Chapter+3

=__10/12/10__= toc //HW: Physics Classroom on Vectors - Lesson 1(ab)// Topic Sentence: Vectors, quantities with magnitude or direction, are represented on a diagram when they have a clear tail and head, when they have a scale, and when the direction is listed. A study of motion will involve the introduction of a variety of quantities that are used to describe the physical world. All these quantities can by divided into two categories - vectors and scalars. A vector quantity is a quantity that is fully described by both magnitude and direction. On the other hand, a scalar quantity is a quantity that is fully described by its magnitude. Each of these vector quantities are unique in that a full description of the quantity demands that both a magnitude and a direction are listed. Vector quantities are not fully described unless both magnitude and direction are listed. Vector quantities are often represented by scaled vector diagrams. Vector diagrams depict a vector by use of an arrow drawn to scale in a specific direction. Such diagrams are commonly called as free-body diagrams. An example of a scaled vector diagram is shown in the diagram at the right. The vector diagram depicts a displacement vector. Observe that there are several characteristics of this diagram that make it an appropriately drawn vector diagram. **Conventions for Describing Directions of Vectors** Vectors can be directed due East, due West, due South, and due North. Thus, there is a clear need for some form of a convention for identifying the direction of a vector that is __not__ due East, due West, due South, or due North. There are a variety of conventions for describing the direction of any vector. The two conventions that will be discussed and used in this unit are described below: Two illustrations of the second convention (discussed above) for identifying the direction of a vector are shown below.
 * A: Vectors and Direction**
 * Vectors and Direction**
 * a scale is clearly listed
 * a vector arrow (with arrowhead) is drawn in a specified direction. The vector arrow has a //head// and a //tail//.
 * the magnitude and direction of the vector is clearly labeled. In this case, the diagram shows the magnitude is 20 m and the direction is (30 degrees West of North).
 * 1) The direction of a vector is often expressed as an angle of rotation of the vector about its "tail" from east, west, north, or south. For example, a vector can be said to have a direction of 40 degrees North of West (meaning a vector pointing West has been rotated 40 degrees towards the northerly direction) of 65 degrees East of South (meaning a vector pointing South has been rotated 65 degrees towards the easterly direction).
 * 2) The direction of a vector is often expressed as a counterclockwise angle of rotation of the vector about its "tail" from due East. Using this convention, a vector with a direction of 30 degrees is a vector that has been rotated 30 degrees in a counterclockwise direction relative to due east. A vector with a direction of 160 degrees is a vector that has been rotated 160 degrees in a counterclockwise direction relative to due east. A vector with a direction of 270 degrees is a vector that has been rotated 270 degrees in a counterclockwise direction relative to due east. This is one of the most common conventions for the direction of a vector and will be utilized throughout this unit.

**Representing the Magnitude of a Vector** The magnitude of a vector in a scaled vector diagram is depicted by the length of the arrow. The arrow is drawn a precise length in accordance with a chosen scale. Using the same scale (__1 cm = 5 miles__), a displacement vector that is 15 miles will be represented by a vector arrow that is 3 cm in length. Similarly, a 25-mile displacement vector is represented by a 5-cm long vector arrow. And finally, an 18-mile displacement vector is represented by a 3.6-cm long arrow. See the examples shown below.



In conclusion, vectors can be represented by use of a scaled vector diagram. On such a diagram, a vector arrow is drawn to represent the vector. The arrow has an obvious tail and arrowhead. The magnitude of a vector is represented by the length of the arrow. A scale is indicated (such as, 1 cm = 5 miles) and the arrow is drawn the proper length according to the chosen scale. The arrow points in the precise direction. Directions are described by the use of some convention. The most common convention is that the direction of a vector is the counterclockwise angle of rotation which that vector makes with respect to due East.

Topic Sentence: There are three ways to add vectors: through the pythagorean theorem, through trigonometry, and through the head to tail method. A variety of mathematical operations can be performed with and upon vectors. One such operation is the addition of vectors. Two vectors can be added together to determine the result (or resultant). Observe the following summations of two force vectors:
 * B: Vector Addition**
 * Vector Addition**

These rules for summing vectors were applied to free-body diagrams in order to determine the net force (i.e., the vector sum of all the individual forces). Sample applications are shown in the diagram below. In this unit, the task of summing vectors will be extended to more complicated cases in which the vectors are directed in directions other than purely vertical and horizontal directions. There are a variety of methods for determining the magnitude and direction of the result. The two methods that will be discussed in this lesson and used throughout the entire unit are: **The Pythagorean Theorem** The Pythagorean theorem is a useful method for determining the result of adding two (and only two) vectors __that make a right angle__ to each other. The Pythagorean theorem is a mathematical equation that relates the length of the sides of a right triangle to the length of the hypotenuse of a right triangle. To see how the method works, consider the following problem: >> Eric leaves the base camp and hikes 11 km, north and then hikes 11 km east. Determine Eric's resulting displacement. This problem asks to determine the result of adding two displacement vectors that are at right angles to each other. The result (or resultant) of walking 11 km north and 11 km east is a vector directed northeast as shown in the diagram to the right. The Pythagorean theorem can be used to determine the resultant. The result of adding 11 km, north plus 11 km, east is a vector with a magnitude of 15.6 km. [|Later], the method of determining the direction of the vector will be discussed **Using Trigonometry to Determine a Vector's Direction** The direction of a //resultant// vector can often be determined by use of trigonometric functions. These three functions, sine, cosine, and tangent, relate an acute angle in a right triangle to the ratio of the lengths of two of the sides of the right triangle. The **sine function** relates the measure of an acute angle to the ratio of the length of the side opposite the angle to the length of the hypotenuse. The **cosine function** relates the measure of an acute angle to the ratio of the length of the side adjacent the angle to the length of the hypotenuse. The **tangent function** relates the measure of an angle to the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle. The three equations below summarize these three functions in equation form. These three trigonometric functions can be applied to the hiker problem in order to determine the direction of the hiker's overall displacement. The process begins by the selection of one of the two angles (other than the right angle) of the triangle. Once the angle is selected, any of the three functions can be used to find the measure of the angle. Once the measure of the angle is determined, the direction of the vector can be found. In this case the vector makes an angle of 45 degrees with due East. The measure of an angle as determined through use of SOH CAH TOA is __not__ always the direction of the vector. The following vector addition diagram is an example of such a situation. Observe that the angle within the triangle is determined to be 26.6 degrees using SOH CAH TOA. Yet the direction of the vector as expressed with the CCW (counterclockwise from East) convention is 206.6 degrees.
 * the Pythagorean theorem and trigonometric methods
 * the head-to-tail method using a scaled vector diagram

In the above problems, the magnitude and direction of the sum of two vectors is determined using the Pythagorean theorem and trigonometric methods (SOH CAH TOA). The procedure is restricted to the addition of __two vectors that make right angles to each other__. When the two vectors that are to be added do not make right angles to one another, or when there are more than two vectors to add together, we will employ a method known as the head-to-tail vector addition method. **Use of Scaled Vector Diagrams to Determine a Resultant** The magnitude and direction of the sum of two or more vectors can also be determined by use of an accurately drawn scaled vector diagram. Using a scaled diagram, the **head-to-tail method** is employed to determine the vector sum or resultant. A common Physics lab involves a //vector walk//. Either using centimeter-sized displacements upon a map or meter-sized displacements in a large open area, a student makes several consecutive displacements beginning from a designated starting position. Suppose that you were given a map of your local area and a set of 18 directions to follow. Starting at //home base//, these 18 displacement vectors could be //added together// in consecutive fashion to determine the result of adding the set of 18 directions. Perhaps the first vector is measured 5 cm, East. Where this measurement ended, the next measurement would begin. The process would be repeated for all 18 directions. Each time one measurement ended, the next measurement would begin. In essence, you would be using the head-to-tail method of vector addition.

A step-by-step method for applying the head-to-tail method to determine the sum of two or more vectors is given below.
 * 1) Choose a scale and indicate it on a sheet of paper. The best choice of scale is one that will result in a diagram that is as large as possible, yet fits on the sheet of paper.
 * 2) Pick a starting location and draw the first vector //to scale// in the indicated direction. Label the magnitude and direction of the scale on the diagram (e.g., SCALE: 1 cm = 20 m).
 * 3) Starting from where the head of the first vector ends, draw the second vector //to scale// in the indicated direction. Label the magnitude and direction of this vector on the diagram.
 * 4) Repeat steps 2 and 3 for all vectors that are to be added
 * 5) Draw the resultant from the tail of the first vector to the head of the last vector. Label this vector as **Resultant** or simply **R**.
 * 6) Using a ruler, measure the length of the resultant and determine its magnitude by converting to real units using the scale (4.4 cm x 20 m/1 cm = 88 m).
 * 7) Measure the direction of the resultant using the counterclockwise convention

Interestingly enough, the order in which three vectors are added has no affect upon either the magnitude or the direction of the resultant. The resultant will still have the same magnitude and direction.

=__10/13/11__= //HW: Physics Classroom on Vectors - Lesson 1(cd)// Topic Sentence: The resultant vector is the sum of two or more vectors of the same vector quantity. The **resultant** is the vector sum of two or more vectors. It is //the result// of adding two or more vectors together. If displacement vectors A, B, and C are added together, the result will be vector R. As shown in the diagram, vector R can be determined by the use of an [|accurately drawn, scaled, vector addition diagram]. **A + B + C = R**  When displacement vectors are added, the result is a //resultant displacement//. But any two vectors can be added as long as they are the same vector quantity. In all such cases, the resultant vector (whether a displacement vector, force vector, velocity vector, etc.) is the result of adding the individual vectors. It is the same thing as adding A + B + C + ... . As an example, consider a football player who gets hit simultaneously by three players on the opposing team (players A, B, and C). The football player experiences three different applied forces. Each applied force contributes to a total or resulting force. If the three forces are added together using methods of vector addition ([|discussed earlier]), then the resultant vector R can be determined. In this case, to experience the three forces A, B and C is the same as experiencing force R. Topic Sentence: A vector that is directed along an angle can be broken up into two separate vectors going along the axes. A [|vector] is a quantity that has both magnitude and direction. [|Displacement], [|velocity], [|acceleration], and [|force] are the vector quantities that we have discussed thus far in the Physics Classroom Tutorial. In the first couple of units, all vectors that we discussed were simply directed up, down, left or right. Now in this unit, we begin to see examples of vectors that are directed in //two dimensions// - upward and rightward, northward and westward, eastward and southward, etc.
 * C: Resultants**
 * Resultants**
 * D: Vector Components**
 * Vector Components**

In situations in which vectors are directed at angles to the customary coordinate axes, a useful mathematical trick will be employed to //transform// the vector into two parts with each part being directed along the coordinate axes. For example, a vector that is directed northwest can be thought of as having two parts - a northward part and a westward part. A vector that is directed upward and rightward can be thought of as having two parts - an upward part and a rightward part. Any vector directed in two dimensions can be thought of as having an influence in two different directions. That is, it can be thought of as having two parts. Each part of a two-dimensional vector is known as a **component**. The components of a vector depict the influence of that vector in a given direction. The combined influence of the two components is equivalent to the influence of the single two-dimensional vector.

Consider a picture that is hung to a wall by means of two wires that are stretched vertically and horizontally. Each wire exerts a tension force upon the picture to support its weight. Since each wire is stretched in two dimensions (both vertically and horizontally), the tension force of each wire has two components - a vertical component and a horizontal component. Focusing on the wire on the left, we could say that the wire has a leftward and an upward component. This is to say that the wire on the left could be replaced by two wires, one pulling leftward and the other pulling upward. If the single wire were replaced by two wires (each one having the magnitude and direction of the components), then there would be no affect upon the stability of the picture. The combined influence of the two components is equivalent to the influence of the single two-dimensional vector.



 Any vector directed in two dimensions can be thought of as having two different components. The component of a single vector describes the influence of that vector in a given direction. In the [|next part of this lesson], we will investigate two methods for determining the magnitude of the components. That is, we will investigate //how much// influence a vector exerts in a given direction.

=__10/14/10__= //Class Notes:// __Collinear__- Velocities that are just added or subtracted to find the total force. Whether they are added or subtracted depends on their direction. __Right Angles__- It is added or subtracted as normal, depending on direction __Resultant__- The answer to a vector addition problem Steps to solve Right Angle Vector Problem: __Estimating__:
 * 1) Draw a sketch of the vector __head-to-tail__ (Draw vector 1. At the head of Vector 1, start drawing Vector 2)
 * 2) Draw the resultant from the tail of the first vector to the head of the second vector (Where you began to where you ended)
 * 3) Do Pythagorean Theorem to find R (resultant)
 * 4) Do Trig to find ø (theta, angle)
 * 1) Size must be between minimum (subtracting) and maximum (adding)
 * 2) Angles must be between the angles of the vectors given

Lab: Orienteering Part 1
//Measurements:// R physical = 2689 cm
 * Legs || Distance (cm) || Direction ||
 * Raymond Tree --> Mickey Tree || 786.8 || N ||
 * Mickey Tree --> Steve Tree || 700.0 || W ||
 * Steve Tree --> Edge of Cement Towards Garden || 533.5 || W ||
 * Edge of Cement Towards Garden --> Edge of Cement Away from Garden || 351.9 || W ||
 * Edge of Cement Away from Garden --> First Pillar || 135.4 || S ||
 * First Pillar --> Largest Tree || 1025.5 || S ||

//__Analysis (Part 1)__//
 * Draw to scale
 * Find resultant magnitude by measuring and by calculating
 * % error measured and actual b/w calculated and actual

R actual = 2869 cm R measured (graph) = 2640 cm
 * Measured Resultant Magnitude**

786.8 N + 700.0 W + 533.5 W + 359.1 W + 135.4 S + 1025.5 W (786.8 N + -135.4 S), (700.0 W + 533.5 W + 359.1 W + 1025.5 W) = 651.4 N, 2618.10 W
 * Calculated Resultant Magnitude**

[(Theoretical - Experimental) / Theoretical] * 100 [(2640 - 2689) / 2640] * 100 = __1.86%__
 * % Error: R Measured**

[(Theoretical - Experimental) / Theoretical] * 100 [(2640 - 2697.92) / 2640] * 100 = __2.19%__
 * % Error: R Calculated**

=__10/17/11__= //HW: Physics Classroom on Vectors - Lesson 1(e)// **Vector Resolution** As mentioned [|earlier in this lesson], any vector directed in two dimensions can be thought of as having two components. For example, if a chain pulls upward at an angle on the collar of a dog, then there is a tension force directed in two dimensions. This tension force has two components: an upward component and a rightward component.

In this unit, we learn two basic methods for determining the magnitudes of the components of a vector directed in two dimensions. The process of determining the magnitude of a vector is known as **vector resolution**. The two methods of vector resolution that we will examine are
 * the parallelogram method
 * [|the trigonometric method]

**Parallelogram Method of Vector Resolution** The parallelogram method of vector resolution involves using an accurately drawn, scaled vector diagram to determine the components of the vector. If one desires to determine the components as directed along the traditional x- and y-coordinate axes, then the parallelogram is a rectangle with sides that stretch vertically and horizontally. A step-by-step procedure for using the parallelogram method of vector resolution is:
 * 1) Select a scale and accurately draw the vector to scale in the indicated direction.
 * 2) Sketch a parallelogram around the vector: beginning at the [|tail] of the vector, sketch vertical and horizontal lines; then sketch horizontal and vertical lines at the [|head] of the vector
 * 3) Draw the components of the vector. The components are the //sides// of the parallelogram.
 * 4) Meaningfully label the components of the vectors with symbols to indicate which component represents which side. A northward force component might be labeled Fnorth.
 * 5) Measure the length of the sides of the parallelogram and [|use the scale to determine the magnitude] of the components in //real// units.

The step-by-step procedure above is illustrated in the diagram below to show how a velocity vector with a magnitude of 50 m/s and a direction of 60 degrees above the horizontal may be resolved into two components.

**Trigonometric Method of Vector Resolution** The trigonometric method of vector resolution involves using trigonometric functions to determine the components of the vector.[|Recall from the earlier discussion] that trigonometric functions relate the ratio of the lengths of the sides of a right triangle to the measure of an acute angle within the right triangle. The method of employing trigonometric functions to determine the components of a vector are as follows:
 * 1) Construct a //rough// sketch (no scale needed) of the vector in the indicated direction. Label its magnitude and the angle that it makes with the horizontal.
 * 2) Draw a rectangle about the vector such that the vector is the diagonal of the rectangle.
 * 3) Draw the components of the vector. The components are the //sides// of the rectangle.
 * 4) Meaningfully label the components of the vectors with symbols to indicate which component represents which side. A northward force component might be labeled Fnorth.
 * 5) To determine the length of the side opposite the indicated angle, use the sine function. Substitute the magnitude of the vector for the length of the hypotenuse.
 * 6) Repeat the above step using the cosine function to determine the length of the side adjacent to the indicated angle.

The above method is illustrated below for determining the components of the force acting upon Fido. As the 60-Newton tension force acts upward and rightward on Fido at an angle of 40 degrees, the components of this force can be determined using trigonometric functions.

In conclusion, a vector directed in two dimensions has two components - that is, an influence in two separate directions. The amount of influence in a given direction can be determined using methods of vector resolution.

=__10/18/11__= //HW: Physics Classroom on Vectors - Lesson 1(gh)// Topic Sentence: Velocity is different depending on the point of view, and when solving riverboat problems, use the correct velocity for the different distances. On occasion objects move within a medium that is moving with respect to an observer. In such instances as this, the magnitude of the velocity of the moving object (whether it be a plane or a motorboat) with respect to the observer on land will not be the same as the speedometer reading of the vehicle. That is to say, the speedometer on the motorboat might read 20 mi/hr; yet the motorboat might be moving relative to the observer on shore at a speed of 25 mi/hr. Motion is relative to the observer
 * G: Relative Velocity and Riverboat Problems**

**Analysis of a Riverboat's Motion** The affect of the wind upon the plane is similar to the affect of the river current upon the motorboat. If a motorboat were to head straight across a river (that is, if the boat were to point its bow straight towards the other side), it would not reach the shore directly across from its starting point. The river current influences the motion of the boat and carries it downstream.

 Motorboat problems such as these are typically accompanied by three separate questions: The first of these three questions was answered above; the resultant velocity of the boat can be determined using the Pythagorean theorem (magnitude) and a trigonometric function (direction). The second and third of these questions can be answered using the average speed equation (and a lot of logic).
 * 1) What is the resultant velocity (both magnitude and direction) of the boat?
 * 2) If the width of the river is //X// meters wide, then how much time does it take the boat to travel shore to shore?
 * 3) What distance downstream does the boat reach the opposite shore?

Consider the following example.  We will start in on the second question. The river is 80-meters wide. The time to cross this 80-meter wide river can be determined by rearranging and substituting into the average speed equation. The distance of 80 m can be substituted into the numerator. But what about the denominator? What value should be used for average speed? Should 3 m/s (the current velocity), 4 m/s (the boat velocity), or 5 m/s (the resultant velocity) be used as the average speed value for covering the 80 meters? With what average speed is the boat traversing the 80 meter wide river? Most students want to use the resultant velocity in the equation since that is the actual velocity of the boat with respect to the shore. Yet the value of 5 m/s is the speed at which the boat covers the diagonal dimension of the river. If one knew the distance C in the diagram below, then the average speed C could be used to calculate the time to reach the opposite shore. If one knew the distance B in the diagram below, then the average speed B could be used to calculate the time to reach the opposite shore. And finally, if one knew the distance A in the diagram below, then the average speed A could be used to calculate the time to reach the opposite shore. In our problem, the 80 m corresponds to the distance A, and so the average speed of 4 m/s (average speed in the direction straight across the river) should be substituted into the equation to determine the time. **time = (80 m)/(4 m/s) = 20 s** It requires 20 s for the boat to travel across the river. During this 20 s of crossing the river, the boat also drifts downstream. Part c of the problem asks "What distance downstream does the boat reach the opposite shore?" The same equation must be used to calculate this //downstream distance//. And once more, the question arises, which one of the three average speed values must be used in the equation to calculate the distance downstream? The distance downstream corresponds to Distance B on the above diagram. The speed at which the boat covers this distance corresponds to Average Speed B on the diagram above. And so the average speed of 3 m/s (average speed in the downstream direction) should be substituted into the equation to determine the distance. **distance = ave. speed * time = (3 m/s) * (20 s)** **distance = 60 m** The boat is carried 60 meters downstream during the 20 seconds it takes to cross the river
 * Example 1 A motorboat traveling 4 m/s, East encounters a current traveling 3.0 m/s, North.
 * 1) What is the resultant velocity of the motorboat?
 * 2) If the width of the river is 80 meters wide, then how much time does it take the boat to travel shore to shore?
 * 3) What distance downstream does the boat reach the opposite shore? ||

Topic Sentence: Perpendicular components of Motion are independent of each other - a change in one doesn't affect the other. Any vector - whether it is a force vector, displacement vector, velocity vector, etc. - directed at an angle can be thought of as being composed of two perpendicular components. These two components can be represented as legs of a right triangle formed by projecting the vector onto the x- and y-axis. The two perpendicular parts or components of a vector are independent of each other. A change in the horizontal component does not affect the vertical component.
 * H: Independence of Perpendicular Components of Motion**

= __10/19/11__ = //HW: Physics Classroom on Vectors - Lesson 2 (ab)//


 * A: What is projectile motion?**
 * 1) Like the title says, what are projectiles?
 * 2) Are projectiles the same thing as free-falling objects?
 * 3) How are projectiles different from free-falling objects?
 * 4) What does inertia have to do with projectile motion?
 * 5) Does gravity change the path of a projectile significantly?

__Central Theme: Gravity is the only force acting upon a projectile, and it affects only the vertical motion of the object.__
 * 1) A projectile is an object that is only acted upon by gravity
 * 2) According to each of their definitions, projectiles and free-falling objects are the same.
 * 3) They are not different
 * 4) Inertia allows an object to stay at a certain speed as long as there is nothing acting upon it. Thus, even in the presence of gravity, a projectile will still have the same speed as one not in gravity.
 * 5) Gravity does change the path of a projectile by making it fall, though it does nothing to slow it down.


 * B: Characteristics of a Projectile's trajectory**
 * 1) What defines a projectile's trajectory?
 * 2) How does it differ from other things?
 * 3) Are the paths parabolic, like free falling objects?
 * 4) Do they have similar characteristics as free falling objects? (t1,v1,y1, t2,v2,y2....)
 * 5) How does gravity affect the path of a projectile?

__Central Theme:__
 * 1) Projectiles move horizontally and vertically at the same time, though these two are independent of each other. The path is parabolic.
 * 2) It differs because the only force acting upon it is gravity.
 * 3) Yes, the path is parabolic.
 * 4) Yes, it does have these same characteristics.
 * 5) Gravity affects the path of a projectile by making it parabolic, making it fall. However, it will stay at the same horizontal speed.

=__10/20/11__= //Class Notes:// //HW: Physics Classroom on Vectors - Lesson 2(c)// Central Theme: Because vector components are independent of each other, gravity will not affect the horizontal path. This means that the horizontal velocity will stay the same while it will accelerate downwards due to gravity.
 * 1) How do you calculate the horizontal and vertical velocities of a projectile?
 * 2) Is there a way to calculate the exact path of a projectile?
 * 3) What are the equations, if necessary, to find these?
 * 4) How are they similar to other equations we have used in the past?
 * 5) How does the sign of the upward/downward velocity change as the projectile goes in its path?
 * 1) The horizontal velocity will not change, so it is constant throughout. The vertical velocity can be found by using an equation.
 * 2) Not that we have learned yet. However, we can make rough estimates.
 * 3) The three equations we need to know are useful for displacement. For vertical displacement, the equation is y = .5gt^2. Horizontal displacement is x = vit. To find the vertical displacement for an angled projectile, the equation is y=(viy)(t) + .5gt^2.
 * 4) The vertical displacement equation is similar to d = vit + 1/2at^2. Because vi is 0 when it is essentially dropped in projectile problems, this makes sense. The x=vit equation is similar to a rewritten v = d/t equation. The vertical displacement for an angled projectile is exactly the same as d = vit +1/2at^2.
 * 5) For an angled projectile, the acceleration will change from negative to positive as it goes towards its max height and then away.

=__10/21/11__= //Class Notes://

Lab: Orienteering Part 2
//Other Group's Data:// Legs Resultant = 6311 cm Starting Position: Bottom Left of Football Field (at 0)
 * Distance || Direction || Direction ||
 * 1 || 3000 cm || East ||
 * 2 || 3000 cm || South ||
 * 3 || 1500 cm || East ||
 * 4 || 1500 cm || South ||
 * 5 || 960 cm || East ||
 * 6 || 276 cm || South ||
 * R Graphical = 7530cm**
 * R Analytical = 7256cm**

//Percent Error:// [(Theoretical - Experimental) / Theoretical] x 100 [(7530 - 6311) / 7530] x 100 =
 * Measured**
 * 16.19 %**

[(Theoretical - Experimental) / Theoretical] x 100 [(7256 - 6311) / 7256] x 100 =
 * Calculated**
 * 13.02 %**

=__10/24/11__=

**Ball in Cup (Part 1):**
//Objective://
 * How fast does the launcher shoot the ball at "medium range"? (Be sure to launch horizontally.)
 * After changing the initial height of the launcher, calculate where to place the cup on the floor so that ball lands inside of the cup 3 times in a row. (You may adjust the position of the cup.)
 * Calculate the %error of the theoretical position of the cup with the actual position of the cup.

//Lab://

=__10/28/11__=

Gourd-O-Rama
__Partner(s): Max Llewellyn__ //Objective:// - Create a cart, with the least possible weight, that will hold a gourd, which weighs at least 1kg, as it travles down a 2 foot wooden ramp and over a distance with the least acceleration possible.

//Materials://
 * Acorn Squash
 * Assorted Lego pieces
 * Duct tape

//Data:// Weight of Cart With Gourd: 1.09 kg Weight of Gourd: 1.01 kg Weight of Cart: .08 kg (80g)

//Results://
 * Best Run: 2m traveled, 1.69s**
 * Initial Velocity: 2.366 m/s**
 * Velocity at the Bottom of the Ramp: 3.45 m/s**
 * Total Acceleration: -1.30 m/s^2**

//Conclusion:// The biggest problem that my groups design had was that the wheels were too small, which caused it fall apart easily when it transferred the energy from vertical to horizontal motion. If they had been bigger, it would have survived the transfer and gotten some actual significant distance. Also, with such a small platform for the gourd to sit upon, it was hard to get a balanced center of gravity. Eventually, we got it to not fall apart and move some distance, although it was not much. However, the rest of our design was good. We had slick, straight axels that would have carried it farther if we had good sized wheels, and our design weighed almost nothing, doing a good job of satisfying the "with the least possible weight" part of the objective.

=__11/8/11__=

Shoot Your Grade Lab
//Objective: // Use your knowledge about projectiles to launch, using a launcher, a plastic ball through strategically placed rings and finally into a cup, with basically no background knowledge except what angle your launcher is at.

//Rationale:// As found out from our previous launcher activity, it is possible to get a reasonably accurate initial vertical velocity for our launcher. From here, we set up the rings of tape to be equal distances apart, allowing us to use a distance equation (∆d = vit + 1/2at^2) to find out the amount of time for the ball to pass through the center of each individual ring, as well as the vertical height needed for the ball to go through each of the tape rings.

//Materials and Methods:// Before we set up the rings, we were given an starting angle of 20 degrees, and had to find our launchers initial velocity. For this, we used regular paper and carbon paper, with the carbon paper being on top of the regular paper. What this does is when the carbon paper is struck, it leaves a clear mark on the regular paper that we can use for measurements. With this, we launch the ball and have it land on the two layers of paper. After five trials, we had five clear points to measure with. We found the distance from the launcher to the dots, and averaged out these distances. The average we calculate is our horizontal distance. Also, we measured the vertical distance from the launcher to the ground. With both vertical and horizontal distances, we were able to calculate the initial velocity. We were now ready to set up our rings. How we did it was we took two pieces of string and tied them onto the tape ring. From there, we ran the other end through the ceiling tiles, making it possible for us to change the vertical distance at will. The distance between rings was as evenly split as possible, with one or two exceptions since there was a light fixture that we could not move. From there, we continued with the experiment, finding what our exact vertical distances were. After much trial and error, adjusting our tape rings ever so slightly when necessary, we got the ball to go through the rings. Finally, we measured one more time since we had to move a few of our tape rings. Although our group never got to it because of time constraints and complications, we would have placed a cup at the horizontal distance that we expected it to land. After all the experimentation, we calculated our percent error, based off of our actual and theoretical values.

//Data and Observations (Initial Velocity)://
 * Dot || Distance (m) ||
 * 1 || 5.024 ||
 * 2 || 5.023 ||
 * 3 || 5.001 ||
 * 4 || 5.034 ||
 * 5 || 5.065 ||
 * Average || 5.03 ||
 * We found the vertical distance by measuring from the launcher to the ground, and we found the horizontal distance by averaging the dots on carbon paper. From there, using our knowledge of projectiles, we separated it into an x component and y component and preset the acceleration of the vertical component to -9.8m/s^s and the horizontal component to zero.
 * Vi = 6.73m/s
 * When we first did the launcher activity, we found that our initial velocity with the same launcher was 7.10 m/s. When we did it this time, we got a slower initial velocity because we are now at an angle, where the force of gravity is more of a presence. Although, when taking matters of percent error in hand, the values were relatively close to each other. Therefore, we knew we had a valid initial velocity.

//Performace:// media type="file" key="physics best trial.m4v"
 * Best Run: 4 rings**

//Calculations://
 * Days Performance:**
 * Trail Performance || Frequency of Trial Occurring ||
 * 4 rings (Best) || 2 times ||
 * 3 rings + hit cup || 2 times ||
 * 3 rings || 5 times ||
 * 1.27m was added to each distance to account for the distance from the launcher to the edge of the cabinets

//Data://
 * Rings || Horizontal Distance (x) || Time || Theoretical Vertical (y) || Actual Vertical (Y) || Percent Error ||
 * 1 || .88m || .14s || 1.49m || 1.40m || 6.00% ||
 * 2 || 1.455m || .23s || 1.54m || 1.46m || 5.20% ||
 * 3 || 2.172m || .34s || 1.48m || 1.35m || 8.80% ||
 * 4 || 2.865m || .45s || 1.31m || 1.26m || 3.80% ||
 * Our percent error was all under 10%, meaning that they are all relatively small. Many of the reasons why this happened will be explained in the conclusion below.

//Conclusion://  Although we did not fulfill every part of the hypothesis, the majority is still correct. We made it through 4 hoops, but did not get it in a cup as we stated we would. Yet our approach was quite successful. What stopped us from completing the lab more thoroughly, as in getting the ball through 5 hoops and into the cup, was a lack of time. This lack of time was due to the fact that we not only had to put each of the tape hoops up ourselves, but also because at the begging of every class, we had to readjust the vertical distances because people would walk into them and change the measurements. Therefore, it was not our approach that was faulty, but a lack of time that caused our procedure to be more difficult.  We were successfully able to find the vertical heights of our tape rings by using our knowledge of projectiles. We had small percent error values, all of which were under 10%. This shows that our theoretical calculations were correct. The reason we had percent error at all, and our experimental calculations were different from our theoretical calculations, was due to a number of reasons. Firstly, the launcher was inconsistent. The launcher works by using tension on a spring. Therefore, many problems that a spring-load design has are: if the spring is hot or cold, if the person using the device pulled the stinger softer or harder, or if the person did not pull the string straight up. If any of these occurred, the launch would be altered. The best way to deal with this problem would be to ensure we were being as consistent as possible with the launcher, as well as using multiple trials before changing the tape at all. Another issue was as we ended up firing the launcher some 50 times, the angle would sometimes be moved after a launch. It could move as severe a degree as 2 degrees. This, again, is a problem with consistency of the launcher that could not be fixed permanently. We would just have to make sure it was clamped down as hard as possible, hold it down while we launched it, and try to be as precise as possible. Finally, we ran into issues with our tape rings falling when other groups moved their tape. This was also a factor in our time constraint, as we had to make sure everything was consistent for our measurements to mean anything. For sure, we could have used a better system of attaching our tape rings to the ceiling. If we had multiple clips, we could have rigged a pulley system instead of just looping the string through. This system would have made our tape basically stay in the same place and not move, despite other groups. All of this would have lowered our percent error.  Projectile motion as discussed through this lab is applicable to circus performers. There is a very popular stunt performed where a person is loaded into a cannon, launched out of it, flung through flaming hoops, and then lands in a small pool of water. As we did in the lab, to do this properly and safely, the performer must be sure to know how fast they are going and at what angle they are being launched at. From there, they need to know how high each of their flaming hoops are in order for the performer to safely fly through them. Finally, they need to know how far they will go and with that, where to place the pool of water. For the performer and for us alike, the measurements have to be almost exact. For us, exactness means better results, for the performer, it means a spectacular stunt or being severely injured.