Ch3_KostmanL

=__**Chapter 3- Vectors: Motion and Forces in Two Dimensions**__= toc Lauren Kostman October 12, 2011 Homework

Lesson 1: a, b
(Online Physics Classroom) __Topic Sentence: vectors can be portrayed by a scaled vector diagram, including vector arrows (for magnitude, direction).__ Study of motion involves introduction of a variety of quantities (distance, displacement, speed, velocity, force, mass, momentum, energy, work, power, etc.) that are used to describe the physical world. These quantities can by divided into two categories - [|vectors and scalars]. A vector quantity is fully described by both magnitude and direction. A scalar quantity is fully described by its magnitude. Vector quantities are often represented by scaled [|vector diagrams], which 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]. The vector diagram above depicts a displacement vector. There are several characteristics of this diagram that make it an appropriately drawn vector diagram. Vectors can be directed due: East, West, South, and North. Some vectors are directed //northeast// (at a 45 degree angle); some are directed //northeast//, yet more north than east. There is a 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 ways for describing the direction of any vector, such as: 2. Direction of a vector is often expressed as a counterclockwise angle of rotation of the vector about its "__ tail __" from due East. Ex: vector with a direction of 30 degrees has been rotated 30 degrees in a counterclockwise direction relative to due east. Illustration of the second convention: 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. Ex: diagram below shows a vector with a magnitude of 20 miles. Since scale is __1 cm = 5 miles__, the vector arrow is drawn with a length of 4 cm (4 cm x (5 miles/1 cm) = 20 mi). Two methods for determining magnitude and direction of the result of adding vectors:
 * 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. The diagram shows the magnitude is 20 m and the direction is (30 degrees West of North).
 * Conventions for Describing Directions of Vectors **
 * 1) Direction of a vector is an angle of rotation of the vector about its "__ tail __" from N, S, E, W. Ex: vector has direction of 40 degrees N of W (vector pointing W is rotated 40 degrees towards N)
 * Representing the Magnitude of a Vector **
 * Pythagorean theorem and trigonometric methods
 * __ Head-to-tail method using a scaled vector diagram __

Used to determine the result of adding **__only__** two vectors __that make a right angle (90 degrees)__. The direction of a //resultant// vector can often be determined by use of trigonometric functions (SOH CAH TOA). These three trigonometric functions can be applied to the __ hiker problem __ in order to determine the direction of the hiker's overall displacement. Chose one of the two angles and apply the functions.
 * The Pythagorean Theorem **
 * Using Trigonometry to Determine a Vector's Direction **

Use of Scaled Vector Diagrams to Determine a Resultant The magnitude and direction of the sum of two or more vectors (that don’t make right angles) can also be determined by use of an accurately drawn scaled vector diagram. Using this, the head-to-tail method determines the vector sum or resultant. This method involves [|drawing a vector to scale] beginning at a designated starting position. Where the head of this first vector ends, the tail of the second vector begins. The process is repeated for all vectors that are being added. Once all vectors are added head-to-tail, the resultant is drawn from the tail of the first vector to the head of the last vector (start to finish). Then, length can be measured and converted to //real// units using the given scale. [|Direction] of the resultant can be determined by using a protractor and measuring its counterclockwise angle of rotation from due East. Example: 20 m, 45 deg. + 25 m, 300 deg. + 15 m, 210 deg. SCALE: 1 cm = 5 m

Lesson 1: c, d
Lauren Kostman October 13, 2011 Homework Resultants __ Topic Sentence: The sum of 2 or more vectors is equal to the resultant. __ Resultant= the vector sum (result) of two or more vectors. If displacement vectors A, B, and C are added together, the result will be vector R. As seen below, vector R can be determined by the use of a scaled vector addition diagram.

Example: a football player gets hit simultaneously by three players on the opposing team (players A, B, and C). He experiences three different applied forces, each contributing to a total or resulting force. If the three forces are added together using methods of vector addition, then the resultant vector R can be determined. Here, to experience the three forces A, B and C is the same as experiencing force R. Vector Components __ Topic Sentence: Vectors that move in both horizontal and vertical directions have components, or parts, which when separated are equal to that vector; each individual component describes the impact that part has on the vector. __ Vector= quantity that has both magnitude and direction. Some vectors are directed in //two dimensions// - upward and rightward, northward and westward, eastward and southward, etc. Any vector directed in two dimensions can be thought of as having an influence in two different directions (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. The single two-dimensional vector could be replaced by the two components.

Orienteering Lab Lauren Kostman October 14, 2011 Lab Partners: Maxx, Garrett *Note: starting position was at the first pillar to the right of the doorway. Actual Displacement: 3.05 m, West. Vector Displacement Map: - Resultant magnitude based on measuring Resultant magnitude based on calculations (analytical): 10.53 m east + 24.75 m west + 10.14 m east= 4.08 m west 13.29 m north + 14.48 m south=1.19 m south Error || + Actual || 28.2% || + Actual || 1.67% || Sample Calculations: Analysis: Upon reviewing the percent errors between both the calculated vs actual, and measured vs actual, it is clear that the measured data is extremely more accurate than the calculated. This can be due to the fact that there was a lack of precision for each measurement; this can be due to several mistakes. It's possible that upon bringing in the measuring tape from one point to another, the position (ending from one segment to another) changed several centimeters, which ultimately could have damaged the results. Additionally, although perpendicular angles were attempted, it's possible that each turn was not exactly 90 degrees, and thus using the pythagorean theorem to calculate the magnitude of the resultant would be an incorrect method.
 * Legs || Distance || Direction ||
 * 0 (starting position) || 0 m || north ||
 * 1 || 10.53 m || east ||
 * 2 || 13.29 m || north ||
 * 3 || 24.75 m || west ||
 * 4 || 14.48 m || south ||
 * 5 || 10.14 m || east ||
 * || Percent
 * Graphical
 * Analytical

Lesson 1: e
Lauren Kostman October 17, 2011 Homework

__ Topic Sentence: vector magnitudes (resolution) can be determined using the parallelogram or trigonometric methods. __ Process of determining magnitude of a vector is known as ** vector resolution **. - Make scale and draw vector in indicated direction. - Sketch 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 meet to form a rectangle - Draw vector components (//sides// of the parallelogram). Tail of the components start at the tail of the vector and stretches along the axes to the nearest corner of parallelogram. Arrowheads for direction! d.Label components of vectors to indicate which one represents which side. - Measure length of sides of parallelogram and [|use scale to determine magnitude] of components in //real// units. Used to determine side lengths of right triangles if an angle measure and length of side are known. Steps: - Sketch without scale the vector in given direction. Label magnitude and angle that it makes with the horizontal. - Draw rectangle around vector so it’s the diagonal of the rectangle. Beginning at [|tail] of vector, sketch vertical and horizontal lines. Sketch horizontal and vertical lines at [|head] of the vector (lines will meet to form rectangle). - Draw vector components (//sides// of rectangle). Tail of each component begins at tail of vector and stretches along axes to nearest corner of rectangle. Use arrowheads! - Label components of vectors.
 * Vector Resolution **
 * Parallelogram Method of Vector Resolution **
 * Trigonometric Method of Vector Resolution **
 * 1) For length of side opposite indicated angle, use sine. Substitute magnitude of vector for length of hypotenuse. Use algebra to solve equation for length of side opposite indicated angle.
 * 2) Repeat using cosine to determine length of side adjacent to indicated angle.

Lauren Kostman October 18, 2011 Homework

Lesson 1: g, h
__ Topic Sentence: Riverboat problems generally have 3 parts, all of which can be answered using simple equations and calculations. __ Motorboat problems usually have three separate questions: b.If the width of the river is //X// meters wide, then how much time does it take the boat to travel shore to shore? Questions b and c can be answered using: ave. speed = distance/time. Motorboat traveling 4 m/s, East encounters a current traveling 3.0 m/s, North. - What is the resultant velocity of the motorboat? - If the width of the river is 80 meters wide, then how much time does it take the boat to travel shore to shore? - What distance downstream does the boat reach the opposite shore? || a.) Resultant velocity= 5 m/s at 36.9 degrees. b.) River is 80-meters wide. Time to cross river can be determined by rearranging and substituting into the average speed equation.** time = distance /(ave. speed) ** Distance of 80 m can be substituted into the numerator. If knew any of the distances (A, B, C), then the average speed of that distance could be used to calculate the time to reach the shore. 80 m corresponds to distance A: average speed of 4 m/s is substituted into equation to determine time. c.) Same equation is used to calculate //downstream distance//. Distance downstream corresponds to ** Distance B **, with ** Average Speed B ** (3 m/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. __ Topic Sentence: perpendicular parts of vectors are independent, and thus don’t affect one another (if changed). __ Any vector directed at an angle can be thought of as being composed of two perpendicular components, represented as legs of a right triangle formed by projecting the vector onto the x- and y-axis. The two perpendicular components of vectors are independent of each other; a change on one component doesn't affect the other. It would require 15 seconds to cross the 60-meter wide river. **d=vt, so t=d/v= (60 m)/(4m/s)= 15 seconds**
 * Relative Velocity and Riverboat Problems **
 * 1) What is the resultant velocity (both magnitude and direction) of the boat?
 * 1) What distance downstream does the boat reach the opposite shore?
 * ** Example: **
 * ** Answers: ** ||
 * time = (80 m)/(4 m/s) = 20 s **
 * Independence of Perpendicular Components of Motion **

Lauren Kostman October 19, 2011 Homework

Lesson 2: a and b
__Lesson a: What is a Projectile__
 * Central Idea/ Theme:** Projectiles are objects that have gravity as the only force acting upon them, and gravity effects the vertical motion of the object, causing it to fall downwards with a downward acceleration.

1.) What is a projectile? A projectile is an object upon which the only force is gravity. It's anything that was once //projected// or dropped, continues in motion by its own inertia, and is only influenced by the downward force of gravity. An example of this is an object dropped from rest (influence of air resistance is negligible). 2.) What would a free-body diagram of a projectile look like? the direction of a projectile (up, down, up and to the right, down and to the left, etc.) doesn't matter; the free-body diagram of a projectile is always shown as a downward arrow (gravity is pulling the object down). 3.) What does Newton's First Law of Motion state (what is a force required to do and not to do)? A force is **__NOT__** required to keep an object moving, it's only required to maintian an acceleration For example: if a projectile is moving upward, there's a downward force and a downward acceleration (the object is moving up but slowing down). 4.) What is Newton's Law of Inertia? Without gravity, an object in motion wil continue to move with the same speed and in the same direction. Similarly, an object at rest would stay at rest. 5.) How does gravity impact a projectile? Gravity causes a projectile to fall below it's inertial path (effects **vertical motion** of projectile). It acts downward to cause a downward acceleration. There are no horizontal forces needed to maintain the horizontal motion (this is consistent with the concept of inertia).
 * 5 Questions and Answers**:
 * His laws are the opposite of the common misconception that a force is needed to keep an object moving!

__Lesson b: Characteristics of a Projectile's Trajectory__


 * Central Idea/Theme:** Projectiles move with a parabolic path since the downward force of gravity accelerates them downwards, but since there are no horizontal forces acting upon projectiles, a constant horizontal velocity is maintained.

1.) What is the vertical velocity of a projectile? Gravity causes a projectile to have a downward velocity (and thus a downward acceleration). 2.) What is the horizontal velocity of a projectile? Since there is only a vertical force acting upon projectiles (gravity), there must be a horizontal force to cause a horizontal acceleration. The vertical force acts perpendicular to the horizontal motion and will not impact it (since perpendicular components of motion are independent of each other). So, the projectile travels with a __constant horizontal velocity.__ 3.) How do the vertical versus horizontal motions affect the velocity of a projectile? Vertical motion causes the velocity to change at a rate of 9.8m/s (acceleration of gravity-g), whereas horizontal motion causes the velocity to remain constant. - Vertical versus Horizontal Velocities of Projectiles: 4.) When do projectiles travel with a parabolic trajectory? When launched non-horizontally (example: upward at an angle to the horizontal from the starting position), projectiles would free-fall below the straight line of it's inertial path, creating a parabolic path. 5.) What causes projectiles to move with a parabolic trajectory? Since the downward force of gravity accelerates projectiles downward from their otherwise straight line (gravity-free trajectory).
 * 5 Questions and Answers:**

Lesson 2: c (parts 1 and 2)
Lauren Kostman October 20, 2011 Homework

__**Describing Projectiles with Numbers**__ __1.) Horizontal and Vertical Components of Velocity__ 1.) What would a vector diagram of a projectile (ex: launched cannanball from top of cliff) look like? 2.) How do the x and y components of the velocity change over time (for projectiles)?
 * Central Idea/Theme:** The values of the x and y components of the velocity and displacement of projectiles changes, or remains constant, depending on the situation, which can be portrayed using vector diagrams.
 * Questions and Answers:**

3.) What would the vector diagram of a projectile look like if the object was launched upward (at an angle) to the horizontal? 4.) In this case, how would the x and y components change over time? 5.) What does the vector diagram for a //non-horizontally// launched projectile that lands at the same height as which it was launched look like? __2.) Horizontal and Vertical Components of Displacement__ 1.) How can the vertical displacement of a projectile be predicted? One can use the equation d=vit+.5at^2; a=g=-9.8 m/s/s; y=vertical displacement 2.) How can the horizontal displacement of a projectile be represented by an equation? 3.) What would a diagram of the trajectory of a projectile (red) look like if it was released from rest with no horizontal velocity (blue), and the path of it when there's no gravity (green)? 4.) What information can be gained from the diagram above? - vertical distance fallen from rest during each second is increasing (vertical acceleration) - vertical displacement coincides with the equation y=.5gt^2 - since there's no horizontal acceleration, horizontal distance traveled each second is a constant value 5.) What is the equation for vertical displacement of an angled-launched projectile?
 * Central Idea/ Theme:** Both equations and diagrams can be used to describe the vertical and horizontal displacement of a projectile.
 * Questions and Answers:**
 * vi=0 if falling from rest and a=g=-9.8 m/s/s, thus simplified equation is:

Lauren Kostman Group: Maxx, Garrett October 21, 2011

Orienteering Lab (part 2)
__Start Position:__ 0 m E
 * Legs || Distance || Direction ||
 * 1 || 9.28 m || S ||
 * 2 || 16.31 m || S ||
 * 3 || 18.36 m || S ||
 * 4 || 16.45 m || W ||
 * 5 || 27.49 m || S ||
 * 6 || 16.34 m || W ||
 * Resultant || 78.55 m || SW ||

__Sample Calculations:__ __Analytical:__ 9.28 m S + 16.31 m S + 18.37 m S + 27.49 m S= 71.5 m S 16.45 m W + 16.34 m W= 32.8 m W
 * || Percent Error ||
 * Analytical + Experimental || .19 % ||
 * Graphical + Experimental || 3.35 % ||





__Graphical:__

Lauren Kostman October 24, 2011 Class Activity

Launcher Activity (ball in cup)
PART I Height: 92 cm
 * Trial || Distance(cm) ||
 * 1 || 203.1 ||
 * 2 || 205.4 ||
 * 3 || 206.4 ||
 * 4 || 206.8 ||
 * 5 || 208.2 ||
 * 6 || 208.4 ||
 * 7 || 209.2 ||
 * 8 || 210.2 ||
 * Average || 207.2 ||

Sample Calculations: __Cup__: Height= 67 cm Time= .369 s Distance: 177 cm Procedure: media type="file" key="ball in cup.mov" width="300" height="300" Percent Error: 4.5% Sample Calculations: PART II: Procedure: Begin by solving for the initial velocity, which includes a similar procedure to that of part I in which carbon paper is placed on the ground, the ball is launched, the positions are averaged, and a (x) distance is measured. By equating this value to cos25 (the angle)*vt, one can plug this value into the equation d=vit+.5at^2, and find the time. After this, the velocity can be calculated. Then, calculations are done to create theoretical values as to where the x and y positions of the rings should be (based on the velocity and the given launcher). Upon shooting the ball, one tries to get it through all five rings, plus have it land inside a cup.
 * || x || y ||
 * vi || 4.79 m/s || 0 m/s ||
 * a || 0 || -9.8 m/s/s ||
 * t || .433 sec || .433 sec ||
 * d || 2.07 m || .920 m ||

Table for horizontal distance: Velocity at 25 degree angle: 4.23 m/s Vx= 4.23 m/s Vy=1.97 m/s
 * Trial || Distance(m) ||
 * 1 || 2.95 ||
 * 2 || 2.94 ||
 * 3 || 2.96 ||
 * 4 || 2.96 ||
 * 5 || 2.97 ||
 * Average || 2.96 ||

Video: media type="file" key="through 5 yay!.mov" Results:
 * Ring: || x-position (m) || time (s) || y-position (m) || experimental value (for y) (m) || percent error ||
 * 1 || .500 || .118 || 1.20 || 1.26 || 5.00% ||
 * 2 || .960 || .227 || 1.24 || 1.28 || 3.23% ||
 * 3 || 1.25 || .296 || 1.18 || 1.20 || 1.69% ||
 * 4 || 1.50 || .355 || 1.11 || 1.09 || 1.80% ||
 * 5 || 1.75 || .414 || .926 || .850 || 8.21% ||
 * cup || 2.96 || .700 || .090 || N/A || N/A ||

Sample Calculations: Analysis: Overall, the experimental values for this lab activity were pretty close to the theoretical values that were calculated for each ring. The fifth ring has the highest percent error. This can be because it's the farthest distance away from any other ring; for example, ring 3 is very close to both ring 2 and ring 4, and it has a very low percent error. So, it's possible that because ring 5 is the farthest from any other ring, it has the highest percent error. All of the other rings, however, come within a percent error of five or below, which is very good.

Lauren Kostman

Gourd-o-Rama Project
**Picture of Pumpkin Mobile:** t=2.75 s d=6.75 m Upon completing the pumpkin project, I was overall pleased with the results of my project. I disassembled an old skateboard and screwed in the the wheels from that onto the bottom of a cardboard box, which was formed (using string) to tightly fit around the pumpkin. When the pumpkin was pushed down the ramp, it encountered several issues along the way, the main one being that it fell out of the box. Therefore, it was then taped in using duct-tape. In the future, nonetheless, I would create a top to the cart so that it would be impossible for the pumpkin to slip out. Furthermore, the cart was almost a perfect square, yet I think a rectangular cart would travel a further distance, thus for future instances I would use a rectangular box, as opposed to forming it into a square. Regardless of these issues, the pumpkin was able to travel a distance of 6.75 m in 2.75 seconds, resulting in an initial celocity of 4.91 m/s, with an acceleration of -1.79 m/s/s.
 * Calculations:**
 * vi=4.91 m/s**
 * a=-1.79 m/s/s**
 * Summary:**