Ch3+LambertM

=__Chapter 3__ = toc 10/11/11

Lab--Measuring Angles
10/12/11

Homework Assignment #1--Vectors Lesson 1 (A,B) (Method 1)
Topic Sentence-Vector quantity requires magnitude and direction. A vector quantity is a quantity that is fully described by both magnitude and direction. Vector quantities that have been include displacement, velocity, acceleration, and force. Vector quantities are represented by scaled vector diagrams that depict a vector by use of an arrow drawn to scale in a specific direction.The vector diagram depicts a displacement vector. Characteristics of this diagram that make it an appropriately drawn vector diagram:
 * 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. above magnitude is 20 m and the direction is 30 degrees West

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. Two conventions used are: > Two illustrations of the second convention (discussed above) for identifying the direction of a vector are shown below. **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.
 * 1) The direction of a vector is expressed as an angle of rotation of the vector from east, west, north, or south.
 * 2) The direction of a vector is expressed as a counterclockwise angle of rotation of the vector from due East.

Topic Sentence-Two vectors can be added together to determine the result. Topic Sentence-The Pythagorean theorem is a used for determining the result of adding two vectors __that make a right angle__ to each other. Topic Sentence- Sine, Cosine, and Tangent can be used to determine the length of a segment or the angle measurement. The direction of a //resultant// vector can often be determined by use of trigonometric functions. SOH CAH TOA in trigonometry-sine, cosine, and tangent functions. The measure of an angle as determined through use of SOH CAH TOA is __not__ always the direction of the vector. Topic Sentence-The magnitude and direction of the sum of two or more vectors can also be determined by use of a drawn scaled vector diagram.
 * Vector Addition**
 * The Pythagorean Theorem**
 * Using Trigonometry to Determine a Vector's Direction**
 * Use of Scaled Vector Diagrams to Determine a Resultant**

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
 * 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.
 * 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 discussed

An example of the use of the head-to-tail method is illustrated below. The problem involves the addition of three vectors: 20 m, 45 deg. + 25 m, 300 deg. + 15 m, 210 deg.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. SCALE: 1 cm = 5 m 10/13/11 Class Notes

Measuring Angles


10/13/11

Homework Assignment #2--Vectors Lesson 1 (C,D) (Method 1)
**Resultants** Topic Sentence-The resultant is the vector sum of all the individual vectors T he resultant is t he vector sum of two or more vectors. 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. **Vector Components** Topic Sentence-Vectors can be directed in two dimensions and can be determined by separating it into two parts. 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. Any vector directed in two dimensions can be thought of as having an influence in two different directions or two parts called components.

10/18/11

Homework Assignment #3--Vectors Lesson 1 (G,H) (Method 1)
** Relative Velocity and Riverboat Problems ** Topic Sentence-Motion is relative to the observer. The magnitude of the velocity of the moving object with respect to the observer on land will not be the same as the speedometer reading of the vehicle.

The angle between the resultant vector and the southward vector can be determined using the sine, cosine, or tangent functions. tan (theta) = (opposite/adjacent) theta = invtan (25/100) **theta = 14.0 degrees**

**ave. speed = distance/time** **distance = ave. speed * time** The motion of a riverboat can be divided into two simultaneous parts - a motion in the direction straight across the river and a motion in the downstream direction. These two parts (or components) of the motion occur simultaneously for the same time duration.

**Independence of Perpendicular Components of Motion** A **component** describes the affect of a single vector in a given direction. The vector sum of these components is always equal to the force at the given angle. A change in the horizontal component does not affect the vertical component. A boat on a river often heads straight across the river, perpendicular to its banks. Yet because of the flow of water (i.e., the current) moving parallel to the river banks, the time required for the boat to cross the river from one side to the other side is dependent upon the boat velocity and the width of the river.

**d = v • t** So **t = d / v** (60 m) / (4 m/s)=**15 seconds**
 * Example**: 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.

10/19/11

Homework Assignment #4--Vectors Lesson 2 (A,B) (Method 3)
What is a projectile? --Any object upon which the only force is gravity. What types of projectiles are there? --Any object that once projected or dropped continues in motion by its own inertia and is influenced only by the downward force of gravity is considered a type of projectile What is inertia's affect of projectile motion? --The projectiles continue in motion by their own inertia until acted upon by gravity. What is a free body diagram? --Shows a single force acting downwards and labeled force of gravity. Is force required to keep an object in motion? --No a force is only required to maintain acceleration What are the characteristics of a projectile's trajectory? --Horizontal and vertical motion
 * Questions**
 * Central Idea**--A projectile is an object upon which the only force acting is gravity.

10/20/11

Homework Assignment #5--Vectors Lesson 2 (C) (Method 3)
Does horizontal velocity ever change in a projectile? --The horizontal velocity remains constant during the course of the trajectory and the vertical velocity changes by 9.8 m/s every second How does gravity affect the projectile? --Gravity causes the cannonball to accelerate downwards at a rate of 9.8 m/s/s How do you find the vertical displacement of a projectile? --y=0.5*g*t 2 it is the same equation used to find the vertical displacement of a free falling object undergoing one dimensional motion, g=-9.8 m/s/s How do you find horizontal displacement? --x=v ix *t, The horizontal displacement of a projectile is only influenced by the speed at which it moves horizontally ( vix ) and the amount of time ( t ) that it has been moving horizontally What is the difference between horizontal and vertical displacement? --The vertical displacement of a projectile is dependent only upon the acceleration of gravity and not dependent upon the horizontal velocity What does this equation mean?-->y = viy • t + 0.5 • g • t 2 The gravity-free path is no longer a horizontal line since the projectile is not launched horizontally. In the absence of gravity, a projectile would rise a vertical distance equivalent to the time multiplied by the vertical component of the initial velocity (viy• t). In the presence of gravity, it will fall a distance of 0.5 • g • t2. Combining these two influences upon the vertical displacement yields this equation**.**
 * Central idea**--The horizontal and vertical components of the velocity vector change with time during the course of projectile's trajectory.

=Halloween Lab= Partner: Rachel Knapel
 * Results:**
 * vehicle mass (g) -
 * time (s) - 5.93 seconds
 * distance traveled (m) - 8.7 meters
 * acceleration value - -.495 m/s/s
 * velocity value - 2.934 m/s


 * Improvements**:
 * To improve our project, we would make sure the axels were more perfectly aligned and that they didn't alter when the car crashed. Our biggest issue was that after the first run, the car crashed into the wall and the axels became lose. Our car then continued to flip over on the ramp, which caused it to not be able to reach its maximum distance. Everything else seemed to work well, so everything else would have been kept the same.

Target Practice

 * Objectives:**
 * 1) Measure the initial velocity of a ball.
 * 2) Apply concepts from two-dimensional kinematics to predict the impact point of a ball in projectile motion.
 * 3) Take into account trial-to-trial variations in the velocity measurement when calculating the impact point.

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 * Pre-Lab Questions:**
 * 1) If you were to drop a ball, releasing it from rest, what information would be needed to predict how much time it would take for the ball to hit the floor? What assumptions must you make?
 * In order to solve for time, the distance from the starting to the ending point is necessary. We would also assume that the initial velocity is 0 m/s and that the acceleration is -9.8 m/s^2.
 * 1) If the ball in Question 1 is traveling at a known horizontal velocity when it starts to fall, explain how you would calculate how far it will travel before it hits the ground.
 * By knowing 3 out of the 4 y-component variables, you could solve for time using these variables in the equation d = (vi)(t) + ½(a)(t^2). Because the ball is traveling horizontally (theta = 0 degrees), the initial y-velocity would be zero. The y-acceleration would be -9.8 m/s^2 (acceleration of gravity), and the y-change in distance would be the distance from the shooter to the floor. You could then solve for time, and use the result to solve for the x-change in distance (range). This is possible since the initial velocity is known, the acceleration is 0 m/s, and the time is known.
 * 1) A single Photogate can be used to accurately measure the time interval for an object to break the beam of the Photogate. If you wanted to know the velocity of the object, what additional information would you need?
 * To solve for velocity, it is necessary to know the hang time (which is calculated by the Photogate timer) and the object’s range (x-change in distance).
 * 1) What data will you need to collect? Remember that you must run multiple trials. Keep in mind your end goal!
 * We will need to collect the distance from the shooter (point where ball is launched) to the ground, and the distance from the shooter to each dot on the carbon paper. This is information is necessary to solve for the shooter's initial velocity at "medium range."
 * 1) How will you analyze your results in terms of precision and/or in terms of accuracy?
 * For the carbon paper test, I would analyze my results in terms of precision. To do so, I would shoot the ball many times to see where it lands on the carbon paper each time it is shot. If the dots are close together, it means that my results are precise. If the dots are very spread out and random, it means that my results are not.
 * For the "ball in the cup" experiment, I would analyze my results in terms of precision and accuracy. To analyze my results in terms of accuracy, I would compare where the ball actually lands to the theoretical position (theoretical position = avg. distance of all dots on carbon paper). I will do this using the percent error formula. To analyze my results in terms of precision, I would see if the ball landed in the same area each time. I would do this by implementing the percent difference formula.

"Shoot Your Grade"
November 5, 2011

Lab Partners: Brianna Behrens & Lerna Girgin

**Purpose and Rationale:** The purpose of this lab was to determine the coordinates of each hoop, through which the ball would theoretically be shot successfully and land in the cup. The angle of our launcher was 20 degrees, which during the process of verifying the initial velocity, allowed us to simply each component to v i x = .94v i and v i y = .34v i. Because the ball was then going to be launched at an angle, we had already known that the velocity would no longer be the same, for either component. We planned our procedure by calculating the initial velocity for the x and y components, measuring the height of the counter top, and then determining the coordinates at which we would hang each hoop by dividing the entire horizontal distance into five segments and calculating the vertical heights.


 * Hypothesis:** The ball will go successfully through the hoops and land in the cup if all our calculations were correct. The ball will have a parabolic trajectory.


 * Methods and Materials:** We were able to calculate the height that the ring should be hung by using the off the cliff method for projectiles. With our calculations were able to set up the rings with clamps and tape from the ceiling tiles. We measured the heights we got from the launcher, both horizontally and vertically, to line up the rings. We set up one ring at a time and tested it. Each required a little bit of changing, but they were all in the correct general area. The materials used in this lab were rings of masking tape, string, shooter, yellow ball, measuring tape, right angle clamps, plumb bomb, and carbon paper.

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Actual Measurements of each ring: (x, y) 1- 4.3 m, 1.4m 2- 3.55m, 1.54m 3- 2.43m, 1.52m 4- 1.94m, 1.32m 5- 1.05m, 1.04m cup- 0m, 0m




 * Conclusion:** The main sources of error we encountered were the air pressure, angle slightly changing, inconsistency in shooter (shown with carbon paper when finding how far it shot), moving of the hoops, ventilation system above our hoops could have caused them to move slightly, and the hoops moving from different classes. We were only able to pass through the first 3 hoops, and for these hoops the highest percent error we recorded was 11 %. The following percents were much higher. The cause of these high percent on the last two hoops were because we were unable to continually test them. In order to get a more accurate experimental value we needed to consistently pass through the beginning hoops, which we were unable to do.