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Full text of "Transparencies Unit 3 - The Triumph of Mechanics: Project Physics"

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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 



.,<trTTTk 



^wrrfTTk 



I- 



.^^ffink 



^^^rTink 



T-25 



-^rrngmv. 




jUtftk. 



jflJWOL 



T25 




A 
B 
C 






ir-^ 



-o^p^ 



-♦- 
-(0^ 



nfljUl^ 



T^§ 









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 



K 



T'26 




T26 




T'J« 




T^ 




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^^T i ' • ' ■ ' I ' . i I 1 1 i 1 1 M ! 1 1 ' 1 ! M (1 




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1 


, . ^ -._^ - I ■ 1 p-| r-p-| p-. 


1 1 1 1 M 1 M ! 1 


1 1 




! ! 


' ■./ -^^ 1 \ 


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-4—^^=^ H^^T^— 


=^^T[TTr^^^ — 


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-^ 1 1 , , p^- y-^^- -\ H- 




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-:__.„-t :-:_:. m^iiii-^iiiiiiiii 


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Agr_:=^^^^iiii±ii mSm, 


=:^_^g^=^^iij:^^^^^ ^Zz 


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^ff^ 




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 



iff''.'.".: 



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; 




^ 



iraa 



A3 N3 A2 N2 A, N, 4 





Q 



(^