Ch_3_tonuzire

toc CHAPTER 3

=Homework:= = =

10/12 Lesson 1 Part a & b - Method 1
Distance, displacement, speed, velocity, acceleration, force, mass, momentum, energy, work, power, etc. are divided into vectors and scalars. A vector quantity is a quantity that is fully described by both magnitude (quantity/amount) and direction and a scalar quantity is a quantity that is described by its magnitude (quantity/amount). ∆D, Velocity, Acceleration, and force describe magnitude. In order to fully describe vectors, must include magnitude and direction.
 * Vectors and Direction **

Scaled vector diagrams often represent vector quantities. ^Vector Diagram. a scale is clearly listed. Vector has a head and a tail (direction – 30º West of North) and the magnitude (20m).

Vectors can be directed North, East, South West, NE, NW, SE, SW, East of North, East of South, West of North and West of South.
 * Conventions for Describing Directions of Vectors **

The first example has a vector that is 40º counterclockwise of East. In the second example the vector has a direction of 240 degrees counterclockwise of East.

To add vectors, simply add the values of the net force to find the resultant. Examples:



To find the result of adding two vectors, the Pythagorean theorem and trigonometric methods can be used. The Pythagorean theorem is used when the result of adding two vectors makes a right angle.





The result of adding 11 km, north plus 11 km, east is a vector with a magnitude of 15.6 km. Use trig to solve for the angle measure. Sin ø, Cos ø, Tan ø. .







The **head-to-tail** method is used to determine the vector sum or resultant.

The head to tail method involves drawing a vector to scale. Then, start with a starting position, and where the head of the first vector ends is where the second vector begins, etc. Then a line is drawn from the starting point to the end point, and the real length is found by using the scale, this is labeled R. Using a protractor and measuring it counterclockwise of East determine the direction of the resultant.



10/13/11 Lesson 1 Part c & d - Method 1
Part C - Resultants The **resultant** is the vector sum of two or more vectors. (It is the result of adding two or more vectors together) In this example, A, B, and C are added together and R is the resultant. **A + B + C = R**

If displacement vectors are added, then the resultant is the overall displacement (the displacement resultant). Vectors can be added if they have the same quantity - displacement, velocity, etc. If two or more velocity vectors are added, then the resultant is a velocity vector.

The Resultant vector is the result of adding each individual vector together. A+B+C is just like doing

Part D -

Vector Components Vectors can be directed in //two dimensions// - upward and rightward, northward and westward, eastward and southward, etc.

When vectors are directed in angles, they can be broken down into two positions. \/ North West - North + West. Each part of a two-dimensional vector is called a **component**. Components help to show the influence of a vector on another. A two dimensional vector is the combined influence of the components of a two dimensional vector. They can be replaced by their components. One component is going up and the other is going across.

The two components make up the same influence as a single 2-D vector, as shown in the pictures with the dog.

The combined influence of the two components is equivalent to the influence of the single two-dimensional vector. There is no effect on the stability of the picture.



Resulting displacement vector is northwest. Displacement of the plane has two components - a component in the northward direction and a component going westward.

10/17/11 Lesson 1 Part e - Method 1
Vector Resolution

The process of determining the magnitude of a vector is called **vector resolution**. There are two methods of vector resolution: the parallelogram method and the trigonometric method.

**Parallelogram Method of Vector Resolution** This requires an accurately draw to scale vector diagram to find out the components of the vector.
 * 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.
 * 3) Draw the components of the vector. The components are the //sides// of the parallelogram. The tail of a component starts 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 then use the scale to determine the magnitude of the components in //real// units. Label the magnitude of the vector on the diagram.



**Trigonometric Method of Vector Resolution** The trigonometric method of vector resolution involves using trigonometric functions to determine the components of the vector. The method of using trigonometric functions to find the components of a vector is:
 * 1) Draw a //rough// sketch (no scale needed) of the vector in the indicated direction. Label the magnitude and the angle that it makes with the horizontal.
 * 2) Draw a rectangle around the vector so that it is pointing diagonally from one corner to another. 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 closest corner of the rectangle. Be sure to place arrowheads on these components to indicate their direction (up, down, left, right).
 * 4) Label the components of the vectors with symbols to indicate which component represents which side. A force going north would be F-north. A rightward force velocity component would be labeled vx; and so on.
 * 5) To find the length of the side opposite of the given 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 step 5 using the cosine function to find the length of the side adjacent to the given angle.


 * Trigonometric Method of Vector Resolution**



10/18/11 Lesson 1 Part g & h - Method 1
G -

Relative Velocity and Riverboat Problems Observers see mediums affecting an airplane - wind, and a boat in a river - current. The velocity on the speedometer is different than the actual speed when being acted upon by mediums. Motion is relative to the observer who is on the ground. **Tailwind is** 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 the sum is 125 km/hr. Headwind - wind approaching head of plane - opposite direction. 100km/hr + -25 km/h = 75 km/h Example: Use Pythagorean theorem when side wind approaches a plane. R = √Rx^2 + Ry^2. Use Vector addition and to find resultant use the Pythagorean theorem. (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**

The angle/direction at which R is going is determined by using sin-1, cos-1, or tan-1. Set Tan-1(Ry/Rx) The tangent function can be used like this:

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. Every vector's direction/angle is measured counter-clockwise of East.

**Analysis of a Riverboat's Motion** Similar to the plane examples, the river current vector velocity influences the boat's velocity. Vector addition and Pythagorean Theorem. (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**

To find ø (theta). Use Tan-1: tan (theta) = (opposite/adjacent) tan (theta) = (3/4) theta = invtan (3/4) **theta = 36.9 degrees**


 * H -**

Independence of Perpendicular Components of Motion

Vectors have components of the X and Y axes. A vector is represented as a right triangle. The X and Y components are independent of each other. If the X component is changed than the Y component, it won't influence the Y component at all. It will only change the direction. The resulting motion of a plane flying in a crosswind is the sum of two velocity vectors traveling at the same time that are perpendicular to each other. Perpendicular components of motion do not affect each other. 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. **d = v • t** So **t = d / v** (60 m) / (3 m/s) **20 seconds**

Use the time of the equation and apply it to the velocity or distance of the x component.

10/19/11 Lesson 2 Part a & b - Method 3
Part A - P = Q = R = S = Projectiles are any object that is dropped/thrown in any direction. It’s own inertia keeps it moving and the force of gravity is the only force influencing it. (Air resistance is not taken into account). Any object that is acted upon by a force other than gravity is not a projectile. Newton’s laws only suggest that forces are only required to cause acceleration and not a motion. (BIG MISCONCEPTION) a force is required to keep an object in motion. (WRONG) A force is only required to maintain acceleration (CORRECT) Absence of gravity will cause an object to stay in the same direction at the same speed. With gravity, the projective will fall downward due to downward acceleration.
 * A projectile is an object that is only affected by gravity.
 * Newton’s law of inertia and first law of motion.
 * How does Newton’s law of inertia apply to projectiles?
 * How does Newton’s first law of motion apply to projectiles?
 * How do projectiles apply to vectors?
 * Is there any other force acting upon projectiles other than gravity?
 * What is a projectile?

T =
 * 1. A projectile is kept going by inertia.
 * 2. An object that is not affected by gravity will keep going in the same direction at the same speed.
 * 3. Projectiles apply to vectors because a force causes them to change their direction and or velocity.
 * 4. No. Only gravity
 * 5. A projectile is an object moving in any direction only acted upon by the force of gravity and is kept moving by its own inertia.





Part B -

P – Many projectiles not only undergo a vertical motion, but also undergo a horizontal motion. The X and Y components are independent of each other.

Q – How does trajectory relate to projectiles? Does gravity affect the horizontal component of a projectile? Is there horizontal acceleration? Is there a horizontal force? Do projectiles in motion relate to free-fall?

R –

S – Projectiles travel with a parabolic trajectory due to downward acceleration from gravity. The Y component and X component are independent of each other – they don’t have influence upon each other. Horizontal acceleration is zero and velocity is constant.

T – They travel with a parabolic trajectory. No, it affects the rate at which it falls, not the horizontal velocity. No, horizontal acceleration is zero because velocity is constant. No, no horizontal acceleration means no horizontal force. Yes, acceleration is -9.8 m/s^2 downward.



10/21/11 Lesson 2 Part c - Method 3
Parts 1 and 2

//Lesson 2 Section C Part 1//

P - to use the numerical factors of projectile trajectory and explain its conceptual features. Q - How is a projectile motion affected by horizontal movement? What are some good things to know about projectiles in motion? What happens when an object is launched horizontally? What happens when a projectile is launched at an angle? How does the vertical component of the projectile change during its launch?

R -

S - Projectiles have different ways of being launched. In some cases, Vy = 0 and in others it has a greater value. They can be launched at angles or horizontally. Time is the same for the x and y component.

T - 1) Horizontal velocity is constant and doesn't change. No force affects it.

2)
 * A projectile is any object upon which the only force is gravity,
 * 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 (a never changing in value),
 * There is a vertical acceleration caused by gravity; its value is 9.8 m/s/s, down,
 * The vertical velocity of a projectile changes by 9.8 m/s each second,
 * The horizontal motion of a projectile is independent of its vertical motion.

3) An object is launched horizontally; it is also like being dropped vertically except it travels a certain distance along the x axis (ground)

4) This is when an object is launched at an angle from the horizontal. Therefore, the vertical velocity would be different from vertical velocity during a horizontal launch.

5) It changes at a rate of 9.8 m/s^2 downward (towards ground). Every second the vertical velocity is increasing by -9.8 m/s.

Lesson 2 Section C Part 2

P - The only force acting upon a projectile is gravity. Vertical acceleration is -9.8 m/s^2 due to gravity on earth.

R -

Q - - Is there a relationship between the horizontal and vertical magnitudes of velocity? - Which of the component’s velocities is constant? Why? - How can the vertical component’s ∆d be found? - What is the best equation to find the horizontal displacement? - What happens if gravity is absent in a projectile launch?

S - The horizontal velocity is always constant due to the fact that there is no horizontal acceleration and because the horizontal and vertical components are independent of each other. Vertical velocity is increasing by -9.8 m/s^2 de to gravity on earth.

T - - Yes, horizontal and vertical components are independent of each other, so therefore it applies to the values as well. - The x component’s velocity is constant. This is so because there is no force or acceleration acting upon the x component. - The equation ∆dy = vit + ½ at^2 is used and the component’s other variables are plugged in to solve for the ∆d. - ∆dx = Vx(t) + 0t. This is the best equation, or v = d/t, because there is no acceleration, therefore no change in velocity. - The Projectile will continue at a constant speed without hitting the ground or speeding up, since there is no horizontal or vertical acceleration =Lab - Orienteering Vector Addition=

Data:

Starting Point is door closest to the building.


 * Leg || Distance || Direction ||
 * 1 || 7.05 m || East ||
 * 2 || 5.72 m || South ||
 * 3 || 17.34 m || East ||
 * 4 || 5.68 m || South ||
 * 5 || 2.95 || West ||

Analytical on top, graphical on bottom



Percent Error -

25.30 meters was the resultant vector that was measured, the graphical resultant was 23.80 meters and the analytical resultant was 23.40 meters. The analytical method was more accurate than the graphical method because the percent error for the analytical was 4.12% vs. the graphical percent error of 6.30%. Also, the analytical method was closer to the measured resultant. A source of error was the fact that I rounded two decimal places, which could have distorted my resultant of the analytical method. I can fix this problem by rounding to four or 5 decimal places. This will lead to a more accurate result.
 * Conclusion: **

= Lab: 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.

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?
 * Pre-Lab Questions: **

To be able to find time, I would need to the distance of the starting point to the ending point on the ground. If it is dropped, the initial velocity is assumed to be 0 m/s and the acceleration due to gravity is -9.8 m/s^2.

2.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 would travel before it hits the ground.

Since three of four y component variables are known, I would use them to solve for time using ∆d = vi(t) + ½a(t^2). The ball is traveling 0º counterclockwise from the horizontal, thus making the y-initial velocity zero. The acceleration due to gravity is 9.8 m/s^2 downward (negative). ∆d for y would be from the shooter’s ball chamber to the ground. I would solve for the ∆dx using time. The vi, horizontal acceleration, and time are known so it is possible to solve for this.

This is made possible since the initial velocity is known, the acceleration is 0 m/s, and the time is known.

3.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?

I would need to know the hang time and the object’s range. The hang time is calculated by the Photogate. 4.What data will you need to collect? Remember that you must run multiple trials. Keep in mind your end goal!

The data we need to collect is the distance from the launcher to the ground where the ball hits. This is the distance from the shooter to the carbon paper. This will give us enough information to solve for the Vi at the medium range of the launcher.

5.How will you analyze your results in terms of precision and/or in terms of accuracy?

I will analyze my results pertaining to precision for the carbon paper testing. I will shoot the ball several times and record the distances that they traveled. This will tell if the distance is precise because if the dots are far apart, it will not be considered precise. If they are close together, that is a sign of precision.

Using both precision and accuracy, I will analyze my results for the ball in the cup experiment. To analyze my results with precision, I will see if the ball landed in the same general area after each time it is launched. Using the percent difference equation I will be able to find the precision. To analyze the results for accuracy, I will use the theoretical position, which is the average of the dots on the paper, and compare them to where the ball actually lands on the carbon paper.

** Procedure: ** 1. Shoot ball at "medium range" 2. Put carbon paper on floor where ball seems to be landing 3. Shoot the ball onto the carbon paper several times. 4. Measure the distance from shooter to each dot on the carbon paper. 5. Take the average of the distances. b) Ball in cup (video)
 * a) Carbon paper measurement (written)

media type="file" key="Ball+in+cup+movie.mov" width="300" height="300"

**Calculations:** a) How fast does the launcher shoot the ball at medium range? A: 6.85 m/s

b) Change initial height, calculate where to place cup on floor so that the ball lands inside. Position calculated - 2.80 meters away from the launcher.



Analysis -

**Actual Position** of the cup where the ball landed in - 2.55 meters away from the launching point. **Theoretical Position** of the cup - 2.80 meters away from the launching point.



Conclusion -

This experiment was meant to show students how to figure out a projectile's initial velocity by meeting the objective's requirements. The initial velocity at the medium range setting of my launcher was 6.85 m/s.

=Lab - Shoot Your Grade=

** Hypothesis: ** If the 5 hoops hanging from the ceiling are aligned correctly, the ball should go through all of them and land in the cup at the end.

** Purpose with Rationale: **
 * My goal for this lab is to shoot a ball through 5 hoops and try to get it to land in a cup at the end. I plan to solve for the exact position each hoop. I plan to do this arithmetically because this will solve for the theoretical positions of the hoops. This is a rational idea because theoretically it will go through all hoops without hitting them at all.

** Materials and Methods: ** Materials – Ball, launcher, masking tape to serve as hoops, tape measure, black string to hang the “hoops”.
 * Materials:

Methods – The launcher is set to medium range. The hoops were rolls of masking tape that were hung from the ceiling. I used the tape measure to measure the distance correctly from the ground to the middle of the hoop so that it matched my calculations. I used a string with a weight at the end to take accurate measurements. This determined the exact horizontal position of the hoop.

Must confirm initial velocity Choose a horizontal position for each hoop, relative to the launcher. Use the known x-variables to solve for time with this equation: d=(vi)(t)+(1/2)(a)(t^2) for each hoop. Use the time to solve for y-distance with the same equation: d=(vi)(t)+(1/2)(a)(t^2) for each hoop. Hang each hoop from the ceiling at its calculated horizontal and vertical position. Place a cup at the ending position that was calculated. Shoot the ball from the launcher at the medium launch setting.
 * Procedure:

** Observations and Data from Initial Velocity: ** To confirm the initial velocity, I shot the ball at 20 degrees at the medium range setting onto a sheet of carbon paper several times. I took the measurements of the dots on the carbon paper to the launcher. I then took the average of these distances.

Confirming Vi

I used the time that I solved for (0.45 s) by using the known y variables: vi = 0, a = -9.8, and d = -1.1582 and the known x variables: a = 0 and d = 4.931 to solve for initial velocity. My initial velocity was 6.8 m/s, which was different from my earlier determined initial velocity of 6.85 m/s. Verifying the initial velocity was important, because all of the following calculations were dependent on the initial velocity.


 * Observations and Data from Performance: **

Trial (video)

media type="file" key="Shooter_FINAL.m4v" width="300" height="300"

Trials -



** Physics Calculations - **

The assumptions that can be made are that vertical accelerations is always -9.8m/s^2 due to gravity. Also, I can assume that the horizontal acceleration is always zero, because projectiles have no horizontal force acting upon them.

Calc. Table for each individual hoop



Calc. Table and Sample Calculation: Hoop 1

Initial Velocity -



**Error Analysis -**

The hoops were measured carefully (from the ground to the middle). I put the data in a data table.

Percent Error -

Percent Error for the First Hoop -

Percent Error = (theoretical - experimental) / theoretical x 100 Percent Error = (0.1961-0.1982)/0.1961 x 100 = 1.07%

Chart of Percent Error for Each Hoop -



We only solved for the y components' % error because the hoops were suspended from the ceiling at a certain distance from the floor. They were based off of the x components' distance from the launcher. There was a large percent error on the fourth hoop, which is the hoop we missed. We weren't able to calculate the Cup's percent error since it didn't land in the cup. This experiment was pretty successful.

**Conclusion:** Was the purpose satisfied?
 * Was your hypothesis correct? Provide __specific__ evidence from the experiment to justify your claims.
 * My hypothesis was correct. Although we didn’t accomplish the goal of getting the ball through all of the five hoops, as well as not getting it into the cup. Our fourth hoop wasn’t properly aligned because my group was in a time crunch and had little time to align all five. My groups launch successfully made it though the first, second, third, and fifth hoops, but not through the fourth. If we had aligned the fourth hoop well, the ball would have went through all of the hoops. This shows my hypothesis was true.

**Conclusion:** Experimental Errors Our first source of error occurred while we were hanging up the hoops to the ceiling. To adjust the hoop, the strings had to be pulled, which were attached to the hoop. It was difficult to adjust. It was impossible to adjust it to the perfect position, vertically and horizontally. The strings were never even and one was always slightly higher than the other. It didn’t cause too much of a problem since the ball was much smaller than the hoop, so it still had room to enter through it.
 * How much? Where did the error occur? Why did the error occur? 2-3 sources of error should be described thoroughly.

The second source of error was when the launcher shot the ball, but each time it was set, it moved back to the left or right slightly, throwing off the alignment. Pushing the ball back against the spring caused this. This had a great effect on our trials because if it moved to one side, it shot to that side, which meant it had a greater chance of hitting the hoop rather than going through it.

A third source of error was when pushing the ball against the spring; it moved the angle of the launcher. This is significant because hoops were based on the angle that we were assigned. Although it didn’t all that much from our given angle (from 20º to about 21.2º), it did cause issues, such as hitting the hoops once launched or missing them completely.

**Conclusion:** Implications for further discussion A way to fix the string issue is to measure them to make sure that they are the same distance from the ceiling and that they are the same length to avoid them being at different heights. I would be able to see exactly how much I need to adjust the hoops. To avoid the launcher from shifting its position, I would use two clamps to keep it steady. I could also tape around the launcher so that I can easily adjust it to its original position. To amend the launcher’s shift in angle, I would make sure to check and adjust the angle to its correct position.
 * How would you change the lab to address the error? What is a relevant real-life application of this concept? (Why is this important to know/understand?)

This concept applies to archery. The arrow is a projectile being launched at a certain velocity. It has to have a certain angle when launched to get bulls-eye. The angle and bow are sometimes shifted, like the launcher. A bow has a low, medium, and high range, depending on how far back it is pulled. Bulls-eye is considered spot on and correct. It takes certain time to reach the bulls-eye, and there is a distance between the target and bow. Calculating correctly depends on whether the arrow hits the bulls-eye or not.

=Gourd-o-rama=

Remzi Tonuzi and Lindsay Marella

Materials- Wood, screws, plastic washers (spacers), roller-blade wheels, pumpkin, Halloween basket, shoelaces

We made a cart out of wood, using screws to assemble it. A Halloween basket was screwed on. We attached roller-blade wheels to the cart so that it would glide smoothly across the hallway. The plastic washers were used to decrease the amount of friction between the wood and the wheels. We used the shoelaces to secure the pumpkin so that it wouldn't fall out due to contact with the ground once it reaches the end of the ramp. We intended it to go much further than 8 meters, but two of the wheels were loose and caused it to veer off into the wall.

Pictures -





Calculations -

Mass of Vehicle: 1.91 kg ∆D: 8 meters Time: 2.73 seconds Initial Velocity: - 5.9 m/s Acceleration: -2.2 m/s^2