Ch3_AronsonJ

= Physics Classroom: Vectors--Motion and Forces = toc
 * Lesson 1-a.,b. **

//Topic Sentences//: This section teaches about vectors and direction. In the previous chapter, we represented motion (for the most part) without indicating direction. However, we will now concern ourselves with direction, as represented in vectors by many ways, including angle about the origin and straightforward direction. We will also use new methods to calculate and represent distance and direction.

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.
 * Vectors and Direction **

Examples of vector quantities that have been [|previously discussed] include [|displacement], [|velocity] , [|acceleration] , and [|force]. 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. Vector diagrams were introduced and used in earlier units to depict the forces acting upon an object. Such diagrams are commonly called as [|free-body diagrams]. There are several characteristics of this diagram that make it an appropriately drawn vector diagram. Vectors can be directed due East, due West, due South, and due North. But some vectors are directed //northeast// (at a 45 degree angle); and some vectors are even directed //northeast//, yet more north than east. The two conventions for describing the direction of a vector that will be discussed and used in this unit are described 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.
 * a scale
 * a vector arrow with //head// and a //tail//
 * the magnitude and direction
 * Conventions for Describing Directions of Vectors **
 * 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).
 * 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.
 * Representing the Magnitude of a Vector **

Two vectors can be added together to determine the result (or resultant). Recall in our discussion of Newton's laws of motion, that the //net force// experienced by an object was determined by computing the vector sum of all the individual forces acting upon that object. That is the [|net force] was the result (or [|resultant] ) of adding up all the force vectors. 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). 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 method is not applicable for adding more than two vectors or for adding vectors that are __not__ at 90-degrees to each other. The measure of an angle as determined through use of SOH CAH TOA is __not__ always the direction of the vector. Using a scaled diagram, the ** head-to-tail method ** is employed to determine the vector sum or resultant.
 * Vector Addition **
 * The Pythagorean Theorem **
 * Using Trigonometry to Determine a Vector's Direction **
 * Use of Scaled Vector Diagrams to Determine a Resultant **

A step-by-step method for applying the head-to-tail method to determine the sum of two or more vectors is given below.
 * 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.
 * 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).
 * 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.
 * Repeat steps 2 and 3 for all vectors that are to be added
 * 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**.
 * Using a ruler, measure the length of the resultant and determine its magnitude by converting to real units using the scale
 * Measure the direction of the resultant using the counterclockwise convention discussed [|earlier in this lesson].


 * Lesson 1-c.,d.**

//Topic Sentences:// This lesson teaches about adding vectors to achieve resultant vectors. We covered most of this material in class, basically reaching the following conclusion: the components of a vector can be added to determine the resultant vector.

The ** resultant ** is the vector sum of two or more vectors. If displacement vectors A, B, and C are added together, the result will be vector R. To say that vector R is the //resultant displacement// of displacement vectors A, B, and C is to say 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. That is why it can be said that ** A + B + C = R. **
 * Resultants **


 * Lesson 1-e.**

//Topic Sentence:// This lesson teaches the two methods, both of which we discussed and practiced in class, of determining vector direction and magnitude: the parallelogram (or graphical) method and trigonometric method.

The process of determining the magnitude of a vector is known as ** vector resolution **. The parallelogram method of vector resolution 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. A step-by-step procedure for using the parallelogram method of vector resolution is: The trigonometric method of vector resolution involves using trigonometric functions to determine the components of the vector. The method of employing trigonometric functions to determine the components of a vector are as follows:
 * Vector Resolution **
 * Parallelogram Method of Vector Resolution **
 * 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.
 * 4) Meaningfully label the components of the vectors with symbols to indicate which component represents which side. A northward force component might be labeled F north.
 * 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.
 * 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 tail of each component begins at the tail of the vector and stretches along the axes to the nearest corner of the rectangle.
 * 4) Meaningfully label the components of the vectors with symbols to indicate which component represents which side.
 * 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.


 * Lesson 1-g.,h.**

//Topic Sentence:// This lesson taught how to solve a vector problem with two moving forces, such as a plane flying with tailwinds.

The effect of the wind upon the plane is similar to the affect of the river current upon the motorboat. The resultant velocity of the boat is the vector sum of the boat velocity and the river velocity. 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. Thus, the [|Pythagorean theorem] can be used to determine the resultant velocity. The [|direction] of the resultant is the counterclockwise angle of rotation that the resultant vector makes with due East. Motorboat problems are typically accompanied by three 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). ** ave. speed = distance/time **  ** time = distance /(ave. speed) ** 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. Perpendicular components of motion do not affect each other.A change in the horizontal component does not affect the vertical component. This is what is meant by the phrase "perpendicular components of vectors are independent of each other." Changing a component will affect the motion in that specific direction. 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.
 * Analysis of a Riverboat's Motion **
 * 1) What are the resultant 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?
 * Independence of Perpendicular Components of Motion **


 * Lesson 2-a.,b.**

// Central Idea //** : ** A projectile is an object affected by gravity and no other forces. No matter the direction or speed in which a projectile travels, gravity is the only acting force (just ask Newton!).
 * 1) What is a projectile?
 * 2) A projectile is an object that is acted on by gravity only.
 * 3) What are the different kinds of projectiles?
 * 1) How is a free-body diagram represented?
 * 1) If an object is thrown into the air, then why is momentum not an acting force?
 * 2) According to Newton’s laws, forces are only required to cause acceleration, but not to keep an object in motion. After the object is released, it is acted upon by gravity only, and is a projectile.
 * 3) How is horizontal motion explained?
 * 4) According to Newton’s first law of motion, a projectile in motion (such as horizontal motion) will stay in motion unless acted upon by an unbalanced force. Gravity does not affect a projectile’s horizontal motion.

// Central Idea: // Projectiles can travel horizontally and vertically at the same time, each force being independent of the other. Gravity acts downward on projectiles, whereas constant velocity acts horizontally on projectiles.


 * 1) Why do projectiles travel in a parabolic shape?
 * 2) The downward force of gravity accelerates projectiles downward from their gravity-free trajectories, causing an increasing curve towards the ground.
 * 3) Do perpendicular components of motion affect each other?
 * 4) No, perpendicular components of motion are independent of each other; therefore, perpendicular components are discussed individually.
 * 5) If gravity does not act upon a horizontal projectile, then what force does act on it?
 * 6) Gravity acts downward on a projectile and cannot change its horizontal motion; therefore, a horizontal force causes horizontal acceleration. There is also vertical force acting upon projectiles. A projectile is acted upon by constant horizontal velocity and downward vertical acceleration.
 * || ** Horizontal Motion **  ||  ** Vertical Motion **  ||
 * ** Forces **  (Present? What direction?)  ||  No  ||  Yes; Gravity acts downward  ||
 * ** Acceleration **  (Present? What direction?)  ||  No  ||  Yes; “g” acts downward at 9.8 m/s2  ||
 * ** Velocity **  (Constant or changing?)  ||  Constant  ||  Changing by 9.8 m/s every second  ||
 * 1) What is the displacement of a projectile trajectory?
 * 2) The downward force and acceleration of a projectile result in a downward displacement from the gravity-free position.
 * 3) Does the vertical force affect the horizontal force?
 * 4) No, because the two forces are perpendicular; therefore, the vertical force is independent of the horizontal force.


 * Lesson 2-c.**
 * Central Idea: ** This section teaches about how to pictorially represent projectiles with vectors.


 * 1) In review, what are the eight concepts to memorize?
 * 2) A projectile is any object upon which the only force is gravity.
 * 3) Projectiles travel with a parabolic trajectory due to gravity.
 * 4) There are no horizontal forces acting upon projectiles, so also no horizontal acceleration.
 * 5) The horizontal velocity of a projectile is constant
 * 6) There is a vertical acceleration from gravity; its value is -9.8 m/s2.
 * 7) The vertical velocity of a projectile changes by 9.8 m/s each second.
 * 8) The horizontal motion of a projectile is independent of its vertical motion.
 * 9) How are projectiles represented pictorially, using vectors?
 * 1) How are the changes in vertical and horizontal motion represented?
 * 2) The horizontal motion of a projectile remains constant and unchanging, while the vertical motion changes at a rate of 9.8 m/s.
 * 3) What forces are acting on a projectile with both horizontal and vertical motion?
 * 4) Gravity is the only force acting upon projectiles, and it affects vertical motion. No horizontal force is present.
 * 5) Do projectiles indicate direction of velocity?
 * 6) Yes, projectiles indicate direction with + and – signs for magnitudes. If a magnitude has a positive (+) value, such as 10 m/s, then it is traveling upwards; if a magnitude has a negative (–) value, such as -10 m/s, then it is traveling downwards. The horizontal magnitude is constant.


 * Central Idea: ** This section teaches how to use the kinematics equations that we learned in the previous chapter to solve projectile problems.

Δdy=vit+½gt2 g=force of gravity (-9.8 m/s2) Δdx=vixt vix=initial horizontal velocity Δdy=vit+½gt2 -78.4=(0)(t)+(½)(-9.8)(t)2 -78.4=-4.9t2 16=t2
 * 1) Because vertical motion is dependent upon the force of gravity, can any of the kinematics equations describe vertical motion?
 * 1) Because horizontal movement is constant, can any equation describe it?
 * 1) How are these concepts represented graphically?
 * 1) What is the shape of a parabolic projectile graph?
 * 2) Projectiles are symmetrical parabolas. For example, if the maximum is at x(t=5) seconds, then the x(t=4)=x(t=6), x(t=3)=x(t=7), etc.
 * 3) Example problem
 * 4) A cannonball is launched horizontally from the top of a 78.4-meter high cliff. How much time will it take for the ball to reach the ground?
 * t=4 seconds **
 * 1) NOTE: the cannon ball’s horizontal speed does not affect the time to fall a vertical distance of 78.4 meters.

=**Displacement Lab**= //By Jake Aronson, Kaila Solomon, Ali Cantor and Jessica Smith// //October 19, 2011// Possible sources of error were imprecise measurement, due to slack in the measuring tape and/or inexact positions. In addition, the directions that we received may have been imprecise.
 * % Error**

=Ball in Cup Class Activity= //By Jake Aronson, Kaila Solomon, Ali Cantor and Jessica Smith//

media type="file" key="New Project - Medium.m4v" width="300" height="300" =
 * Video**
 * Medium-Range Ball Launch Calculations**
 * Ball in Cup Calculations**

= =Shoot Your Grade Lab= //By Jake Aronson, Kaila Solomon, Ali Cantor and Jessica Smith// //November 13, 2011//

Launcher, ball, carbon paper, masking tape, right angle clamps, plumb bomb, string
 * Materials**

Using kinematics equations, our calculations will tell us exactly where to hang each ring. The rings will be hung in a horizontally parabolic shape, because our angle of elevation is only 10 degrees.
 * Hypothesis**

We used the "Big 5" kinematics equations to calculate predicted heights off of the ground and distances from the launcher where we should hang the rings. We then measured those distances along the floor and down from the ceiling and hung one hoop at each designated point. However, sources of error caused the theoretical values to be slightly imprecise, so we had to slightly adjust each hoop as the ball did not clear its hole. Trial and error assumed responsibility for where to hang the rings! After manipulating each ring, we finally achieved positions that allowed the ball to shoot through all five rings (our video shows only four, but we scored a fifth that was out of the camera's line of sight!). Enjoy the video!
 * Procedure**

media type="file" key="4 hoops lol.m4v" width="300" height="300"
 * Video**


 * Calculations**

//Data:// //% Error:// Initially, our group hypothesized that our calculations would measure the exact horizontal and vertical distances for the locations to hang our rings. However, after the launches revealed that some error had affected our theoretical distances, we had to readjust. As is evident from the above table, the percent error was 0.06% for the first ring, 0.21% for the second, 0.14% for the third, 0% for the fourth and 3.4% for the fifth. All told, the sources of error affected our calculations only slightly, and not at all for the fourth ring. The small percents error suggest that our calculations were fairly accurate and precise, which allowed us to launch our ball through all five rings (though Ms. Burns witnessed it go through only four).
 * Analysis**
 * Conclusion**

A few possible sources of error were imprecise measurement with the flimsy measuring tape, movement of the rings due to the ventilation, shifting of the string out of its place in the ceiling, change in the angle of elevation of the launcher and movement of the launcher along the counter. If we had the opportunity to perform the lab again, we could use a more taught measuring tape; turn off the air conditioning in the classroom; staple the string into the ceiling, or just have enough space for every group in all of the classes to perform its own lab; check the angle of elevation more precisely; and adjust the clamp to make sure that the launcher would not shift. We learned that theoretical results are not always achieved, so trial and error is often useful to obtain the best possible results. In addition, we learned the value of attention to detail.

=Gourd-o-Rama Project= //By Jake Aronson and Michael Solimano//

Our project did not travel as far as we had hoped that it would have, but we believe that we can improve our design. First, rather than use a flimsy axel with tennis balls, we would use a smoother axel with wheels that would cause less friction with the ground. For example, we could use roller blade wheels, which are specifically designed to minimize friction. Next, we would make sure that our project was balanced after rolling down the incline. Because our pumpkin's weight made the project "backheavy," we would try to put enough weight on the front axel so as to keep the project balanced and momentum forward. Finally, we would decorate our project with images of the "Batmobile"...after all, who better wears Halloween costumes and masters the laws of physics than Batman? toc
 * Final Project**
 * Mass:** 2.0 g
 * Calculations:**