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HARVARD PROJECT PHYSICS
Overhead Projection Transparencies
UNITS
Harvard Project Physics
Overhead Projection Transparencies
Unit 3
T19 One Dimensional Collisions
T20 Equal Mass Two-Dimensional Collisions
T21 Unequal Mass Two-Dimensional Collisions
T22 Inelastic Two-Dimensional Collisions
T23 Slow Collisions
T24 The Watt Engine
T25 Superposition
T26 Square Wave Analysis
T27 Standing Waves
T28 Two-Slit Interference
T29 Interference Pattern Analysis
T19
T-19 One-Dimensional Collisions
Facsimiles of stroboscopic photographs involving two-body collisions in one dimension are provided
in this transparency. Two events which are useful in discussing the principle of conservation of momen-
tum are presented. The first event (T19-A) shows ball A coming in from the right with ball B initially
at rest. Both move off to the left after the collision. Event 2 (T19-B) shows ball B coming in from the
left with ball A initially at rest. Overlay T19-C shows the directions after collision: ball B moves to
the left and ball A moves to the right.
You may make measurements directly on the transparency or you may have students make measure-
ments from images projected onto a chalkboard or wall.
Ball A
d = 4.05 cm (from photo)
4.05 cm X
1 m
= 0.348 m
11.65 cm
V = 3.48 m/sec
mv"= 1.85 kg-m/sec
1/2 mv= =^ 3.22 joules
d' ~ 1.1 cm (from photo)
d' = 0.0945 m (actual)
V' = 0.945 m/sec
mv' = 0.503 kg-m/sec
Vomv'^ = 0.237 joules
EVENT 1
Before
I mT=
I
After
I d' =
I
Ball B
4.5 cm (from photo)
0.386 m (actual)
I V' = 3.86 m/sec
mv' = 1.35 kg-m/sec
I i/jmv'^ = 2.61 joules
d' = 3.4 cm (from photo)
d' = 0.245 (actual)
T' = 2.45 m/sec
mv* = 1.31 kg-m/sec
V^ mv'^ = 1.61 joules
EVENT 2
Before
I d = 4.5 cm (from photo)
d = 0.325 m (actual)
I ~v = 3.25 m/sec
rnv = 1.14 kg-m/sec
I V^ mv- — 1.85 joules
After
I d' = 0.6 cm (from photo)
j d' = 0.0434 m (actual)
I V = 0.434 m/sec
nw' = 0.152 kg-m/sec
I 1/2 niiv'^ = 0.033 joules
T19
Event I
At = 0.1 sec
mg^
0.350 kg
mA =
0.532 kg
6
5
7 6 5 4
' 123
3
2
AFTER
BEFORE
4
1 t>-* /->-|-/->r»
^
1 lllcLcl
►
«
Tl»
Event I
At = 0.1 sec
mg = 0350 kg
111^= 0.532 k
t= V,
6
5
^ 123
BEFORE COLLISION^
AFTER m
B
3
2
I meter
meter -
BEFORE
#
Event 2
At = 0.1 sec
mg = 0.350 kg
m^= 0.532 kg
BEFORE COLLISION^
^ AFTER ^
^ COLLISION ~
mg- m^
I meter
TW
^.
#
6 5 4 3 2 1 I
2
3
4
T20
T-20 Equal Mass Two-Dimensional Collisions
This transparency and T21 and T22 are intended to accompany the film loops and stroboscopic photo-
graphs dealing with the two-dimensional collisions discussed in Chapter 9 of the text. You may wish to
distribute stroboscopic photograph No. 69 to the students and proceed to explain the technique required
in determining momentum before and after the collision as your students follow with their own photo-
graphs. Transparencies T21 and T22 may be used in a similar manner, or you may wish to assign stro-
boscopic photographs 73 and 74 for homework. You may then use T21 and T22 on the following day
to show the students the correct solution. Alternately you may have students work independently or in
groups directly on the transparencies or on a projected image on a chalkboard.
Overlay A Shows a facsimile of photograph 69. The impinging ball (mass = 0.367 kg) approaches
from the upper left of the picture while the target ball (mass = 0.367 kg) which was orig-
inally stationary proceeds to the right after the collision. The distance between the two
marks at the lower portion of the transparency represents one meter.
Overlay B Shows the first step in computing the momentum quantities. Determine the velocities by
measuring the distances in the indicated places and divide these by the time interval (At
= 0.05 sec). The momentum is then computed by multiplying these velocity values by the
respective masses. You may now remove Overlay A and proceed with the others.
Overlay C Shows momentum vectors before and after collision drawn to a scale of 7cm = 1 kg-m/sec.
Overlay D Shows the resultant momentum vector P' after the collision. It can be seen to be approx-
imately equal to the momentum before collision P.
BEFORE COLLISION
Impinging Ball
m = 0.367 kg
d = 3.20 cm (from photo)
1 m
d=: 3.20 cm X
22.8 cm
d = 0.141 m (actual)
V = 2.81 m/sec
mv = 1.03 kg-m/sec
14 mv= = 1.45 joules
d' = 2.40 cm (from photo)
d' = 0.105 m (actual)
V = 2.10 m/sec
mv' = 0.771 kg-m/sec
1/2 mv'2 = 0.810 joules
I
I
Scale : 1 meter = 22.8 cm
I
I
I
Target Ball
m = 0.367 kg
At = 0.05
1
1
1
1
1
sec
T=0
niv"=
1/2 mv^ =
AFTER COLLISION
1 d =
1
= 1.65 cm
(from photo)
d' =
= 0.0725 m (actual)
I V' = 1.45 m/sec
mv' = 0.532 kg-m/sec
I 1/2 mv'- = 0.386 joules
(niv" scale: 7 cm = 1 kg-m/sec)
T20
T-20
m
At
0367kg
O.OSsec
e
^ ® © © ^
© m = 0.367kg
^d
T20
m =
At = 0.05$ec
-tT
'MS-
s.
d =
■#•••#••-_.
L±
^■■■0-0-
m = 0.367kg
^^
^^
^S
T^
B
C
TIO
I
T-21 Unequal Mass Two-Dimensional Collisions
'21
This transparency is facsimile of stroboscopic photograph 74 which shows the collision of spheres
of unequal mass both of which are moving before collision.
Overlay A Shows a sphere of mass 1.80 kg approaching from the upper left and a sphere of mass 4.29
kg approaching from the upjier right. After the collision the less massive ball proceeds to
the lower left while the larger ball travels toward the lower central portion of the picture.
The time interval shown is 0.10 sec; the scale is indicated by the one meter markers at the
bottom of the overlay.
Overlay B Shows the first step in computing the momentum quantities. Determine the four velocities
by measuring the distances in the indicated places and divide these by the time interval
(At = 0.10 sec). The momentum is then computed by multiplying these velocity values
by their respective masses. You may now remove Overlay A and proceed with the others.
Overlay C Shows momentum vectors drawn to a scale of 1 cm = 1 kg-m/sec.
Overlay D Shows the resultant jnomentum vector before collision P and the resultant momentum
vector after collision P' to be equal.
BEFORE COLLISION
Small Ball Large Ball
m = 1.80 kg \ m — 4.29 kg
d = 6.40 cm (from photo) | d = 4.25 cm (from photo)
Scale : 1 meter = 22.8 cm
1 m I
d = 6.40 cm X |
22.8 cm I
d = 0.281 m (actual) I d = 0.187 m (actual)
At = 0.10 sec
"v = 2.81 m/sec i ~v"= 1.87 m/sec
nfiv'= 5.06 kg-m/sec I mv = 8.03 kg-m/sec
V2 mv= =: 7.11 joules I 1/2 rnv"- = 7.50 joules
(rniv scale: 1 cm =: 1 kg-m/sec)
AFTER COLLISION
d' = 5.30 cm (from photo) I d' = 3.15 cm (from photo)
d' = 0.232 m (actual) j d' = 0.138 m (actual)
T' =: 2.32 m/sec | "v' = 1.38 m/sec
mv' = 4.18 kg-m/sec j mv"' = 5.79 kg-m/sec Z9.97
1/2 mv'== = 4.85 joules | 1/2 mv'^ = 3.92 joules 18.77 ;
A profitable discussion might focus on:
1. What became of the "lost" momentum?
2. Are we likely to observe perfectly elastic collisions between large objects?
3. What became of the "lost" kinetic energy?
4. If we add the two "before" momentum vectors and compare them to the sum of the "after" vec-
tors, is momentum conserved?
5. Can the kinetic energies be added vectorially?
T21
T-21
€
m = I
SO kg
•
•
•
•
•
»
«
m = 4.29 kg
At = 0.10 sec
T-21
••• \j
m = 4.29 kg
At = 0.10 sec
irai
..••' B
T^
D
T-22 Inelastic Two-Dimensional Collisions
T22
This transparency is a facsimile of stroboscopic photograph 73. It depicts an inelastic collision
between spheres of equal mass (0.50 kg). The spheres are covered with plasticene, a clay-like material.
Overlay A Shows one sphere approaching from the left and one approaching from the right. After the
collision they proceed as one body to the lower right. The time interval is 0.10 sec
Overlay B Shows the first step in computing the momentum quantities. Determine the three velocities
by measuring the distances in the indicated places and divide these by the time interval
(At = 0.10 sec). The momentum is then computed by multiplying these velocity values by
their respective masses. You may now remove Overlay A and proceed with the others.
Overlay C Shows momentum vectors drawn to a scale of 7 cm = 1 kg-m/sec.
Overlay D Shows the resultant momentum vector before collision to be equal to the momentum vec-
tor after collision.
BEFORE COLLISION
Left Ball
m = 0.500 kg
d = 6.30 cm (from photo)
d = 0.276 m (actual)
"v=: 2.76m/sec
mv = 1.38 kg-m/sec
Vi mv= = 1.91 joules
Right Ball
I m = 0.500 kg
I d = 4.55 cm (from photo)
I d = 0.199 m (actual)
At = 0.10 sec
[ ~v^=z 1.99 m/sec
I mv = 0.999 kg-m/sec
I 1/2 rrw= = 0.995 joules
AFTER COLLISION
m' = 1.00 kg
d' = 1.90 cm (from photo)
d' = 0.0834 m (actual)
T' = 0.834 m/sec
mv' = 0.834 kg-m/sec
1/2 riiv'^ = 0.348 joules
(mv" scale: 7 cm = 1 kg-m/sec)
\
T22
T-22
^^
9
m = 0.5kg
At = 0.10 sec
9 o
^
te
^k
m
m = 0.5kg
T^»
^^
m = 0.5kg
At = 0.10 sec
^•
m = 0.5kg
T22
1^%
■I
I
T-23 Slow Collisions
T23
This transparency is helpful in analyzing the stroboscopic photographs which your students will
take of colliding carts in the Slow Collisions Experiment. It may also serve as a source of data if the
student photographs are inadequate.
The events depicted in 3 separate overlays are as follows :
1- 4 : 4.0 kg cart moving to right before collision
4-12: Same cart during collision
12-15: Same cart after collision
4-12: 2.0 kg target cart during collision
12-15: Same cart after collision
Image Number 12 is lighter in each level since one slit on the strobe disc is taped to produce a narrower
aperture. This image represents the same instant of time for each cart. Data to be taken from this
photograph include values for mv and mv- before, during, and after the collision.
>
T23
1
o
2
O
3
O
m
before
collision
T-23
4
o
m
= 4.0 kg
At - g^ sec
I cm = .05 m
m2 = 2.0 kg
Tt2^
■4
d
CO
uring
llision
456 7 8
9
10 11 12
ooo o
O
OOO
1 2 3
4
5
6 7 8
9
10 11 12 13 14 15
o o o
4 rn
O
O
e o o
O
o ooooo
^ 1 1 ii
rrii
before
during
collision
^
collision
^
m,
= 4.0 kg
1
m2 = 1.0 kg
'^• = 60
sec
I cm = .05 m
1F^
A
B
C
1
o
2
O
3
O
m
4
o
1
before
collision
5
O
6
O
7
o
m,
8
O
during
collision
9
O
— m2 —
during
collision
456 7 8 9
ooo e o
10 11 12 13 14 15
O O0O O O
m^^
after
collision
10
o
11
o
12
o
— rrij —
after
collision
13
O
14
o
m, = 4.0 kg
At = g^ sec
m2 = 2.0 kg
I cm = .05 m
T-24 The Watt Engine
T24
This transparency will aid in discussing the contributions of Watt to the design of the steam en-
gine. With the introduction of a separate external condenser the necessity of reheating the piston cham-
ber is avoided. Also the engine's efficiency is increased and its fuel consumption is markedly decreased.
Overlay A Shows the general components of the Watt steam engine in a diagrammatic form.
Overlay B Shows the piston in the steam-expansion phase with the steam valve open and the condens-
er valves off. Remove this ovei'lay and introduce Overlay C.
Overlay C Shows the piston in the condensation phase of the cycle. The piston has fallen as the open
condenser valve allows steam to condense in the outer chamber.
T24
T-24
The Watt Engine
X
Cylinder
Boiler
v6
Source of Heat
Condenser
T-24
T24
The Watt Engine
Valve A
Cooling
Water
Cylinder
A ^
d i
'Boiler
Condenser
Source of Heat
Condensed Steam
T25
>v
T-25 Superposition
This transparency is intended to expedite the presentation of the principle of superposition. Se-
quences of traveling pulses on the same side and on opposite sides of the equilibrium line are shown.
Amplitudes can be conveniently added with the aid of a mechanical drawing dividers and a wax pencil.
Overlay A Shows a pulse moving to the right at four successive time intervals. The vertical bars pro-
vide convenient mark-off points for the algebraic summations of the amplitudes.
Overlay B Shows a pulse moving to the left. Sum up the two pulses by adding one amplitude to the
other (see diagram). When you have generated a number of points with this method intro-
duce Overlay C to illustrate the completed superposed wave.
Overlay C Shows the complete resultant wave forms. Remove Overlays B and C and introduce Over-
lay D.
Overlay D Shows a pulse moving to the left on the reverse side of the equilibrium line. Now with the
aid of the dividers algebraically add the two pulses by subtracting the smaller amplitude
from the larger (see diagram).
Overlay E Shows the complete resultant wave forms. Notice how the approaching pulse.^ interact with
each other as they "pass through" each other.
T25
T-25
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T-25
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T25
A
B
C
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T26
T-26 Square Wave Analysis
You may employ this transparency to illustrate the effectiveness of the superposition principle in
analyzing a complex wave form into a series of simpler component waves. The graphic procedure par-
allels a Fourier analysis. Additionally, the various wave forms will afford ample practice work for
students in applying the superposition principle. (The formation of a square wave is important in all
pulsing radar transmitters.)
Overlay A Shows a grid with a heavy central line from which all amplitude measurements are deter-
mined.
Overlay B Shows a wave which approximates a square wave. You may suggest the possibility of con-
structing this complex wave by superposing a number of simpler waves and proceed to show
how it is done. (The formation of a square wave is important in all pulsing radar trans-
mitters.)
Overlay C Shows a component wave with a wavelength equal to that of the square wave but with some
regions of excess and defect which require smoothing out by the addition of more compo-
nents.
Overlay D Shows the next component with a wavelength one-third the main component. Remove Over-
lay B and proceed to sum up the two waves of Overlays C and D as indicated in T23. Draw
in the resultant wave form. The residual irregularities can be largely removed with addi-
tional components of wavelengths 1/5, 1/7, 1/9 . . . times that of the original.
Overlay E Shows a component wave with a wavelength 1/5 the original. Remove Overlay C
superpose the resultant wave of Overlays C and D with the wave on Overlay E.
and
Overlay F Shows a component wave with a wavelength 1/7 the original. Remove Overlay D and super-
pose wave F with the resultant wave appearing on Overlay E. The final wave should approxi-
mate the original square wave. Return Overlay B to verify this.
\
T26
T-26
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T'26
T26
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T-28 Two-Slit Interference
T28
This transparency uses the moire pattern technique to simulate the regions of constructive and
destructive interference as produced in a two-source two dimensional interference pattern. A sliding
overlay shows the dependence of the nodal line pattern upon the source separation.
Overlay A Shows circular waves emanating from a single slit or from a circular wave source.
Overlay B Shows another set of circular waves emanating from another slit or circular wave source.
Match this overlay carefully with the dot on Overlay A. A distinct nodal line pattern will
be seen where these two sources overlap.
To illustrate the dependence of this pattern on the separation of the sources, carefully move
this loose overlay above or below the point on Overlay A. Notice how the pattern changes
as the sources are moved apart.
Overlay C Shows the fringe pattern which might be obser\-ed at a distance from the sources in any
number of wave phenomena (light, sound, microwaves, etc.). Be sure Overlay B is lined up
with the dot on Overlav A.
\
T28
T-28
T28
T-28
r>** !•• •* •
weak
intense
weak
intense
weak
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intense
weak
T-29 Interference Pattern Analysis
129
This transparency will be useful in analyzing the interference pattern produced by a double slit
barrier. You may also employ it to derive the wavelength equation.
Overlay A Shows a set of crests (lines) and troughs (space between lines) emanating from source S.
Overlay B Shows a similar set of waves emanating from source Sj. You may discuss the formation of
constructive and destructive interference points by indicating the crest-crest, trough-trough
and crest-trough overlaps. Place a series of points with a wax pencil directly on the overlay
to indicate interference points. Then introduce Overlay C to show the complete pattern.
Overlay C Shows the complete pattern with nodal and anti-nodal lines. Remove Overlays A and B and
introduce Overlay D.
Overlay D Shows the geometry needed to derive the wavelength equation. First choose a point P on a
anti-nodal line and show that the relationship PS,-PS, == n A holds, where n is the number
of the anti-nodal line.
Now draw a line from S, to A such that PS. = PA (see diagram). Then
PS, — PS, — ASi. Assume that P is far enough so that PS, and PS., are
nearly parallel and that angle S,AS, is a right angle. Then Sin = AS,/d
(Identify angle AS,S, as $ )■
Since AS, = PS, - PS, then PS, - PS, = d sin d . When P is on the nth
anti-nodal line PS, — PS, = nA, therefore n A = d sin as long as P is
far from S, and S,.
Now show that = 0^ by observing that line L and PS, are practically par-
allel to each other and both are perpendicular to AS,. The center line Aq is
perpendicular to d and $ = Q
Then sin^n^
d(Xn/L)
Xjj/L or
I
T29
T-29
I-
T-29
T^»
A
C
A3 N3 A2 N2 A, N, Ao N, Ai N2 A2 N3 A;
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A3 N3 A2 N2 A, N, 4
Q
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