Ch3_Doomank

=Chapter 3 :) = toc

Honors Physics Rocks Wiki
= Homework 10/12/11 =
 * Vectors Lessons 1**

**Part A** Vectors and Direction
 * A vector quantity is a quantity that is fully described by both magnitude and direction
 * Examples of vector quantities include [|displacement], [|velocity], [|acceleration], and [|force].
 * Each of these quantities are unique in that a full description of the quantity demands that both a magnitude and a direction are listed.
 * Ex: Your teacher tells you "A bag of gold is located outside the classroom. To find it, displace yourself 20 meters."
 * This statement may provide yourself enough information to pique your interest; yet, there is not enough information included in the statement to find the bag of gold.
 * The displacement required to find the bag of gold has not been fully described.
 * On the other hand, suppose your teacher tells you "A bag of gold is located outside the classroom. To find it, displace yourself from the center of the classroom door 20 meters in a direction 30 degrees to the west of north."
 * This statement now provides a complete description of the displacement vector - it lists both magnitude (20 meters) and direction (30 degrees to the west of north) relative to a reference or starting position (the center of the classroom door). Vector quantities are not fully described unless both magnitude and direction are listed.
 * [[image:http://www.physicsclassroom.com/Class/vectors/u3l1a3.gif width="223" height="234" align="right"]]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]: depict forces acting on an object.
 * An example of a scaled vector diagram is shown in the diagram at the right.
 * The vector diagram depicts a displacement vector.
 * A good vector diagram has:
 * 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).

**Conventions for Describing Directions of Vectors** Two illustrations of the second convention (discussed above) for identifying the direction of a vector are shown below.
 * [[image:http://www.physicsclassroom.com/Class/vectors/u3l1a5.gif width="160" height="158" align="right"]]Vectors can be directed due East, due West, due South, and due North.
 * Some vectors are directed //northeast// (at a 45 degree angle); and some vectors are even directed //northeast//, yet more north than east.
 * There are a variety of conventions for describing the direction of any vector.
 * 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.
 * 2) 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).
 * 3) The direction of a vector is often expressed as a counterclockwise angle of rotation of the vector about its "[|tail]" from due East.
 * 4) 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.
 * 5) 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.
 * 6) 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.

Observe in the first example that the vector is said to have a direction of 40 degrees.
 * You can think of this direction as follows: suppose a vector pointing East had its [|tail] pinned down and then the vector was rotated an angle of 40 degrees in the counterclockwise direction.
 * Observe in the second example that the vector is said to have a direction of 240 degrees. This means that the tail of the vector was pinned down and the vector was rotated an angle of 240 degrees in the counterclockwise direction beginning from due east.
 * A rotation of 240 degrees is equivalent to rotating the vector through two quadrants (180 degrees) and then an additional 60 degrees //into the// third quadrant.

**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. For example, the diagram at the right shows a vector with a magnitude of 20 miles.
 * Since the scale used for constructing the diagram is __1 cm = 5 miles__, the vector arrow is drawn with a length of 4 cm. That is, 4 cm x (5 miles/1 cm) = 20 miles.
 * 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.
 * A 25-mile displacement vector is represented by a 5-cm long vector arrow.
 * An 18-mile displacement vector is represented by a 3.6-cm long arrow.



**Part B**
 * Two vectors can be added together to determine the result (or resultant)
 * Ex:

**The Pythagorean Theorem** >> //Eric leaves the base camp and hikes 11 km, north and then hikes 11 km east. Determine Eric's resulting displacement.// **Using Trigonometry to Determine a Vector's Direction**
 * These rules for summing vectors were applied to [|free-body diagrams] in order to determine the net force (the vector sum of all the individual forces).
 * Examples of Applications:
 * A vector directed up and to the right will be added to a vector directed up and to the left.
 * The //vector sum// will be determined for the more complicated cases shown in the diagrams below.
 * The two methods that will be used throughout the entire unit are:
 * the Pythagorean theorem and trigonometric methods
 * [|the head-to-tail method using a scaled vector diagram]
 * 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:
 * 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.
 * Since the northward displacement and the eastward displacement are at right angles to each other, the Pythagorean theorem can be used to determine the resultant (i.e., the hypotenuse of the right triangle).
 * The result of adding 11 km, north plus 11 km, east is a vector with a magnitude of 15.6 km.
 * The direction of a //resultant// vector can often be determined by use of trigonometric functions.
 * SOH CAH TOA (sine, cosine, and tangent functions)
 * These three functions relate an acute angle in a right triangle to the ratio of the lengths of two of the sides of the right triangle.
 * **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
 * **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
 * **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
 * These three trigonometric functions can be applied to the [|hiker problem] in order to determine the direction of the hiker's overall displacement.
 * 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. Thus, the direction of this vector is written as 45 degrees. (Recall from [|earlier in this lesson] that the direction of a vector is the counterclockwise angle of rotation that the vector makes with due East.)
 * **The measure of an angle as determined through use of SOH CAH TOA is __not__ always the direction of the vector.**
 * Ex: The angle within the triangle is determined to be 26.6 degrees using SOH CAH TOA. This angle is the southward angle of rotation that the vector R makes with respect to West. Yet the direction of the vector as expressed with the CCW (counterclockwise from East) convention is 206.6 degrees.

**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.
 * Example:
 * 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.


 * head-to-tail method involves [|drawing a vector to scale] on a sheet of paper beginning at a designated starting position
 * Where the head of this first vector ends, the tail of the second vector begins (thus, //head-to-tail// method)
 * The process is repeated for all vectors that are being added
 * Once all the vectors have been added head-to-tail, the resultant is then drawn from the tail of the first vector to the head of the last vector; i.e., from start to finish
 * Once the resultant is drawn, its length can be measured and converted to //real// units using the given scale
 * The [|direction] of the resultant can be determined by using a protractor and measuring its counterclockwise angle of rotation from due East.

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.

The problem shown involves the addition of three vectors: 20 m, 45 deg. + 25 m, 300 deg. + 15 m, 210 deg.SCALE: 1 cm = 5 m SCALE: 1 cm = 5 m
 * The head-to-tail method is employed as described above and the resultant is determined (drawn in red). Its magnitude and direction is labeled on the diagram.


 * 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.
 * For example, consider the addition of the same three vectors in a different order.
 * 15 m, 210 deg. + 25 m, 300 deg. + 20 m, 45 deg.****SCALE: 1 cm = 5 m**

SCALE: 1 cm = 5 m
 * When added together in this different order, these same three vectors still produce a resultant with the same magnitude and direction as before (20. m, 312 degrees).
 * The order in which vectors are added using the head-to-tail method is insignificant.

= Homework 10/13/11 =


 * Part C **
 * Resultants**


 * **resultant** is the vector sum of two or more vectors
 * //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].


 * vector R is the //resultant displacement// of displacement vectors A, B, and C= saying that a person who walked with displacements A, then B, and then C would be displaced by the same amount as a person who walked with displacement R.
 * Displacement vector R gives the same //result// as displacement vectors A + B + C.


 * 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.
 * If two or more velocity vectors are added, then the result is a //resultant velocity//.
 * If two or more force vectors are added, then the result is a //resultant force//.
 * If two or more momentum vectors are added, then the result is //momentum force//


 * In all such cases, the resultant vector (whether a displacement vector, force vector, velocity vector, etc.) is the result of adding the individual vectors.
 * [[image:http://www.physicsclassroom.com/Class/vectors/u3l1c5.gif width="148" height="107" align="right"]]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.
 * To experience the three forces A, B and C is the same as experiencing force R.
 * To be hit by players A, B, and C would result in the same force as being hit by one player applying force R.

**Lesson D**


 * 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.
 * 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.
 * 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.
 * Each part of a two-dimensional vector is known as a **component**.
 * components of a vector depict the influence of that vector in a given direction
 * combined influence of the two components is equivalent to the influence of the single two-dimensional vector
 * The single two-dimensional vector could be replaced by the two components.


 * If Fido's dog chain is stretched upward and rightward and pulled tight by his master, then the tension force in the chain has two components - an upward component and a rightward component.
 * To Fido, the influence of the chain on his body is equivalent to the influence of two chains on his body - one pulling upward and the other pulling rightward.
 * If the single chain were replaced by two chains. with each chain having the magnitude and direction of the components, then Fido would not know the difference.
 * The combined influence of the two components is equivalent to the influence of the single two-dimensional vector.




 * 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.
 * The wire on the left has a leftward and an upward component. (it 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.

= Homework 10/17/11 =
 * 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**
 * method involves drawing the vector to scale in the indicated direction, sketching a parallelogram around the vector such that the vector is the diagonal of the parallelogram, and determining the magnitude of the components (the sides of the parallelogram) using the scale
 * 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.


 * 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; the sketched lines will meet to form a rectangle (a special case of a parallelogram).
 * 3) Draw the components of the vector. The components are the //sides// of the parallelogram. The tail of the components start at the tail of the vector and stretches along the axes to the nearest corner of the parallelogram. Be sure to place arrowheads on these components to indicate their direction (up, down, left, right).
 * 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. A rightward velocity component might be labeled vx; etc.
 * 5) Measure the length of the sides of the parallelogram and [|use the scale to determine the magnitude] of the components in //real// units. Label the magnitude on the diagram.

*The diagram shows that the vector is first [|drawn to scale] in the indicated direction

**Trigonometric Method of Vector Resolution**
 * 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. Beginning at the [|tail] of the vector, sketch vertical and horizontal lines. Then sketch horizontal and vertical lines at the [|head] of the vector. The sketched lines will meet to form a rectangle.
 * 3) Draw the components of the vector. The components are the //sides// of the rectangle. The tail of each component begins at the tail of the vector and stretches along the axes to the nearest corner of the rectangle. Be sure to place arrowheads on these components to indicate their direction (up, down, left, right).
 * 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. A rightward force velocity component might be labeled vx; etc.
 * 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. Use some algebra to solve the equation for the length of the side opposite the indicated angle.
 * 6) Repeat the above step using the cosine function to determine the length of the side adjacent to the indicated angle.

Vector Addition Lab/Activity
Lab Partners: Julie and Lindsey

Percent Error Calculations: The percent error for the theta calculations was actually very good, meaning that it was low, only 2.84% error. However, the calculations for the R (resultant) value had a percent error of almost 12% which is not as good. However, it also is not absolutely terrible either. The mistakes could have come from simple human error (not drawing accurately enough, misreading the ruler, or even misreading data). However, the data was also confusing, as the group who mapped the path did not go at right angles, giving us directions to go west three times in a row. Though this simply meant that we continued in a straight line, it also meant that the line was longer and therefore it was a bit more difficult to find a scale that would fit all the data, which could have also led to mistakes in calculations, causing a higher percent error. Next time, I would make sure to get a set of directions going at right angles, and have someone else double check measurements just in case someone misread them.

= Homework 10/18/11 = 
 * Part G **
 * Relative Velocity and Riverboat Problems**
 * On occasion objects move within a medium that is moving with respect to an observer.
 * Ex: an airplane usually encounters a wind - air that is moving with respect to an observer on the ground below
 * 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.
 * 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
 * The observer on land, often named (or misnamed) the "stationary observer" would measure the speed to be different than that of the person on the boat.
 * // The observed speed of the boat must always be described relative to who the observer is. //

Consider a plane flying amidst a **tailwind**. The algebraic steps are as follows: (100 km/hr)2 + (25 km/hr)2 = R2 10 000 km2/hr2 + 625 km2/hr2 = R2  10 625 km2/hr2 = R2  SQRT(10 625 km2/hr2) = R  **103.1 km/hr = R**
 * A tailwind is merely a wind that approaches the plane from behind, thus increasing its resulting velocity.
 * If the plane is traveling at a velocity of 100 km/hr with respect to the air, and if the wind velocity is 25 km/hr, then what is the velocity of the plane relative to an observer on the ground below?
 * The resultant velocity of the plane (that is, the result of the wind velocity contributing to the velocity due to the plane's motor) is the vector sum of the velocity of the plane and the velocity of the wind.
 * This resultant velocity is quite easily determined if the wind approaches the plane directly from behind.
 * The plane travels with a resulting velocity of 125 km/hr relative to the ground.
 * If the plane encounters a headwind, the resulting velocity will be less than 100 km/hr.
 * Since a headwind is a wind that approaches the plane from the front, such a wind would decrease the plane's resulting velocity.
 * Suppose a plane traveling with a velocity of 100 km/hr with respect to the air meets a headwind with a velocity of 25 km/hr.
 * The resultant velocity would be 75 km/hr; this is the velocity of the plane relative to an observer on the ground.
 * A plane traveling with a velocity of 100 km/hr, South that encounters a **side wind** of 25 km/hr, West.
 * The resulting velocity of the plane is the vector sum of the two individual velocities.
 * To determine the resultant velocity, the plane velocity (relative to the air) must be added to the wind velocity.

The tangent function can be used; this is shown below: tan (theta) = (opposite/adjacent) tan (theta) = (25/100) theta = invtan (25/100) **theta = 14.0 degrees** If the resultant velocity of the plane makes a 14.0 degree angle with the southward direction (theta in the above diagram), then the direction of the resultant is 256 degrees.

**Analysis of a Riverboat's Motion** (4.0 m/s)2 + (3.0 m/s)2 = R2 16 m2/s2 + 9 m2/s2 = R2  25 m2/s2 = R2  SQRT (25 m2/s2) = R  **5.0 m/s = R**
 * 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.
 * The motorboat may be moving with a velocity of 4 m/s directly across the river, yet the resultant velocity of the boat will be greater than 4 m/s and at an angle in the downstream direction.
 * While the speedometer of the boat may read 4 m/s, its speed with respect to an observer on the shore will be greater than 4 m/s.
 * Since the boat heads straight across the river and since the current is always directed straight downstream, the two vectors are at right angles to each other.
 * The [|Pythagorean theorem] can be used to determine the resultant velocity.
 * Suppose that the river was moving with a velocity of 3 m/s, North and the motorboat was moving with a velocity of 4 m/s, East. What would be the resultant velocity of the motorboat (i.e., the velocity relative to an observer on the shore)?

tan (theta) = (opposite/adjacent) tan (theta) = (3/4) theta = invtan (3/4) **theta = 36.9 degrees**
 * [[image:http://www.physicsclassroom.com/Class/vectors/u3l1f7.gif width="105" height="53" align="right"]]The [|direction] of the resultant is the counterclockwise angle of rotation that the resultant vector makes with due East. This angle can be determined using a trigonometric function as shown below.

Given a boat velocity of 4 m/s, East and a river velocity of 3 m/s, North, the resultant velocity of the boat will be 5 m/s at 36.9 degrees.

Motorboat problems such as these are typically accompanied by three separate questions: The second and third of these questions can be answered using the [|average speed equation]. **ave. speed = distance/time** Consider the following example.  The solution to the first question has already been; the resultant velocity of the boat is 5 m/s at 36.9 degree
 * 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?
 * 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? ||

**time = distance /(ave. speed)** 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** **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.
 * 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.
 * 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?
 * The value of 5 m/s is the speed at which the boat covers the diagonal dimension of the river.
 * And the diagonal distance across the river is not known in this case.
 * 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
 * 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.
 * During this 20 s of crossing the river, the boat also drifts downstream.
 * "What distance downstream does the boat reach the opposite shore?"
 * The same equation must be used to calculate this //downstream distance//.
 * 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 (i.e., the speed at which the current moves - 3 m/s).
 * The average speed of 3 m/s (average speed in the downstream direction) should be substituted into the equation to determine the distance.

**Part H**
 * Independence of Perpendicular Components of Motion**
 * A force vector that is directed upward and rightward has two parts - an upward part and a rightward part.
 * These two parts of the two-dimensional vector are referred to as [|components].
 * A **component** describes the affect of a single vector in a given direction.
 * Any force vector that is exerted at an angle to the horizontal can be considered as having two parts or components.
 * // The vector sum of these two components is always equal to the force at the given angle. //

**d = v • t** So **t = d / v** (60 m) / (3 m/s) **20 seconds**
 * 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.
 * Ex: If the horizontal pull upon Fido increases, then Fido would be accelerated at a greater rate to the right; yet this greater horizontal pull would not exert any vertical influence upon Fido.
 * Pulling horizontally with more force does not lift Fido vertically off the ground.
 * //A change in the horizontal component does not affect the vertical component//--> "perpendicular components of vectors are independent of each other."
 * A change in one component does not affect the other component.
 * While the change in one of the components will alter the magnitude of the resulting force, it does not alter the magnitude of the other component.
 * The resulting motion of a plane flying in the presence of a crosswind is the combination (or sum) of two simultaneous velocity vectors that are perpendicular to each other.
 * Suppose that a plane is attempting to fly northward from Chicago to the Canada border by simply directing the plane due northward.
 * If the plane encounters a crosswind directed towards the west, then the resulting velocity of the plane would be northwest.
 * The northwest velocity vector consists of two components - a north component resulting from the plane's motor (the //plane velocity//) and a westward component resulting from the crosswind (the //wind velocity//).
 * These two components are independent of each other.
 * An alteration in one of the components will not affect the other component.
 * Ex: if the wind velocity increased, then the plane would still be covering ground in the northerly direction at the same rate. It is true that the alteration of the wind velocity would cause the plane to travel more westward; however, the plane still flies northward at the same speed.
 * Air balloon descending through the air toward the ground in the presence of a wind that blows eastward.
 * Suppose that the downward velocity of the balloon is 3 m/s and that the wind is blowing east with a velocity of 4 m/s.
 * The resulting velocity of the air balloon would be the combination (i.e., the vector sum) of these two simultaneous and independent velocity vectors.
 * The air balloon would be moving downward and eastward.
 * If the wind velocity increased, the air balloon would begin moving faster in the eastward direction, but its downward velocity would not be altered.
 * If the balloon were located 60 meters above the ground and was moving downward at 3 m/s, then it would take a time of 20 seconds to travel this vertical distance.
 * With a wind speed of 4 m/s, the distance traveled eastward in 20 seconds would be 80 meters.
 * If the wind speed increased from the value of 4 m/s to a value of 6 m/s, then it would still take 20 seconds for the balloon to fall the 60 meters of downward distance.
 * A motion in the downward direction is affected only by downward components of motion.
 * An alteration in a horizontal component of motion (such as the eastward wind velocity) will have no affect upon vertical motion.
 * A variation of the eastward wind speed from a value of 4 m/s to a value of 6 m/s would only cause the balloon to be blown eastward a distance of 120 meters instead of the original 80 meters.

**d = v • t** So **t = d / v** (60 m) / (4 m/s) **15 seconds**
 * Suppose that the boat velocity is 4 m/s and the river velocity is 3 m/s.
 * The magnitude of the resultant velocity could be determined to be 5 m/s using the Pythagorean Theorem.
 * The time required for the boat to cross a 60-meter wide river would be dependent upon the boat velocity of 4 m/s. It would require 15 seconds to cross the 60-meter wide river.
 * The southward river velocity will not affect the time required for the boat to travel in the eastward direction.
 * If the current increased such that the river velocity became 5 m/s, then it would still require 15 seconds to cross the river.
 * An increase in the river velocity would simply cause the boat to travel further in the southward direction during these 15 seconds of motion.
 * An alteration in a southward component of motion only affects the southward motion.

= Homework 10/19/11 = Lesson 2 **Part A** 1. What is a projectile? 2. How do projected objects travel? 3. How can you draw diagrams of motion of a projectile? >
 * A projectile is an object upon which the only force acting is gravity.
 * Ex: An object that is thrown vertically upward (as long as air-resistance is negligible)
 * They move in two dimensions.
 * When dropped or throw, etc., it continues in motion by its own inertia
 * A free-body diagram f a projectile would show a single force acting downwards and labeled force of gravity (Fgrav)
 * Ex:

4. If an object is moving upwards then isn't there an upward force? 5. Does gravity affect how motion/how far projectiles travel? **Theme:** A projectile is any object that once //projected// or dropped continues in motion by its own inertia and is influenced only by the downward force of gravity.
 * No; forces are only required to cause/maintain acceleration. A force is __not__ required to keep an object in motion. In the case of a projectile that is moving upward, there is a downward force and a downward acceleration. That is, the object is moving upward and slowing down
 * No; gravity causes a vertical acceleration. An object will drop vertically below its otherwise straight-line, inertial path. Gravity is the downward force upon a projectile that influences its vertical motion and causes the "parabolic trajectory" that is typical of projectiles.

** Part B ** 1. What kinds of motion do projectiles have? 2. Can the types of motion be discussed together? 3. Does acceleration due to gravity (-9.8 m/s 2 ) apply to projectiles? 4. What happens when an object is shot upwards instead of just dropped from a height? 5. How is the horizontal distance affected when a projectile is shot upward? > **Theme:** Projectiles have upward/downward AND horizontal motion, but they must be discussed separately.
 * They move upward or downward they are also moving horizontally; there are the two components of the projectile's motion - horizontal and vertical motion.
 * Perpendicular components of motion are independent of each other; the horizontal and vertical components of motion MUST be discussed separately.
 * Yes it does. However, it does not affect the horizontal motion (as mentioned in the question above).
 * The projectile would travel with a //parabolic// //trajectory// (the shape of the path would look like a parabola due to the downward force of gravity). The downward force of gravity would act upon the cannonball to cause the same vertical motion as before - a downward acceleration
 * The projectile still moves the same horizontal distance in each second of travel as it would if it were not under the influence of gravity. The force of gravity is a vertical force and does not affect horizontal motion; perpendicular components of motion are independent of each other

= Homework 10/20/11 =

1. How do you describe a projectile with numbers? 2. How would you draw a diagram to represent this? >
 * Part C **
 * You use vectors to represent the motion. Make sure to remember that the two motions (horizontal and vertical) are independent and therefore need to be discussed separately.
 * Ex: A cannonball is launched horizontally with no upward angle whatsoever and with an initial speed of 20 m/s. Gravity causes the cannonball to accelerate downwards at a rate of 9.8 m/s/s; the vertical velocity is changing by 9.8 m/s every second.
 * If a vector diagram (showing the velocity of the cannonball at 1-second intervals of time) is used to represent how the x- and y-components of the velocity of the cannonball is changing with time, then x- and y- velocity vectors could be drawn and label the magnitudes

3. How would you represent horizontal velocity? 4. What if the projectile was shot upwards? How would that be represented? 5. How can we calculate vertical and horizontal displacement?
 * Since the horizontal velocity remains constant during the travel of the projectile, you can just write it multiple times, like shown in the diagram above. If you are representing it in a table, you would just write the same horizontal velocity for each change in vertical velocity.
 * The horizontal velocity would remain the same. The vertical velocity would have some initial velocity that would slowly decrease to 0 when it reached the maximum height, before falling back down towards the ground. In a table, you could write this by showing the initial velocity and the direction (up), then writing the next (decreasing) velocities in the same direction (up). Once the velocity reached 0, a direction would not be included, but after that, you would write the (now decreasing) velocities and their new direction (down).
 * Vertical displacement can be found using this equation:
 * y = 0.5 • g • t2
 * g is acceleration due to gravity (-9.8 m/s 2 )
 * t is the time the projectile has been traveling vertically
 * Horizontal displacement can be found using this equation:
 * x = vix • t
 * vix is the speed at which the projectile moves horizontally
 * t is time that it has been moving horizontally
 * Theme:** There are multiple ways in which horizontal and vertical motion of projectiles can be represented, whether that is through a diagram or through equations.

Ball In Cup Lab/Activity
Lab Partners: Julie and Lindsey

__Launch | Position (cm)__ __1 | 275.7__ __2 | 280.2__ __3 | 281.6__ __4 | 283.4__ __5 | 285.2__ __6 | 290.2__
 * Objective:** Find the position to place a cup, so that a ball can be launched directly into it 3 times.
 * Materials:** Launcher, ball, cup, tape measure
 * Data:**

The percent error, 8.03% was not amazing, but it was not terrible either. The sources of error came from mainly human mistakes. For example, someone could have pushed the launcher so that it was turned in a different direction, or pushed it forward or backward. The angle at which the launcher was positioned could also have been messed up by another group using the launcher. Additionally we simply could have measured the distance wrong, forgetting to take 10 cm off the total (we started the measuring tape at 10 cm instead of at 0) or just misreading the numbers. To fix this, we could be more careful, have two people check the distance measured, and we could have re-checked the alignment/placement of the launcher to make sure that it was consistent with what we had measured the day we started the activity. media type="file" key="Movie on 2011-10-25 at 13.07.mov" width="300" height="300" media type="file" key="Movie on 2011-10-24 at 12.42.mov" width="300" height="300"
 * Calculations:**

Gourd-O-Rama Lab
Lab Partner: Maggie L. Front View:

Back View:

Side-View:


 * Calculations**:



If I were to re-do this experiment, I would make sure to glue the wheels on straight. Also, I would use lighter materials, such as a shoebox instead of a piece of plywood, which is much heavier. Additionally, I would secure the pumpkin (or gourd) onto the cart so that its falling off or instability would not cause the cart to crash. This could be achieved by either taping, gluing, or even nailing the pumpkin onto the cart.
 * Analysis:** Our cart went a fairly good distance (5 meters), considering the minimum distance it needed to travel was 1 meter. However, it kept veering off the the right. After repositioning the ramp, we ruled out ramp position as the cause of this. Instead, we believe that the axels of the wheels were not glued on completely straight, which would cause the cart to veer off. Our acceleration was also fairly low, which was consistent with the goals of this experiment.

Shoot Your Grade Lab
Lab Partners: Julie and Lindsey


 * Objective**: To get a ball launched from a 15 degree angle (on setting 2, medium range) through a series of five hoops placed relatively equal distances apart, then get the ball into a cup after passing through the hoops.


 * Materials and Methods**: The purpose of this lab was to gain more experience and apply the physics behind the projectile launch class problems to a real situation. In our experiment, the launcher we used was set at a 15 degree angle. The purpose of the launcher was to shoot a small yellow ball through 5 hoops made of tape, then at the end, the ball should (hopefully) land in a plastic cup. To set up the hoops so that the ball would go through them, it was necessary to use calculations from the initial velocity components were. These were used to determine the exact position of the projectile at 6 time intervals (hoops 1-5 and the cup). These measurements were then used to place the hoops at their specific locations and than place the cup in its designated position so that the ball would fall into it. After a series of "test-runs" we were able to adjust the positions of the cups so that we would be able to successfully get the ball through.

Assigned Angle- 15 degrees Measured Vertical Distance- -1.16m Average Horizontal Distance- 2.74m

media type="file" key="physics+vid2.mov" width="300" height="300" This video shows one of the trials for the lab.

__**Calculations**__

To fine the (vertical) y-distance, our group measured the distance from the launcher to the floor. Then we found the x-distance by using the average horizontal distance, which we calculated after doing a few test launches. Considering that this lab is deals with projectiles and their motion, we know that the x- acceleration will always be 0 m/s 2 while the y-acceleration will be -9.8 m/s 2. The angle of the launcher, theta, was given to us; 15 degrees. Using the information that we already knew, we were able to solve for the total time of the projectile. Then, we used the total time to find the initial velocity.
 * Initial Velocity**:


 * Example Calculations for Hoop Position**:

**Results from Calculations for Actual Ring Positions** Ring 1- (.45m, 1.3m) Ring 2- (.81m, 1.24m) Ring 3- (1.38m, 1.09m ) Ring 4- (1.67m, .99m ) Ring 5- (2.24m, .52m )

**Percent Error Sample Calculation:** __ l theoretical - experimental l __ x 100 theoretical

__ l 1.23 - 1.3 l __ x 100= 5.70% 1.23

**Percent Error:** If we were to do this lab again I would consider doing things differently. I would definitely use a different launcher because the angle of the launcher changed every time we shot the projectile. Additionally, I would fine a better way to secure the string to the ceiling, making sure that it would not fall or change its angle.
 * Conclusion**: The theoretical calculations for the hoops' position were not correct. We had to move the hoops multiple times to get the ball to go through each. This can be seen when looking in the percent error which shows that our theoretical calculations were wrong when looked at in relation to the actual positions. One possible source of error could be due to the inconsistency of the launcher; it tended to shift left or right every time we shot it. This means that the ball wasn't being shot straight ahead, but instead to one side. Human error should also be considered; it is possible that when calculating the positions, or even the initial velocity, we may have read the numbers on the measuring tape wrong. Additionally, the tape tended to come off, making the hoops drop, messing up our calculations. If a group had used clamps to keep the string in position, it is possible that they may have shifted and changed angles, again altering the hoops' position.