Ch3_llewellynm

Chapter 3 Honors Physics Max Llewellyn toc

__**10/12/2011 Homework**__ Vectors Lesson 1 a, b

__**Vectors and Direction**__
 * Topic Sentence:Vectors have magnitude and direction, there are many ways to describe that direction**

Distance, displacement, speed, velocity, acceleration, force, mass, momentum, energy, work, power, etc are quantities used to describe the physical world. All 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 that is fully described by its magnitude.

Vectors eg: [|displacement], [|velocity], [|acceleration], and [|force]. 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 is 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 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 can be used to depict the forces acting upon an object. Such diagrams are commonly called as [|free-body diagrams]. Look a scaled vector diagram -->. The vector diagram depicts a displacement vector. Why it's a good vector diagram.


 * a scale is clearly listed
 * a vector arrow (with arrowhead) is drawn in a specified direction.
 * the magnitude and direction of the vector is clearly labeled.

**Conventions for Describing Directions of Vectors**

northeast does not always mean 45º, so there must be other ways to describe vectors. There are many ways, two of them are:

>
 * 1) an angle of rotation of the vector about its "[|tail]" from east, west, north, or south. a direction of 40 degrees North of West = west + 40 more north
 * 2) a counterclockwise angle of rotation from due East. a direction of 30 degrees is 30 degrees more counterclockwise then due east. This is used a lot

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 to scale



__**Vector Addition**__
 * Topic Sentence: There are a few ways to add vectors, they all make sense.**

See below diagram, remember how to add.

When vectors aren't pointing striaght one way or another there is some more math required to add them. Here are two ways to do that.
 * the Pythagorean theorem and trigonometric methods
 * [|the head-to-tail method using a scaled vector diagram]

**The Pythagorean Theorem** If the vectors are at a right angle to each other they can be aligned in a right triangle and math can be done to find the resultant.



**Using Trigonometry to Determine a Vector's Direction** Hey remember SOA-CHA-TOA? Yes, good then you'll be able to figure out everything in this next section

Sometimes the angle you get from trig is not the angle you need, be sure to apply logic to your numbers. 

T

**Use of Scaled Vector Diagrams to Determine a Resultant** You can also draw out all the vectors from head to tail and then draw a line from the start to the end of the last vector. Measure with ruler and protractor, and be happy.

\ 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

The order doesn't matter because addition is like that.
 * 15 m, 210 deg. + 25 m, 300 deg. + 20 m, 45 deg.****SCALE: 1 cm = 5 m**

SCALE: 1 cm = 5 m

__**10/13/2011 Classnotes**__ Naming angles
East is 0º, North 90º, and so on East is also X+, North Y+, and so on  Can name angles based on either amount CCW from E, or in a given direction from a given point.

__**10/13/2011 Homework**__ Vectors Lesson 1 c, d
The **resultant** is the vector sum of two or more vectors. "doing" vector A B and C in the above diagram is the same as doing vector R. A resultant is a sub and behaves in a logical way and has logical attributes similar to how any other sum behaves.
 * Resultants**
 * Topic Sentence:the sum of vectors is called a resultant, and hey it's a sum just like any other sum in the math you know and love**


 * Vector Components**
 * Topic Sentence: A vector that's pointed at an angle can be though of as having 2 parts. Think of it like the hypotenuse of a right triangle.**

Vectors can be 2 dimensional, eg point a little to the left and a little up.

Each part of a two-dimensional vector is known as a **component**. eg: 10 m/s N and 5 m/s E. The single two-dimensional vector could be replaced by the two components and vis versa.

A place is going along a northwest vector. If this is the case, then the displacement of the plane has two components - a component in the northward direction and a component in the westward direction.

__**10/13/2011 Classnotes**__ Naming angles
Collinear- on the same line, or angle

Right angles- to add two vectors at right angles you can use pithag. 1. Draw a sketch of the vector head to talk (draw vector 1. at the head of vector 1 start drowing vector 2) 2. draw resultant, from tail of vector 1 to head of vector 2 (from begining to end) 3. do pythag to find the R. 4. do trig to find the ange.

Estimating 1. size must be between minimum (subtracting) and maximum (adding) 2. Angles must be between the angels of the vectors given

__**10/17/2011 Homework**__ Vectors Lesson 1 e

 * Vector Resolution**
 * Topic Sentence:There are two ways to combine vectors, using a diagram drawn to scale and using trig.**

Here are two basic methods for determining the magnitudes of the components of a vector directed in two dimensions. The process of determining the magnitude of a vector is known as **vector resolution**. These methods are:
 * the parallelogram method
 * [|the trigonometric method]

**Parallelogram Method of Vector Resolution** The parallelogram method of vector resolution involves using an accurately drawn, scaled vector diagram to determine the components of the vector. Briefly put, the 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.

More briefly put draw the first then the second vector head to tail and the other way round. The result is a parallelogram and geometry can be done to determine the various lengths.

**Trigonometric Method of Vector Resolution** The trigonometric method of vector resolution involves using trigonometric functions to determine the components of the vector. Treat each vector as the hypotonuse of a right triangle and do trig to determine the X and Y components of that triangle. Then add the respective componets to form a bigger triangle and use trig to determine that the hypotnuse is.

__**10/18/2011 Homework**__ Vectors Lesson 1 g & h

 * Topic Sentence:Sometimes due to forces like headwind or current the speed of one object relative to an observer won't be the same as the speed of the object relative to the current or wind. All these vectors can be added and determining what's going where and how fast is simple math.**

On occasion objects move within a medium that is moving with respect to an observer. For example, an airplane usually encounters a wind - air that is moving with respect to an observer on the ground below. As another example, a motorboat in a river is moving amidst a river current - water that is moving with respect to an observer on dry land. 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.
 * Relative Velocity and Riverboat Problems**

For example, consider a plane flying amidst a **tailwind**. Or a headwind. The same principle applies to more then one axis of motion. See below. In this situation of a side wind, the southward vector can be added to the westward vector using the [|usual methods of vector addition]. The magnitude of the resultant velocity is determined using Pythagorean theorem. 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**

Crikey! If you look closely you can see math in it's natural habitat.

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. Like any vector, the resultant's [|direction] is measured as a counterclockwise angle of rotation from due East.

**Analysis of a Riverboat's Motion** The affect of the wind upon the plane is similar to the affect of the river current upon the motorboat. The resultant velocity of the motorboat can be determined in the same manner as was done for the plane. The math is below the concept is above. (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**

tan (theta) = (opposite/adjacent) tan (theta) = (3/4) theta = invtan (3/4) **theta = 36.9 degrees**

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 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** Consider the following example.
 * 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?

The solution to the first question has already been shown in the [|above discussion]. The resultant velocity of the boat is 5 m/s at 36.9 degrees.
 * 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? ||

Using the vector components calculated above it is simple division to determine how much time will pass before the boat travels the total river width. After the time is know it is simple multiplication to determine how much distance the boat covered. **time = (80 m)/(4 m/s) = 20 s** **distance = ave. speed * time = (3 m/s) * (20 s)** **distance = 60 m**


 * Independence of Perpendicular Components of Motion**


 * Topic Sentence:**

Do you remember this? Don't worry it hasn't changed since Relativity. Good then lets continue. The two perpendicular parts or components of a vector are independent of each other. If you (talking to myself) though they weren't hang your head in shame for the rest of this chapter. Perpendicular components of motion do not affect each other.

Remember those rules about vectors that we covered earlier? Well math is still math so using the ideas you learned earlier on this you might not have seen specifically but have seen similar to will work because all physics is math and all math can be physics.

__**10/14/2011 Activity**__ Vector Mapping

 * Partners: Gabby, Madie**

Our starting was the set of doors on the right on the front of the building side of the courtyard. We started measuring in the middle of the doors on the ground.

Measured displacements
 * Legs || Distance (m) || Direction ||
 * 0 || 0 || E ||
 * 1 || 7.05 || E ||
 * 2 || 5.72 || S ||
 * 3 || 17.34 || E ||
 * 4 || 5.68 || S ||
 * 5 || 2.95 || W ||

Resultant calculation via graphical method

Analytical method: 7.05 m E - 17.34 m E - 2.95 m E = 21.44 m E 5.72 m S - 5.68 m S = 11.40 S

Magnitude of resultant = Sqrt(A^2 + B^2) = Sqrt(21.44^2 + 11.40^2) = 24.282372207014703 m

Percent Error
 * Analysis:**


 * Alert! *NULL*** I was out during the day that this lab was preformed so all of my data is simply copied from my group. I preformed the calculations in the interest of completing as much of the lab as possible but anything else would simply be rewording and plagiarizing their work.

Graphical Method (24.6 - 25.69) / 24.6 = 4.4% error

Analytical Method (24.3 - 25.69) / 24.3 = 5.72% error

__**10/19/2011 Homework**__ Vectors Lesson 2 a & b
**a)What is a Projectile?**
 * Main Idea:** To introduce the fundamentals of projectile motion. Projectile motion is is comprised of both vertical motion and horizontal motion.

1)What is a projectile? An object that once released is only carried forward by its own inertia, influenced only by the downward force of gravity.

2)Why is inertia important an important concept? This is the force that makes projectiles project, the only other force is gravity so any X wise motion is the result of inertia.

3)What is a common misconception of projectile motion? That there is a force other then gravity affecting a projectile's motion, or that some how gravity has an effect X wise. The only force that ever plays a role in projectile motion in the X direction is inertia.

4)What are the types of projectile motion? A projectile can either be dropped, or can be launched at an angle. That angle can be 90 or 0 degrees, which makes the math much easier. However formulas developed for non right angle launches will still work when fed an angle of 0, 90, 180, ect.

5)Why is this not free fall? Free fall is the same as projectile motion when dropped straight down. In face projectile motion can be thought of as simply free fall with inertia.


 * b)Characteristics of a Projectile's Trajectory **


 * Main Idea:**To provide examples of various types of projectile motion, and to give a clear understanding of it.

Annoying things we have to memorize.
 * A projectile is any object which the only force action upon it is gravity
 * (Unless launched straight up or down) Projectiles travel with a parabolic trajectory due to the influence of gravity
 * There are no horizontal forces acting upon projectiles and thus no horizontal acceleration
 * The horizontal velocity of a projectile is constant. (Hey I already knew that!)
 * These is vertical acceleration, this is caused by gravity and has the value g
 * The vertical velocity of a projectile changes by 9.8 m/s each second (Hey I already knew that!)
 * The horizontal motion of a projectile is independent of its vertical motion (Well duh)

1)What would happen to projectiles without gravity? They would have a constant vertical motion and travel in a straight line indefinitely.

2)What about with gravity? The vertical motion would no longer be at a constant velocity and the projectile would eventually come down to earth, faster and faster.

3) (Rhetorical question) Does the X wise have any effect on the Y wise motion and vis versa? No. *dope slap*

__**10/20/2011 Homework**__ Vectors Lesson 2 c
**c)Describing Projectiles With Numbers: (Horizontal and Vertical Velocity)**
 * Main Idea:** To describe projectile motion in terms of X and Y velocity.

1) Does gravity effect X wise motion? No. *dope slap*

2) If the object is launched at a 45 degree angle how do the velocities change? The X wise velocity is constant. End of story. The Y wise velocity changes at g. Yup still works. Vf = Vi + AT

3) What equations represent the X and Y velocity?
 * Vertical: Y = .5g*T^2 + ViyT + H**
 * Horizontal: X = VixT**

5)How are those velocities calculated when the projectile is launched at an angle? Break the velocity in to X and Y components like in vector addition. There you have it.

__**10/21/2011 Activity**__ Target Practice (Ball in a cup)
**Objective**
 * Measure the initial velocity of a ball.
 * Apply concepts from two-dimensional kinematics to predict the impact point of a ball in projectile motion.
 * Take into account trial-to-trial variations in the velocity measurement when calculating the impact point.

**Pre Lab Questions**

**Procedure**media type="file" key="Movie on 2011-10-25 at 11.01.mov" width="300" height="300"
 * 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?
 * 2) The initial height, that the ball would be in free fall [on earth] (no wind resistance, ect.)
 * 3) 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.
 * 4) If you know how long the ball will take to fall, which can be calculated by knowing the height above the ground from which the ball was launched, the only calculation necessary would be to multiply the time the ball spends in the air by the horizontal velocity.
 * 5) Write your procedure and get approval from Mrs. Burns before you proceed any further!
 * 6) What data will you need to collect? Remember that you must run multiple trials. Keep in mind your end goal!
 * 7) The height of the launcher.
 * 8) The distance from the launcher that the ball landed.
 * 9) How will you analyze your results in terms of precision and/or in terms of accuracy?
 * 10) We will look for where the most consistent landing spot of the ball is, taking in to account that factoring in outliers to our calculations might contribute to inaccuracy in the the average.

**Data** Data of ball impact point and accompanying calculations**.**

If you throw any velocity and angle in to the left side there are functions on this workbook that will, calculate nearly everything you could want to know, create a data table, make an accurate position vs height graph of the object's path, and has spaces where you can plug in multiple X values and find out what the height at that point is.



**Analysis** This activity was solely to measure the Vi of the launcher. As the theoretical data was calculated using the experimenter data there is nothing to compare our results to calculate percent error or anything of that nature. However we could have gathered better data if the ball was in the same place inside the launcher every time or if the launcher was more consistent.

__**10/28/2011 Lab**__ Gourd-O-Rama

 * Partner(s): Noah Pardes**

-Create a cart, with the least possible weight, that will hold a 1kg+ gourd as it travles down a ramp and over a distance with the least acceleration possible.
 * Objective**

-none
 * Pre-lab Questions**

-Place the cart on the ramp and release it. -When the cart reaches the bottom of the ramp start a timer. -When the cart either stops moving or hits a wall stop the timer and measure the distance traveled. -Record the above data and make calculations.
 * Procedure**

Cart-


 * Data**

Cart weight with gourd- 1.09 kg Cart weight without gourd- Weight of gourd-

Distance traveled- 2.00 m Time taken- 1.69 s

Velocity at bottom of ramp- 3.45 m/s Acceleration- -1.40 m/s/s

The only errors we had were related to the engineering of our cart, not the physics (at least not the physics in two chapters in any direction) of it. We had a small wheel base and a high center of gravity, that's a really good way to tip over, which we did.
 * Analysis**

__**11/8/2011 Lab**__ Shoot Your Grade

 * Group Members: Gabby, Madie**


 * Purpose with Rationale**: In this lab you have to use your knowledge about projectiles to launch, using a launcher, a plastic ball through five strategically placed rings and finally into a cup.


 * Procedure**
 * 1) Measure the height from the ground of the launcher and it's angle relative to the horizon.
 * 2) Fire the launcher so that it hits a piece of charcoal paper, repeat this process at least 5 times for more accuracy.
 * 3) Measure the points of impact and their X distance from the launcher.
 * 4) Use the above data to calculate the initial velocity of the projectile.
 * 5) Use the above data to calculate the path the ball will take and it's Y position at given X points.
 * 6) Hang rings at those X points.
 * 7) Measure the height of the cup and calculate it's optimal location.
 * 8) Fire the launcher and watch it go through the mathematically perfect rings and in to the cup every time.

media type="file" key="Shoot+Your+Grade.m4v" width="300" height="300"


 * Data**

X Distance of ball from launcher
 * Calculations**

Y = -4.9 Time^2 + Vi * Sin(Angle) * TIme + Height X = Time * Vi * Cos(Angle) Time = (X / Vi * Cos(Angle)) Y = -4.9 * (X / Vi * Cos(Angle))^2 + Vi * Sin(Angle) * (X / Vi * Cos(Angle)) + Height At ground Y = 0

0 = -4.9 * (X / Vi * Cos(Angle))^2 + Vi * Sin(Angle) * (X / Vi * Cos(Angle)) + Height 0 = -4.9 * (X / Vi * Cos(Angle))^2 + Tan(Angle) * X + Height Tan(Angle) * X + Height = 4.9 * (X / Vi * Cos(Angle))^2 (Tan(Angle) * X + Height) / 4.9 = (X / Vi * Cos(Angle))^2 (Tan(Angle) * X + Height) / 4.9 = X^2 / (Vi * Cos(Angle))^2 4.9 / (Tan(Angle) * X + Height) = (Vi * Cos(Angle))^2 / X^2 4.9 * X^2 / (Tan(Angle) * X + Height) = (Vi * Cos(Angle))^2
 * Sqrt ( 4.9 * X^2 / (Tan(Angle) * X + Height)) / Cos(Angle) = Vi**

Sqrt ( 4.9 * 5.0575^2 / (Tan(21) * 5.0575 + 1.175)) / Cos(21) = 6.79291419
 * Ring Height Calculations**
 * Ring Height Calculations**

Y = -4.9 * (X / Vi * Cos(Angle))^2 + Vi * Sin(Angle) * (X / Vi * Cos(Angle)) + Height f(x) = Y = -4.9 * (X / 6.79291419 * Cos(21))^2 + 6.79291419 * Sin(21) * (X / 6.79291419 * Cos(21)) + 1.175 f(x) where X is the ball's position relative to the X axis will return the height of the ball at that point. Use this function to hang the rings at the optimal height.

Really cool launch simulator!! --> Check it out! Can be used for any type of 2D launch!


 * Analysis**

When the ball when through one of the rings that we hung at the height we calculated it also went through the rest of the rings hung at the height we calculated. This means that, when the launcher launched at the angle we though it would and with the velocity we thought it would, all our calculations were right. However when the launcher was inconsistent, it was inconsistent. That all said when we measured the rings after the launch succeed they were slightly off the position predicted. Personally my ego would like to believe that this was due to two wrongs (the launcher and the little bit off rings) making a right but in case my arrogance is getting ahead of me here are the calculations.

(1.37 - 1.32) / 1.37 = 0.03649 = 3.649%
 * Ring || Predicted height (m) || Actual height (m) || % Error ||
 * 1 || 1.37 || 1.32 || 3.65% ||
 * 2 || 1.47 || 1.41 || 4.08% ||
 * 3 || 1.39 || 1.34 || 2.88% ||
 * 4 || 1.24 || 1.19 || 3.23% ||
 * 5 || 1.23 || did not achieve || null ||
 * Sample**


 * Conclusion**

While I am proud to say that my math was, in my humble opinion, mathematically perfect, and our calculations were accurate to a tenth of a millimeter our launcher did not find it within itself to step up to the same standards. Out launcher, in addition to firing the ball differently when the angle was consistent, had the habit of moving and changing it's angle. Due to this no matter how good out calculations, or how well we hung the rings, or placed the cup based on our math the launcher chose to use different, randomly chosen, numbers, resulting in something less then total success.Aside form the infernal launcher other sources of error could be shifting of the rings, and the string holding the rings. This would have changed the height and Z offset of the rings, throwing them out of the path of the projectile. This could have been solved by standing the rings on top of tripods, or other similar ground based, more rugged, structures.