Practicums
For
Physics
Teachers
(Second Edition)
Edited by Michael Crofton
Central Theme Developed
by
Henry J. Ryan and Jon E. Barber
Fall 08
2
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
3
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
4
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
5
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!”
6
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
7
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.
8
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.
9
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
Screen With
Image
Concave
Mirror
Show Case
Bulb
Paper Tape
Optical Bench
Focal
Point
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.
10
Sample Calculations
Distance to the Image measured to the focal point:
SoSi = f 2 ; Si =
Si =
f2
So
(21.2 cm) 2
= 41.4cm
10.85 cm
Measured value was 41.4cm
! Height of image:
Hi f
( f )(Ho)
=
; Hi =
Ho So
So
Hi =
(21.2 cm)(8.4 cm)
= 16.4cm
10.85 cm
Measured value was 16.2cm
! Notes:
11
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
Screen
Lens
Focal length
f = 25.0 cm
Light Source with
Hood
Height of object
Ho = 7.9 cm
Object to focal point distance
So = 35.0 cm
Calipers
Lab Jack
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.
12
Sample Calculations
Distance to the image measured from the focal point:
SoSi = f 2 ; Si =
Si =
f2
So
(25.0 cm) 2
= 17.9cm
35.0 cm
Measured value was 18.0cm
! Height of the image:
Hi f
f (Ho)
=
; Hi =
Ho So
So
Hi =
25.0 cm (7.9 cm)
= 5.6 cm
35.0 cm
Measured value was 5.7cm
! Notes:
13
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
Number of lights in each set
The fact that all the lights are of equal power (same lights)
Standard intensity of a single lamp (optional)
86.7 cm
2 and 4
150 candles
Sample Calculations
I1 =
K1
r2
I2 =
K2
r'2
For the paraffin photometer to receive equal illumination from each set of lights it has to be
! placed so that I1 = I2.
!
Therefore:
K1 K 2
=
r 2 r' 2
!
14
In addition:
r + r'= 86.7 cm
r'= 86.7 cm " r
By substitution:
!
K1
K2
=
2
r
(86.7 cm " r) 2
r2
K1
=
2
(86.7 cm " r)
K2
Taking the square root of both sides:
!
r2
(86.7 cm " r)
2
=
r
= 1.41;
86.7cm " r
!
k1
4
=
= 1.41
k2
2
r = (86.7 cm " r)(1.41); r + 1.41r = 122 cm
r = 50.7 cm
r'= 86.7 cm " 50.7 cm = 36.0 cm
!
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:
15
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
Wire Leads
Diameter of circle
D = 20.0 cm
Index of refraction of water
nw= 1.33
Bucket showing the
circle of light
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.
16
Sample Calculations
Radius
The critical angle for water:
1
sin "c =
1.33
Wire Leads
Normal Line
n w = 1.33
Light Bulb
sin " = .75
Critical Angle
" = 48.6°
!
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.
!
tan 48.6° =
10.0 cm
10.0 cm
; Depth =
Depth
tan 48.6°
Depth = 8.81cm
! Measured results 8.8cm
! Notes:
17
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
Angle of incidence and refraction for the liquid
40.0°; 25.4o
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
18
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:
(n w )(sin " w ) = (n g )(sin " g ) ; sin " g =
(1.33)(sin 40°)
= .57
1.50
" g = 34.7°
sin " g =
Measured value was 34.5°.
! Notes:
(n w )(sin " w )
ng
19
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
Laser
Beam
Slide held
in clamp
Screen
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
Length from slide to paper screen
Wavelength of light
Nodal line assigned
d = .011 cm
L = 452 cm
" = 6.328 x 10-5 cm
n=2
!
20
Additional Comments
Screen
The set up for this practicum is similar to
d
pictures in texts for a two slit apparatus
producing interference patterns on screens.
The screen can be white paper taped to a
X
L
black board. The laser beam is allowed to
hit the paper screen with no slide in place
to establish the center point for the
2nd Nodal Line
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:
1
(n " ) #L
x2
1
2
d = (n " ) # ; x 2 =
L
2
d
1
(2 " )(6.328x10"5 cm )(425 cm )
2
x2 =
= 36.7cm
0.0011cm
Measured value was 37.2cm
! Notes:
21
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)
(T) ; where Vo = 331m /s
°C
V = 331m /s +
!
(0.60m /s)
(20°C) = 343m /s
°C
22
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:
" = 4(slide movement) = 4(0.182m) = 0.728m
The frequency is therefore:
!
V = f" ;
f =
V 343m /s
=
= 471 Hertz
" 0.728m
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:
23
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:
va =
x 1.27m
=
= 0.127m /s
t 10.0sec
vb =
1.38m
= 0.138m /s
10.0sec
Distance traveled by each buggy before they collide:
!
!
xa = va t
x a = 0.127 m / s(t)
xb = vb t
x b = 0.138 m / s(t " 5.0sec)
24
The two buggies will travel a total distance of 4.00 meters before they collide therefore:
0.127 m / s(t) + 0.138 m / s(t " 5sec) = 4.00m
0.127 m / s(t) + 0.138 m / s(t) " 0.690 m = 4.00m
t=
!
4.69 m
= 17.7sec
0.265 m / s
Using buggy “A” to determine the collision point:
X a = v a t = 0.127 m / s(17.7 s) = 2.25m
!
Checking the total distance:
!
x b = v b (t) = 0.138 m / s(17.7 s " 5.0 s) = 1.75m
2.25m + 1.75m = 4.00m
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:
25
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
Top View
Air Table
Thread
Side View
Thread
Puck
Puck
Air Table
Students need to measure the length of the pendulum thread and the angle between horizontal and
the table surface.
Measurements
Length of pendulum thread
Angle between horizontal and the table surface
L = 0.72 m
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.
26
Sample Calculations
The acceleration down the ramp due to gravity is calculated using the following relation:
sin" =
a(down the ramp )
g
a(down the ramp ) = sin19.4°(9.8m /s2 ) = 3.25m /s2
The period of the pendulum is calculated using the following relation:
!
T = 2"
L
0.72m
= 2"
= 2.96sec
g
3.25m /s2
Measured values were 2.93 sec and 2.91 sec.
! Notes:
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
Radius of inner beaker
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.
Ro = 0.80 m
Ri = 0.40 m
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:
4" 2 R
ac =
T2
T=
4 " 2 (0.80m)
9.8m /s =
T2
2
4 " 2 (0.80m)
= 1.80sec
9.8m /s2
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:
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
Ramp length
from the experiment
that are not near the
Formica board
average value for a
run down the ramp.
Ramp height
Measure the length
of the ramp and the
Drilled center
height used during
hole to fit the
the timed runs.
turntable
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
Ramp height
Average time for pucks descending the ramp
Period of spin for the board on the turntable
Mass of the pucks
Gravitational field strength of the earth
L = 1.22 m
H = 0.437 m
t = 1.5 s
T = 2.1 s
Mp = 0.22 kg
g = 9.8 m/s2
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.
sin " =
0.437m
= 0.358
1.22m
The force acting down the ramp is therefore:
!
Fr = sin " (m p g) = 0.358(0.22kg)(9.8N /kg) = 0.77N
The actual acceleration of the pucks down the ramp can be calculated from:
!
!
1
d = at 2
2
a=
2d 2(1.22m)
=
= 1.08m /s2
2
2
t
(1.50s)
31
The unbalanced force acting on the pucks down the ramp:
Fnet = ma = (0.22kg)1.08m /s2 = 0.24N
Determination of the friction force:
!
F friction = 0.77N " 0.24N = 0.53N
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:
!
Fc =
4 " 2 Rm
Fc t 2
0.53N(2.1s) 2
;
R
=
=
= 0.27m
t2
4 " 2 m 4 " 2 (0.22kg)
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:
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.
2.
3.
4.
Toy helicopter, or plane, battery operated
Heavy duty table clamp
Lab rod, around one meter long
Stopwatch
Ø
Ø
Diagram #1
h
Measurements
Length of string (from point of connection to the center of helicopter)
Distance from table level to helicopter level
Distances from table level to top of lab rod
1.02 m
0.19 m
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.
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:
Tup " W = 0
Ts
Diagram #2
Fup = W
Ø
Tup = Ts (sin # )
Tx
W = mg
mg
Ts =
sin #
Ts (sin # ) = mg
W
Diagram #3
!
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.
b 2 = c 2 " a 2 = (1.02m) 2 " (0.90m) 2
b = 0.48m
c =1.02m
a =0.90m
Ø
b =?m
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:
Tx = Fc =
!
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.
cos" =
!
!
Tx
Ts
Tx = Ts (cos" )
4 # 2 Rm
Ts (cos" ) =
t2
Ts =
!
4 " 2 Rm
t2
mg
sin "
From diagram number 2
Substitute Ts (cos" ) for Tx
From diagram number 2
!
34
mg(cos " ) 4 # 2 Rm
=
sin "
t2
Substitute
mg
for Ts
sin "
Solve for t
!
t=
4 " 2 Rm(sin # )
=
mg(cos # )
tan " =
!
t=
!
0.90m
= 1.875
0.48m
4 " 2 R(tan # )
g
!
From diagram number 3
4 " 2 (0.48m)(1.875)
= 1.89sec or 56.7 sec for 30 revolutions
9.8m /s2
Measured value was 1.9 s/rev or 56.9 sec for 30 revolutions.
! Notes:
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.
PASCO
projectile
launcher
Carbon
Paper
Measurements
Distance from the barrel to the floor
Horizontal range
1.14 meters
1.78 meters
Sample Calculations
Time of fall when launched horizontally is given by:
y=
gt 2
2
1.14m =
(9.8m /s2 )t 2
2
t=
2(1.14m)
9.8m /s2
The horizontal range then yields the muzzle velocity from:
!
vm =
!
x
t
vm =
1.78m
= 3.69m /s
.482sec
t = 0.482sec
36
To find the range when launched at an angle requires that the muzzle velocity be broken into
vertical and horizontal components, therefore:
v y = v m sin "
v y = 3.69m /s(sin 30°) = 1.85m /s
v x = vm cos "
v x = 3.69m /s(cos 30°) = 3.20m /s
Use the time of flight equation and solve the quadratic:
!
y = vy t +
1
gt2
2
1
"1.14m = (1.85m /s)(t) + ( )("9.8m /s2 )(t 2 )
2
4.9t 2 "1.85t "1.14 = 0
t = 0.71sec
The horizontal range is therefore:
x = vx t
!
x = 3.20m /s(0.71s) = 2.26m
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:
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.
2.
3.
4.
5.
6.
Blow dart gun and quality darts
Level
Sheet of fine grained Styrofoam insulation
Ring stands, clamps, etc.
Plastic bucket
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
Horizontal distance of the shot
Earth's gravitational field strength
h =1.70 m
r = 7.00 m
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:
h = 1/2gt 2 ; t 2 =
2h
g
; t=
2h
2(1.70m)
=
= 0.59sec
g
9.8m /s2
Muzzle velocity:
Vmuzzle =
!
range 7.00m
=
= 11.9m /s
t
0.59s
X and Y components of the impact velocity:
Vx = 11.9m /s
!
Vy = gt = 9.8m /s2 (0.59s) = 5.8m /s
Impact angle:
!
tan " impact =
Vx 5.8m /s
=
= 0.48
Vy 11.9m /s
" impact = 26°
Measured value: 26o
! Notes:
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.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
Air track
Photo-gates and timer
Plumb bob
Paper strip
Carbon Paper
Meter sticks
Ball with a hole for projectile
Coat hanger wire
Clay
Wood block
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.
Photogates
Air Sled
(Spring Bumpers Removed)
Projectile
Path
Carbon Paper
Target
Air
Track
Measurements
Vertical fall
Horizontal distance
y = 1.130 m
x = 0.642 m
40
Sample Calculations
Calculate the time of fall:
1
2y
2(1.130m)
y = gt 2 ; t =
=
= 0.48sec
2
g
9.8m /s2
Calculate the horizontal velocity:
!
vx =
x 0.642m
=
= 1.338m /s
t
0.48s
Calculate time between photo-gates:
!
vx =
xp
t
; t=
xp
0.800m
=
= 0.60sec
t 1.338m /s
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:
Wire
Clay Counter Weight
Wood Block
Ball
Air Sled
(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
Muzzle velocity allowed by the window
Muzzle to ceiling distance
Height of muzzle from the floor
Muzzle to back wall of the classroom
Ø = 17.8°
Vm= 19.2m/s
Hc = 2.29m
Hm = 0.87m
Xw = 13.67m
Sample Calculations
X and Y components of the muzzle velocity
sin " =
!
vy
; Vy = sin(17.8°)(19.2m /s) = 5.87m /s
19.2m /s
42
cos" =
Vx
; Vx = (cos17.8°)(19.2m /s) = 18.3m /s
19.2m /s
Muzzle Velocity
19.2 m/sec
!
17.8o
v = 18.3 m/sec
x
Vy = 5.87 m/sec
Horizontal
Flight time for the shot as it returns to muzzle level:
T=
2Vy 2(5.87m /s)
=
= 1.20sec
g
9.8m /s2
Maximum height of the dart during which occurs at half the flight time:
!
1
1
y max = v y t + gt 2 = 5.87m /s(.60s) + ("9.8m /s2 )(.60s) 2 = 1.76m
2
2
Note: The dart will not hit the ceiling.
! Range for the shot when it falls back to muzzle level:
X range = Vx (t) = 18.3m /s(1.20s) = 21.9m
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:
x w = Vx t ; T =
x w 13.67m
=
= 0.75sec
Vx 18.3m /s
The height of the dart 0.75 seconds into the flight relative to the gun muzzle:
!
1
1
y 0.75s = v y t + gt 2 = 5.87m /s(0.75s) + ("9.8m /s)(0.75) 2 = 1.65m
2
2
The component velocities (Vx, Vy) at the moment of impact:
!
!
Vx = 18.3m /s ; Vy = 5.87m /s + ("9.8m /s2 )(0.75s) = "1.48m /s
43
The impact angle of the dart with the Styrofoam:
V = 18.3 m/s
x
V
dart
= 18.4 m/.s
Horizontal
V = -1.48 m/s
y
The Sine of the angle below the horizontal:
sin " =
1.48m /s
= 0.081 ; " = 4.6°
18.3m /s
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 smallgrained 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.
Tw =
!
2Yd
g
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"
Yd = 0.68m ; Yb = 2.22m ; X w = 6.87m
Tw =
!
2(0.68m)
6.87m
2.22m
= 0.373sec ; Vx =
= 18.4m /s ; Vy =
= 5.95m /s
2
9.8m /s
0.373s
0.373s
The angle of the gun barrel:
!
tan " =
5.95m /s
= 0.232 = 17.9°
18.4m /s
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.
a 2 + b 2 = c 2 ; c = a 2 + b 2 = (18.4m /s) 2 + (5.95m /s) 2 = 19.2m /s
Notes:
!
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.
2.
3.
4.
5.
Two low friction pulleys
Assorted masses
Fish line
Two cylinders to hide the masses
Protractor
θ1
m1 (known)
m2
θ2
m3
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.
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:
Cos"1 =
left pull
150gr
Cos" 2 =
right pull
m3
150gr(Cos25°) = m3 (Cos48°)
m3 = 203gr
The upward pull of m1 plus the upward pull of m3 must equal the downward pull of m2 therefore:
!
Sin25° =
upward m1
150gr
Sin48° =
upward m3
203gr
Sin25°(150gr) + Sin48°(203gr) = m2
m2 = 214gr
Notes:
!
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
Mass at the 30cm mark
Mass at the 90cm mark
Mass of the meter stick
Center of mass of the meter stick
100.0 grams
500.0 grams
50.0 grams
140.0 grams
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
Clip locations (right to left)
Mass of meter the stick
Mass of the clips
Mass of suspended weight
60.0o
20.0 cm, 60.0 cm, 90.0 cm marks
0.13823 kg
0.0211 kg
0.5000 kg
50
Sample Calculations
∑ Torques clockwise = ∑ Torques counterclockwise, therefore starting with the clip on the right:
Counterclockwise(ccw) " = [(mass and clip)" + (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=
2.59Nm
= 3.71N
0.700m
Taking into account the angle to get the scale reading:
!
sin60° =
3.71N
scaler
; Scaler =
3.71N
3.71
=
= 4.28N
sin60° .8660
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:
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.
2.
3.
4.
5.
Two low friction pulleys.
Assorted masses
Fish line
Three cylinders to hide the masses
Protractor
Knot “a”
Knot “b”
θ
1
θ
3
θ
2
m1 (known)
m2
m3
m4
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
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:
Cos25°(150gr) = 136gr
The pull to the right is produced only by m4 and must be equal to 136 grams therefore:
!
Cos48°( m4 ) = 136gr
m4 =
136gr
= 203gr
Cos48°
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”:
!
Knot
“a”
136gr
136gr
Downward vector at"a"
Tan10° =
136gr
Downward vector at"a"= Tan10°(136gr) = 24gr
10°
Up and down forces at point “a” are equal and therefore:
!
Sin25°(150gr) = m2 + 24.0gr
m2 = Sin25°(150gr) " 24.0gr = 39.4gr
Up and down forces at point “b” are equal and therefore:
!
m3 = Sin48°(203gr) + 24.0gr = 175gr
To check our answers we can calculate the total upward pull and the total downward pull:
!
upward = Sin25°(159gr) + Sin48°(203gr) = 218gr
downward = 39.4gr + 175gr = 214gr
!
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.
2.
3.
4.
5.
6.
7.
Dynamics cart and flag
Photo-gates (2)
Low friction pulley
String
Weight
Ring stand and some clamps
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,
Photo-gates
Clamped
but under tension
ring-stand
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”
Mass on the end of the string
Distance between photo-gates
0.521 kg.
0.100 kg.
0.600 meters
55
Sample Calculations
Force that causes the system to accelerate:
F = Weight = mg = (0.1kg)(9.8m /s2 ) = 0.98N
Acceleration of the cart and weight:
!
F = ma; a =
F
0.98N
=
= 1.60m /s2
M 0.0512kg + 0.100kg
1
2x
2(0.600m)
x = at 2 ; t =
=
= 0.750s2 = 0.866s
2
a
1.60m /s2
!
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:
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.
2.
3.
4.
5.
6.
Level air track with a pulley on one end
Photo-gate used in gate and pulse mode
String
Film canister with hook on the lid
Balance
Bottle of small lead shot
Photo-gates
glider with flag
pulley
Film-canister
with a hook in
the lid
Air-track
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
57
Sample Calculations
Equation for the glider:
Fstring = mg a
Equation for the canister:
!
W canister " Fstring = mc a
Adding the equations:
!
W c " mg a = mc a ; W c = mc a + mg a ; mc g = mc a + mg a ; mc g " mc a = mg a
mc (g " a) = mg a ; mc =
mg a 0.210kg(1.8m /s2 )
=
= .0472kg
g " a 9.8m /s2 "1.8m /s2
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:
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.
2.
3.
4.
Ramp (PASCO 2 meter track works well)
Cart with friction (PASCO dynamics cart with friction plate works well)
String
Masses, hooks, clamps, etc.
Measurements
1.
2.
3.
4.
Mass of the cart (m1)
Mass of the pulling weight
Distance above the floor (pulling weight)
Ramp angle at which the cart will descend
with a constant speed
5. Angle of the incline for the practicum
510 grams
500 grams
1.00 meters
9.0°
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:
µ = tan9.0° = 0.158
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:
!
m1g " sin #mc g " µ cos #mc g = (m1 + mc )a
a=
m1g " sin #mc g " µ cos #mc g
(m1 + mc )
a=
.500kg(9.8nt /kg) " (sin28°)(.510kg(9.8nt /kg) " .158(cos28°)(.510kg(9.8nt /kg)
(.500kg + .510kg)
a = 1.83m /s2
The time for m1 to reach the floor starting from rest is:
1
x = at 2
2
!
t2 =
2x
a
; t=
2x
2(1.00m)
=
= 1.05s
a
1.83m /s2
Generally there is no problem getting results within 5% (.05s).
! Notes:
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.
2.
3.
4.
5.
6.
7.
8.
Air track with pulley on end
Glider and flag
Photo-gates used in the gate and pulse mode
String
Two low friction double pulleys
Low friction single pulley
Two ring stands
Clamps
Photo-gates
Pulley
Glider and flag
Air-track
Suspended
pulley
Ring-stand
Stool or
table with a
ring-stand
clamped on
top
Measurements
Mass of the glider
Mass of the suspended pulley
Width of the flag on top of glider
200.0 grams
80.0 grams
10.0 cm (needed only for the photo gate
program)
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:
Fs = mgl agl
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:
!
Fs =
W p m p ap
"
2
2
The acceleration of the glider coming forward will be twice that of the pulley downward,
therefore:
!
agl = 2a p
!
ap =
agl
2
Equating the force in the string and replacing acceleration of pulley with the acceleration
of glider:
mgl agl =
w p m p a p w p m p (agl /2) w p m p agl
"
=
"
=
"
2
2
2
2
2
4
Solve for the acceleration of the glider:
!
4mgl agl = 2w p " m p agl ; 4mgl agl + m p agl = 2w p ; ag ( 4mgl + m p ) = 2w p
ag =
2w p
(2)(0.0800kg)(9.80N /kg)
=
= 1.78m /s2
4mgl + m p 4(0.200kg) + (0.0800kg)
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.
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
Mass of sled
Compression distance
Force on spring
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.
xa = 1.00 m
m = 0.23462 kg
xc = 0.0250 m
F = 10.0 N
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:
F = "kx s ; k = "
F
10.0N
="
= 400N /m
xs
"0.0250m
Using the spring constant, calculate the potential energy stored in the spring:
!
1
1
E p = kx 2 = 400N /m(0.0250m) 2 = .125J
2
2
Set the potential energy before the expansion equal to the kinetic energy of the sled after the
expansion and calculate the sleds velocity:
!
2(E p )
1
2(0.125Nm)
E p = E k = mv 2 ; v =
=
= 1.066m /s
2
m
0.2346kg
At this velocity the time required to span the distance between the timing gates:
!
V=
"x
"x
1.00m
= ; "T =
=
= 0.94 sec
"T
V 1.066m /s
Three measured values were, 0.90, 0.93 and 0.93 sec.
!
Notes:
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.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
Spring
Ring Stand, clamps and rods
Meter sticks
Mass with hook
Thread
Scissors
Lab jack (may substitute blocks)
Egg
Carbon paper
Clamp type clothespins
Clay
Jaw type clothespins will
produce a right angle when
clipped to the ring-stand which
allows you to mark the unSpring with thread in the center
stretched length of the spring,
the release position (r) and
Mass
later the calculated egg
location. Cover the bottom of
Carbon Paper
the mass with carbon paper and
tape in place. Next raise the
Clothespins
mass to the release position
and tie it to the upper rod
supporting the spring. Place
Egg
the egg directly below the mass
and raise it with the lab jack or
Lab Jack/Blocks
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).
66
Measurements
Mass
Period of the harmonic motion
m = 0.877 kg
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:
T = 2"
m
k
; k=
4 " 2 (m) 39.48(.877kg)
=
= 36.5kg /s2 = 36.5N /m
2
2
T
(.974s)
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:
1 2 1
mgh + kx r = k(x r + h) 2
2
2
h=
2(mg " x r k) 2(0.877kg(9.8N /kg) " (0.100m)(36.5N /m)
=
= .271m
k
36.5N /m
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.
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.
2.
3.
4.
5.
Ballistic Pendulum apparatus
Tape measure
Ruler, Meter sticks
Clay
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
Rise of center of mass of the pendulum
Mass of the ball (projectile)
Mass of the ballistic pendulum
h1 = 1.02 m
h2 = 0.069 m
mb = 0.0675 kg
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:
1
(mb + mp )(V p +b ) 2 = (mb + mp )gh
2
1
9.8m
(0.0675Kg + 0.2761kg)(V p +b ) 2 = (0.0675Kg + 0.2761Kg)( 2 )(0.069m)
2
s
(0.1718kg)(V p +b ) 2 = 3.367kgm /s2
V p +b =
.2323kgm 2 /s2
= 1.163m /s
0.1718kg
Calculate the velocity of the projectile using the Law of Conservation of Momentum:
!
mbVb = (mb + m p )(V p +b ) ; Vb =
Vb =
(mb + m p )(V p +b )
mb
(0.0675kg + 0.2761kg)(1.163m /s)
= 5.92m /s
0.0675kg
The time the projectile is in flight:
!
1
2h
2(1.02m)
h = gt 2 ; t =
=
= .456sec
2
g
9.8m /s2
The range of the shot:
!
Vav =
"x r
"t
; "x r = Vav ("t ) = 5.92m /s(0.456s) = 2.70m
Measured values were 2.71m and 2.67m.
!
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
Mass of cart
Mass of cart + brick
Distance between carts and the frame
Mf = 0.535 kg
Mc = 1.181 kg
Mb = 3.492 kg
∆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.
M bVb = "M cVc ; 3.492kgVb = "1.181kgVc
Neither velocity is known but the velocity of the brick and cart can be found in terms of the
velocity of the cart alone.
!
Vb = "
1.181kg
(Vc ) = "0.338Vc
3.492kg
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.
M bVb = "M cVc
#X b = "
therefore : M b
#X b
#X c
= "M c
#t
#t
M c #X c
1.181kg(0.16m)
="
= "0.054 M
Mb
3.492Kg
The distance between the massive cart and the frame is now:
!
!
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.
M cVc = (M c + M f )Vc + f
Vc + f =
M cVc
1.18kg(Vc )
=
= 0.688Vc
M c + M f 1.181kg + 0.535kg
The speed at which the cart + frame approaches the bricked cart is:
Speed of approch = Vc + f + Vb = .688Vc + 0.338Vc = 1.026Vc
!
Of the distance to be covered by the two approaching bodies, the cart carrying the frame will
travel:
!
0.688Vc
(0.106m) = 0.071m
1.026Vc
Measured values were 0.068 m and 0.070 m.
! Notes:
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
Fall distance of pendulum bob after being cut free
Earth's gravitational field strength
h = 0.711 m
f = 0.759 m
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:
1
E p = E k ; mgh = mv 2
2
v = 2gh = 2(9.8m /s2)(0.711m) = 3.73m /s
Using the fall distance solve for the time of the fall.
!
1
2(0.759m)
f = gt 2 ; t =
= 0.394 sec
2
9.8m /s2
Determine the horizontal distance the bob moves during the fall:
x = vt = 3.73m /s(0.394s) = 1.46m
!
Measured value was 1.43 m.
! Notes:
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
Mass of ball #2
Horizontal distance to ball #1
Horizontal distance to ball #2
Angle off the centerline to ball #1
Angle off the centerline to ball #2
The vertical drop of the balls (optional)
m1 = 0.1616 kg
m2 = 0.0455 kg
∆x1 = 0.900 m
∆x2 = 1.323 m
Ø1 = 15.0°
Ø2 = 40.5°
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:
!
V '1 =
"x1 0.900m
=
= 0.900m /tick
"T 1.0ticks
V '2 =
"x 2 1.323m
=
= 1.323m /tick
"T 1.0ticks
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:
p'1 = m1V '1 = 0.1616kg(0.900m /tick) = 0.1454kgm /tick
p'2 = m2V '2 = 0.0455kg(1.323m /tick) = 0.0602kgm /tick
Vector addition of p'1 and p'2:
!
P ' = 0.0602
2
40.5
'
'
P + P =
1
2
o
15
0.184
center line
o
'
P = 0.0602
2
'
P = 0.1454
1
m
Therefore the horizontal momentum of ball #1 before collision must have been 0.184 kg tick .
The horizontal velocity of ball #1 before collision:
p1 = m1V1 ; V1 =
p1 0.184kgm /tick
=
= 1.14m /tick
V1
0.1616kg
The horizontal distance ball #1 will travel when no collision occurs:
!
"x h = Vh T = 1.14m /tick(1.00ticks) = 1.14m
The actual result in this experiment was 1.15 m.
!
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
Height of weight off the floor
Height of table from floor
Mass of weights
Earth's gravitational field strength
mp = 0.087 kg
h1 = 0.20 m
h2 = 1.245 m
mw = 0.250 kg
g = 9.8 m/s2 or N/kg
80
Sample Calculations
Potential energy of the weights:
E p = mgh = 0.250kg(9.8N /kg)(0.20m) = 0.49J
This potential energy becomes kinetic energy in the puck and the weights:
!
E pw = E kw + E kp =
1
1
2
2
M wVw + M pV p
2
2
1
1
0.49J = (0.250kg)(Vw 2 ) + (0.087kg)(V p 2 )
2
2
!
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):
2(0.49J) = (0.250kg)(V 2 ) + (0.087kg)(V 2 ) ; 2(0.49J) = 0.337kg(V 2 )
V=
2(0.49J)
= 1.70m /s
0.337kg
The time of fall of the puck:
!
1
2(1.245m)
h2 = gt 2 ; t =
= 0.50sec
2
9.8m /s2
Calculate the range (x) using the horizontal velocity of the puck and the time of fall:
!
x = Vave"t = 1.70m /s(0.50s) = 0.85m
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.
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
Power source
Ammeter
Voltmeter
Calculated section of Chromel
A wire clipped off for testing
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
Radius of the wire
Voltage (for internal resistance)
Amperage (for internal resistance)
Assigned Resistance of wire section
Handbook value for wire resistivity
d = 0.65 x 10-3 meter
r = 0.325 x 10-3 meter
V = 0.325 V
A = 2.0 A
R = 2.0 ohms
r = 1.00 x 10-6 ohm meters
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.
Power
supply
Voltmeter
Ammeter
Sample Calculations
Cross sectional area of the wire:
A = "r 2 = (3.14)(0.325x10#3 m) 2 = 3.3x10#7 m 2
Calculation of proper wire length:
!
resistance(R) = (resistivity)
L=
!
Length
L
=r
Cross sec tional area
A
AR (3.3x10"7 m 2 )(2.0ohms)
=
= 0.643 meters
r
1.00x10"6 ohm meters
Calculation of internal resistance of the system:
!
V = IR ; R =
V 0.325V
=
= 0.1625ohms
I
2.0A
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:
!
V = IR ; I =
V
6.0V
=
= 2.77A
R 2.1625ohms
Measured value was 2.70 A.
!
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)
Fuse
6.3 Volts
Bulbs
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 =
!
V
6.3V
=
= 0.25A for each bulb
R 25.2ohms
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.
2.
3.
4.
5.
6.
7.
One 30 Ω resistor (device 1)
One 60 Ω resistor (device 2)
Two variable resistors (up to 100 Ω) for the unknown resistors
Two digital DC voltmeters
Two digital DC Ammeters
Power supply for up to 20 Volts
Wires to complete the circuit shown below
Measurement
Power supply
Device #1 (upper branch)
Device #2 (lower branch)
20.0Volts
15.0 Volts @ 0.50 Amps
12.0 Volts @ 0.20 Amps
Sample Calculations
Total resistance for the top branch:
Total resistance for the bottom branch:
V 20.0Volts
=
= 40.0"
I 0.50Amps
R = 40.0" total resistance
V 20.0Volts
=
= 100.0"
I 0.20Amps
R = 100.0" total resistance
R=
!
R=
!
88
Resistance in device #1:
Resistance in device #2:
R=
V 15.0Volts
R= =
= 30.0"
I 0.50Amps
Therefore for variable resistor #1:
!
40.0" # 30.0" = 10.0"
V 12.0Volts
=
= 60.0"
I 0.20Amps
Therefore for variable resistor #2:
!
100.0" # 60.0" = 40.0"
Measured results when variable resistors were set to 10.0 and 40.0 ohms:
!
Notes:
!
V1 = 15.1Volts
V2 = 12.0Volts
I1 = 0.50Amps!
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
1000Ω
500Ω
500Ω
200Ω
2000Ω
1000Ω
1000Ω
500Ω
(Unknown)
500Ω
90
1000Ω
1000Ω
500Ω
(Unknown)
350Ω
Tower
Top View
Measurements
Potential difference across circuit
Current leaving power supply
8.94 V
9.40 mA
Sample Calculations:
Total circuit resistance:
R=
V
8.94V
=
= 951"
I 0.0940A
Resistance in tower series:
!
1000" + 500" + 500" = 2000"
Equivalent resistance in the tower:
!
1
1
1
1
1
1
1
=
+
+
+
+
+
Req 2000" 200" 2000 1000" 500" 1000"
Req =
2000"
= 100"
20
Unknown resistance must be:
!
500" + 100" + Runknown = 951"; Runknown = 951" # 600" = 351"
Measured value was 350.0 ohms
! Potential difference across the tower:
V = IR = (0.00940A)(100") = 0.940V
!
91
Current in the top series of resistors:
I=
V 0.940V
=
= 0.00047A
R 2000"
Potential difference across the 500 ohms resistor in the series:
!
V = RI = 500"(0.00047A) = 0.235V
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.
92
Notes: