Ch3_BarinskyL

toc Chapter 3.

Lesson 1 a-h
10/12/11, 10/13/11, 10/17/11, 10/18/11

Vectors and Direction Quantities that describe the physical world include distance, displacement, speed, velocity, acceleration, force, mass, momentum, energy, work, power, etc. All these quantities can by divided into two categories, vectors and scalars.

__Vector__: a quantity that is fully described by both magnitude and direction. ex: displacement, velocity, acceleration, force __Scalar__: a quantity that is fully described by its magnitude.

Vector diagrams depict a vector by use of an arrow drawn to scale in a specific direction. The vector diagram on the right depicts a displacement vector. There are several characteristics of this diagram that make it an appropriately drawn vector diagram.
 * 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. In this case, the diagram shows the magnitude is 20 m and the direction is (30 degrees West of North).

Conventions for Describing Directions of Vectors There is a clear 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. Two conventions are described 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. The diagrams below use the scale 1 cm =5 miles. The length of each vector is 1 cm for every 5 miles being represented in each diagram.
 * 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.
 * 2) Ex: 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) of 65 degrees East of South (meaning a vector pointing South has been rotated 65 degrees towards the easterly direction).
 * 3) The direction of a vector is often expressed as a counterclockwise angle of rotation of the vector about its "tail" from due East.
 * 4) Ex: 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. A vector with a direction of 270 degrees is a vector that has been rotated 270 degrees in a counterclockwise direction relative to due east. This is one of the most common conventions for the direction of a vector and will be utilized throughout this unit.

Vector Addition Two vectors can be added together to determine the result (or resultant). The //net force// experienced by an object is determined by computing the vector sum of all the individual forces acting upon that object. That is the net force was the result of adding up all the force vectors. Observe the following summations of two force vectors:

Two methods for determining the magnitude and direction of the result of adding two or more vectors: 1) Pythagorean theorem and trigonometric methods 2) Head-to-tail method using a scaled vector diagram

The Pythagorean Theorem 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. __Example problem:__ Eric leaves the base camp and hikes 11 km, north and then hikes 11 km east. Determine Eric's resulting displacement.

This problem asks to determine the result of adding two displacement vectors that are at right angles to each other. The result of walking 11 km north and 11 km east is a vector directed northeast as shown in the diagram below. Find the hypotenuse of the triangle. The result of adding 11 km, north plus 11 km, east is a vector with a magnitude of 15.6 km.

Using Trigonometry to Determine a Vector's Direction The direction of a //resultant// vector can often be determined by use of trigonometric functions. These three trigonometric functions can be applied to the previous hiker problem in order to determine the direction of the hiker's overall displacement. The process begins by the selection of one of the two angles (other than the right angle) of the triangle. Once the angle is selected, any of the three functions can be used to find the measure of the angle. Once the measure of the angle is determined, the direction of the vector can be found. In this case the vector makes an angle of 45 degrees with due East. Thus, the direction of this vector is written as 45 degrees.

The measure of an angle as determined through use of SOH CAH TOA is __not__ always the direction of the vector. The following vector addition diagram is an example of such a situation. Observe that the angle within the triangle is determined to be 26.6 degrees using SOH CAH TOA. This angle is the southward angle of rotation that the vector R makes with respect to West. Yet the direction of the vector as expressed with the CCW (counterclockwise from East) convention is 206.6 degrees.

Use of Scaled Vector Diagrams to Determine a Resultant The magnitude and direction of the sum of two or more vectors can also be determined by use of an accurately drawn scaled vector diagram. Using a scaled diagram, the __head-to-tail__ is employed to determine the vector sum or resultant.

The head-to-tail method involves drawing a vector to scale on a sheet of paper beginning at a designated starting position. Where the head of this first vector ends, the tail of the second vector begins (thus, //head-to-tail// method). The process is repeated for all vectors that are being added. Once all the vectors have been added head-to-tail, the resultant is then drawn from the tail of the first vector to the head of the last vector; i.e., from start to finish. 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.

__Ex:__ Add the 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

The order in which three vectors are added has no affect upon either the magnitude or the direction of the resultant. The resultant will still have the same magnitude and direction.

Consider the addition of the same three vectors in a different order. 15 m, 210 deg. + 25 m, 300 deg. + 20 m, 45 deg. SCALE: 1 cm = 5 m

When added together in this different order, these same three vectors still produce a resultant with the same magnitude and direction as before (20. m, 312 degrees). The order in which vectors are added using the head-to-tail method is insignificant. SCALE: 1 cm = 5 m

Resultants __Resultant__: the vector sum of two or more vectors. If displacement vectors A, B, and C are added together, the result will be vector R. As shown in the diagram, vector R can be determined by the use of an accurately drawn, scaled, vector addition diagram. 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

The resultant can be determined by adding the individual forces together using vector addition methods.

Vector Components A vector is a quantity that has both magnitude and direction. We begin to see examples of vectors that are directed in //two dimensions// - upward and rightward, northward and westward, eastward and southward, etc.

When vectors are directed at angles to the customary coordinate axes, a useful mathematical trick will be employed to //transform// the vector into two parts (or influences) with each part being directed along the coordinate axes (a vector that is directed northwest can be thought of as having northward part and a westward part). __Component__: each part of a vector The components of a vector depict the influence of that vector in a given direction. The component of a single vector describes the influence of that vector in a given direction. The combined influence of two components is equivalent to the influence of a single two-dimensional vector. The single two-dimensional vector could be replaced by the two components.

Example: If a dog's chain is stretched upward and rightward and pulled tight by his master, then the tension force in the chain has two components - an upward component and a rightward component. To the dog, the influence of the chain on his body is equivalent to the influence of two chains on his body - one pulling upward and the other pulling rightward. If the single chain were replaced by two chains with each chain having the magnitude and direction of the components, the dog would not know the difference. This is because the combined influence of the two components is equivalent to the influence of the single two-dimensional vector.



Vector Resolution Any vector directed in two dimensions can be thought of as having two components. Determining the magnitude of a vector is known as __vector resolution__. The two methods of vector resolution are:
 * the parallelogram method
 * the trigonometric method

Parallelogram Method of Vector Resolution This 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.
 * 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. 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 use the scale to determine the magnitude of the components in //real// units. Label the magnitude on the diagram.

__Example:__ Determining Vx and Vy of a velocity vector with a magnitude of 50 m/s and a direction of 60 degrees above the horizontal: Trigonometric Method of Vector Resolution The trigonometric method of vector resolution involves using trigonometric functions to determine the components of the vector. This method can be used to determine the length of the sides of a right triangle if an angle measure and the length of one side are known.
 * 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 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 nearest corner of the rectangle. 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 force velocity component might be labeled vx; etc.
 * 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. Use some algebra to solve the equation for the length of the side opposite the indicated angle.
 * 6) Repeat the above step using the cosine function to determine the length of the side adjacent to the indicated angle.

__Example:__ Determining the components of a 60-Newton tension force acting upward and rightward on a dog at an angle of 40 degrees:

Relative Velocity and Riverboat Problems Objects may 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. In such instances as this, 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. Motion is relative to the observer.

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. If a motorboat were to head straight across a river, it would not reach the shore directly across from its starting point. The river current influences the motion of the boat and carries it downstream. The motorboat may be moving with a velocity of 4 m/s directly across the river, yet the resultant velocity of the boat will be greater than 4 m/s and at an angle in the downstream direction. While the speedometer of the boat may read 4 m/s, its speed with respect to an observer on the shore will be greater than 4 m/s. 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. Suppose that the river was moving with a velocity of 3 m/s, North and the motorboat was moving with a velocity of 4 m/s, East. The magnitude of the resultant can be found as follows: (4.0 m/s)2 + (3.0 m/s)2 = R2 25 m2/s2 = R2 SQRT (25 m2/s2) = R **5.0 m/s = R** The direction of the resultant is the counterclockwise angle of rotation that the resultant vector makes with due East. This angle can be determined using a trigonometric function as shown below. tan (theta) = (opposite/adjacent) tan (theta) = (3/4) theta = invtan (3/4) **theta = 36.9 degrees** The resultant velocity of the boat will be 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? ||

The second and third of these questions can be answered using the average speed equation. ave. speed = distance/time Most students want to use the resultant velocity in the equation since that is the actual velocity of the boat with respect to the shore. Yet the value of 5 m/s is the speed at which the boat covers the diagonal dimension of the river, and the diagonal distance across the river is not known in this case. If we know Distance A in the diagram below, then the Avg. Speed A could be used to calculate the time to reach the opposite shore. 80 m corresponds to the distance A, and so the average speed of 4 m/s (average speed in the direction straight across the river) should be substituted into the equation to determine the time. time = (80 m)/(4 m/s) = 20 s  It requires 20 s for the boat to travel across the river. During this 20 s of crossing the river, the boat also drifts downstream. Part C of the problem asks "What distance downstream does the boat reach the opposite shore?" The same equation must be used to calculate this //downstream distance//. The distance downstream corresponds to Distance Bon the above diagram. The speed at which the boat covers this distance corresponds to Avg. Speed Bon the diagram above, so the average speed of 3 m/s (average speed in the downstream direction) should be substituted into the equation to determine the distance. 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.

The current velocity itself has no affect upon the time required for a boat to cross the river. The river moves downstream parallel to the banks of the river. As such, there is no way that the current is capable of assisting a boat in crossing a river. While the increased current may affect the resultant velocity - making the boat travel with a greater speed with respect to an observer on the ground - it does not increase the speed in the direction across the river.

Independence of Perpendicular Components of Motion A component describes the affect of a single vector in a given direction. Any force vector that is exerted at an angle to the horizontal can be considered as having two parts or components. The vector sum of these two components is always equal to the force at the given angle. This is depicted in the diagram below. The two perpendicular parts or components of a vector are independent of each other. A change in one component does not affect the other component. 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.

__Example:__ If a plane is flying North at 100 km/hr and the wind is moving East at 25 km/hr, a change in the velocity of the wind will not change the Northward component (the plane's velocity according to the speedometer). If the wind's velocity increases to 35 km/hr, the plane will move Eastward faster, but it's Northward velocity (Northward component) will not be altered. The plane will travel X km in the Northward direction in the same amount of time no matter the wind's velocity. The only difference will be the km travelled in the Eastward direction.

Lab 1: Orienteering
10/14/11

//PART 1://


 * Leg || Actual distance (m) || Direction || Measured distance (m) || Calculated distance (m) ||
 * 1 || 2.0 || S ||  ||   ||
 * 2 || 6.8 || W ||  ||   ||
 * 3 || 14.6 || S ||  ||   ||
 * 4 || 15.1 || W ||  ||   ||
 * 5 || 2.8 || N ||  ||   ||
 * Resultant || 25.30 || - || 25.75 || 25.89 ||

__Analysis__:

Actual resultant: 25.30 m

Finding the resultant by drawing a scaled diagram: Measured resultant: 25.75 m

Finding the resultant by calculating: Calculated resultant: 25.89 m

Percent error between __measured__ (experimental) resultant & __actual__ (theoretical) resultant:

Percent error between __calculated__ (experimental) resultant & __actual__ (theoretical) resultant:

Both percent errors are very low, showing that both methods were fairly accurate. However, the drawing-to-scale-and-measuring method proved to be more accurate in this case because its percent error is lower. I find this to be interesting because I would assume that calculating a resultant would be more dependable and accurate than drawing and measuring one by hand.

//PART 2://


 * Leg || Actual distance (m) || Direction || Measured distance (m) || Calculated distance (m) ||
 * 1 || 7.00 || E ||  ||   ||
 * 2 || 9.25 || S ||  ||   ||
 * 3 || 17.48 || E ||  ||   ||
 * 4 || 9.25 || S ||  ||   ||
 * 5 || 24.50 || E ||  ||   ||
 * 6 || 18.30 || S ||  ||   ||
 * Resultant || 61.35 || - || 60.81 || 61.26 ||

__Analysis:__

Actual resultant: 61.35 m

Finding the resultant by drawing a scaled diagram: Measured resultant: 60.81

Finding the resultant by calculating: Calculated resultant: 61.26

Percent error between __measured__ (experimental) resultant & __actual__ (theoretical) resultant:

Percent error between __calculated__ (experimental) resultant & __actual__ (theoretical) resultant:

Both percent errors are very low, showing the success of our two experimental methods. The calculation method proved to be more accurate in this case because its percent error is lower. Using the measuring method might not be as accurate because it is easy to make small mistakes when drawing and measuring by hand.

Lesson 2 a-c
10/19/11, 10/20/11

A: 1) What is a projectile? -//A projectile is an object that has gravity as the only force acting on it.// 2) If an projectile's horizontal speed is constant, what keeps it going? //-Inertia says that an object will remain at the speed// //it is going until another force acts upon it.// 3) What does the free-body diagram of a projectile look like? -//It shows a single downward force (gravity).// 4) What is the difference between projectiles and freely falling objects? -//According to their definitions, they are the same. However, freely falling objects move in one dimension, while projectiles move in either one or two dimensions.// 5) What would happen if there was no gravity? //-The object would continue moving at the same speed in the same direction because of inertia.//

__Central theme__: A projectile can move in two dimensions, only has gravity acting upon it, and continues moving horizontally because of inertia.

B: 1) What does gravity affect? -//Gravity affects the vertical motion of a projectile but has no influence on the horizontal motion (until the object reaches the ground of course, but then the object stops moving horizontally because of the friction between it and the ground, not directly because of gravity).// 2) Why doesn't gravity influence the horizontal motion? -//The vertical downward force (gravity) acts perpendicular to the horizontal motion and will not affect it because perpendicular components of motion are independent of each other.// 3) Is there an acceleration on projectiles? -//There is no horizontal acceleration because the horizontal motion remains constant, but there is a vertical acceleration (-9.8 m/s/s) due to the downward pull of gravity.// 4) What path does a projectile follow? //-Projectiles travel with a parabolic trajectory resulting from gravity's influence.// 5) How does a projectile's position change? -//The downward force of gravity results in a downward displacement from the position that the object would be if there were no gravity.//

__Central theme__: Gravity has no influence over the horizontal motion of a projectile (horizontal velocity remains constant) but does influence the vertical motion (negative acceleration).

C: 1) How do the vertical and horizontal velocity of a horizontally launched projectile change over time? //-The horizontal velocity remains constant, and the vertical velocity changes by 9.8 m/s every second.// 2) Do the vertical and horizontal velocities act differently over time if the projectile is launched upward at an angle? //-No. The// //horizontal velocity still remains constant, and the vertical velocity still changes by 9.8 m/s every second.// 3) Does a projectile ever have the same speed at two points in time? //- Yes. The projectile's speed is symmetrical about the peak of its position. If an object is thrown at 45 degree angle, its speed at 2 seconds before the peak is the same as it is 2 seconds after the peak.// 4) How does the displacement of a horizontally launched projectile change over time? //-The horizontal displacement changes by the same amount every second, but the vertical displacement changes by a larger amount every second.// 5) How can you determine the vertical displacement of a projectile? //-You can use the equation y = (V iy )(t) + .5(g)(t 2 )//

__Central theme__: A projectile's trajectory has a symmetrical nature. Vertical displacement changes by a different amount every second due to acceleration, while horizontal displacement changes at a constant rate due to the 0 acceleration.

Activity: Ball in Cup
10/24/11

__Part 1: Find the initial velocity of the ball when it is launched.__

Measured vertical height: 0.986 m Measured horizontal distance traveled by ball: 2.85 m (see table below)
 * Trial || Distance (m) ||
 * 1 || 2.75 ||
 * 2 || 2.77 ||
 * 3 || 2.78 ||
 * 4 || 2.78 ||
 * 5 || 2.83 ||
 * 6 || 2.94 ||
 * 7 || 2.92 ||
 * 8 || 2.96 ||
 * 9 || 2.97 ||
 * **Avg** || **2.85** ||

//Calculating initial velocity://

Y: d = Vit + .5at 2 -0.986 = 0t + .5(-9.8)(t 2 ) t= 0.45 s
 * =  ||= **X** ||= **Y** ||
 * = **Vi** ||= ? ||= 0 ||
 * = **a** ||= 0 ||= -9.8 m/s/s ||
 * = **t** ||= ? ||= ? ||
 * = **d** ||= 2.85 m ||= -0.986 m ||

X: d = Vit + .5at 2 2.85 = Vi(0.45) + .5(0)(0.45) 2 **Vi = 6.35 m/s**

__Part 2: Using the initial velocity, find where to place a cup so that the ball lands in it when launched from a different vertical height.__

New vertical height: 0.775 m

//Calculating horizontal distance://

Y: d = Vit + .5at 2 -0.775 = 0(t) + .5(-9.8)(t 2 ) t= 0.398 s
 * =  ||= **X** ||= **Y** ||
 * = **Vi** ||= 6.35 m/s ||= 0 ||
 * = **a** ||= 0 ||= -9.8 m/s/s ||
 * = **t** ||= ? ||= ? ||
 * = **d** ||= ? ||= -0.775 m ||

X: d = Vit + .5at 2 d = 6.35(0.398) + .5(0)(0.398) 2


 * d = 2.53 m**

Theoretical distance the cup should be placed at: 2.53 m Actual distance we placed the cup at for success: 2.47 m

One trial (miss): media type="file" key="ballincuptrialshort.mov" width="270" height="270"

//Calculating percent error://



__Discussion__ Our experiment yielded a percent error of 2.37. We were only off by .06 m. One main source of error came from the inconsistency of our launcher. We noticed that the ball would land in different spots all over the piece of paper when we kept everything the same. To fix this source of error, we would either have to get a different, consistent launcher (maybe a new one with a better spring) or use a ramp instead to ensure accuracy. Another source of error came from the movement of the ball within the launcher. We noticed that when we would load it, sometimes the ball would move to the end of the launcher where we could see it. When we did see it, we pushed it back into the launcher. However, I'm sure there were times at which we launched the ball while it was not all the way in, which changed the results from time to time. This may have been happening if the surface we attached the launcher to was not perfectly level. To minimize this source of error, we should make sure the the surface is level and check that the ball stays all the way back when loaded before we start the experiment. We could also make sure to launch the ball as quickly as possible after loading it so it does not have time to roll to the end instead of loading it and waiting to launch it. If we did an experiment with the launcher set at an angle, the ball would probably stay back. Also, the spring in the launcher takes some some launches to warm up. Launches done with a cold spring probably yield a shorter horizontal displacement compared to launches done with a warmed up, looser spring. To minimize error from this source, multiple launches should be done prior to the lab trials that will be recorded so that the spring can warm up.

Lab 2: Shoot Your Grade
Lab Partners: Timothy Hwang, Jonathan Itskovitch 10/28/11

__Purpose:__ The aim of this lab was to measure the initial velocity of a ball launched at a certain angle, to predict the impact point of the ball, and predict the ball's position at five different points mid-flight. A projectile is only influenced by gravity. When launched at an angle, it will follow a parabolic trajectory due to the pull of gravity. The accuracy of the predicted five positions of the ball will be tested by suspending rings in those places and observing whether or not the ball launches through them. Our launcher was set at an angle of 20 degrees, and the initial height of the ball was measured to be 1.17 m.

__Materials and Method:__ Launcher, ball, paper, carbon paper, 5 rolls of masking tape, string, cup, measuring tape

We first find the initial velocity of the ball shot from our particular launcher by doing a sample launch, taping a piece of paper to the ground in the region that the ball landed, placing carbon paper over it, and launching the ball 7 times. Each launch yields a spot on the paper where the ball landed, which enables us to measure the horizontal displacement of the ball. This part of the procedure was previously done in the Ball in Cup activity. The average horizontal displacement of the ball, along with the measured vertical displacement (initial height), is used to find the initial velocity of the ball.

Once the initial velocity is found, positions of where to put the rings and cup are calculated. The rings are rolls of masking tape suspended from ceiling tiles using string. Two lengths of string are used for each ring to minimize movement. The first ring is positioned; the ball is launched and the ring position is adjusted until the ball successfully launches through it multiple times. The same is repeated for the next ring, adding rings along the ball's trajectory until the ball successfully launches through 5 rings and lands in a cup at the end. In the picture on the right, a cup is shown at the predicted impact point and five rings are suspended along the ball's trajectory. This view is looking straight at the launcher.

__Observations and Data:__

After seven trials, the following horizontal displacements of the ball were measured.
 * Trial || Horizontal Displacement (m) ||
 * 1 || 2.96 ||
 * 2 || 2.96 ||
 * 3 || 2.90 ||
 * 4 || 2.93 ||
 * 5 || 2.93 ||
 * 6 || 2.94 ||
 * 7 || 2.94 ||
 * **Avg** || **2.937** ||

//Finding initial velocity:// Initial velocity of ball: 4.623 m/s

//Calculating ring positions://

To find the horizontal (x) and vertical (y) displacement of the ball, we used five 0.10 second intervals. One ring will be placed at each of the five positions along the ball's trajectory corresponding to the calculated displacements at each interval.

Sample calculation, Ring #1:

Distance from ground = (initial height, 1.17 m) + (vertical displacement) Distance from launcher = horizontal displacement

The following table shows the theoretical values of the ring positions: The theoretical position of the cup was 2.937 m away from the launcher because this was the average horizontal displacement of the ball.
 * Ring || Time (s) || Horizontal Displacement (m) || Vertical Displacement (m) || Distance from ground (m) || Distance from launcher (m) ||
 * 1 || 0.1 || 0.43 || 0.11 || 1.28 || 0.43 ||
 * 2 || 0.2 || 0.87 || 0.12 || 1.29 || 0.87 ||
 * 3 || 0.3 || 1.30 || 0.03 || 1.20 || 1.30 ||
 * 4 || 0.4 || 1.74 || -0.15 || 1.02 || 1.74 ||
 * 5 || 0.5 || 2.17 || -0.43 || 0.74 || 2.17 ||

//Performance//:

The ball successfully went through five rings. Although it did not land inside of the cup, it did hit it. media type="file" key="ballthroughrings.mov" width="293" height="293"

__Analysis:__

We did not adjust the horizontal positions of the rings, only the vertical positions. The distance of each ring from the ground was measured when we successfully shot the ball through all five rings. The table below compares these vertical heights to the ones we calculated earlier and originally placed the rings at. Since the ball did not land in the cup, we could not evaluate our placing of it at 2.937 m.
 * Ring || Calculated/Theoretical Vertical Height (m) || Actual/Experimental Vertical Height (m) || Percent Error ||
 * 1 || 1.28 || 1.26 || 1.56 ||
 * 2 || 1.29 || 1.27 || 1.55 ||
 * 3 || 1.20 || 1.21 || 0.83 ||
 * 4 || 1.02 || 1.02 || 0.00 ||
 * 5 || 0.74 || 0.70 || 5.41 ||

Example calculation of percent error, Ring #1:

__Conclusion:__

The purpose was satisfied. The other lab groups used the same launcher they used in the Ball in Cup activity and just verified the initial velocity they had already calculated while our group switched to a more consistent launcher and found the initial velocity for the first time. We found the positions along the ball's trajectory to place five rings and adjusted them until the ball successfully went through all of them. We found where to place the cup, but we did not get the ball to land in it.

Our experiment yielded minimum error. The percent errors for the five rings ranged from 0 to 5.41, although the majority were within the 0 to 1.56 range. The fourth ring yielded a percent error of 0, meaning that we had correctly calculated the position of the ball at 0.4 seconds. The fifth ring yielded a percent error of 5.41, the highest one of the five. This is probably because the inconsistency of the launcher is usually seen more apparently farther along the launch. Even so, the position of the fifth ring was only off by 0.04 m. The rest of the positions of the rings were off by 0.02 m or less.

One source of error could be that the launcher shoots the ball differently depending on how much it is used. If it is not used for a few days, the internal spring is cold and may be weaker when launching the ball than when it has been used already and the spring is warm and loose. Because we switched to a launcher that no one had been using previously, this factor probably caused our horizontal displacements measured in the first part of the lab to be slightly different than what they actually ended up being during the rest of the trials throughout the lab. This means that we had calculated positions according to the inaccurate, lesser horizontal displacement, and the ball was actually being launched slightly farther when we were attempting to shoot it through the rings. This source of error could be minimized by doing multiple launches to warm up the spring before doing any actual trials for the lab.

Another source of error came from the movement of the launcher head. Even when the launcher is held down by someone in addition to the clamp keeping it steady, the part of the launcher that actually projects the ball was found to change its launching angle very easily. We needed to keep the angle at 20 degrees to get consistent results, but this was difficult because the screws keeping the angle stable would constantly loosen. After careful observing, we found that even when we would check the angle and tighten the screws after each trial, the angle would often change mid-launch. Even the slightest change in angle could throw off the results, sending the ball off of its normal path. This source of error could be minimized by (1) replacing the screws and (2) using some type of sturdy fixture independent from the launcher to hold the launcher head in its desired place. This would prevent the launcher head from moving up and down and undesirably changing the angle.

Applications of these concepts can be seen anywhere. Any object that is projected through the air is a projectile (ignoring air resistance). All of these objects will follow a parabolic trajectory and act similarly to the ball launched in this lab. Examples of these objects include a baseball thrown across a field, a package dropped from an aircraft, and a snowboarder projected off of a jump. Given certain information, such as initial velocity and launch angle, the position of the object can be predicted at different points in time.

Gourd-o-Rama
Partner: Garrett Almedia 11/1/11



__Results:__ Out best trial was the third one, and this is the one we submitted to be graded.
 * Trial || Distance (m) || Time (s) || Initial Velocity (m/s) || Acceleration (m/s/s) ||
 * 1 || 7.3 || 2.57 || 5.68 || -2.21 ||
 * 2 || 7.35 || 1.97 || 7.46 || -3.79 ||
 * 3 || 10.5 || 5.49 || 3.83 || -0.70 ||

__Calculations:__

__Analysis:__ The main problem we had with our cart was making it go straight. It travelled fast and kept its speed well, but it would hit the side walls before it had a chance to roll to a stop. To fix this issue, the ends of the axes should be kept an equal distance apart from each other using some type of solid frame. This would prevent the axes from moving back and forth in our cardboard cart, thus enabling a straight travel.