it^
The Project Physics Course
Programmed Instruction
Waves 1
Waves 2
#
The Kinetic-Molecular Theory of Gases
INTRODUCTION
You are about to use a programmed text.
You should try to use this booklet where there
are no distractions — a quiet classroom or o study
area at home, for instance. Do not hesitate to
seek help if you do not understand some problem.
Programmed texts require your active participa-
tion and are designed to challenge you to some
degree. Their sole purpose is to teach, not to
quiz you.
In this book there are three separate pro-
grams. The first. Waves 1, proceeds from left to
right across the top port of the book. Waves 2
parallels it, starting at the front of the book
again, and using the middle portion of each
page. The third program, Kinetic-Moleculor
Theory, takes up the bottom part of each page.
It con be studied either before or after the two
wave programs.
This publication is one of the many instructional material*
developed for fh<! Project Physics Course. These mo-
terials include Texts, Handbooks, Teocher Resource Books,
Readers, Progrommed Instruction Booklets, Film Loops,
Transparencies, 16mm films ond laboratory equipment.
Development of the course has profited from the help of
many colleagues listed in the text units.
Directors of Project Physics
Gerald Holton, Department of Physics, Harvard University
F. James Rutherford, Chairman, Department of Science
Education, New York University
Fletcher G. Watson, Harvard Graduate School of Education
Copyright© 1974, Project Physics
Copyright (C) 1971, Project Physics
All Rights Reserved
ISBN: 0-03-089643-6
(H2 (!()« 9K76
Project Physics is a registered trademark
A Component of the
Proiect Physics Course
Distributed by
Holt, Rinehort ond Winston, Inc.
New York — Toronto
Cover Art by Andrew Ahlgren
Waves 1 The Superposition Principle
Knowledge of the behavior of waves is
of basic importance in physics, in this pro-
gram you will learn something about brief
wave disturbances — pulses: how they travel
and what happens when two pulses pass
through the same region at the same time.
Waves 2 Periodic Waves
When the seme wave shape is repeated
over and over again, the wove is called a
periodic wave. In this program you will learn
the relationships among the frequency, period/
wavelength, and speed of a periodic wove.
Kinetic-Molecular Theory of Gases
This program consists of a series of problems
that will help you to understand the kinetic mole-
cular theory end the behavior of gases. Following
most problems is a hint (printed upside down below
the dashed line). In the answer frames the solutions
to the problems are worked out in detail. To derive
the most benefit from this program, make a reason-
able effort to solve each problem without help. If
success does not come, read the hint. If you still
hove trouble, look at the solution, then go back
and work through the problem yourself.
The problem sequence used in this
program is adapted from:
Physics For the Enquiring Mind
Eric M. Rogers
Princeton University Press, 1960
INSTRUCTIONS
1. Frames: Each frame contains a question. Answer the question by writing in the blank space next to the frome.
Fromes ore numbered 1, 2, 3, ...
2. Answer Blocks: To find an answer to o frame, turn the page. Answer blocks are numbered Al, A2, A3, . . .
This booklet is designed so that you can compare your answer with the given answer by folding
back the page, like this:
1
"Tu^
2
;__
1
— -■■■
3. Always write your answer before you look at the given answer.
4. If you get the right answers to the sample questions, you do not hove to complete the program.
INSTRUCTIONS: Same as for Waves 1 aboN
INSTRUCTIONS: Same as for Waves 1 abov
Sample Question A
The disturbance travels from P^ to P2 in a time At.
If At = 1.0 seconds, what is the speed ot wl, h the
disturbance travels?
I ■ I H
Lmetfit.-
I I I li
i ! r- i, II I ^
-U»- ' ♦
,,,:'P.:;!
^
^■
t-^
1 1 ' 1 1
I M I I I
£
:^
t > ■ I I — . . I . . . i I I I I H ' ! M ! T )
' I I ) I I ' ' — ' ' t t r p t ' ■ ' I '
Sample Question A
The waves on the diagram were produced in 1.5 seconds.
What is the frequency f in cycles per second?
Turn page to begin the program
The Kinetic-Molecular Theory of Gases
Answer to A
Ad
" = a7
\6 = 3.0 m
At = 1.0 sec
3.0 m
3.0 m, sec
Answer to A
6.0 cycles
1.5 sec
4.0 cycles sec
In developing the kinetic-molecular theory of gases, we use elastic collisions of
bolls as a model, and find results that correspond to the observed behavior of gases.
So, for that set of phenomeno, we consider invisible molecules as though they were per-
fectly elastic spheres. The theory may seem difficult at first, but if you carry it through
once, thinking about it corefully, it will soon begin to moke good sense.
Sample Question B
The diagram shows two pulses traveling towards each other
along a rope. An instant later the pulses arrive at Point P, the
center of the rope. What is the maximum displacement of the
rope at that instant?
Sample Question B
The frequency of a wave is the number of
(i) that pass a point
per (ii) , and is equal
to the inverse of the (iii) .
A ball of moss 2.0 kg moving 12 m/sec to the right hits
a massive wall head-on and stops dead.
The momentum of the
ball before impact is .
("nits) t^vV
'X|i30|3A sauji; ssoiu sjonba ujn;u9LuoiiJ joij; |io33>j uutu
An!
to B
a + b
Maximum displacement is a > b.
If you missed this or Somple
Question A, you can profit by
working through this progrom
booklet, beginning with Question
1. if you got Sample Questions
A and B right, go on to Sample
Question C.
Anss
to B
(i) cycles
(complete waves)
(ii) second
(iii) period
A1
p = mv
m = 2.C kg
w « -^12 w sec
T = (2.0 kg; (♦12 m sec)
1 p - *24-3' —
Sample Question C
A rope is hung as in the diagram, one end fixed
and the other end free. At point A the rope is moved
sideways and back suddenly, creating similar pulses
that travel towards the ends.
Sketch what the pulses will look like after re-
flection from the ends.
Fixe<i
aaj-j
Sample Question C
If waves of frequency f = 20 cycles per
second travel with speed v = 40 meters per
second in a given medium,
(i) what is the wavelength of the waves in
the medium?
(ii) what is the period of the waves?
Since the ball stops dead when it strikes the wall, the
momentum of the boll after impact is
(units)
'0J3Z SI IDOduJI J3|^D
{|Dq 3L|4 jO Xil30|3A 9L|4 'UOISI{|OD DliSD|3UI Sll^4 U| ■iU}\-{
Answer to C
^^-
^
>^
l( you got thrs question right also, you
need not complete Waves 1. Turn instead
to the Proiect Physics Programmed lr>-
struction booklet Waves 2. If you missed
Somple Question C (but got A and B right)
begin this progrom at question 16.
Answer to C
(i) A =1
40
20 cycles sec
A = 2.0 m cycle
(ii) T = — = — sec cycle
f 20 ^
T = .05 sec cycle
A2
p = m V
m' = 2.0 kg
v' = 0.0 m sec
■p - (2.0 kg) (0.0 m sec)
T = 0.0
kg ■ m
Two people hold opposite ends of a rope. The sender,
s, snaps his end of the rope, creating a disturbance that
travels along the rope towards the receiver, r.
1
T
—A
A short time (At) later, the disturbance passes point P.
Draw the disturbance as it passes P.
Sample Question D
The figure shows an attentuated (domped) periodic wave.
Of the following, which property is changing?
speed of wave, wavelength, amplitude,
frequency, period.
If the momentum of the ball before impact was
-^24 leg • m/ sec, and the momentum after impact was
0.0 kg- m/ sec, then the change o{ momentum that
the ball underwent is .
(units)
•d - ,d = d\7
'looduji sjoisq luntusujouj snujiu ^Doduii
j^^D ujoiudujouj Sjonba ujniusujouj ^o sSuoi^ij) -^^!H
Al
A
P
Answer to D
Only amplitude is changing.
If you were able to answer al
of the sample questions cor-
rectly, the rest of the prograr
is optionol.
A3
Ap" = p' - "p
P -- 0.0 - —
sec
— ^ »9' •"
_-. _ kg • rr
Ap (0.0 - 24 ;-
The disturbance below travels through the rope from P^
to Pj in time At.
If At = 1.0 seconds, what is the speed at which the dis-
turbance travels in the rope?
^^ 1 mete-- ^ ^ ■
»' < » t I — I — I — I — ( — I — I— ♦• •
-• r
P^2
-•r
Waves are shaken on a rope by moving point P up and down.
Crests are generated when the displacement of P is positive,
and troughs are made by negative displacements of P.
Label the crests and troughs on the diagram.
Newton's third law applies to this case, therefore we can
say that the change of momentum of the wall must be
(units;
•Das/iu.6)| t^j -
SI {{0<^ am 10 uin^usuiouj ^o 36uoij3 dij; ;ol|| mou)|
3M puo 'OJ3Z s: mn^usujouj lo s36uol|3 9L|4 ^o luns 3L| i
';33|ao J3L|io 9L|; Xq p3u;o6 ujn(U3UJoiu 91^4 0{ spn;
-luboLLi ui lonoa si uois||iod d ui i33iqo suo Xq jsoj oin;
-U3UJOUJ joq; S3|0|S ujn^udiuouj jO uojiOAjasuoD ^o mdi 9i^j_ -t^lH
A2
Ad
At
Ad = 3.0 m
At = 1.0 sec.
3.0
1.0 sec
= 3.0 m/sec
A1
c
r«
;s
t
s
' 1
%
r
\
i
i
S-T..
i
'
\
1
1
J
Ik
I
i ,
- n
I
r
JK-- J
1
/
1 TL
\
J
-W^
^
f
-3^
\
*i
K
^ .
^
t
troughs
A4
ATwoII -^ -^boll =
i5iPwoll = --^Pbo
= - (-24
leg* m
APwoii » +24
kg- m
3 [
A brief disturbance that moves through a medium is |
called a pulse. |
In the cose of the previous two frames, the medium '
is (i) , and the source of '
the pulse is (ii)
2 1
A complete waveform is a cycle. | ^
■ I 1
Draw 3 cycles between points P and P'. . " •
5 1
Now suppose that the wall is hit by a stream of such bolls, '
each of mass 2.0 kg and velocity of +12 m sec. Suppose that '
1000 balls hit the wall head-on during 10 seconds. '
The total change of momentum of the wall in that 10- second l
Deriod is i
(units) j
•uBis uj aijsoddo inq 'S||Dq qqq[ ai^; 40 '
96u0LfD UJn4U3UJ0UJ 31^; 04 |Dnb3 SI IJOM 3lj4 ^0 36U0U3 1
ujniusiuoiu am 3jo^3J3ij; 'p3aj3suo3 5 1 ujn4U3UJoyy '
•S||D<| QQQ[ JO^ luniUSUJOlU 40 36uDgD 31^4 34D|nD|03 :*ui|^ 1
A3
(i) the rope
(ii) the hand movenient of the sender'
A2
A5
~m. kg • m
APwoll = -24——
ATiooo - -24000
eg • m
ball)
APwoll • Apiooo =
balls
ATwoII " - Apiooo
bolU
= - ( - 24000 ^^^-^ )
ATwoII - * 24000
kg- m
A small object is fastened to the rope.
Draw the object and pulse when the pulse is
centered at point P.
^
In one second., point P makes 6 cycles, and each cycle
contains a crest and a trough.
(i) How many cycles per second pass an observer at point S?
(11) How many troughs pass the observer at S in one second?
crest
trough
Note: Actually, *he changes of momentum occur in bumps, one
bump when each boil strikes the wall. You can still cal-
culate the total change ot momentum, then use this total
change to calculate an overoge force. Just forget that
the momenta changes occur in bumps, and divide the
total change cf momentum by the total time (ten seconds
in this case) and obtain overage force, F
^ ' av
The following frame is en illustration of this point.
A4
A3
(i) 6 cycles
(ii) 6 troughs
^ 1
We know that the water wove below contains energy '
because |
^ 1
^ ^^ ^"^ 1
J^ i
4 1
The number of cycles per second is the frequency •
of the wave. (In the last question, the frequency was •
6 cycles per second.) j
If 12 cycles are produced in 3.0 ■
seconds, the frequency of the i
waves is 1
(units) 1
6 j
The average force on the wall, during the ten second period, '
due to all 1000 bolls losing momentum, is given by applying '
Newton's second law to the whole collection of balls. '
^ _ A^ 1
F = ma and a = "T" • 1
At 1
Hence '
— Av" m A v" A (niv) Ap"
~ "" At " At " At " At ■ !
Average force, Favf o" ^he wall must be . ,
(units) 1
A5
For one thing, the wave lifts
the boat. The increase in the
boots gravitotiorvjl potential
energy must hove cor»>e from
the wove.
A4
f =
12 cycles
3.0 sec
= 4.0 cycles sec
A6
Fw
^w
oil
oil =
\t
A^Pwall = ^24000
9 • m
At - 10 sec
F'woII • 24000 kg • m/sec
2400
10 sec
kg • m
sec^
Fqv " +2400n«wton$
If you lift a weight against gravity, you do work, and
to do work requires energy.
A pulse passes a small weight attached to the rope.
JX
-•r
s«-
-i-
5*-
/M
From this experiment it can be seen that the pulse
transmits , because it lifted the
weight as it passed.
The waves on the diagram were produced in 1.5 seconds.
What is the frequency f in cycles per second?
In the previous frames we dealt with inelastic collisions;
the balls struck the wall and stopped dead. Suppose 1000 hard
steel balls, each of mass 2.0 kg, hit a massive wall head-on in
the course of 10 seconds; but this time they arrive with a velo-
city of +12 m/sec, bounce straight back with equal speed,
12 m/sec, but in the opposite (negative) direction.
The momentum of each ball before impact is
vunits)
A6
energy
A5
6 cycles
1.5 sec
= 4.0 cycles sec
A7
p = m V
m = 2.0 kg
V = ♦12m sec
P = (2.0 leg) (+12 m sec)
kg- m
Notice that momentum is o vector
quantity and the oirection ot the
momentum vector i$ the direction
of the velocity vector.
A pulse is a disturbance moving through
(i) , and
transmitting (ii) .
The quantity T represents the time required to
generate one cycle (complete wave). The time interval
(T) is called the period of the wave.
When f = 1 cycle/sec; T = 1 sec/cycle
f = 2 cycles/sec; T = 1/2 sec/cycle
f = 3 cycles/sec; T = 1/3 sec/cycle
In general:
f = X cycles/sec; T = .
8
The momentum of each ball after impact is
(units)
i^
•43od
-lUI (a) Jdi^o puD
(a) SJO^Sq X4j30|3A
31^; ^O UO!;33Jip 31^4
SMOl^S UIOjSdjP 31^^
Hint: This is easy, but watch the sign indicating the direction
of the vector.
A7
(i) a medium
(ii) energy
A6
1
T - — sec/cycle.
A8
p = mv
m = 2.0 kg
v" = - 12 m sec
?' = (2.0 kg) (-12 m sec)
■p' = -24
The minus sign before the
onswer meons thof the mo-
mentum of each ball after
collision IS directed in the
opposite direction from the
momentum before collision.
A disturbance can be described as a displacement of
particles in a medium from their normal positions.
Using the information below, complete the table.
t-rr+-
Displacement
Point
A
B
C
D
From frame 6 we see that the period is related to the
frequency of a wave.
1 , t ^
T - — , and t = — .
What is the period of a wave whose frequency is
10 cycles/sec?
The change of momentum of one ball is
(units)
-OJ3Z 40U SI J3MSUD 9l^J_ -^"!|-|
A8
Displacement
2
1
Point
A
B
C
D
A7
T = r
= — sec/cycle
10
= 0.1 sec/cycle
A9
Ap = "p • - "p
P" = -24
■p = *24
kg
sec
kg- m
AT = -24
kg
kg- m
_^ kg* m
Ap=-48^^
Draw a possible pulse that has the following displacements:
Displacement
-1
+2
.2
At Point
A
B
C
D
E
F
G
8
The frequency of a wave is the number
of (i) that pass a point
per (ii) , and is equal to
the inverse of the (iii) .
10
For each collision the change of momentum of the wall
(units)
-|u;od SjLji uo ^uaiunSjD aiji
MO{|o^ u|o6d puo p siuDjj o; ujn^s^ ^pssn^uo^
•Q = "°''dv + ll^^dy
'00| SUOISIIJOS 3I|SD|3 Ul pSAJSSUOD Sj LUn;U3lUO^ -^'^!H
A9
one possibility:
A8
(i) cycles
(ii) second
(iii) period
A10
Apwoii + Ap^boii =
APwoll = "Apboll
APbol. = -48^
APwoii = -(-48 '^ll^)
sec
APwoll - -^48
Kg- m
10
Two pulses are shaken onto the rope from opposite ends,
and the pulses pass at mid-rope. A "movie camera" series
shows the interaction of the two pulses.
Picture 12 3 4 S
^W
•■--^
ii:
-i-^-i—
:^3g;
Has there been any change in the shape of the pulses as
a result of the interaction?
Points marked P' on the diagram correspond to the
point P because they are in the same part of a complete
cycle.
On the diagram, mark points H that correspond to
the point H.
11
When each ball collides with the wall, the wal
undergoes a momentum change of +48 kg- m/sec.
If, during 10 seconds, 1000 balls strike the wall
and rebound thus, the total change in momentum of the
wa 1 1 is .
(units)
A10
A9
-^ Ul.l . ■
tiUtil^ln. t-'fmt^
All
Ap, - lOOOAp
kg
Sp, =- (1000) (.48 ^^ )
^ sec
^p^ - > 48000
kg ■ m
11
When two waves pass through the same region of a medium, the displacement of each point is
the sum of the displacements that each wave would cause by itself. For example, as two pulses
of displacements a and b pass a point P on a rope, the displacement of point P will be a + b:
a*b
10
On the wave diagram, mark as A', B' and C
all points which correspond respectively to points
marked A, B and C.
12
Average force on the wall is
(units)
IV +V
*V
iv {i^)s/ ^v"" -i
:md| puooas s^uo+MatNj uioj^ uoi^onba 9^ jsqiuauiay :iu\y\
AID
A12
F -^P
' av - -7—
At
Ap = ♦ 48000
kg- m
At = 10 sec
h 48000 kgm/s«c
h4800
10 sec
kg- m
sec2
^4800 rtewtons
12
Sketch what the displacement of the rope will be
as the two pulses arrive at the center of the rope:
n
The Greek letter lambda (A) is used to represent
wavelength.
The wavelength is the distance between corres-
ponding points of adjacent cycles. For example,
points A and A' are one A apart.
On the diagram mark points whose separation is
A from points B and C.
4l . |A . I A'
Mi
13
The balls are directed at a patch of wall whose
dimensions are 2.0 m high by 3.0 m wide.
The area in which the balls impact is
(units)
A12
All
A13
areo » height x width
height < 2.0
width . 3.0
area = (2.0m) (3.0 m)
orea = 6.0 m*
13
Sketch what the displacement of the rope will be
as the two pulses arrive at the center of the rope.
12
Higher frequencies in the same medium
produce shorter wavelengths.
Mathematically speaking, the wavelength are
frequency are proportional.
14
Pressure is usually defined as the ratio of perpendicular
force to the area, and written as
force
pressure = -^ .
area
The pressure on the patch of wall due to the impocting
balls is , ^.
(units)
pUD 3UJ Q'9 = D3JD 'SUJ3|qOJd SnoiAdjd lUOJ^ ■^''!H
A13
I 1
A12
inversely
A14
pressure =
for
pressure -
force = + 4800 N
orea ^ 6.0 m'
♦ 4800N
6.0 ni2
pressure - 800 N/m'
14
1
in
5
"
^
'
1
i
1
-5
1 1
j
.1. .
e
The two pulses
in the top dia-
gram are shown
superposed in
the lower dia-
gram. Complete
the displace-
ment table for
the labeled
points.
Displacement
Point
A
B
C
D
E
F
13
The top diagram shows a wave moving away from source s, and
the bottom diagram shows the wave one complete cycle later.
(i)
(ii)
During the time interval from one diagram to the other, the
elapsed time is equal to the (i) of the wave, and the
wave had moved a distance of (ii) .
Before we moke the change from bouncing balls to bouncing
molecules, we must look at the behavior of things moving inside
a closed box. Suppose we have an oblong box, 4.0 meters long
from end to end, with only one ball in it. The ball moves from
end to end with a speed of 12 m/sec. The ball hits head-on and
rebounds elastically with a speed of 12 m/sec toward the other
end. The same ball will hit the front end of the box many times
in 10 seconds. Instead of
using the number of balls
hitting the end, we must
calculate and use the num-
ber of hits made by this one
ball. To find the force on
one end, we use the hits on
that end only.
A14
Displacement
5
8
2
5
Point
A
B
C
D
E
F
A13
(i) period (T)
(ii) one wovelength (A)
15
Complete the drawings, showing superposition of the
two pulses when the pulse centers coincide.
n
u
u
n
14
If a wave moves one wavelength A in one period of time T,
wave speed can be calculated.
V (wave speed) =
distar
time
Find the speed of a wave whose wavelength A is 1.5 cm,
and whose period T is 0.1 seconds.
15
Between successive hits on the front end of the box, the
ball travels one "round trip." It travels the whole length of
the box from the front end to the opposite end and back to the
front end again.
The distance traveled by the ball during one round-trip
is meters.
A15
tn
net =
At the exoct moment they cross
the center the displacement is
zero.
A14
" = T
A = 1.5 cm
T = 0.1 sec
1 .5 cm
0.1 sec
= 15 cm sec
A15
8.0 meters
16
A pulse on a rope approaches a point that is
attached to a wall.
7\
m
Experiments show thot the wave does not continue
past the fixed point, but is reflected back in the oppo-
site direction, and the puise appears on the opposite
side of the rope.
Draw the pulse after reflection.
15
1 A
Recall that f = — (frame 7) and v = — (frame 14).
What is fhe speed of o wove written in terms of wavelength A and
frequency f?
16
The speed of the boll is 12 m/sec. The total distance
traveled by the ball in 10 seconds is meters.
-9UJ{4 S8UJJ4 p33dS 51 aDU04SIQ UUJI^
A16
A15
fA
A16
d = vf
V - 12 m sec
t = 10 sec
d = (12 nv sec)(10 sec)
d - 120 m
17
Draw the pulse after reflection.
A
\ A_
16
!f waves of frequency f = 20 cycles per second travel in a
particular medium with speed v = 40 meters per second,
(i) what is the wavelength of the waves in the medium,
(ii) what is the period of the waves?
17
The number of round trips made by the ball in 10
seconds is .
SDUD^Sip dlJ| punOJ 9L|4 pUO 'lU Q3[ S| D9S Q[
Ul p3|dADJ| 93UD4SJP 'SlUOJ^ snoiASJd 9L|4 UIOJ^ -^"IH
A17
A16
(i) A = -
= 40 m/sec
20 cycles/sec
= 2.0 m/cycle
Since a cycle is always implied,
wavelength is usually written in
terms of distance units only. In
this example the answer would
be reported os A = 2.0 m.
(..)T = -j- = 20
T = .05 sec (per cycle)
A17
120
8.0
15
15 round trips
18
A rope is suspended in a stairwell, and a
pulse is shaken on the rope from the top.
Experiments show that the wave is reflected
from the free end but on the same side of the
rope.
Draw the pulse after reflection.
>\
17
Actually, it is very difficult to produce a perfectly periodic
wave. One reason is the dissipation of energy which causes
waves to be "damped" or "attenuated."
The figures shows an attenuated (damped) periodic wave.
Of the following, which property is changing?
speed of wave, wavelength, amplitude, frequency, period
18
The number of times that the ball strikes the front end of
the box in 10 seconds is (times).
•diJi i^Doa aDuo xoq aij; jo pua ^uoj^ aL|; sajjiJ^s ||Dq aL|j^ :;u!|_|
A18
i=i
A17
On Only amplitude is changing.
Amplitude refers to the heights of
the crests and troughs above end
below the undisturbed position.
As the figure indicates, the ompli-
tude of the damped wove is
decreasing.
A18
15 times
19
A rope is hung in the diagram, one end fixed and
the other end free. At point A the rope is moved side-
ways and back suddenly, creating similar pulses that
travel towards the ends.
Sketch what the pulses will look like after reflec-
tion from the ends.
38J^
This ends the two programs on waves. To further your knowledge of waves
you should refer to some of the other Project Physics materials, especially
the Unit 3 laboratory activities, some of the articles in Reader 3 (as for
example, "What is a Wave?" by Albert Einstein and Leopold Infeld), and
the following Film Loops:
Superposition
Standing Waves on a String
Standing Waves in a Gas
Vibrations of a Rubber Hose
Vibrations of a Wire
Vibrations of a Drum
Vibrations of a Metal Plate
19
1
1
1
1
If we
want to have 1000 hits
on the front wall in 10
1
seconds, '
as
we had before w
ith the stream
of bolls, we would have to
hove '
more than
one boll
in the box. In
fact, we would need a
aout
ba
Is, all
moving back and forth between the ends.
■M
30|3A
>UD SSDUJ
3UJDS am 3AOL|
•xoq
s||Dq {|D 4DL|; SLunssy
3L|| ui S||oq jo jaqiunu
am Oi
ouoi;jodojd si S|il| ^o jaqiunu 3L|; ;di^; siunssy
uui
H [
A19
r
A19
1000
15
66.7
67 bails
20 [
What will the rope look like when the two pulses cross? 1
Pressure i
s g
ven as fo
rce/area; if yo
1
u know the area of i
the
front of the
bo
K you can
calcula
te the average pressure on | |
the
end caused
by
repeated
mpacts
of the
balls. 1
If we have
on
y 67 ball.
in the
box, a
1
calculation of the |
"pressure" does not seem to
be use
ful. But if the box contains | |
thousands of m
1 lions of mol
ecules
moving
at high speeds, it l
bee
omes worthwhi
e to compute on
average
result for the in- i
vis
ble particles.
A20
They cancel
21 j
After onof/ier reflection the pulses again cross. What will |
the rope look like at that instant? 1
20 j
So let's now do a similar calculation using gas molecules in a box.'
A metal box, 4.0 meters long, with ends 3.0 meters wide by 2.0 |
meters high, contains one gas molecule which moves to and fro along |
the length of the box with a speed of 500 meters per second. The |
molecule bounces elasticolly from each end, so the speed remains |
constant at 500 m/sec. The mass of the molecule is approximately 1
5.0 X 10-26 kilograms.* 1
The momentum of the molecule before impact with the front end ,
is 1
(units) j
*This is approximately the moss of on average molecule of air. '
A21
Both pulses will be on the left
side of the rope and so the dis-
placement will be twice that of
either pulse alone.
A20
m = 5.0 . 10-^« kg
V = 500 m sec
- 5.0 ^ 10' m sec
p - (5.0 X 10-" kg) (5.0 y 10'
p - *25 X io-»* i^i:^ I
sec)
This is the end of Waves 1. Now that you understand the nature of wave |
pulses, you can study the behavior of trains of wave pulses. This is |
dealt with in Waves 2 Periodic Waves which begins just below this pro- |
gram in the front of the book. |
21
The momentum of the molecule after impact (p') with
the front end of the box is
(units)
A21
p' = mv'
m . 5.0 . 10-" leg
v' = -500 m sec
r -5.0 . 10^ msec
p' = (5.0 . 10-" kg) (-5.0 . 10' m sec)
p' = -25 - 10
-24 ^JJl
22
The total change of momentum of the molecule, therefore
IS
(units)
•d - ,d = dy 'ajo^aq sy uui^
A22
Ap p' -"p
p = .25 . 10
P = -25. 10
.2.^
-2. ^i:
Ap = (-25 . 10-2*) _ (.25 . 10
-J4v ^g
Ap -50 - 10
-24 ^9-
23 j
In 10 seconds the molecule can make trips to 1
end fro, and so can moke this number of impacts on the front end. I
•sja^aiu o'8 ^' 33UD4Sip dui-punoj a^ \
puD 'sjaiauj OOOS ^^°^ p|noM a|noa|ouj am spuooas q[ u| 4ui|_| |
A23
5000
8.0 m per trip
625 trips
24 j
In 10 seconds, the molecule mokes impacts |
with the front end of the box. |
-aiuoj^ snoiAdid 3i|; u\ uousanb 3L|4 poe^ -^"!H 1
A24
625 impacts
(one per trip)
25 1
The change of momentum of the molecule per impact was |
calculated to be -50 x 10~^* kg • m/sec. Therefore, the total |
change of momentum of the front wall of the box in 10 seconds 1
is '
(units) 1
•a|nDa|ouj 3^\ jo lun^uaoiouj ^o aBuoi^D 31^4 ot uou |
-DSJip ui a;isoddo si ||dm 3l|4 ^0 luniusuiouj ^0 aSuoi^D 3l|j_ ^^^IH I
A25
A?wall ' -50 . 10-" ^l- per ,mpoct
\Pwoii = '625) (.50 . 10
.24 kg- m
sec
31250. 10-" ^1—
APwoii - -3.12 . 10-
kg
26 j
The average force on the front of the box, during the 10 1
second period, is . 1
(units) 1
•Xj^ jslhoud sil^ aAi6 '41 Buimsia .
-aj ja+^D puD 9 aujojj o; >|3Dq oq ^a|qnoj^ 6u|adl| \\\l^ 1
•jDAjaiui 3UJU 31^; Aq papiAip oiniusiuouj ui sBudijd '
3L|; so p9;Din3|Do som 3djo^ aSojaAO aij; |DL|4 jsqiusiud^ '^^!H
A26
P _ ACmv) ^ Ap
" At ° aT
Ap^„„ = .3.12.10-
At = 10 sec
* ♦3.12-10-"
r = newtons
10
F = .3.12 . 10-*' newtons
27
The front end wall has dimensions of 2.0 meters by 3.0
The nvernge pressure on the end wall is
meters. 1
(units)
-sSjOj 9<^ o^ djnsssjd silj^ psdxs i^uoQ
•pa;DejD poij sm ^d^ mnnooA ai^; ^o pnojd a^inb aq p|noM sm
'it ui 3|nos|OLU auo ^snl i^;im xoq d a>jDLu o; 3\<^d sjsm sm j|
•D3JD ijun jad aojoj si ajnssajj
UUJH
A27
pressure = force orea
area = 6.0
pressure
Torce = 3.12 . 10-'' N
3.12 > 10-'' N
6.0 m2
! oressore = 5.0 - 10"'' N m* i
Now suppose that this box contains 6.0 x 10^^ molecules (that is, 600,000,000,000,000,000,
000,000,000 molecules). This figure is roughly the number of molecules in such o box if it were
filled with air at atmospheric pressure. In reality these molecules would be moving about in all
directions at random; but in order to simplify the calculation, suppose that they are sorted out
into three groups. One lot moves up and down, another lot moves to and fro along the length,
and the third lot moves back and forth across the width. Symmetry considerations suggest we
should have the molecules equally divided between the three groups.
(If you wish a more rigorous explanation, think of dividing the velocity of each molecule
into three components Vx, Vy and v^. This cannot be done individually for each molecule, so
it is done statistically for the sample of randomly-moving molecules. The results are the same
as given in the first paragraph.)
ouDuose we nove three
groups, 2.0 . IC^' in
each, having some ove*'-
oge velocity parallel tc
one edge of the box.
There ore 6.0 ■ 10^^ molecules
in the box moving in random
directions.
iTaj
I
^
^.
The pressure on an end of the box will be due only to impacts
of molecules moving to and fro along the length. We will now cal-
culote the pressure on the end of the box. The pressure is caused
by 2.0 X 10^^ molecules, moving ot a speed of 500 m sec, and col-
liding elostically with the front end of the box.
Assume that the pressure
on the end face is aue to the
impacts of the molecules that
are moving parallel to the
length of the box.
2m
28
The average pressure on the end of the box will be
^units;
-XOO 3U4 p pu3 9^
L|4iM 5u!pi||0D sainD3joui g^Ql * 3 ^^°^ ^'^ asoD sjlji u:
z^ H 22-01 " C? 5DM a|nDa|oui
9U0 04 9np djnssajd dSojdAO 3lj; ;di^| ^3 auioj^ ujoj^ jiooa^ -^"!H
A28
(5.0 ■ 10-" N m2) (2.0 . 10^*)
= 10 - 10* N fn2
pressure = 1.0 • 10^ N m^
The values used for the moss
of a molecule and the number of
molecules m a box of that sue
ore roughly correct for ordirxiry
oir in o room.
Standard atmospheric pres-
sure IS 1.0132 ■ lO'^ N m'.
Thus your calculated resT;lt
approximates the standard atmo-
spheric pressure quite closely.
29
Suppose that we reduce the size of the box slowly to one-half
of the original length (that is, to 2.0 meters). Assume that we do
not change the speed of the molecules or the size of the end wall.
The average pressure of the end of the box will now be
(units)
-SJOjaq so ujouisj suoi;o|n3|03 am |o 4S9J aL|j[
'spodiui uasM^sq jsadj^ 04 33UDj
-sip JdjJOLjS D aAOL| S3|no9|oui am 9DUIS '||0M pus am uo
S||L| jo jaquinu 9i{; 3^0|nD|0D3j noX loiji sjmbaj ||im sii|j[_ '^"!H
A29
number of impacts per molecule
in 10 seconds is 1250
change of momentum per irrtpact
is .50 . 10-'* ''?■"' (per molecule)
sec
2.0 « 10^' molecules strike the face
pressure^^ = 2.0 - 10*K m'
This is what you should expect
since twice as many impacts per
unit of time should result m twice
the pressure.
30 1
If the box is reduced in size to one-half the original |
length, what effect does this have on the volume of the 1
box? 1
A30
The volume is reduced to
half its originol size.
31 1
How did this change in volume of one-half affect the |
pressure exerted by the gas? |
A31
The pressure of the gas
was doubled.
32
From the calculations made in this program, how
pressure of a gas related to its volume?
is the j
-S3|n39|OUJ ^O SUOISJIIOD OU pUO 'S3|n03|OUJ pUD ||DM
uddMjaq suo|Si||OD 3uso|3 XjiosjJdd puo s3|n33|ouj
jo jdqiunu slugs 'spssds JO|n33|Oui jonbs diunssy
:+"!H
A32
The pressure of a gcs is
inversely prcportionol to
the volume it occupies.
Having now completed this program, it should be much easier to follow
the more complete discussion of The Kinetic-Molecular Theory of Gases
to be found in Chapter II of the Project Physics Text, and in other
physics books. Also you will now be in a position to enjoy many of the
Project Physics Reader 3 articles, such as:
The Great Molecular Theory of Cases by Eric M. Rogers
Entropy and the Second Low of Thermodynamics by Kenneth W. Ford
The Low of Disorder by George Gamow
The Law by Robert M. Coates
The Arrow of Time by Jacob Bronowski
Also in the same Reader, you will find a brief biography of James Clerk
Maxwell, the great scientist, who was a key figure in the development of
kinetic theory.
0-03-089643-6