Practicums For

Physics Teachers

(Second Edition)

Edited by Michael Crofton

Central Theme Developed

by

Henry J. Ryan and Jon E. Barber

08 Fall  

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Table of Contents

History and Motivation 4

Introduction 6

Light and Optics Image With a Mirror 9

Image With a Lens 11

How Intense 13

It is Critical 15

Relative Index of Refraction 17

Interference Call 19

Waves and Sound Interference Trombone 21

Kinematics Colliding buggies 23

Circular Motion Lazy Pendulum 25

Centripetal Acceleration 27

Centripetal Force 29

The Helicopter Ride 32

Projectiles Shot Heard Round the Lab 35

Projectile Impact Angle 37

Arkansas Traveler 39

Non-horizontal Projectile 41

Static Equilibrium Static Equilibrium for Beginners 45

Center the Mass 47

Beams in Equilibrium 49

Static Equilibrium for Nerds 51

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Dynamics The Runaway Cart 54

Canister and Accelerating Glider 56

An Uphill Climb 58

Suspended Pulley 60

Momentum/Energy

Spring Potential Energy 63

Humpty Dumpty 65

Ballistic Pendulum 68

Jumping Frame 71

Tarzan 74

Vector Conservation of Momentum 76

Gravitational Potential Energy 79

Electricity Resistivity 82

Blow the Circuit 85

Resistance Will Vary 87

Resistance is Futile 89

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History and Motivation (2006)

If frustration can be the mother of invention, then that’s how Physics Practicums were born. Hank Ryan, then a new physics teacher at Minnesota’s Mounds View High School, simply hated chapter test reviews. “These chapter reviews took exciting concepts and just flattened them,” Hank says. “I wanted kids to understand that physics concepts are a real live part of their world—and that they describe how the world behaves. I wanted to show them that physics is experienced. That it’s something they can see with their eyes, feel with their hands—and something they could learn to predict.” The idea of a class practicum came to him whole one day, in a flash of inspiration. Well, anyway, that’s Hank’s story and he is sticking to it. He worked out his first physics practicum on centripetal force and gave it a try in his class. Though it went over well, and he was excited about it, but the idea got lost in the whirlwind of school life. The next year, when Hank got back to circular motion, he decided to use the practicum again. This time, he invited his colleague Jon Barber to come see the practicum review session. Jon was impressed with the results. And being a natural organizer and motivator, he dug into the project with Hank. In fact, Jon was so good at inventing practicums that soon this team of two had eight practicums worked out in their heads, though not much on paper. In 1985 at a summer workshop for teachers held at the University of Minnesota, Hank presented the practicum concept to 40 physics teachers. The university liked the idea and made funds available for creating a written and somewhat illustrated version of the first eight practicums for the forty. That winter the University of Minnesota, through the efforts of Dr. Roger Jones, also made a video of Hank and his class doing a practicum. Hank claims to have perhaps the only remaining copy of the video—if he could only lay hands on it. In any case, with the video under their belts, Jon began applying to any granting authority that might have funds available to help further develop the idea. After many rejections, he secured a grant from the American Institute of Physics (AIP) for a few thousand dollars to develop new practicums and publish a book on how to use them. The summer of 1990 saw Hank and Jon spending most every day in the physics lab at Mounds View High School reading physics books, playing around with equipment, writing and illustrating practicums both new and old. It was a great summer. The work was exciting, and Jon was so funny, Hank says he often returned home with sore tummy muscles from all the laughing. In 1991, to fulfill a requirement of the grant, Hank and Jon went to Vancouver, British Columbia, to give a paper about their practicums at the summer meeting of the American Association of Physics Teachers (AAPT). With 25 draft copies of their book, Practicums for Physics Teachers, in hand they ventured forth. “It was an early session in fact we were the first in the day. So, I hoped that maybe ten people would come to hear us,” Jon remembers. Needing plenty of set-up time the two were up and working by 6:30 a.m. At 7:30, some teachers began to arrive. Hank asked why they were coming so early. “We wanted to be sure to get a

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seat!” they responded. Hank and Jon were mystified, as the room could seat 120 people. But by 8:00, the house was packed, with people sitting on the stairs to boot. Hank and Jon were truly blown away by the response. Needless to say, they were a tad short on draft copies of their practicum book. After the great reception in Vancouver, Jon and Hank gave workshops and sold books around the United States and Canada. They visited Boise, Idaho; Bangor, Maine; St. Louis, Missouri; Winnipeg, Canada; and beyond, as they spread the word about the value of these exercises. Physics teachers were thrilled and teachers from around the country, and as far away as Japan, sent the duo new practicums to be added to the publication.

If this is your first look at the concept, Hank and Jon hope you become a user and an advocate of the pedagogy. They have certainly seen positive results with their students. Jon and Hank would like to thank everyone who gave them encouragement along the way even if just by showing interest and a special thanks to those who made a contribution. They would also like to give a very special thanks to Michael Crofton for his work in editing the second edition. Both now retired, Hank and Jon have had a great journey with the practicum idea. To physics teachers and students everywhere, they say, “It’s been great fun, thanks for the memories, and good luck in developing your own physics practicums!”

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Introduction THE PHYSICS PRACTICUM The practicum is started by the teacher setting out laboratory equipment that allows data to be measured and then posing a problem for the class to solve. Students organize themselves and make the proper measurements and calculations in an attempt to find the answer to the question posed. After all students agree that the proper solution has been attained, the apparatus is assembled and the experiment performed. If the class prediction is correct, the students receive a pass, and if incorrect, all receive a fail. Much enthusiasm and satisfaction is generated because successful solutions are almost always achieved by the students INSTRUCTOR'S RESPONSIBILITY It is not a simple project to develop a problem that is equipment centered for use in a class practicum. All the needed data must be measurable in a relatively short time in front of the class, and the whole project must be designed to fit a single period. It complicates the situation if data needed cannot be measured and therefore must be announced to the class. The goal is to present a word problem without the words, and we suggest not having any unmeasured mystery numbers. However, the method of taking data is one way to control time during the hour. If the calculations are long and complex, the teacher makes most of the data measurements leaving much of the hour for students to work. Conversely, if the calculations are short with less complexity, which is often the case early in the year, the students can make most of their own data measurements. The practicum is always a culminating activity. The problem posed to the class should contain as many concepts that have been covered in the unit as possible. It should not require the use of relationships yet to be studied but can make use of concepts studied in the distant past. The problem should not be simply a reconstruction of a word problem already studied in class, but a new twist on such a problem. The new twist introduced need not be dramatic. Students are often very proud of making a small, new step on their own. There is a temptation to use the practicum to teach a new concept. The introduction of a new concept on the day of the practicum is inadvisable because it will dampen much of the positive effects of the exercise. It is absolutely essential that the equipment determine if the class solution is correct. As the instructor you should set the experiment in motion. The students' attention, hopes, fears, credibility, grade and the like should be glued to the equipment. If all goes well the reward to the students will be right from nature laws and their own savvy. We often discuss "add on situations" for extra credit. Students who wish the extra credit turn in their calculations the next day and a few moments are taken for the additional run of the equipment. STUDENTS' RESPONSIBILITY Students can organize in any manner they wish to accomplish the task. We do not give directions as to what might work best as far as organization is concerned. Regardless of organization, students must take data, make calculations, reach consensus on the solution to the problem and make sure all class members understand. The work, including all units, drawings, etc. must be placed on the blackboard and the prediction clearly stated. This must be accomplished with a minimum of five minutes left in the hour to allow time for the following steps. First, the instructor needs time to quiz a few students on their understanding. This can include erasing part

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of the blackboard work and asking any student to replace it without notes or coaching. Failure at this stage stops the practicum and results in the class failing the activity. This is a key point and is needed to make sure all students are involved. One must not yield to the temptation of helping out if things are going poorly. Students often will ask for help from classmates when they are uncomfortable with part or all of the calculations. Students will break into coaching groups to give the needed help so all are ready and able to answer the teacher's questions. This is the moment of truth! You will find more discussion on this topic in the section ADDITIONAL COMMENTS AND SUGGESTIONS THE BENEFITS OF THE CLASS PRACTICUM The debate during a class practicum is authentic. No student wishes to "own the solution". Students’ respect others rights to disagree and also learn to listen in order not to miss a bit of wisdom. Everyone is a little humble because when "Mother Nature" does her thing there will be no one to argue with or to blame. The high ability student often takes a leadership role and is looked to as a real asset in class, instead of a rival. Often when a practicum is announced students "check with the talent" to make sure they will be present on the appointed day. Asking these students to cancel doctor appointments etc. is not uncommon. It's hard for the class to depend on these students one day and dislike the same people the next day just because they get good marks. However, students of great physics ability must learn to explain what they know so well, or all is still in danger of being lost. This ability to articulate clearly is not always held by those who have great insight into physics, but is valued by the class. Therefore, other students have opportunities to gain the respect and appreciation of their classmates by being good explainers or for coming up with the one good idea that allows the class to surge forward on a practicum problem. Organization of the class is up to the students, but the following systems are the most common. One system is the whole class working as a single unit, debating each idea in a free for all setting. The success of this situation depends greatly on the skill of the leader. It is interesting to see that the leader need not be a great physics student, but must exercise good people skills. This is the least productive method of organizing, but is the most interesting to watch. A second method is many small groups working independently with debate postponed until the groups are pulled back together. Several groups will often come to the same correct results. The more groups with the same results, the more weight on their side during any debate. This brings about consensus quickly. A third method is again in small groups, but with runners moving around keeping everyone informed about what others have done and are thinking. This system usually results in no large debate. The consensus is reached by uniformly shared information and thought. No matter what style of organization is used, if a major conflict arises and time gets short students must figure out how they are going to decide. Should they go with the experts? Take a class vote? Not decide and fail? It can be a very lively discussion. Students are forced to think, debate, change their minds, yield to better ideas and cooperate during a class practicum. Class tension always builds during the questioning session and when nature is asked if the student prediction is true. When the results do match the class prediction, the students cheer and applaud. Many students shake hands and congratulate each other. This generates a team spirit, not unlike what we observe in sports. A team of students is a more powerful organization for learning physics than a class of students.

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ADDITIONAL COMMENTS AND SUGGESTIONS: Once the class has the problem and starts working, it might seem that as the instructor, you are free to correct papers etc. However, there are many things to be done that will insure success of the practicum concept all year. Look for students who are not involved, though this is not common, and talk with them. Most often a gentle reminder of the questioning session coming up is enough. If not, inform them that you will be calling on them today and everyone's success depends directly on their understanding. If the non-involvement problem does not exist, you need to start looking for the students you are going to call on. Always pick students that will be successful but are as low on the grading ladder as possible. This gives the student some encouragement and sends the signal to others that the teacher will pick anyone. Picking the right students to ask questions requires careful observation and good judgment. Make yourself available for students to ask questions or to ask for information. Don't give out information other than making the data clear and then, only if you made the measurements. Don't answer questions unless a student has thought of some perceived flaw in the set up and, based on topics covered in class, has no way to judge its importance. Always answer these questions using concepts studied with extrapolations that will lead the student to the correct conclusion. Check after the discussion to make sure he or she understands. You do not want to be responsible for a wrong impression, going back for class discussion that has your name and weight attached to it. During class debate it is tempting to give just "a little signal" about what might be a good idea, but don't do it. Students will begin to look to you during future debates instead of to each other. Organize the points for the practicum in such a way that it will not influence any student's individual grade very much. One can include a quiz or test item, based on the practicum, to increase the grade significance. Students are very jealous of their solutions to a practicum problem and do not share them with other classes. In fact, sometimes classes will deliberately plant false solutions on the blackboard to try to throw off another class. In addition, most of our practicums can be altered each hour with little difficulty if needed. In conclusion, class practicums are a useful method to culminate a chapter by allowing students to organize and solve a complex problem. No other technique produces such enthusiasm. Students ask, "When is the next practicum?" and they truly look forward to the next opportunity to do battle with Nature's Laws. Students rarely fail a practicum because they are putting a lot of talent and organizational energy to work in an effort to achieve a single goal. The number of practicums is limited only by the ingenuity to conceive of situations that fit the curriculum, where all the variables can be measured and one has, or can build, the equipment.

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Light and Optics Image With a Mirror

Developed by Hank Ryan and Jon Barber, Mounds View High School, Arden Hills, MN Problem With the object filament a given distance from a concave mirror, you are to place a screen to produce a real and in focus image. You must also predict the size of the image. Equipment and Setup

1. Concave mirror 2. Show case light bulb 3. Meter sticks 4. Modeling clay 5. Paper screen, 8 1/2" by 11" sheet of paper 6. Ring stand, clamps and rods 7. Sticky tape 8. Adding machine tape for optical bench markings

Measurements

Focal length of the mirror f = 21.2 cm

Height of object Hi = 8.4 cm

Object to focal point distance So = 10.85 cm Additional Comments Emphasize that success in this practicum depends on careful measurement. Encourage students to obtain the focal length of mirrors by finding the position where the object and real image are side by side. This occurs at 2f and this position of "f" can then be marked on the paper tape. When students have completed their work on the focal length determination, you should cover the mirror and place the object light where you want it for the practicum. Placing the object inside of 2f a short distance will cause the image to form a small distance outside of 2f. With this arrangement the image distance does not change dramatically with small errors made in measuring the object distance. This will help students obtain satisfactory results. The precise position of the object may be found if you place a meter stick along the paper optical bench and move a small card back and forth next to the light to observe where the shadow cast by the card is perpendicular to the ruler. When this step is completed, turn out the object light, uncover the mirror and leave students to work through the final calculations for the practicum.

Focal

Point

Show CaseBulb

Screen With

Image

Concave

Mirror

Paper TapeOptical Bench

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Sample Calculations Distance to the Image measured to the focal point:

Measured value was 41.4cm Height of image:

Measured value was 16.2cm Notes:

!

SoSi = f 2 ; Si = f 2

So

Si = (21.2cm)2

10.85cm= 41.4cm

!

Hi

Ho=fSo

; Hi =( f )(Ho)So

Hi =(21.2cm)(8.4 cm)10.85cm

=16.4cm

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Light and Optics Image With a Lens

Developed by Jon Barber, Mounds View High School, Arden Hills, MN Problem Given an object at a given distance from a convex lens, place a screen so a real and in focus image will appear. Also, predict the height of the image produced on the screen. Equipment and Setup

1. Converging lens 2. Show case bulb 3. Paper screen (8.5" x 11" sheet of white paper) 4. Ring stands, clamps, rods, etc. 5. Meter stick

6. Show case lamp cover with an arrow cut out for an image 7. Caliper (optional)

8. Lab jack (optional) An arrow is cut into the cardboard hood and can be covered with colored plastic to diffuse the light and produce a sharp image. Measurements Focal length

f = 25.0 cm

Height of object Ho = 7.9 cm

Object to focal point distance

So = 35.0 cm Additional Comments The So distance selected should be around 2f. Under this condition small errors in placing the object will not produce large changes in the position of the image. We encourage students to measure the focal length of lenses by finding the point where object and image are equal distance from the lens and the same size; this occurs at 2f. After the students have measured the focal length, the teacher turns off the light and places it where desired for the problem. Students complete their measurements, position the screen and set the calipers so the real images height will fall between the caliper arms. You can alter the practicum by stating only the size of the image (Hi) required. Students must now position the object light as well as the screen and the lens.

Lens

Light Source withHood

Screen

Lab JackCalipers

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Sample Calculations Distance to the image measured from the focal point:

Measured value was 18.0cm Height of the image:

Measured value was 5.7cm Notes:

!

SoSi = f 2 ; Si = f 2

So

Si = (25.0cm)2

35.0cm=17.9cm

!

Hi

Ho=fSo

; Hi =f (Ho)So

Hi =25.0cm(7.9cm)35.0cm

= 5.6cm

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Light and Optics How Intense

Developed by Hank Ryan and Jon Barber, Mounds View High School, Arden Hills, MN Problem Place the paraffin photometer between the sets of showcase lamps in such a way that both sides will be equally illuminated. Equipment and Setup 1. Six, or more, showcase lamps that are of equal intensity 2. A large Joly (paraffin-aluminum foil) photometer. 3. Meter stick. 4. A room that can be substantially dark. 5. A dark, non-reflecting surface to work on (flat black art board works well). The lights are not turned on until the class has placed the paraffin photometer. Students are assured that all the lamps are equal in their ability to produce light. Students may want a more concrete starting point such as a measurement of the intensity of a single lamp. The intensity of one showcase lamp can be measured with any light meter and expressed in foot-candles or lumens. The value you measure does not have any effect on the outcome of the calculations. Measurements Distance between the two sets of lights 86.7 cm Number of lights in each set 2 and 4 The fact that all the lights are of equal power (same lights) Standard intensity of a single lamp (optional) 150 candles Sample Calculations

For the paraffin photometer to receive equal illumination from each set of lights it has to be placed so that I1 = I2. Therefore:

!

I1 =K 1

r2

!

I 2 =K 2

r'2

!

K 1

r2=K 2

r' 2

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In addition: By substitution:

Taking the square root of both sides:

Additional Comments The practicum can be varied by changing the distance between the banks of lights or by changing the number of lights on each side. You could even offset one light on one side of the photometer to produce a more difficult problem suitable for advanced students. In addition, one bank of lights can be set a certain distance from the photometer and the class asked where to place the other bank to produce equal illumination. Use strong light sources in order to make any ambient light in the room a small factor in the outcome of the experiment. After the test has been run, move the Joly photometer a centimeter or two each way to demonstrate how quickly the intensity changes and that the sensitivity of the photometer is such that the change is easy to detect. Notes:

!

r + r'= 86.7cm r'= 86.7cm " r

!

K 1

r2=

K 2

(86.7cm " r)2r2

(86.7cm " r)2=K 1

K 2

!

r2

(86.7cm " r)2 =

k1k2

=42

=1.41

!

r86.7cm " r

=1.41; r = (86.7cm " r)(1.41); r +1.41r =122cm

!

r = 50.7cmr'= 86.7cm " 50.7cm = 36.0cm

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Light and Optics Is it Critical?

Developed by Ted Hale, Blake High School, Minneapolis MN. Problem You are to determine the depth of a light bulb under water by observing the circle of light produced on the water surface. Equipment and Setup

1. Large bucket 2. Light bulb and base (1.5 V) 3. Voltage source 4. Wire 5. Lycopodium powder (or powdered milk coffee creamer) 6. Ruler 7. Clamps 8. Dividers 9. Battery jar 10. Table of indices of refraction

Students need to measure the diameter of the circle of light on the surface. Dividers will allow this measurement to be made accurately without disturbing the water surface. Measurements Diameter of circle D = 20.0 cm Index of refraction of water nw= 1.33 Additional Comments Painting the inside of the bucket flat black and sprinkling the surface of the water with lycopodium powder will increase the visibility of the circle of light. It should be emphasized that measurements are to be made carefully. The socket and the wire leads to the bulb at the bottom of the bucket are dipped in paraffin to prevent the system from shorting out. The wires leading to the bulb should also be attached to the side of the bucket with clamps to prevent any unwanted movement of the light source. When checking the student prediction of the depth make sure to measure to the filament of the bulb and not just to the top of the bulb. A second bulb that the students can see and handle helps the class understand this measurement.

Wire Leads

Bucket showing the

circle of light

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Sample Calculations The critical angle for water:

Use the tangent of the critical angle and the radius of the circle on the water surface to determine the depth of the light source.

Measured results 8.8cm Notes:

!

sin"c =11.33

nw =1.33

!

sin" = .75 " = 48.6°

!

tan48.6° =10.0cmDepth

; Depth =10.0cmtan48.6°

!

Depth = 8.81cm

Radius

Wire Leads

Normal Line

Critical AngleLight Bulb

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Light and Optics

Relative Index of Refraction Developed by Jon Barber, Mounds View High School, Arden Hills, MN

Problem Place a small paper target on the opposite side of the apparatus from the light source (laser) in such a way as to catch the beam after it has passed through two different refracting mediums. Equipment and Setup 1. Semicircle of glass 2. Semicircular thin walled plastic tub 3. Transparent liquid such as water, alcohol etc. 4. Optical Disk (can be done with polar graph paper) 5. Small paper target 6. Blocking paper (paper card placed between the two mediums)

7. Low energy laser (any laboratory beam of light will suffice) Measurements Angle of incidence in the water Øw = 40.0° Index of refraction for the glass semicircle Øg=1.50 Index of refraction for the transparent liquid nw =1.33

To have your students calculate the index of refraction for the materials used in the setup, you will need to include at least one angle of incidence/angle of refraction pair for each material. Angle of incidence and refraction for the glass 40.0°; 25.4o Angle of incidence and refraction for the liquid 40.0°; 28.9° Comments The paper target is constructed to be ±2° which allows for small measurement errors. Measure the angles for each substance on the rotating platform as the setup is assembled. When you have completed this process put the blocking paper into position between the glass and the plastic tub and rotate the system to the angle of incidence you wish to have for the final problem. Your students can measure this angle in the same manner you measured each angle for the individual substances. Students often get into a debate about how many times the light will bend. Some feel

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it bends several times due to the thin layer of air (caused by the removal of the blocking paper) and plastic that exists between the two substances. Because of the parallel sides and thinness of these materials, their influence on the final direction of the beam is minimal. Students generally come to this correct understanding by themselves during the practicum. However, due to the presence of the air a critical angle can be experienced. You can have this situation as the outcome of the practicum if you wish, but do it intentionally and not by accident. Sample Calculations Using Snell’s Law:

!

(nw )(sin"w ) = (ng )(sin"g ) ; sin"g =(nw )(sin"w )

ng

sin"g =(1.33)(sin40°)

1.50= .57

"g = 34.7°

Measured value was 34.5°. Notes:

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Light and Optics Interference Call

Developed by Jon Barber, Mounds View High School, Arden Hills, MN Problem Given a HeNe laser and a double slit slide predict the distance from the center to an assigned nodal line when the laser light forms an interference pattern produced by passing through the slide. Equipment and Setup

1. HeNe laser 2. Painted microscope slide 3. Razor blades 4. Meter stick 5. Micrometer 6. Ring stand 7. Clamps 8. Paper Screen 9. Sticky tape

Students will need to measure the distance between slits on the microscope slide. To do this use a micrometer to measure the thickness of the razor blade used to make the slide. The PSSC Laboratory Guide, Sixth Edition explains how to prepare slides. Measure the distance from the slide to a sheet of paper taped on a wall. The student must also look up the wavelength of light emitted from the laser. Measurements

Distance between slits d = .011 cm Length from slide to paper screen L = 452 cm Wavelength of light = 6.328 x 10-5 cm Nodal line assigned n = 2

LaserBeam

Slide held

Screenin clamp

!

"

20

Additional Comments The set up for this practicum is similar to pictures in texts for a two slit apparatus producing interference patterns on screens. The screen can be white paper taped to a black board. The laser beam is allowed to hit the paper screen with no slide in place to establish the center point for the students. Then close the gate for the laser, put the slide in place and complete final measurements. After the students have made their prediction they should mark the value for the distance from the center maximum to the second nodal line, on the paper. Then open the gate and compare the interference pattern directly. Sample Calculations The distance to the second nodal line (x2) from the center maximum:

Measured value was 37.2cm Notes: !

d x2L

= (n " 12)# ; x2 =

(n " 12)#L

d

x2 =(2 " 1

2)(6.328x10"5cm )(425cm)

0.0011cm= 36.7cm

d Screen

2nd Nodal Line

L X

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Waves and Sound Interference Trombone

Developed by George Amann, Franklin D. Roosevelt High School, Hyde Park, NY Problem Calculate the frequency of the sound being played into an interference producing slide trombone apparatus. Equipment and Setup

1. Interference slide trombone apparatus (shown below) 2. Oscilloscope and microphone or decibel meter to measure sound 3. Meter stick 4. Frequency generator with the dial covered

A speaker connected to a frequency generator plays sound into the top of the “trombone” with the slide arm adjusted so the two sides are of equal length. The frequency generator is kept at the same frequency and the moveable arm of the trombone is shifted so that the sound coming out of the bottom end is at a minimum. The movement of the slide is measured, as is the room temperature. Measurements

Room temperature = 20 ° C Slide movement = 0.182 meter

Sample Calculations Velocity of sound at this temperature:

!

V =Vo +(0.60m /s)

°C(T) ; where Vo = 331m /s

V = 331m /s+(0.60m /s)

°C(20°C) = 343m /s

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When the trombone slide arm is shifted from this maximum to the first minimum, the path difference between the two halves must be 1/2 λ. Since the sound must go through both the top and bottom, the slide arm would be moved 1/4 λ. The wavelength of the sound wave is therefore:

The frequency is therefore:

Measured value = 466 Hertz Additional Comments Directions for building the “slide trombone” apparatus can be found in Exploring Physics in the Classroom (AAPT/PTRA publication) by George Amann, appendix C. Notes:

!

" = 4(slide movement) = 4(0.182m) = 0.728m

!

V = f" ; f =V"

=343m /s0.728m

= 471Hertz

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Kinematics Colliding Buggies

Suggested by Val Michaels Problem Determine the point where two constant velocity buggies (they do not travel at the same speed) collide if they are released from opposite ends of the lab table at different times. Equipment and Setup

1. Two battery powered constant velocity buggies (science catalog item) 2. Ten-meter metric tape or meter sticks 3. Stopwatches

Each buggy is tested individually in order to collect data that allows their speeds to be determined. They are then set facing each other a known distance apart and one buggy is delayed in starting by a known amount of time. Measurements

Buggy “A” distance 1.27 meters time 10.0 seconds.

Buggy “B” distance 1.38 meters in time 10.0 seconds. In this example the buggies were placed 4.00 meters apart (measured from the front bumpers) and buggy “B” was delayed 5.0 seconds. Sample Calculations The speed of each buggy is calculated:

Distance traveled by each buggy before they collide:

!

va =xt

=1.27m10.0sec

= 0.127m /s vb =1.38m10.0sec

= 0.138m /s

!

xa = vat xa = 0.127m / s(t)

xb = vbt xb = 0.138m / s(t " 5.0sec)

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The two buggies will travel a total distance of 4.00 meters before they collide therefore:

Using buggy “A” to determine the collision point:

Checking the total distance:

Actual distance from buggy A’s starting point when they collided was measured as 2.18 meters. Additional Comments Problems arise from not having the second buggy start at exactly five seconds after the first, so several trials may be necessary, and the results averaged. A possible solution to the release problem is to hold the one buggy with a thread tied to the back and anchored. It could sit and spin for the five seconds and then be released on time (you may have to keep it pointed properly). Notes:

!

0.127m / s(t) + 0.138m / s(t " 5sec) = 4.00m

0.127m / s(t) + 0.138m / s(t) " 0.690m = 4.00m

!

t =4.69m0.265m / s

=17.7sec

!

Xa = vat = 0.127m / s(17.7s) = 2.25m

!

xb = vb (t) = 0.138m / s(17.7s" 5.0s) =1.75m2.25m +1.75m = 4.00m

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Circular Motion Lazy Pendulum

Developed by Hank Ryan and Jon Barber, Mounds View High School, Arden Hills, MN

Problem Given an air table-pendulum apparatus predict the period of the pendulum. Equipment and Set Up

1. Air table 2. Clamps 3. Pucks 4. Meter sticks 5. Level 6. Thread 7. Lab jack 8. Stop watch 9. Protractor

Students need to measure the length of the pendulum thread and the angle between horizontal and the table surface. Measurements

Length of pendulum thread L = 0.72 m Angle between horizontal and the table surface Ø = 19.37o

Additional Comments A variety of methods can be used to measure the angle between horizontal and the air table surface. Use trigonometric functions and right angles or measure directly using a protractor.

Thread

Side View

Puck

Air Table

Top View

Air Table

Puck

Thread

26

Sample Calculations The acceleration down the ramp due to gravity is calculated using the following relation:

The period of the pendulum is calculated using the following relation:

Measured values were 2.93 sec and 2.91 sec. Notes:

!

sin" =a(down the ramp )

g

a(down the ramp ) = sin19.4°(9.8m /s2) = 3.25m /s2

!

T = 2" Lg

= 2" 0.72m3.25m /s2

= 2.96sec

27

Circular Motion Centripetal Acceleration

Developed by Jon Barber and Hank Ryan, Mounds View High School, Arden Hills, MN Problem Turn a propeller blade through a vertical circle with a calculated period so the ball from the inner beaker falls out, but a ball in the outer beaker remains. Equipment and Set Up

1. Two beakers, 100 ml 2. Metronome 3. Propeller blade assembly 4. Small colored balls (ping-pong balls work well) 5. Clamps 6. Tape

The board that acts as the propeller blade is the same length on each side of the pivot point. The upper half is not show here to save space. Measurements Radius to outer beaker Ro = 0.80 m Radius of inner beaker Ri = 0.40 m

Additional Comments The propeller blade assembly consists of two boards nailed to a base. Drill a half-inch hole in each board and insert a half-inch dowel rod. This dowel rod should be long enough to attach the propeller blade board (1" by 4" approximately 5 ft long) on one end and the crank handle on the other.

28

Tape the beakers, each containing a different colored ball, to the propeller blades with masking tape. Starting with a rocking motion, rotate the propeller blade rapidly enough to avoid having either ball drop out. At this point stop the rotation and allow the students to work on the problem. The students set the metronome at the proper period so as to drop the ball from the inner beaker and not drop the ball from the outer beaker. After the students have completed the calculations and set the metronome you may wish to cut the metronome period in half. It is easier to match the propeller period to the metronome period if it "clicks" as each blade of the propeller passes a fixed point. Start up as before keeping the balls in their beakers and then slow the propeller to the rhythm of the metronome. Sample Calculations Using the equation for centripetal acceleration:

Repeating the calculation for the inner beaker you get 1.30sec. Because the beakers have walls that will hold ball the class should choose the 1.80sec and keep the outer ball in place while insuring the inner ball will fall. You can note to the class that the balls start to rattle in the beakers at just about the right period of rotation. Notes:

!

ac =4" 2RT 2

9.8m /s2 =4" 2(0.80m)

T 2

T =4" 2(0.80m)9.8m /s2

=1.80sec

29

Circular Motion Centripetal Force

Developed by Hank Ryan, Mounds View High School, Arden Hills, MN (Note: This was the first practicum)

Problem When a flat horizontal board is rotated on the turntable with pucks arranged on its surface radiating out from the center, determine which pucks will remain in place and which will fly off. Equipment and Set Up 1. Old record player turntable. 2. Formica board with center point drilled to fit the turntable. 3. Ring stands, clamps, etc. 4. Stop clock. 5. Alcohol for cleaning the Formica surface. 6. Metal pucks of consistent mass, surface characteristics, etc. 7. Balance. Mass a few of the pucks and record an average value. Angle the ramp so that the pucks will accelerate down when released. Test different angles and select the smallest angle that still produces a definite smooth acceleration. Releasing the pucks from rest at the top of the ramp make measurements of the time required to complete a run to the bottom. Average the time for several trials using different pucks. Remove any pucks from the experiment that are not near the average value for a run down the ramp. Measure the length of the ramp and the height used during the timed runs. Turn on the record player to rotate the board on the turntable and record the time for completing five to ten turns. Stop the board and place the pucks on its surface radiating out from the center. To maintain balance pucks need to be spaced apart from each other and placed the same distance from the center on each side. Measurements

Ramp length L = 1.22 m Ramp height H = 0.437 m Average time for pucks descending the ramp t = 1.5 s Period of spin for the board on the turntable T = 2.1 s Mass of the pucks Mp = 0.22 kg Gravitational field strength of the earth g = 9.8 m/s2

Ramp height

Ramp length

Formica board

Drilled center hole to fit the turntable

30

Additional Comments The pucks can be large washers purchased at a hardware store as long as they are of equal mass and do not have sharp edges or other features that would alter their frictional characteristics. A child's phonograph player is inexpensive and has the power to spin a rather large board. A Formica surface works well because the frictional characteristics are quite consistent over the entire surface. However, alcohol should to be used to wash the board and the pucks to remove grit and to reduce any static charges that may exist. In this practicum the kinetic friction is measured and used for all the calculations even though the pucks are at rest when the final test is made. The difference in kinetic and static friction between a Formica surface and smooth metal puck is small and also the motor in the record player causes vibrations during the spinning, which further reduces the difference. If students ask how they are to find static friction, explain these factors and assure them that the kinetic friction present can be used to predict the behavior of the pucks. Many classes will not correct the friction calculated for the pucks on the angled ramp and when the board is horizontal. If the smallest angle possible is used for the ramp the change is small and has a minimal effect on the results. Space the pucks on the board in such a way as to have the answer to the problem fall between two pucks and not near the center of mass of any given puck. This gives a reasonable margin of error for small effects not controlled in the setup. Because students may need most of the hour for the calculations in this practicum, the teacher should do the measurements. It takes only about six to eight minutes for the measurements if the equipment is in place and ready to go. You can reduce the difficulty of the practicum by finding the angle at which the pucks proceed down the ramp with a constant velocity. The pucks will need to be given a small initial velocity to find this angle. The calculation of the friction, in this case, is more straightforward and eliminates the need to measure the time for the pucks going down the ramp. Sample Calculations The calculations shown here are not corrected for the increase in friction between the pucks and the board caused by the change from the angled ramp to the horizontal position during the spin.

The force acting down the ramp is therefore:

The actual acceleration of the pucks down the ramp can be calculated from:

!

sin" =0.437m1.22m

= 0.358

!

Fr = sin"(mpg) = 0.358(0.22kg)(9.8N /kg) = 0.77N

!

d =12at 2 a =

2dt 2

=2(1.22m)(1.50s)2

=1.08m /s2

31

The unbalanced force acting on the pucks down the ramp:

Determination of the friction force:

The maximum radius, measured from the center of the board, where there is enough friction to supply the centripetal force required for the pucks to stay in place:

If you correct the friction force for the change from the angled ramp to the horizontal position during the spin, the answer is 28 cm. All the right pucks will fly off because the spacing between their centers of mass is large enough to provide the needed tolerance for any small differences in friction, etc. that may exist. Generally the answer calculated is accurate to about plus or minus 2.5 centimeters. Even if you space the centers of mass of the pucks by as much as ten to fifteen centimeters, the demonstration is still impressive. Notes:

!

Fnet = ma = 0.22kg( )1.08m /s2 = 0.24N

!

Ffriction = 0.77N " 0.24N = 0.53N

!

Fc =4" 2Rmt 2

; R =Fct

2

4" 2m=0.53N(2.1s)2

4" 2(0.22kg)= 0.27m

32

Circular Motion The Helicopter Ride

Developed by Rex Rice, Clayton High School, St. Louis, MO

Problem

Given a toy helicopter attached to a string and traveling in a circle at a constant speed about a vertical lab rod, determine the reading on a stopwatch that has recorded the time to make 30 revolutions. Equipment and Set Up 1. Toy helicopter, or plane, battery operated 2. Heavy duty table clamp

3. Lab rod, around one meter long 4. Stopwatch

Measurements

Length of string (from point of connection to the center of helicopter) 1.02 m Distance from table level to helicopter level 0.19 m Distances from table level to top of lab rod 1.09 m

Additional Comments Start the helicopter and wait until its motion stabilizes. Being careful not to knock the helicopter out of orbit, place a meter stick vertically just outside of the path of the helicopter with the zero mark at table level. Have students watching from the side tell you where to move your fingers so that they are at the level of the center of the helicopter. Have students record the distance from the tabletop to center of the helicopter. Have the students count 30 revolutions while you time with a stopwatch. Put the stopwatch in your pocket. Stop the helicopter. Only after you have put the stopwatch away should you explain to students the problem is to determine the reading on the stopwatch.

Ø

Ø

h

Diagram #1

33

Sample Calculations Draw the force diagram for the helicopter. Since the helicopter stays in a horizontal circle, it is not accelerating in the vertical direction; therefore the net vertical force must be zero. Choosing up as the positive direction:

Calculate the radius of the circle flown by the helicopter by using the diagram above and the Pythagorean theorem. Side b is the radius of the helicopter's flight.

Since the weight and the vertical component of the tension in the string are balanced, the only remaining force acting on the helicopter is the horizontal component of the tension (Ts). The net force acting on the helicopter is therefore the x component of the tension (Tx). This provides the centripetal force required to keep the helicopter in uniform circular motion therefore:

Where R is the radius of flight, m is the mass of the helicopter, t is the period, Fc is the centripetal force and Tx is the component of the Tension in the string in the x direction.

From diagram number 2

Substitute for Tx

From diagram number 2

!

Tup "W = 0 Fup =W

Tup = Ts(sin#)

W = mg

Ts(sin#) = mg Ts =mgsin#

!

b2 = c 2 " a2 = (1.02m)2 " (0.90m)2

b = 0.48m

!

Tx = Fc =4" 2Rmt 2

!

cos" =TxTs

Tx = Ts(cos")

!

Ts(cos") =4# 2Rmt 2

!

Ts(cos")

!

Ts =mgsin"

W

Tx

Ts

Ø

Diagram #2

Ø

c =1.02m

b =?m

a =0.90m

Diagram #3

34

Substitute for Ts

Solve for t

From diagram number 3

or 56.7 sec for 30 revolutions

Measured value was 1.9 s/rev or 56.9 sec for 30 revolutions. Notes:

!

mg(cos")sin"

=4# 2Rmt 2

!

mgsin"

!

t =4" 2Rm(sin#)mg(cos#)

=4" 2R(tan#)

g

!

tan" =0.90m0.48m

=1.875

!

t =4" 2(0.48m)(1.875)

9.8m /s2=1.89sec

35

Projectiles Shot Heard Round the Lab

Developed by George Amann, Franklin D. Roosevelt H.S. Hyde Park, NY Problem After measuring needed data from a horizontal shot fired by a PASCO projectile launcher determine the range of the same projectile fired from the same launcher at an angle above level. Equipment and Set Up

1. PASCO projectile launcher 2. Meter stick 3. Clamp 4. Sheet of Carbon Paper

For the horizontal shot a face down piece of carbon paper placed on the floor will accurately mark the impact point. The launcher is then reset to fire at an angle above level. For this example the height above the floor remains unchanged and the angle above level is 30°. The class is given a “bulls eye” target and the carbon paper to place on the floor to measure their results.

Measurements

Distance from the barrel to the floor 1.14 meters Horizontal range 1.78 meters

Sample Calculations Time of fall when launched horizontally is given by:

The horizontal range then yields the muzzle velocity from:

!

y =gt 2

21.14m =

(9.8m /s2)t 2

2t =

2(1.14m)9.8m /s2

t = 0.482sec

!

vm =xt

vm =1.78m.482sec

= 3.69m /s

Carbon Paper

PASCO projectile launcher

36

To find the range when launched at an angle requires that the muzzle velocity be broken into vertical and horizontal components, therefore:

Use the time of flight equation and solve the quadratic:

The horizontal range is therefore:

Measured value was 2.23 meters Additional Comments If students are asked to measure the horizontal shot some care must be taken to measure the range from a point on the floor directly below the end of the barrel on the launcher. The PASCO launcher gives very good reproducible results. Notes:

!

vy = vm sin" vy = 3.69m /s(sin30°) =1.85m /s

vx = vmcos" vx = 3.69m /s(cos30°) = 3.20m /s

!

y = vyt +12gt2 "1.14m = (1.85m /s)(t) + (1

2)("9.8m /s2)(t 2) 4.9t 2 "1.85t "1.14 = 0

t = 0.71sec

!

x = vxt x = 3.20m /s(0.71s) = 2.26m

37

Projectiles Projectile Impact Angle

Developed by Jon Barber, Mounds View High School, Arden Hills, MN Problem Predict the angle a dart enters a horizontal target after being fired from a level blowgun. Equipment and Set Up 1. Blow dart gun and quality darts 2. Level 3. Sheet of fine grained Styrofoam insulation 4. Ring stands, clamps, etc. 5. Plastic bucket 6. Small propane torch (optional) Shoot a dart across the classroom that strikes a sheet of Styrofoam insulation that has been placed on the floor. Measure the horizontal distance from the dart to the blowgun and cover the dart with a bucket. After the students have completed their calculations to predict the impact angle, remove the bucket for the comparison. Measurements Height of barrel from the floor h =1.70 m Horizontal distance of the shot r = 7.00 m Earth's gravitational field strength ge = 9.8 m/sec2 Additional Comments The blowgun consists of a piece of half inch electrical metal tubing about 1.0 m long. This is clamped horizontally and leveled using a spirit level. The darts are made from number five or six box nails fitted with a paper cone made from ordinary writing paper. The paper cone is held together and first attached to the nail with scotch tape. Once you have the cone attached (make sure it is properly aligned with the nail) put a few drops of glue inside on the nail head and let dry. To trim the paper cone of the dart to the proper diameter to fit the tube, you first drop the dart into the tube and mark the cone with a pencil. Trim off excess paper with a scissors. The cone should be kept short as accuracy decreases when the cone is long and must have a snug fit with the barrel. To check the student's prediction, cut a piece of paper at an angle to match the class prediction. Burn the paper cone off the dart leaving only the nail. Then, slide the paper angle up to the nail for the check.

38

Sample Calculations Time of flight:

Muzzle velocity:

X and Y components of the impact velocity:

Impact angle:

Measured value: 26o Notes:

!

h =1/2gt 2 ; t 2 =2hg

; t =2hg

=2(1.70m)9.8m /s2

= 0.59sec

!

Vmuzzle =ranget

=7.00m0.59s

=11.9m /s

!

Vx =11.9m /s

Vy = gt = 9.8m /s2(0.59s) = 5.8m /s

!

tan"impact =VxVy

=5.8m /s11.9m /s

= 0.48

"impact = 26°

39

Projectiles Arkansas Traveler

Developed by Richard L. Picard

Problem Predict the time for an air sled to travel a measured distance on an air track. Equipment and Set Up 1. Air track 2. Photo-gates and timer 3. Plumb bob 4. Paper strip 5. Carbon Paper 6. Meter sticks 7. Ball with a hole for projectile 8. Coat hanger wire 9. Clay 10. Wood block 11. Tape Set up an air track with one end over the edge of a table. Attach a section of coat hanger wire to the air sled in such a way that a metal ball with a hole in it will slide off when the sled strikes the end of the air track. The wire is attached to the sled using a wood block with a hole drilled into the front and the block taped to the sled (see below). Place a strip of paper on the floor and tape in place. Place carbon paper face down on the paper. Use the plumb bob to mark the position on the paper where the ball begins to fall. Push the glider along the track, releasing it before it enters the first photo-gate and have the timer covered. Measurements Vertical fall y = 1.130 m Horizontal distance x = 0.642 m

Photo-gates

Air Track

Air Sled (Spring Bumpers Removed)

Projectile Path

Carbon Paper Target

40

Sample Calculations Calculate the time of fall:

Calculate the horizontal velocity:

Calculate time between photo-gates:

Photo-gate time reading was 0.59 s Addition Comments The metal ball from a simultaneous fall apparatus works well for the projectile. The sled may need a counterweight to remain horizontal to the track and glide smoothly. Notes:

!

y =12gt 2 ; t =

2yg

=2(1.130m)9.8m /s2

= 0.48sec

!

vx =xt

=0.642m0.48s

=1.338m /s

!

!

vx =xpt

; t =xpt

=0.800m1.338m /s

= 0.60sec

Wire

Ball

Air Sled

Clay Counter Weight

Wood Block

(Spring Bumpers Removed)

41

Projectiles Non-Horizontal Projectile

Developed by Hank Ryan, Mounds View High School, Arden Hills, MN Problem Place a piece of Styrofoam on the ceiling, wall, or the floor of the classroom so that it will be hit by the dart that is shot through the "velocity window". Predict the angle formed between the dart and the Styrofoam that is produced during this hit. Equipment and Set Up

1. Blow gun 2. Darts 3. Ring stands, clamps, etc. 4. Calibrated "velocity window" 5. Styrofoam (fine grained) approximately 75 cm wide and 50 cm in length 6. Radar device for measuring velocity (optional)

Give a brief demonstration and explanation of the "velocity window". During the demonstration, the window is closed so that the darts will not pass through. If you have a radar device, the muzzle velocity of a shot that would pass through the window can be measured directly. If not, the class will accept your value after an explanation. Measure the angle of the gun, above the horizontal, and inform the class that other measurements such as the distance from the floor to the ceiling, the distance from the gun muzzle to the floor, the distance from the gun muzzle to the back wall, etc. have been measured and are available if and when needed. Measurements

Angle of the gun Ø = 17.8° Muzzle velocity allowed by the window Vm= 19.2m/s Muzzle to ceiling distance Hc = 2.29m Height of muzzle from the floor Hm = 0.87m Muzzle to back wall of the classroom Xw = 13.67m

Sample Calculations X and Y components of the muzzle velocity

!

sin" =vy

19.2m /s; Vy = sin(17.8°)(19.2m /s) = 5.87m /s

42

Flight time for the shot as it returns to muzzle level:

Maximum height of the dart during which occurs at half the flight time:

Note: The dart will not hit the ceiling. Range for the shot when it falls back to muzzle level:

Note: The range is larger than the classroom is long, therefore the dart will hit the back wall. The time it takes for the dart to reach the back wall of the classroom:

The height of the dart 0.75 seconds into the flight relative to the gun muzzle:

The component velocities (Vx, Vy) at the moment of impact:

!

cos" =Vx

19.2m /s; Vx = (cos17.8°)(19.2m /s) =18.3m /s

!

T =2Vy

g=2(5.87m /s)9.8m /s2

=1.20sec

!

ymax = vyt +12gt 2 = 5.87m /s(.60s) +

12("9.8m /s2)(.60s)2 =1.76m

!

Xrange =Vx (t) =18.3m /s(1.20s) = 21.9m

!

xw =Vxt ; T =xwVx

=13.67m18.3m /s

= 0.75sec

!

y0.75s = vyt +12gt 2 = 5.87m /s(0.75s) +

12("9.8m /s)(0.75)2 =1.65m

!

Vx =18.3m /s ; Vy = 5.87m /s+ ("9.8m /s2)(0.75s) = "1.48m /s

v = 18.3 m/secHorizontal

V = 5.87 m/sec

19.2 m/sec

Muzzle Velocity

17.8o

x

y

43

The impact angle of the dart with the Styrofoam:

The Sine of the angle below the horizontal:

The angle that the dart makes with the Styrofoam: 90° - 4.6° = 85.4°. The actual results, picking a shot that went directly through the center of the "velocity window", hit 1.61 m above the gun muzzle on the back wall of the classroom and made an angle of 85.0° with the Styrofoam. The piece of Styrofoam used by the student to catch the dart should be small-grained material because larger grained Styrofoam will, at times, alter the impact angle. Construction of the darts and dart gun are discussed in the practicum titled "Projectile Impact Angle". Additional Comments The construction of a “velocity window” and calibration is not difficult. The velocity window filters out shots from the blowgun that are not at a given velocity. Make the window of heavy cardboard and hang it from the ceiling of the classroom. Cut an opening into the cardboard to produce the actual "velocity widow". Shots that are too fast will stick into the cardboard above the window, and shots that are too slow will hit the cardboard below the opening. The shot you need for this experiment must pass through the center of the window. The blowgun is set on the floor and is "bore sighted" at a target "dot" near the top of the "velocity window". After the blowgun is in place, make the measurements in the above graphic so the window can be calibrated. To compute the time for a shot that passes through the center of the window to reach the opening use the following expression.

V = 18.3 m/sx

= 18.4 m/.sdart

V

Horizontal

V = -1.48 m/sy

!

sin" =1.48m /s18.3m /s

= 0.081 ; " = 4.6°

!

Tw =2Ydg

44

The time value from this calculation divided into Xw gives the horizontal component of the muzzle velocity of the shot and this same time divided into Yb gives the vertical component of the muzzle velocity. These two vector velocities added together at right angles should give the angle of the gun above the horizontal. The actual angle of the gun can be measured and checked against the calculations. If there is good agreement between the predicted angle and the actual angle then the "velocity window" is properly calibrated. Sample Calculation for a "velocity window"

The angle of the gun barrel:

The gun barrel angle should be 17.9° above the horizontal and by actual measurement was 17.8°. You can now calculate the muzzle velocity of the gun, for a shot going through the center of the window, by taking the square root of the sum of the squares of the two component velocities.

Notes:

!

Yd = 0.68m ; Yb = 2.22m ; Xw = 6.87m

!

Tw =2(0.68m)9.8m /s2

= 0.373sec ; Vx =6.87m0.373s

=18.4m /s ; Vy =2.22m0.373s

= 5.95m /s

!

tan" =5.95m /s18.4m /s

= 0.232 =17.9°

!

a2 + b2 = c 2 ; c = a2 + b2 = (18.4m /s)2 + (5.95m /s)2 =19.2m /s

45

Static Equilibrium Static Equilibrium for Beginners

Developed by Michael Crofton, Spring Lake Park High School, Spring Lake Park, MN Problem Given a protractor, determine the amount of mass on the end of each string hidden in the cylinders. Equipment and Setup

1. Two low friction pulleys 2. Assorted masses 3. Fish line 4. Two cylinders to hide the masses 5. Protractor

Measurements Mass of m1 (150 g)

Angle of θ1 (25.0o)

Angle of θ2 (48.0o) Additional Comments The cardboard cylinders are about 3-4 feet tall and 6 inches in diameter. These cylinders are capped by a piece of paper with a hole in the center and a cut made from the edge of the paper to the hole that allows the cap to slide over the string. The pulleys are mounted on the ceiling and the whole setup is four to six meters wide (It can be setup using much less space if desired). The pulleys used must be very low friction, ball bearing type. You want to make sure the masses are not touching the sides when you’re done setting up. This practicum generally yields answers within plus or minus 5 grams and can easily be done in a single class period.

m1 (known)

θ2 θ1

m3 m2

46

Sample Calculations The pull to the left caused by m1 must be equal to the pull to the right caused by m3 (m2 dose not pull right or left) therefore:

The upward pull of m1 plus the upward pull of m3 must equal the downward pull of m2 therefore:

Notes:

!

Cos"1 =left pull150gr

Cos"2 =right pull

m3

150gr Cos25°( ) = m3 Cos48°( )

m3 = 203gr

!

Sin25° =upwardm1150gr

Sin48° =upwardm3

203gr

Sin25° 150gr( ) + Sin48° 203gr( ) = m2

m2 = 214gr

47

Static Equilibrium Center the Mass

Developed by Michael Crofton, Spring Lake Park High School, Spring Lake Park, MN

Problem Determine where a clamp should be placed under a weighted meter stick so that the whole system will balance when placed on a knife-edge support. Equipment and Setup

1. Meter stick with four 1/8" small holes drilled in it. In this example they are at the 10, 30. 60 and 90 cm marks. 2. Various hooked masses from 50g to 500 g as desired. In this case they are 50, 100, 100, and 200g. 3. Clamp with knife edges to rest on the fulcrum 4. Fulcrum type support Measurements

Mass at the 10cm and 60cm marks 100.0 grams Mass at the 30cm mark 500.0 grams Mass at the 90cm mark 50.0 grams Mass of the meter stick 140.0 grams Center of mass of the meter stick 49.8 grams

48

Sample Calculations (100g x 10cm) + (200g x 30cm) + (140g x 49.8cm) + (100g x 60cm) + (50g x 90cm) = 590g x Xcm In this case, X = 41.5 cm Additional Comments This practicum can easily be done in a 40 minute period. You may need to straighten out the ends of the hooks on the hooked masses so they fit into the holes properly. Notes:

49

Beams In Equilibrium Developed by John O' Leary, The American School in Japan, Tokyo

Problem

Given a weighted beam suspended horizontally by two angled strings, compute the tension force (spring scale reading) in one of the strings. Equipment and Setup 1. Two identical meter sticks 2. Three meter stick clips 3. String 4. Two rings stands 5. Weight with a hook 6. Triple beam balance 7. Spring scale 8. Protractor 9. Spirit level

Attach the mass off center. Adjust the clip so the string makes an angle from 60o to 80o with the meter stick adjusted to be horizontal. Cover the scale prior to the arrival of students. Measurements

String angle with the horizontal 60.0o Clip locations (right to left) 20.0 cm, 60.0 cm, 90.0 cm marks Mass of meter the stick 0.13823 kg Mass of the clips 0.0211 kg Mass of suspended weight 0.5000 kg

50

Sample Calculations ∑ Torques clockwise = ∑ Torques counterclockwise, therefore starting with the clip on the right:

Taking into account the angle to get the scale reading:

Results are within 2% are easy to attain. Additional Comments By changing the location of the suspended mass the practicum can be altered. The practicum can also be done using a spring force scale calibrated in grams. Notes:

!

Counterclockwise(ccw) "= (massandclip)" + (meter stick)" + (clip)"[ ] g[ ]

ccw "= (0.5211kg)0.400m + (0.13823kg)0.300m + (0.0211kg)0.700m[ ] 9.8N /kg[ ]= 2.59Nm

Clockwise" = F(0.70m) = 2.59Nm

F =2.59Nm0.700m

= 3.71N

!

sin60° =3.71Nscaler

; Scaler =3.71Nsin60°

=3.71.8660

= 4.28N

51

Static Equilibrium Static Equilibrium for Nerds

Developed by Michael Crofton, Spring Lake Park High School, Spring Lake Park, MN Problem Given a system at equilibrium, a protractor and the mass of the visible weights, determine the amount of mass on the end of each string hidden in the cylinders.

Equipment and Setup

1. Two low friction pulleys. 2. Assorted masses 3. Fish line 4. Three cylinders to hide the masses 5. Protractor

Measurements

Mass of m1 150 g

Angle of θ1 25.0o

Angle of θ2 10.0°

Angle of θ3 48.0o

Additional Comments

The cardboard cylinders are about 3-4 feet tall and 6 inches in diameter. These cylinders are capped by a piece of paper with a hole in the center and a cut made from the edge of the paper to the hole that allows the cap to slide over the string. The pulleys are mounted on the ceiling and the whole setup is four to six meters wide (It can be setup using much less space if desired). The pulleys used must be very low friction, ball bearing type. You want to make sure the masses are

θ3

θ2

θ1

Knot “a” Knot “b”

m1 (known) m2

m3 m4

52

not touching the sides when you’re done setting up. This practicum generally yields answers within plus or minus 5 grams and can easily be done in a single class period.

Sample Calculations The knot at point “a” (where m2 hooks into the system) is motionless and therefore the pull left on point “a” must be equal to the pull towards the right. Keeping in mind that m1 produces the only pull to the left (m2 and m3 pull only down) it can be calculated by:

The pull to the right is produced only by m4 and must be equal to 136 grams therefore:

The tension in the line between points “a” and “b produces an upward pull on point “b” and downward pull on point “a” that are equal. Vector analysis of point “a”:

Up and down forces at point “a” are equal and therefore:

Up and down forces at point “b” are equal and therefore:

To check our answers we can calculate the total upward pull and the total downward pull:

!

Cos25° 150gr( ) =136gr

!

Cos48° m4( ) =136gr

m4 =136grCos48°

= 203gr

!

Tan10° =Downwardvector at"a"

136grDownwardvector at"a"= Tan10° 136gr( ) = 24gr

!

Sin25° 150gr( ) = m2 + 24.0gr

m2 = Sin25° 150gr( ) " 24.0gr = 39.4gr

!

m3 = Sin48° 203gr( ) + 24.0gr =175gr

!

upward = Sin25° 159gr( ) + Sin48° 203gr( ) = 218grdownward = 39.4gr +175gr = 214gr

Knot “a”

136gr 136gr

10°

53

Notes:

54

Dynamics

The Runaway Cart Developed by George Amann, Franklin D. Roosevelt H.S. Hyde Park, NY

Problem Given the mass of a cart, the mass of a falling weight that pulls the cart from rest across a horizontal table, and the distance between two photo-gates, predict the time it will take the cart to pass from the first to the second gate. Equipment and Setup

1. Dynamics cart and flag 2. Photo-gates (2) 3. Low friction pulley 4. String 5. Weight 6. Ring stand and some clamps 7. Level lab table

A dynamics cart with a “flag” is placed on a level, horizontal surface and is attached to a mass on a string, which passes over a low friction pulley at the end of the table and which will not hit the floor during the time the cart passes through the gates. A master and slave photo-gate set up is placed a fixed distance apart on the table and rigged to read the time it takes the cart to go from the first gate to the second. The cart is adjusted so that the “flag” on the dynamics cart is just at the first gate, when the cart is at rest, but under tension from the string attached to the mass. This can be accomplished by holding the cart back with a thread (to be cut with a sharp scissors) and is adjusted so that the first gate is just at the position of the carts flag. Measurements

Mass of the cart and “flag” 0.521 kg. Mass on the end of the string 0.100 kg. Distance between photo-gates 0.600 meters

Photo-gates Clamped ring-stand

55

Sample Calculations Force that causes the system to accelerate:

Acceleration of the cart and weight:

Additional Comments Problems may arise if surface is not level, with friction effects, if the cart does not start from rest and if the start photo gate is not triggered at the very beginning of the carts motion. This practicum can be made more complicated by having the start gate a given distance from the initial position of the cart that starts form rest. Notes:

!

F =Weight = mg = (0.1kg)(9.8m /s2) = 0.98N

!

F = ma; a =FM

=0.98N

0.0512kg+ 0.100kg=1.60m /s2

!

x =12at 2; t =

2xa

=2(0.600m)1.60m /s2

= 0.750s2 = 0.866s

56

Dynamics Canister and Accelerating Glider

Developed by Rex Rice of Clayton HS, Clayton, MO. Problem Determine the amount of lead shot needed in a film canister, that hangs on the end of a string draped over a frictionless pulley at the end of a lab table and gravitationally pulls a frictionless cart of known mass, to produce an assigned acceleration.

Equipment and Setup

1. Level air track with a pulley on one end 2. Photo-gate used in gate and pulse mode 3. String 4. Film canister with hook on the lid 5. Balance 6. Bottle of small lead shot

The glider with flag is massed in front of the students and placed on the air track. Students are informed that when the practicum is checked the air will be turned on under the glider so that friction can be ignored. Students are to take the film canister and place the appropriate amount of lead shot inside of it so the glider will accelerate equal to the assigned value. Assigned acceleration values should fall between 1.0 and 2.5 m/s2. This practicum can be done in lab groups rather than the whole class approach. A piece of paper with each assigned acceleration value can be placed in the canister at each lab station.

Measurements Mass of glider and flag (mg) 210.0 g

Assigned acceleration 1.8 m/s2

Air-track

glider with flag Photo-gates

Film-canister with a hook in the lid

pulley

57

Sample Calculations Equation for the glider:

Equation for the canister:

Adding the equations:

Additional comments

When students have calculated the answer they place the empty canister on a balance and add lead shot until it has the correct mass. They then bring the canister to the air track and hang it on the end of the string. The photo-gate program is started, the glider is held stationary and the air is turned on. When the air is steady the glider is released. The answer is checked by running the photo-gate program in gate and pulse mode with the width of the flag entered in the program. This practicum generally yields answers within 2%. If the teacher wants the answer even closer they can angle the track down slightly. The practicum can easily be done in a 45 minute class period. Depending on the ability of your students it can be run as one problem for the whole class or given to smaller groups, with a different value for each group. The success of the group can be checked in 10 seconds if the photo-gates are used.

Notes:

!

Fstring = mga

!

Wcanister " Fstring = mca

!

Wc "mga = mca ; Wc = mca + mga ; mcg = mca + mga ; mcg "mca = mga

mc (g " a) = mga ; mc =mgag " a

=0.210kg(1.8m /s2)9.8m /s2 "1.8m /s2

= .0472kg

58

Dynamics An Up Hill Climb

Developed by Michael Crofton, Spring Lake Park High School, Spring Lake Park, MN

Problem Predict the time a given mass released from rest at a known height will take to reach the floor, when it is pulling a cart of known mass (not frictionless) up a ramp of a given angle. Equipment and Setup

1. Ramp (PASCO 2 meter track works well) 2. Cart with friction (PASCO dynamics cart with friction plate works well) 3. String 4. Masses, hooks, clamps, etc.

Measurements

1. Mass of the cart (m1) 510 grams 2. Mass of the pulling weight 500 grams 3. Distance above the floor (pulling weight) 1.00 meters 4. Ramp angle at which the cart will descend

with a constant speed 9.0° 5. Angle of the incline for the practicum 28°

Additional Comments Demonstrate to the class, on a second smaller ramp of the same material, the angle at which the cart slides at a constant rate. Providing the values of the masses and the angle of the incline will save needed time for students to work on the calculations. Make sure the string is taut when you release the falling weight. Timing the results can be done with a photo-gate (to start) and a plate with a piezo-electric sensor. The problem can be made a bit more difficult by hooking a friction block to the back of the cart with a horizontal string. The block has a higher coefficient of friction than the cart and the angle at which the block slides down at a constant speed must also be provided.

59

Sample Calculations Coefficient of friction from the ramp data:

The net force (pull of m1 less the force of the cart acting down the ramp and the drag (due to friction) must equal the total mass in the system times the acceleration therefore:

The time for m1 to reach the floor starting from rest is:

Generally there is no problem getting results within 5% (.05s). Notes:

!

µ = tan9.0° = 0.158

!

m1g " sin#mcg "µcos#mcg = (m1 + mc )a

a =m1g " sin#mcg "µcos#mcg

(m1 + mc )

a =.500kg(9.8nt /kg) " (sin28°)(.510kg(9.8nt /kg) " .158(cos28°)(.510kg(9.8nt /kg)

(.500kg+ .510kg)

a =1.83m /s2

!

x =12at 2

t 2 =2xa

; t =2xa

=2(1.00m)1.83m /s2

=1.05s

60

Dynamics Suspended Pulley

Developed by Michael Crofton, Spring Lake Park High School, Spring Lake Park, MN Problem Given the mass of a frictionless glider and a gravitationally falling suspended pulley, determine the acceleration of the glider. Equipment and Setup

1. Air track with pulley on end 2. Glider and flag 3. Photo-gates used in the gate and pulse mode 4. String 5. Two low friction double pulleys 6. Low friction single pulley 7. Two ring stands 8. Clamps

Measurements

Mass of the glider 200.0 grams Mass of the suspended pulley 80.0 grams Width of the flag on top of glider 10.0 cm (needed only for the photo gate

program)

Suspended pulley

Stool or table with a ring-stand clamped on top

Ring-stand Pulley

Air-track

Glider and flag

Photo-gates

61

Additional Comments The pulleys are the low friction ball bearing type. Masses can easily be added to the suspended pulley if wished. The answer is checked by running the photo-gate program in the gate and pulse mode with the width of the flag entered in the program.

Sample Calculations The force in the string is what will cause the glider to accelerate, therefore:

The force in the string is produced by half the weight of the pulley less that part of the weight that causes the acceleration of the pulley downward, therefore:

The acceleration of the glider coming forward will be twice that of the pulley downward, therefore:

Equating the force in the string and replacing acceleration of pulley with the acceleration of glider:

Solve for the acceleration of the glider:

Measured value was 1.74 m/sec2 Note: When I do the practicum I angle the track downward slightly to compensate for friction effects. However, you will get within about two percent of the correct value without making any adjustment.

!

Fs = mglagl

!

Fs =Wp

2"mpap2

!

agl = 2ap ap =agl2

!

mglagl =wp

2"mpap2

=wp

2"mp (agl /2)

2=wp

2"mpagl4

!

4mglagl = 2wp "mpagl; 4mglagl + mpagl = 2wp; ag 4mgl + mp( ) = 2wp

ag =2wp

4mgl + mp

=(2)(0.0800kg)(9.80N /kg)4(0.200kg) + (0.0800kg)

=1.78m /s2

62

Notes:

63

Momentum/Energy Spring Potential Energy

Developed by Jon Barber, Mounds View High School, Arden Hills, MN Problem Predict the time measured on a photo-gate timer for the run of a spring launched bumper sled. Equipment and Set Up 1. Air track with spring bumper sled 2. Ring stands, clamps, etc 3. Meter sticks 4. Newton scale 5. Thread 6. Photo-gate timer 7. Kg mass 8. Balance Compress the spring with a known force, and then release it suddenly by cutting a thread. The photo-gate timing distance is located beyond the point where the spring is fully extended, and the air track is considered frictionless. Measure and record the mass of the sled, the distance the spring is compressed, and the pull applied to the spring. Adjust the photo-gates to be approximately 1.00 m apart. Measurements

Air track timing distance xa = 1.00 m Mass of sled m = 0.23462 kg Compression distance xc = 0.0250 m Force on spring F = 10.0 N

Additional Comments This practicum needs the coordinated efforts of several students. One to turn on the air track, one to pull the spring balance to 10.0 N, one to cut the thread, and one to stop the sled after it has passed through the timing area in order to prevent it from bouncing back and re-triggering the photo-gates.

64

You need to check the air track sled with loop circular steel springs to insure that they obey Hooke's Law. The Newton scale must pull horizontally on the sled and parallel to the air track using a thread. Support the scale at a height equal to the air track with a lab jack or a pile of books and move back and forth to achieve proper alignment. Use a meter stick clamped to a ring stand and placed very close to the sled to read the springs compression. Sample Calculations Calculate the spring tension constant:

Using the spring constant, calculate the potential energy stored in the spring:

Set the potential energy before the expansion equal to the kinetic energy of the sled after the expansion and calculate the sleds velocity:

At this velocity the time required to span the distance between the timing gates:

Three measured values were, 0.90, 0.93 and 0.93 sec. Notes:

!

F = "kxs ; k = "Fxs

= "10.0N

"0.0250m= 400N /m

!

Ep =12kx 2 =

12400N /m(0.0250m)2 = .125J

!

Ep = Ek =12mv 2 ; v =

2(Ep )m

=2(0.125Nm)0.2346kg

=1.066m /s

!

V ="x"T

= ; "T ="xV

=1.00m1.066m /s

= 0.94sec

65

Momentum/Energy Humpty Dumpty

Developed by Jon Barber, Mounds View High School, Arden Hills, MN

Problem Place an egg under a suspended mass that, when released, stretches a spring as it falls, so that the egg is just touched or cracked, but not smashed. Equipment and Setup 1. Spring 2. Ring Stand, clamps and rods 3. Meter sticks 4. Mass with hook 5. Thread 6. Scissors 7. Lab jack (may substitute blocks) 8. Egg 9. Carbon paper 10. Clamp type clothespins 11. Clay Jaw type clothespins will produce a right angle when clipped to the ring-stand which allows you to mark the un-stretched length of the spring, the release position (r) and later the calculated egg location. Cover the bottom of the mass with carbon paper and tape in place. Next raise the mass to the release position and tie it to the upper rod supporting the spring. Place the egg directly below the mass and raise it with the lab jack or blocks to the calculated lowest position the mass is predicted to fall by the class. A small piece of clay will hold the egg in place. Before placing the mass in position for the problem measure the period for the spring and the mass to allow for the calculation of the spring tension constant (k).

Spring with thread in the center

Mass

Carbon Paper

Clothespins

Egg

Lab Jack/Blocks

66

Measurements

Mass m = 0.877 kg Period of the harmonic motion T = 0.974 sec

Spring extension at the thread held position xr = 0.100 m

Additional Comments Place the supporting rod high enough so the mass will not hit the base of the ring stand when it is dropped. When tying off the thread to support the mass, first tie the thread to the mass, then run the thread through the spring and wrap it around the support rod several times. This allows you to make small adjustment to the height. When you have the mass set at the position you want wrap the thread around the rod several more times and secure with a piece of tape. Any value between zero and the rest position can be chosen for the initial spring extension. When cutting the thread, place the scissors at the highest point possible so they won't interfere with the spring. If the mass falls and just touches the egg it will leave a black spot, if it falls a slight bit more then the egg may crack. If the students miscalculate, the egg will either not be touched or it will be smashed. Some old newspapers under the equipment to aid in clean up might be called for. Sample Calculations Calculate the spring constant:

Assuming zero gravitational potential energy at the shell of the egg then all the potential energy in the system (spring plus gravitational) must be conserved in the spring and therefore the distance the mass will fall (h) is calculated:

Trial 1 produced a black spot on the egg and no cracks. Trial 2 produced a black spot on the egg and a small crack. There is an alternative method of calculating the answer that does not use energy considerations. You may wish to inform your class that their solution must use energy as the central thought process. If not, it is interesting to see if students proceed with energy after having been indoctrinated with the concept.

!

T = 2" mk

; k =4" 2(m)T 2

=39.48(.877kg)(.974s)2

= 36.5kg /s2 = 36.5N /m

!

mgh +12kxr

2 =12k(xr + h)2

h =2(mg" xrk)

k=2(0.877kg(9.8N /kg) " (0.100m)(36.5N /m)

36.5N /m= .271m

67

Notes:

68

Momentum/Energy Ballistic Pendulum

Developed by Hank Ryan and Jon Barber, Mounds View High School, Arden Hills, MN Problem Predict the distance a projectile will travel when shot horizontally from a ballistic pendulum. Equipment and Set Up

1. Ballistic Pendulum apparatus 2. Tape measure 3. Ruler, Meter sticks 4. Clay 5. Balance

Additional Comments Use a small amount of clay to help catch the projectile when it strikes the pendulum. Press the clay into the pendulum and form it so as to absorb the projectile when it hits. After students have solved for the velocity of the projectile and predicted the impact point on the floor, move the pendulum out of the line of fire. The projectile is then shot again to determine the range and compare it with the calculated result. Use a "pancake" of clay to mark the target point. Place this clay pancake (15 cm to 20 cm in diameter) over the impact point and mark it to indicate the exact predicted spot. When the projectile hits the clay, it leaves a permanent record of the impact for discussion purposes and also serves to absorb the energy of the projectile. Measure the mass of the projectile and the ballistic pendulum as well as height of spring gun muzzle from the floor. Fire the gun into the ballistic pendulum and measure the rise of the center of mass. Measurements

Height of gun muzzle h1 = 1.02 m Rise of center of mass of the pendulum h2 = 0.069 m Mass of the ball (projectile) mb = 0.0675 kg Mass of the ballistic pendulum mp = 0.2761 kg

69

Sample Calculations The kinetic energy of the pendulum and the projectile after the collision is equal to the potential energy of the pendulum and projectile when the pendulum reaches it's highest point, therefore:

Calculate the velocity of the projectile using the Law of Conservation of Momentum:

The time the projectile is in flight:

The range of the shot:

Measured values were 2.71m and 2.67m.

!

12(mb + mp)(Vp+b )

2 = (mb + mp)gh

12(0.0675Kg + 0.2761kg)(Vp+b )

2 = (0.0675Kg + 0.2761Kg)(9.8ms2

)(0.069m)

(0.1718kg)(Vp+b )2 = 3.367kgm /s2

Vp+b =.2323kgm2 /s2

0.1718kg=1.163m /s

!

mbVb = (mb + mp )(Vp+b ) ; Vb =(mb + mp )(Vp+b )

mb

Vb =(0.0675kg+ 0.2761kg)(1.163m /s)

0.0675kg= 5.92m /s

!

h =12gt 2 ; t =

2hg

=2(1.02m)9.8m /s2

= .456sec

!

Vav ="xr"t

; "xr =Vav "t( ) = 5.92m /s(0.456s) = 2.70m

70

Notes:

71

Momentum/Energy Jumping Frame

Developed by Jon Barber and Hank Ryan, Mounds View High School, Arden Hills, MN Problem Predict how far, and in which direction, a frame will move when the enclosed carts explode apart. Equipment and Setup 1. Light wooden frame with low friction sliders 2. Two carts, one must be spring loaded 3. Velcro 4. Felt tip pen 5. Paper 6. Sticky tape 7. Meter stick 8. Balance 9. Brick

The system consists of a low friction frame with two carts in the middle. The carts and frame have Velcro on them so when the carts spring apart and make contact with the frame they will stick and not rebound. The carts have unequal masses, but equal distances to travel to the frame. Students will need to measure the mass of both carts, one with a brick taped to it, and the mass of the frame with a felt tip pen attached. The distance between the cart and the frame on either end should be equal and measured to the nearest 0.001 m. Sample data is given. Measurements Mass of frame + pen Mf = 0.535 kg Mass of cart Mc = 1.181 kg Mass of cart + brick Mb = 3.492 kg Distance between carts and the frame ∆X = 0.16 m

72

Additional Comments The distance the frame moves is recorded using a felt tip pen inserted through a hole drilled in the frame. The tip of the pen is set to drag over a piece of paper beneath it, leaving a permanent record of the frames trip. If the paper beneath the pen is bent over, but not folded, it has enough spring left in it to press against the pen. After the calculations are complete, explode the carts apart and compare the prediction with the length of line left by the felt tip pen. It is important to tape the brick securely in place on the cart because any motion of the brick introduces an impulse into the system that is not accounted for and can significantly alter the results. The problem can be made more difficult by not having the run distance to the frame for the two carts equal. Sample Calculations Using conservation of momentum where Vc is the velocity of the cart and Vb is the velocity of the more massive cart with the brick.

Neither velocity is known but the velocity of the brick and cart can be found in terms of the velocity of the cart alone.

The momentum equation can also be used to find the distance the brick and cart moves while the empty cart reaches the frame. Let ∆Xb equal the distance the massive cart moves while the lighter cart covers the 0.16 m.

The distance between the massive cart and the frame is now:

!

MbVb = "McVc ; 3.492kgVb = "1.181kgVc

!

Vb = "1.181kg3.492kg

(Vc ) = "0.338Vc

!

MbVb = "McVc therefore : Mb#Xb

#t= "Mc

#Xc

#t

#Xb = "Mc#Xc

Mb

= "1.181kg(0.16m)

3.492Kg= "0.054M

!

0.16m " 0.054m = 0.106m

73

To find the velocity of the cart and frame after they have bumped and stuck together, use conservation of momentum.

The speed at which the cart + frame approaches the bricked cart is: Of the distance to be covered by the two approaching bodies, the cart carrying the frame will travel:

Measured values were 0.068 m and 0.070 m. Notes:

!

McVc = (Mc + Mf )Vc+ f

Vc+ f =McVc

Mc + Mf

=1.18kg(Vc )

1.181kg + 0.535kg= 0.688Vc

!

Speedof approch =Vc+ f +Vb = .688Vc + 0.338Vc =1.026Vc

!

0.688Vc

1.026Vc

(0.106m) = 0.071m

74

Momentum/Energy Tarzan

Developed by Jon Barber and Hank Ryan, Mounds View High School, Arden Hills, MN Problem A pendulum bob swings down through an arc, at the very bottom of the swing a razor blade cuts the bob free, determine the point of impact of the bob with the floor. Equipment and Setup 1. Scissors 2. Ball to be used as pendulum bob, about 1 kg 3. Non-stretch string, linen works well 4. Clamps 5. Sticky tape 6. Meter stick 7. Razor blade 8. Thread 9. Step ladder Measurements

Height of the bob above rest position h = 0.711 m Fall distance of pendulum bob after being cut free f = 0.759 m Earth's gravitational field strength g = 9.8 m/s2

Additional Comments To obtain the needed height for this practicum, mount the pendulum on a stepladder set on a lab table. To avoid introducing unnecessary impulses during the cutting of the string, use a new razor blade and set it at an angle so the string will slide along it as it is cut. To hold the razor blade rigid, clamp it between two rulers with "C" clamps exposing a sliver of the sharp side. While aligning the razor blade to the vertical pendulum string, cover its edge with a piece of tape. Use non-stretch string for the pendulum or some of the energy will go into potential energy in the string and convert to heat after the weight is cut free. While the mass of the pendulum bob does not need to be known, a fairly massive ball of about 1.0 kg is desirable. This helps minimize factors such as impulse from the razor blade, air friction, and release disturbances that would affect the accuracy of the practicum. The pendulum bob is pulled back and held at its release point with a thread. Tape is used on the bob to create

75

loops for the two threads that need to be attached to it. Take care to cut the thread with as little disturbance as possible when releasing the bob. Sample Calculations Set the loss in gravitational potential energy equal to the gain in kinetic energy and solve for v, the horizontal velocity:

Using the fall distance solve for the time of the fall.

Determine the horizontal distance the bob moves during the fall:

Measured value was 1.43 m. Notes:

!

Ep = Ek ; mgh =12mv 2

v = 2gh = 2(9.8m /s2)(0.711m) = 3.73m /s

!

f =12gt 2 ; t =

2(0.759m)9.8m /s2

= 0.394sec

!

x = vt = 3.73m /s(0.394s) =1.46m

76

Momentum/Energy Vector Conservation of Momentum

Developed by Hank Ryan and Jon Barber, Mounds View High School, Arden Hills, MN Problem When a ball is released down an acceleration ramp, and no collision with a second ball occurs, where will it land on the floor? Equipment and Setup 1. Large scale "collision in two dimensions" apparatus. 2. Two balls, sized for the equipment, each of different mass. 3. Plumb line. 4. A flat pancake of clay approximately 10 cm by 15 cm. Release a ball down the ramp from a point below middle and mark where it hits the floor. Draw a line from the end of the plumb line to this mark to establish a centerline on the floor that runs straight out from the end of the ramp. Next, set the pivot holding the front ball to the side in such a way as to produce a glancing collision. Run a collision of the two balls using the full ramp and mark the positions of both balls as they impact on the floor . Point out that both balls hit the floor at the same moment. Place carbon paper face down at these points and run the collision a few more times. The marks produced will allow the class to get a feel for the consistency of the results and to determine an average value for the impact points. Measurements

Mass of ball #1 m1 = 0.1616 kg Mass of ball #2 m2 = 0.0455 kg Horizontal distance to ball #1 ∆x1 = 0.900 m Horizontal distance to ball #2 ∆x2 = 1.323 m Angle off the centerline to ball #1 Ø1 = 15.0° Angle off the centerline to ball #2 Ø2 = 40.5° The vertical drop of the balls (optional) Y = 0.95 m

Sample Calculations Using the vertical drop of the balls you can calculate the results in real time. The calculations shown below refer to the time of fall as a single unit called a "tick". This time of one "tick" is true for all trials and therefore will not cause any complications in the calculations.

77

The horizontal velocity of ball #1 and ball #2 after V’1, V’2) collision:

Ball #1 is directed at 15.0° to the right of the centerline and ball #2 at 40.5° to the left of the centerline. The horizontal momentum of ball #1 and ball #2 after (p’1, p’2) collision:

Vector addition of p'1 and p'2:

Therefore the horizontal momentum of ball #1 before collision must have been 0.184 kg m

tick .

The horizontal velocity of ball #1 before collision:

The horizontal distance ball #1 will travel when no collision occurs:

The actual result in this experiment was 1.15 m.

!

V '1="x1"T

=0.900m1.0ticks

= 0.900m / tick

V '2 ="x2"T

=1.323m1.0ticks

=1.323m / tick

!

p'1= m1V '1= 0.1616kg(0.900m / tick) = 0.1454kgm / tick

p'2 = m2V '2 = 0.0455kg(1.323m / tick) = 0.0602kgm / tick

P + P = 0.1841 2

' '

15o

40.5o

P = 0.06022

'

center line

P = 0.06022

'

P = 0.14541

'

!

p1 = m1V1 ; V1 =p1V1

=0.184kgm / tick0.1616kg

=1.14m / tick

!

"xh =VhT =1.14m / tick(1.00ticks) =1.14m

78

Additional Comments To construct a large-scale "collision in two dimensions" apparatus make the frame of lightwood (pine) and the curved ramp is heavy-duty plastic corner molding. The last few centimeters of the lower end of the ramp must be horizontal. In order to have a consistent start for all trials, the board at the high end of the ramp can go past the plastic molding. For each run the ball is pushed up against this "backstop" and released. On the lower end of the ramp the balls must hit center to center. Because the two balls may not have the same diameter, adjustments will be required. This is easy to achieve if the stationary ball rests on a three way adjustable pedestal. The adjustments needed are up and down, in and out, and swinging from side to side. A threaded electric lamp assembly with a slot in the clamping arm will allow for all the needed adjustments. The clamping arm can swing back and forth on the mounting screw to produce collisions that range from a glancing blow to head on. This adjustment plus switching the two balls around allows for all classes to have a different problem. Check the two balls for collision in two ways: 1. They must meet on center. 2. The ball coming down the ramp must be just clearing end of the ramp at the moment of collision (it cannot already have started to fall). Notes:

79

Momentum/Energy Gravitational Potential Energy

Developed by Jack Netland, Osseo High School, Osseo, MN Problem Calculate where the puck will hit the floor after it is pulled to the edge of an air table by falling weights. Equipment and Set Up 1. Air table 2. String and low friction pulley 3. Weight hanger 4. Weights 5. Clay 6. Meter sticks 7. Air table pucks 8. Plumb bob Friction is negligible when using an air table and a low friction pulley. Height (h1) is set equal to the puck's run distance on the air table so that the weight contacts the floor just as the puck leaves the table. The distance from the table to the falling weights is large enough so the puck falls freely to the clay target on the floor without making any contact with other objects. The clay protects the puck from damage and also gives a permanent mark at the impact point. The small piece of clay near the table can be made the same thickness as the puck target for easy measuring of h2. Hold a plumb bob against the launching edge of the air table and drop it a fraction of a centimeter onto the clay below to produce a convenient mark from which to measure the puck's flight distance. Measurements

Mass of puck mp = 0.087 kg Height of weight off the floor h1 = 0.20 m Height of table from floor h2 = 1.245 m Mass of weights mw = 0.250 kg Earth's gravitational field strength g = 9.8 m/s2 or N/kg

80

Sample Calculations Potential energy of the weights:

This potential energy becomes kinetic energy in the puck and the weights:

When the weights have fallen through h1 and hit the floor the puck and these weights will have equal velocities, therefore Vw = Vp. Solving the above equation for this velocity (V):

The time of fall of the puck:

Calculate the range (x) using the horizontal velocity of the puck and the time of fall:

The measured value was 0.86 m. Additional Comments

Students have come up with equally valid solutions not involving conservation of energy. It is interesting, after teaching momentum and energy, to see if this is now how your students think. If the class proceeds with a forces only approach one of the pitfalls is using the average velocity of the puck as it is pulled across the table in calculating the range. The horizontal velocity for the flying puck is the same for the whole flight and is equal to the final velocity achieved as it was pulled across the table. This physics practicum has also been done successfully using carts and adding the proper weights to adjust for the friction in the system.

!

Ep = mgh = 0.250kg(9.8N /kg)(0.20m) = 0.49J

!

Epw = Ekw + Ekp =12MwVw

2 +12MpVp

2

0.49J =12(0.250kg)(Vw

2) +12(0.087kg)(Vp

2)

!

2(0.49J) = (0.250kg)(V 2) + (0.087kg)(V 2) ; 2(0.49J) = 0.337kg(V 2)

V =2(0.49J)0.337kg

=1.70m /s

!

h2 =12gt 2 ; t =

2(1.245m)9.8m /s2

= 0.50sec

!

x =Vave"t =1.70m /s(0.50s) = 0.85m

81

Notes:

82

Electricity Resistivity

Developed by Jon Barber, Mounds View High School, Arden Hills, MN

Problem First, determine the length of Nichrome wire (Chromel A) needed to have a resistance of 2 ohms, and then predict the amperage in a circuit containing the Nichrome wire, an ammeter, a voltmeter with the power source set at 6.0 volts. Equipment and setup 1. Variable DC power supply 2. Ammeter 3. Voltmeter 4. Meter stick 5. Sponges 6. Wire and alligator clips 7. One meter of Nichrome wire (Chromel A) 8. C clamps Students need to measure the diameter of the Nichrome wire and find its resistivity. Resistivity is given in the Handbook of Chemistry and Physics and most textbooks. Measurements

Diameter of the wire d = 0.65 x 10-3 meter Radius of the wire r = 0.325 x 10-3 meter Voltage (for internal resistance) V = 0.325 V Amperage (for internal resistance) A = 2.0 A Assigned Resistance of wire section R = 2.0 ohms Handbook value for wire resistivity r = 1.00 x 10-6 ohm meters

Ammeter

Calculated section of Chromel A wire clipped off for testing

Voltmeter

Power source

83

Additional Comments Students clamp the wire to a meter stick and clip in with alligator clips to include only the calculated length in the circuit. Control the temperature of the wire as the wire gets very hot and changes its resistance. Wet sponges laid on the wire during the test works well. To obtain accurate results include the internal resistance of the system with the wire resistance. The internal resistance can be calculated by the voltmeter-ammeter method. Sample Calculations Cross sectional area of the wire:

Calculation of proper wire length:

Calculation of internal resistance of the system:

Adding the 0.1625 ohms to the 2.0 ohms of the wire gives the total resistance of the system as 2.1625 ohms. Determine the amperage value for the circuit when 6.0 volts are applied:

Measured value was 2.70 A.

!

A = "r2 = (3.14)(0.325x10#3m)2 = 3.3x10#7m2

!

resistance(R) = (resistivity) LengthCross sec tional area

= r LA

!

L =ARr

=(3.3x10"7m2)(2.0ohms)1.00x10"6ohm meters

= 0.643meters

!

V = IR ; R =VI

=0.325V2.0A

= 0.1625ohms

!

V = IR ; I =VR

=6.0V

2.1625ohms= 2.77A

Ammeter

Voltmeter

Power supply

84

Notes:

85

Electricity

Blow the Circuit Developed by Jon Barber, Mounds View High School, Arden Hills, MN

Problem Prepare a circuit (with light bulbs, wire, clips, and one 0.85 amp fuse) so that when the circuit is initially activated it has one bulb left out of it's receptacle and all other bulbs lit and then when the last bulb is turned into its receptacle the fuse blows and all the lights go out except one. Equipment and Setup 1. Variable DC power source 2. Wires and alligator clips 3. Lamps, 6.3V, 0.25A (# 46 Lamps, Radio Shack) 4. Fuse, 0.50A (I have found that this type of fuse will blow at about 0.85A) The lamps have a resistance of 25.2 ohms and will be connected to a power supply set at 6.3 volts. Because at first several bulbs will be operating at 6.3 volts and after the fuse blows only one bulb will be lit, a parallel circuit is needed to keep the voltage constant. The fuse blows at 0.85 amps, so four bulbs (which use a total of 1.0 amp) will cause the fuse to blow. If one lamp is to stay lit, the fuse must be properly placed. This one bulb must be placed in such a way so that it will not be affected when the fuse blows out. A successful circuit is shown above. Additional Comments Students need to determine the amperage each bulb uses, the type of circuit needed, the number of bulbs needed and the placement of the fuse. It is necessary to set the voltage on the variable DC power source to 6.3 volts with a load of 4 bulbs prior to the practicum. Each bulb will then be using 0.25 amps. After students have completed the calculations and drawn out their circuit, test it out. The one shown in the diagram above will not blow the fuse with any one of the four bulbs on the right not screwed in. When this bulb is turned in, the fuse will blow and all the lamps go out except the one on the left. Sample Calculations Using Ohm's Law:

!

V = IR ; I =VR

=6.3V

25.2ohms= 0.25A for each bulb

6.3 Volts Bulbs

Fuse

86

Notes:

87

Electricity

Resistance Will Vary Developed by George Amann, Franklin D. Roosevelt High School, Hyde Park, NY

Problem

An electric circuit has two devices that must operate at the same time in a parallel circuit with a given power supply voltage. Each device requires a certain voltage and amperage for proper operation. Determine the amount of resistance needed in each circuit to provide the proper current and voltage for each device and set the variable resistors accordingly.

Equipment and Set Up

1. One 30 Ω resistor (device 1) 2. One 60 Ω resistor (device 2) 3. Two variable resistors (up to 100 Ω) for the unknown resistors 4. Two digital DC voltmeters 5. Two digital DC Ammeters 6. Power supply for up to 20 Volts 7. Wires to complete the circuit shown below

Measurement Power supply 20.0Volts Device #1 (upper branch) 15.0 Volts @ 0.50 Amps Device #2 (lower branch) 12.0 Volts @ 0.20 Amps Sample Calculations Total resistance for the top branch: Total resistance for the bottom branch:

!

R =VI

=20.0Volts0.50Amps

= 40.0"

R = 40.0" total resistance

!

R =VI

=20.0Volts0.20Amps

=100.0"

R =100.0" total resistance

88

Resistance in device #1: Resistance in device #2:

Therefore for variable resistor #1: Therefore for variable resistor #2: Measured results when variable resistors were set to 10.0 and 40.0 ohms:

Notes:

!

R =VI

=15.0Volts0.50Amps

= 30.0"

!

R =VI

=12.0Volts0.20Amps

= 60.0"

!

40.0"# 30.0" =10.0"

!

100.0"# 60.0" = 40.0"

!

V1 =15.1Volts I1 = 0.50AmpsV2 =12.0Volts I2 = 0.20Amps

89

Electricity Resistance is Futile

Developed by Michael Crofton, Spring Lake Park High School, Spring Lake Park, MN Problem Determine the resistance of a hidden resistor in a circuit and the potential difference across one resistor in a parallel branch. Equipment and Setup 1. Collection of resistors, wired to a board 2. Unknown resistor, hidden so that students cannot see its resistance 3. Multimeters or voltmeter and ammeter 4. Power supply and patch cables to connect to circuit 5. Wire leads

2000Ω

1000Ω 500Ω 500Ω

500Ω

200Ω

1000Ω

1000Ω

(Unk

now

n)

500Ω

90

Measurements

Potential difference across circuit 8.94 V Current leaving power supply 9.40 mA

Sample Calculations: Total circuit resistance:

Resistance in tower series:

Equivalent resistance in the tower:

Unknown resistance must be:

Measured value was 350.0 ohms Potential difference across the tower:

!

R =VI

=8.94V0.0940A

= 951"

!

1000"+ 500"+ 500" = 2000"

!

1Req

=1

2000"+

1200"

+1

2000+

11000"

+1

500"+

11000"

Req =2000"20

=100"

!

500"+100"+ Runknown = 951"; Runknown = 951"# 600" = 351"

!

V = IR = (0.00940A)(100") = 0.940V

(Unk

now

n)

3

50Ω

1000Ω

Top View 500Ω

Tower

1000Ω

91

Current in the top series of resistors:

Potential difference across the 500 ohms resistor in the series:

Measured value was 0.234V Additional Comments The circuit can of course be constructed using a multitude of designs. The one shown arranges many of the resistors in a vertical tower to optimize space and minimize wires. This circuit has 9 known resistors and the one unknown resistor. Be sure to use a digital multimeter to check the resistance of each of the resistors, rather than use the rated values. In this setup the resistance was written on a piece of wood under each resistor. To get started connect the power supply between the terminals of the circuit and place a digital voltmeter across the power supply and a digital ammeter between the terminal and the first series resistor. Turn on the power supply and allow the students to read the values of the two meters. Disconnect and remove the meters. Remove the unknown resistor and allow the students to examine the circuit with the condition that are not allowed to use any meters to solve the problem. Students should sketch a circuit schematic to help them determine the equivalent resistance of the portion of the circuit that they can see, so that they can determine the value of the hidden resistor. Once they have solved for the unknown values the unknown resistor is checked with a digital ohmmeter. It is then placed back in the circuit. Then the power supply is reconnected and the potential difference across the circuit is checked to see that it is unchanged. Then the voltmeter is placed across the top resistor in the tower to check the value. This practicum can be done in one 45 minute period. For General Physics, the resistor value is probably enough. Honors classes should be able to solve for both values. A predicted value within 1.0 Ω on the resistor and 1% on the potential difference can be obtained.

!

I =VR

=0.940V2000"

= 0.00047A

!

V = RI = 500"(0.00047A) = 0.235V

92

Notes: