questions: 1. Where would the cars meet if their speeds were exactly equal? For the collision part, they would meet exactly at 300 cm (or the midpoint) because if they had the same speeds, they would cover the same distance in the same amount of time. For the catch up part, one would never catch up to the other (theoretically) since they are moving at the same speed in the same direction. They would stay 100cm apart from each other for as long as the batteries had enough power. 2. Sketch position-time graphs to represent the catching up and crashing situations. Show the point where they are at the same place at the same time. Look at the calculations for 1. 3. Sketch velocity-time graphs to represent the catching up situation. Is there any way to find the points when they are at the same place at the same time? Collision: Catch Up: Analysis/Class Discussion Percent Error = (theoretical - experimental) / theoretical x 100 a)percent error of Collision: (204.84 - 205.4) / 204.84 x 100 =- 0.2734%

b)percent error of Catch Up: (207.55 - 206) / 207.55 x 100 =.747%

Percent Difference = (Average experimental value - Individual experimental value) / Average experimental value x 100.

Conclusion: The method we used was incredibly accurate in predicting the points of collision and the point where one CMV would catch up to the other. The points where the two CMVs collided were at an average of 205.4 cm (the slower CMV moved 205.4 cm, the faster car moved 600-205.4=394.6 cm). Compared with our theoretical result, this was only .56 cm off - a percent error of .27%. In the Catch Up experiment, the point where the faster CMV caught up with the slower CMV was at an average of 206 cm from the starting point (the slow CMV moved 106 cm, the faster CMV moved 206 cm). This also was very accurate; our theoretical distance was 207.55 cm. The percent error in this case was only .747%. Using the equation of the speeds (or the slope of the position/time graphs) made it easy to accurately find where the two CMVs collided and caught up. One source of error was probably that two people were turning on the CMVs; both of them could not possibly turn them on at the same time. This would mean that the CMV turned on faster would have had more time to gain more distance than the other one. We could fix this by using a machine to turn on both at the same time, but in all practicality we really can't do much more than our best in the class labs. Another source of error may have been from the curvature of the path of the CMV. Some of the CMVs curved to the left or right, and did not go directly straight; this would make it have to travel more distance in order to reach the point of collision or catch up. This could be fixed by using a track to guide the CMVs on a straight track. One other source of error may have been the CMVs themselves; some people used different CMVs than the ones they used for the initial speed test; this will definitely throw off the results. They could fix this by using the same CMVs, or by calculating the speeds of their new CMVs.

-->Egg Drop Project (9/28/11)
Robert Kwark and Noah Pardes

Final Structure WAS A SUCCESS!!! It was 26.9 grams. (The final design was the one in front) It had the mat of straws to catch air and slow it down. The little jenga tower-like structure was to hold the egg; we made sure to give it a snug fit. The top went over the egg, and we taped it down once the egg was in. As long as nothing hit the device, it would succeed.

Analysis: Finding acceleration d = 8.5 m t = 1.52 s V(initial) = 0 a=? d = vt + 1/2 a(t^2) 8.5 = 0t + 1/2 x (1.52)^2 x a 8.5 = 1.1552a a = 7.36 m/s^2 This result was lower than 9.8 m/s^2, which is maximum acceleration an object can reach in the earth's atmosphere, which makes sense. On the other hand, our result was a higher than the result of many other groups that succeeded; I think that our device had the fastest acceleration without resulting in a broken egg.

Conclusion: Fortunately for us, the 1st prototype worked incredibly well. We did attempt to make a better device for the 2nd prototype, but it didn't work nearly as well, probably because the part where the egg would be held was not directly in the center of the device. The 2nd prototyped kept hitting the wall, which caused the entire device to go head over heels and ultimately end up with egg yolk all over. So we went with the 1st design, hoping for it to work again; it did. The reason it worked was because the large surface area of the mat of straws slowed the device enough for the egg to land safely; also, we researched that the narrow end of the egg is the strongest and designed so that that end would point downwards. The acceleration of our device was 7.36 m/s^2, which was lower than the maximum acceleration 9/8 m/s^2. However, it was still accelerating faster than many other groups' devices, which is surprising; our device probably had the fastest acceleration without resulting in a broken egg. If I could do this again, I would probably do the same thing again. Our design was very good, and weighed very little. The only flaw was that if it hit something, or if the crosswind was very strong, the device would start tumbling around and not land on the designated spot (the padded bottom). Another design I might have made would be a heavily padded cube molded to the egg's shape made with straws with a parachute attached.

-->Free-Fall Lab: What is Acceleration due to Gravity?
Partner: Gabby﻿ Leibowitz

Hypothesis: The acceleration due to gravity is 9.8 m/s. The velocity time graph should be a diagonal line from the origin downwards (if it is being dropped). Using this graph, we can figure out a way to find the gravitational acceleration by finding its slope.

Procedure: Get spark timer and tape, get masking tape, and get an object and measure its mass. Tape the tape to the object. One person holds the timer and the other drops the object; start the timer and drop the object.

Data : Graphs: Position-Time Graph Velocity-Time Graph Analysis and Conclusion : The Position-time graph was a parabolic curve, as expected; since the object was accelerating, it covered more distance per unit of time, leading to a steeper and steeper curve. When we calculated the velocity at each interval (the middle time between each tenth of a second), we found that the rate of acceleration was constant; this is shown in the graph of the v-t graph by the straight line with a slope (m) of 881.46 cm/s^2. This result was below the theoretical yield of 9.8m/s^2 (or 980 cm/s^2). This gave us a percent error of 10.06% and a percent difference from the class of 5.69%. The reason that we had a 10.06% error was due one major factor: Friction. The entire class's results were affected by this; nobody's results were over 900cm/s^2. The friction between the ticker tape and the ticker timer slowed the acceleration down of every single experiment in the class. Thus, any range from 10%-25% would be acceptable due to this factor. The reason my hypothesis about the v-t graph being a diagonal line downwards does not match the v-t graph above is because all of the data/dots on the ticker tape were measured in terms of positive distance on the ticker tape. Thus, the graph of the v-t graph would go from the origin upwards. However, in reality, since the object dropped downwards, all of the values of displacement/distance should have been negative, which would make both the parabolic curve of the x-t graph and the linear v-t graph negative, showing my original hypothesis to be correct. The first part of my hypothesis was not correct in accordance with my data. However, since 9.8 m/s^2 is a scientific constant, the reason I did not get the correct answer of 980 cm/s^2 is because of various factors and errors (which are mentioned in the next paragraph). The other part of my hypothesis was correct. The slope of the v-t graph did gave me the the acceleration of 881.46 cm/s^2. We didn't set the y-intercept to 0 because at 0 time, we were still holding the tape with our hands, which were moving infinitesimally, so we can't really tell precisely the point where the first dot was on the ticker tape (the first dot would be the measurement of distance at 0 time, which should be 0 cm, but we couldn't tell what the first dot was because there were a lot of dots there). According to our trendline on the v-t graph, the y-intercept (b) should be at -13.08, which shows that there was some error; if it was perfect, there would have been no b value (or a b value of 0). One error margin would have come from there being kinks in the tape, which would have led to misprints or no-prints of the dots. Another margin of error would have been the force we dropped the object with; since we cannot hold our hands perfectly still, there was bound to be some initial velocity (either positive/upwards or negative/downwards) at the precise moment we dropped the object, affecting the position-time graph and the velocity-time graph to a certain extent. Another large margin of error could have emerged from the fact that our tape tore into three pieces upon the object's impact with the ground; this would lead to measuring errors, even if we taped it; it would not be the exact results that we would have gotten. The data would be off by a few millimeters for each tear. Analysis: Velocity Mid time = (final time + initial time) / 2 Example Problem: mid time = (0 s + .1 s) / 2 mid time = .5 Velocity = change in position/ change in time Example problem: Velocity = (31.42 cm - 0.0 cm) / (.1 s - 0.0 s) Velocity = 314.2 cm/s Analysis: Percent Error ((Theoretical-Experimental) / (Theoretical)) x 100. ((980 cm/s^2-881.46 cm/s^2) / (980 cm/s^2)) x 100 = 10.06% error Analysis: Percent Difference from the clas s : Class results (in cm/s^2): 5.69% Difference.
 * Avg exp. Value - Indiv. exp. Value| / Avg Exp. value x 100
 * 834.03 cm/s^2 - 881.46 cm/s^2| / 834.03 cm/s^2) x 100

__ Discussion Questions: __ 1. Does the shape of your v-t graph agree with the expected graph? Why or why not? Yes because a V-t graph of a parabolic x-t graph should always be a steadily increasing (or decreasing) line with a slope. 2. Does the shape of your x-t graph agree with the expected graph? Why or why not? Yes because the object is accelerating, meaning that it is covering more and more distance in the same amount of time. This is shown by the parabolic curve. 3. How do your results compare to that of the class? (Use Percent difference to discuss quantitatively.) Our results differ by a 5.69% difference. Our result was 881.46 cm/s^2, the second highest in the class, while the class's result was 834.03 cm/s^2. That means that our result was actually closer to the theoretical yield of 980 cm/s^2, but we differed a bit from other groups. 5.69%, however, is not a large deviation and is relatively small compared to the amount of error that could be made on this experiment. 4. Did the object accelerate uniformly? How do you know? In our experiment, it did not accelerate uniformly. We know this because the r^2 value was not 100%. Also, on the graph, we can see that a few points are NOT on the line of best fit, which means the acceleration at that point was not the same as the acceleration at other points. However, we can tell that the object was very close to accelerating uniformly, shown by the 99.93% r^2 value of our line of best fit. 5. What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be? The only factor I could think of that would cause G to be higher than it should be is that for that experiment, the gravitational pull of the earth was somewhat stronger than usual. But that is probably not the case. There are a range of factors why acceleration due to g would be lower than it should be. One is that there was friction between the ticker tape and the box as the object fell, reducing the acceleration and speed. Another would be air resistance, though this would have a very minute affect on the acceleration and speed because the object in question was very small and was quite massive/heavy for its size. The thickness of the air would determine the extent of the affect of air resistance on the object.