Ch2_GreensteinJ

toc =Honors Physics Chapter 2=

=Class Notes 9/7=



=Speed of CMV Lab 9/07=


 * Purpose**: Understanding speed on a position-time graph


 * Hypothesis**: The CMV will travel 30.48 cm (1 foot) per second.


 * Procedure**: Measure the distance over time of a CMV. Attach ticker tape to a CMV and feed it through a spark timer. Turn the CMV and spark timer on.


 * 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. This lab helped us understand motion and how it translates into a graph. It shows the speed at any given point, and the distances traveled over a period of time. Errors may have come from false measurements. These may have arisen 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. Most of the error would have come from determining the distances on the ticker tape (this proved difficult as the tape moved a bit). Next time, fully securing the tape to the desk and then taking measurements will yield much more accurate results.

=Homework 9/8=
 * 1) I fully understood the difference between scalar and vector. Scalar is a form of measurement, such as speed or distance, that requires only one piece of information. Vector, by comparison, also involves a second measurement, such as direction.
 * 2) I still don't fully understand when we know to use a scalar or vector form of measurement, and what their different applications are.
 * 3) When do we use scalar or vector measurements?
 * 4) We covered everything in class

=Class Notes 9/9=
 * Average speed
 * total distance divided by total time
 * Constant speed
 * going the same speed over a given amount of time
 * Instantaneous speed
 * speed in a given moment
 * Types of motion
 * can be represented by motion diagram
 * arrows to show change in motion
 * velocity (v) over arrows, with v represented by arrows of direction or = 0
 * acceleration (a) under arrows, with an arrow (or = 0) showing direction of change
 * at rest
 * constant
 * increasing
 * decrease
 * decrease

=Class Notes 9/12=

= = =Homework 9/12=
 * 1) I understood the difference between ticker tape diagrams and vector diagrams
 * 2) I still am confused with how to draw a vector diagram.
 * 3) How do I draw a vector diagram accurately?
 * 4) Everything was covered in class

=Activity 9/12= Position stays constant over time Velocity stays at 0 over time Acceleration stays at 0 over time
 * 1) How can you tell that there is no motion on a…
 * 2) position vs. time graph
 * 1) velocity vs. time graph
 * 1) acceleration vs. time graph

Position will increase in a steady line over time Velocity will stay constant over time No acceleration over time
 * 1) How can you tell that your motion is steady on a…
 * 2) position vs. time graph
 * 1) velocity vs. time graph
 * 1) acceleration vs. time graph

Based on increasing intervals of position over time Based on intervals of velocity over time Based on how quickly it accelerates over time
 * 1) How can you tell that your motion is fast vs. slow on a…
 * 2) position vs. time graph
 * 1) velocity vs. time graph
 * 1) acceleration vs. time graph

Graph will not be a line velocity will change over time Acceleration will stop and start again in another direction
 * 1) How can you tell that you changed direction on a…
 * 2) position vs. time graph
 * 1) velocity vs. time graph
 * 1) acceleration vs. time graph

Shows linear distance Shows speed in a direction Shows rate of acceleration
 * 1) What are the advantages of representing motion using a…
 * 2) position vs. time graph
 * 1) velocity vs. time graph
 * 1) acceleration vs. time graph

No directions doesn't show distance traveled doesn't show direction
 * 1) What are the disadvantages of representing motion using a…
 * 2) position vs. time graph
 * 1) velocity vs. time graph
 * 1) acceleration vs. time graph

Object is not moving Object is moving at a speed that does not change
 * 1) Define the following:
 * 2) No motion
 * 1) Constant speed

=Class Notes 9/13= 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=

Purpose: Understand acceleration and velocity

//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.

//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 it 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



Time (s) Distance (cm)

Conclusion: Our hypothesis was correct. The graph showed a curve, starting from a small slope to a big slope, showing acceleration. As the object went down the ramp, each point on the ticker tape was a greater distance from the last one. One thing that may have affected our project was friction; the ticker tape going through the box may have given some resistance, slowing down the object as it went down the ramp. This would make our results precise but not accurate. In real life, judging the speeds of cars is similar; you must remove all other factors in order to judge the true potential of a car (traction, wind resistance, etc.). Next time. we could use a different instrument for measuring distance.

=Class Notes 9/14=



= =

= = = Homework 9/14 =

Acceleration is a [|vector quantity] that is defined as the rate at which an object changes its [|velocity]. An object is accelerating if it is changing its velocity. Change its velocity by the same amount each second is constant acceleration. The direction of the acceleration vector depends on two things: The general RULE OF THUMB is: If an object is slowing down, then its acceleration is in the opposite direction of its motion. (negative acceleration)
 * whether the object is speeding up or slowing down
 * whether the object is moving in the + or - direction

=Homework 9/15= Lesson 3 & 4:
 * 1) I understood that the shape of a velocity time graph indicates speeding up or slowing down, as well as the shape of a position time graph indication positive of negative motion.
 * 2) I understood everything I read
 * 3) What is the meaning of something being towards or away from the origin?
 * 4) We didn't go over finding area of our graphs.

=Lab 9/21= A Crash Course in Velocity Info: A Crash Course, 9/20, Jake Greenstein, Lerna Girgin, Katie Dooman,

Purpose: In this lab, we are trying to demonstrate the relationships between velocities, and provide a visual (graph) explanation for our results.

Yellow (fast): y=27.726x Yellow (slow): y=12.897x

Run 1 (two cars meet) Algebra: 12.897x=-27.726x+600 40.623x=600 x=14.77 (seconds) 12.897(14.77)=190.5 (cm)

Graph:

Cars will meet after 14.77 seconds at 190.5 centimeters

Run 2 (one car passes the other) Algebra: 27.726x-100=12.897x 14.829x=100 x=6.74 (seconds) y= 27.726(6.74) - 100 y= 86.8 (centimeters)

Graph:

Yellow fast will catch up with Yellow slow after 6.74 seconds at 86.8 centimeters (away from the Yellow slow)

Videos: media type="file" key="Movie on 2011-09-20 at 13.33 media type="file" key="Movie on 2011-09-20 at 13.41.mov" width="300" height="300" media type="file" key="Movie on 2011-09-20 at 13.47.mov" width="300" height="300"

Data: Discussion:
 * 1) Where would the cars meet if their speeds were exactly equal?

If the speeds of both CMVs were equal, the overtake would not be possible. The two cars would go at an equal speed parallel to one another and never catch up. On the other experiment, at equal speeds the cars would meet at 300cm. They would travel the same speed and meet halfway at their 600cm distance apart.


 * 1) 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. (above)
 * 2) 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? (above)

Conclusion: Using our previous knowledge of the equations of the lines of our CMVs, we were able to manipulate the data to determine where the two would over-take or meet. Our calculations gave us points where the two would meet, and our graphs gave us a visual representation of the path that each would take until both lines met at one point. Our results were not perfectly accurate with our calculations, but were close enough to show that we had calculated properly. Our first catch up took 7.5 seconds, and they met at 90 cm. The second run took 7.9 seconds and they met at 93 cm. These were close to our theoretical data, which yielded 6.78 seconds to meet at 86.8 cm. Our first crash took 16 seconds and they met at 185 cm. The second took 16.5 seconds and they met at 189 cm. These too were close to our theoretical results, which yielded 14.77 seconds to meet at 190.5 cm. % errors were between 10.35% and 11.2% for the crash, and 8.33% and 8.58% for the catch up. These figures showed our percent error away from the correct answer, making our answers more accurate than not. The catch up had a % difference of .984% and .934%. This shows that our answers were very precise, as they were both close to each other. Our crash lab had percent differences of 8.65% and .462%. This greater range in percent differences shows that our answers were not as precise. The errors in precision could arrive from not taking care to perform each experiment the same, such as starting the cars at the same distances apart, or turning them on at the same time. Accuracy could have been improved with a car that stayed in a straight line. Next time, taking these precautions will give us experimental yields closer to the theoretical.

=Egg Drop Project=

 **Procedure** - The procedure of the egg drop was relatively simple. An egg was placed in or device, and was dropped out a third story window (8.5m).

 **Weight** - On our final device, the total weight of the device and egg was 70.45g, with the egg weighing in at 55.25g. Our device weighed 15.2g.

 **Results**: Cracked egg

**Final Design:**



 Our final project still yielded a cracked egg. The egg was leaking slightly, with a large crack. Our parachute design made our egg take a long time to fall (1.58 and 1.75 seconds, average of 1.67 seconds). Our acceleration was 6.09 m/s^2.

 **Calculations and Analysis:**

 Δd=(Vo)(t)+1/2(a)(t2)

 8.5m=(0)(1.67s)+1/2(a)(1.67s^2)

 8.5=1/2(a)(2.79)

 17=2.79a

 17/2.79=a

 a=6.09m/s^2

 Our acceleration was calculated properly, since we expect a value below 9.8 m/s/s.

 While in concept our design was good, we failed to build one properly to protect the egg. We wanted to use a parachute to slow down the egg enough to reduce the force of impact, allowing our project to use less cushioning. Our first prototype was a double parachute with a paper-skin aluminum foil filled nose cone. This failed because the parachute design did not function properly. Our second prototype used a hexagonal parachute with a similar nose cone, which failed because of the weak thread holding the parachute to the cone. Our final product was a hybrid of the two designs: a larger hexagonal parachute, attached to the cone with strong tape, and a cocoon surrounding the egg made of loose foil. The problem with this design was that there was not enough protection. Extremely light and slow to fall, it only needed a little more protection from impact to make the egg survive. A larger cone with more loose foil would have done the trick.

=Class Notes 9/22=



=Class Notes 10/3=

Free fall- an object moving under the influence of gravity ONLY . V=0
 * ^** **|** v **|** a
 * v **v** **v**
 * ^** **|** v **|** a
 * v **v** **v**

=**Class Work 10/3**=



=Homework 10/3= 5.1

<span style="display: block; font-family: 'Times New Roman',Times,serif; text-align: left;">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: <span style="display: block; font-family: 'Times New Roman',Times,serif; text-align: left;"> 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)

<span style="display: block; font-family: 'Times New Roman',Times,serif; text-align: left;">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.


 * An object in free fall encounters no resistance and accelerates at 9.8m/s/s**

<span style="display: block; font-family: 'Times New Roman',Times,serif; text-align: left;">5.2

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. 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.


 * G represents acceleration of gravity constant, which is 9.8 m/s/s**

<span style="display: block; font-family: 'Times New Roman',Times,serif; text-align: left;">5.3

<span style="display: block; font-family: 'Times New Roman',Times,serif; text-align: left;">A position versus time graph for a free-falling object is shown below. <span style="display: block; font-family: 'Times New Roman',Times,serif; text-align: left;">

<span style="display: block; font-family: 'Times New Roman',Times,serif; text-align: left;"> 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.Object starts with a small velocity (slow) and finishes with a large velocity (fast). Slope of any position vs. time graph is the velocity of the object (<span class="wiki_link_ext">as learned in Lesson 3 ), the small initial slope indicates a small initial velocity and the large final slope indicates a large final velocity. Finally, the negative slope of the line indicates a negative (i.e., downward) velocity. <span style="display: block; font-family: 'Times New Roman',Times,serif; text-align: left;">A velocity versus time graph for a free-falling object is shown below. <span style="display: block; font-family: 'Times New Roman',Times,serif; text-align: left;">

<span style="display: block; font-family: 'Times New Roman',Times,serif; text-align: left;">Observe that the line on the graph is a straight, diagonal line. Since a free-falling object is undergoing an acceleration (g = 9,8 m/s/s, downward), velocity-time graph would be diagonal. Object starts with a zero velocity (as read from the graph) and finishes with a large, negative velocity; that is, the object is moving in the negative direction and speeding up. An object that is moving in the negative direction and speeding up is said to have a negative acceleration .Slope of any velocity versus time graph is the acceleration of the object the constant, negative slope indicates a constant, negative acceleration. An object moving with a constant acceleration of 9.8 m/s/s in the downward direction. <span style="display: block; font-family: 'Times New Roman',Times,serif; text-align: left;">**A position time graph of an object in free fall shows it starting from 0 with no slope, and ending with a large slope in a curved line. A velocity time graph will show a straight line with a slope (acceleration) of -9.8** <span style="display: block; font-family: 'Times New Roman',Times,serif; text-align: left;">5.4

<span style="display: block; font-family: 'Times New Roman',Times,serif; text-align: left;">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. Thus, 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 <span style="display: block; font-family: 'Times New Roman',Times,serif; text-align: left;">t  <span style="display: block; font-family: 'Times New Roman',Times,serif; text-align: left;">seconds is   <span style="display: block; font-family: 'Times New Roman',Times,serif; text-align: left;">vf = g * t  <span style="display: block; font-family: 'Times New Roman',Times,serif; text-align: left;">where 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. <span style="display: block; font-family: 'Times New Roman',Times,serif; text-align: left;">Example Calculations:At t = 6 svf = (9.8 m/s2) * (6 s) = 58.8 m/sAt t = 8 svf = (9.8 m/s2) * (8 s) = 78.4 m/sd = 0.5 * g * t2 <span style="display: block; font-family: 'Times New Roman',Times,serif; text-align: left;">g is the acceleration of gravity (9.8 m/s/s on Earth). <span style="display: block; font-family: 'Times New Roman',Times,serif; text-align: left;">Example Calculations:At t = 1 ad = (0.5) * (9.8 m/s2) * (1 s)2 = 4.9mAt t = 2 sd = (0.5) * (9.8 m/s2) * (2 s)2 = 19.6mAt t = 5 ad = (0.5) * (9.8 m/s2) * (5 s)2 = 123 m(rounded from 122.5 m) <span style="display: block; font-family: 'Times New Roman',Times,serif; text-align: left;"> **<span style="display: block; font-family: 'Times New Roman',Times,serif; text-align: left;">Time * acceleration will yield velocity, since the object travels at constant acceleration. Distance can be calculated through ∆d=1/2(g)t^2 ** <span style="display: block; font-family: 'Times New Roman',Times,serif; text-align: left;">5.5 <span style="display: block; font-family: 'Times New Roman',Times,serif; text-align: left;">Value (known as the acceleration of gravity) 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. <span style="display: block; font-family: 'Times New Roman',Times,serif; text-align: left;">Doesn't a more massive object accelerate at a greater rate than a less massive object - absolutely not! That is, absolutely not if we are considering the specific type of falling motion known as free-fall. Free-fall is the motion of objects that move under the sole influence of gravity; free-falling objects do not encounter air resistance. More massive objects will only fall faster if there is an appreciable amount of air resistance present. <span style="display: block; font-family: 'Times New Roman',Times,serif; text-align: left;">The actual explanation of why all objects accelerate at the same rate involves the concepts of force and mass. Acceleration of an object is directly proportional to force and inversely proportional to mass. Increasing force tends to increase acceleration while increasing mass tends to decrease acceleration. Thus, the greater force on more massive objects is offset by the inverse influence of greater mass. Subsequently, all objects free fall at the same rate of acceleration, regardless of their mass. All free falling objects accelerate at the same rate, since a greater mass is offset by force. Mathematically, all objects in free fall have the same acceleration.

=**Falling Object Lab 10/4**=


 * Purpose**: Determine acceleration of a falling body, comparing our results to free-fall acceleration due to gravity.

<span style="font-family: 'Times New Roman',Times,serif;">**Hypothesis**: Acceleration due to gravity is 9.8 m/s/s.Therefore, a v-t graph would be above the X axis, and have a positive slope, since the velocity is increasing due to acceleration. With the slope, we can find acceleration.

<span style="font-family: 'Times New Roman',Times,serif; font-size: 14.4px;">**Procedure**: Attach ticker tape with masking tape to a mass. With the ticker tape feeding through a spark timer, drop the mass. Be sure to avoid coiling the ticker tape and allow the mass to drop with minimal friction interfering.

<span style="font-family: 'Times New Roman',Times,serif; font-size: 14.4px;">**Data**:

<span style="font-family: 'Times New Roman',Times,serif; font-size: 14.4px;">Class Average = 805.9

<span style="font-family: 'Times New Roman',Times,serif; font-size: 14.4px;">


 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 14.4px;">Sample Calculations: **

<span style="font-family: 'Times New Roman',Times,serif; font-size: 14.4px;">



<span style="font-family: 'Times New Roman',Times,serif; font-size: 14.4px;">


 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 14.4px;">Graphs: **

<span style="font-family: 'Times New Roman',Times,serif; font-size: 14.4px;"> <span style="font-family: 'Times New Roman',Times,serif; font-size: 14.4px;">**Analysis**: This graph shows the object's position over time. The curved line shows how our velocity increased over time (velocity is the slope). The distance traveled between an interval of time steadily increased, since acceleration affected the distance traveled. R^2 is very close to 1, showing our precision.

<span style="font-family: 'Times New Roman',Times,serif; font-size: 14.4px;">

<span style="font-family: 'Times New Roman',Times,serif; font-size: 14.4px;">**Analysis**: The graph represents velocity over time for the falling object. The slope of our line is the acceleration, which theoretically should be 9.8. Our first value is not at 0, because the initial velocity calculated was after the object began moving. Our r^2 value shows our precision.

<span style="font-family: 'Times New Roman',Times,serif; font-size: 14.4px;">Discussion Questions: 1) The shape of the v-t graph agreed with the expected graph, proving my hypothesis correct. This is because the velocity increased at a constant acceleration (slope) over time, giving our points a positive slope and moving away from the x axis. 2) The shape of the x-t graph also agreed with the expected graph. Since velocity (slope) is increasing, the points become progressively farther from the last. The created an expected curved line, showing distance over time as a quadratic function. 3) My results were close o those of the class. The class average was 805.9, and mine was 846.9. After calculating percent difference, I got 5.09%. This means that my results were only 5.09% off from the class average, making my results consistent with those of the class. 4) Our object did not accelerate uniformly. Acceleration is constant, and therefore acceleration should have been the same over each interval. However, our data reflects a varying acceleration, which could have been a result of other factors affecting acceleration, such as friction. 5) Acceleration due to gravity could be higher if the object was being pulled through the air at an acceleration faster than 9.8. To be lower, something would have to slow the object down, such as friction.

My hypothesis was correct. Specifically, the slope of the v-t graph did represent acceleration.Our slope (acceleration) was an average of 8.47 m/s/s, which was the slope of our graph. To have been more accurate, we would have wanted a slope of 9.8. The purpose of the experiment was satisfied by the correlation of our experimental results to the predicted results. Experimental errors were varied low. Our percent error ranged between 16.4% and 155.2%, reflecting the differences in acceleration within the experiment. However, as a whole, our experiment only had a percent difference of 5.09%, meaning that our results were very close to those of other groups. The error (occurring in acceleration) could be due to numerous things. Specifically, the procedure for our experiment allowed many factors to interfere. Air resistance may have contributed slightly, but a greater source of error came from the spark timer. The friction caused could have lowered acceleration greatly, and if the tape did not run through smoothly, our results will not be consistent. To address the error, we could alter the materials used. Since dropping the object in a vacuum (space) is not practical, we would instead use a distance sensor to allow the object to drop relatively friction free and record the position over time.
 * Conclusion:**

=Class Work 10/7=