fri. 6.11, 14-17 visualizing electric and rest energy re 6...
TRANSCRIPT
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Fri. 6.11, 14-17 Visualizing Electric and Rest Energy RE 6.d,e
Mon.
Tues.
Things Engineers and Physicists Do
EP6, HW6: Ch 6 Pr’s 58, 59, 91,
99(a-c), 105(a-c)
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A huge asteroid smacks into the leading edge of the Earth – stopping
the Earth’s orbit. Subsequently, the Earth falls straight into the sun!
With what speed would the Earth hit the Sun’s surface?
System= Earth + Sun
Active environment = none
Em
kgmS
301099.1
0. iEv
?. fEvmr iSE
11
. 105.1
SEfSE RRr .
m81002.7 extsystemWE 0
0,.., SESESrestErestSE UKKEEE
Not changing
0,, SEESE UKE
iEfEE KKK .. 2
.21
fEEvm
iES
sE
fES
sE
ES
sEES
r
mmG
r
mmG
r
mmGU
..
02
.21
, ESi
sE
ESf
sEfEESE
r
mmG
r
mmGvmE
ESiESf
sfErr
Gmv11
2.
smmm
kgkgNm /1014.6105.1
1
1002.7
11099.1/1067.62 5
118
302211
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System: comet + star
Surroundings: negligible
As a comet travels away from a star, how do the
kinetic energy and potential energy of the system
change?
K U
a) increase decrease
b) increase increase
c) decrease increase
d) decrease decrease
e) no change no change
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Force as negative gradient (3-D slope) of
Potential Energy
dzFdyFdxFrdFdU zyx .21.21.2121212,1 small change in potential
Say only moves in the x direction, then
dxFdU x.212,1 so xFdx
dU.21
2,1
x̂
2,1U
x̂
xFdx
dU.21
2,1
repuls
ive
attra
ctive
repuls
ive
Similarly, if only moves in the y direction, then
dyFdU y.212,1 so yFdy
dU.21
2,1
or, if only moves in the z direction, then
dzFdU z.212,1 so zFdz
dU.21
2,1
Moving in all directions,
21
2,1
21
2,1
21
2,1
.21.21.2121 ,,,,
dz
dU
dy
dU
x
UFFFF zyx
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|r|
K+U = Const K=0 K=0
Energ
y
Limits / Turning
Points
E = Wext=0
(K + U + Erest) = 0
Kf+Uf = Ki + Ui = Constant
y
Conceptual Understanding from Energy Diagrams
Roller Coaster
Ugra
v=
mg
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04_potential_energy_well.py
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|r|
K+U = Const
r1
K=0 K=0
Energ
y
Limits / Turning
Points
U(r1)
K(r1)+U(r1)=Const
K(r1)=Const - U(r1)
Conceptual Understanding from Energy Diagrams
Ex. Nucleus - Proton Potential
E = Wext=0
(K + U + Erest) = 0
Kf+Uf = Ki + Ui = Constant
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|r| r1
K(r1)
K=0 K=0
Energ
y
Limits / Turning
Points
U(r1)
K+U = Const
K(r1)+U(r1)=Const
K(r1)=Const - U(r1)
Conceptual Understanding from Energy Diagrams
Ex. Nucleus - Proton Potential
E = Wext=0
(K + U + Erest) = 0
Kf+Uf = Ki + Ui = Constant
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|r|
K+U = Const
r1
K(r1)
K=0 K=0
Energ
y
Limits / Turning
Points
U(r1)
K(r1)+U(r1)=Const = K(r2)+U(r2)
K(r1)=Const - U(r1)
K(r1)
r2
U(r2)
Conceptual Understanding from Energy Diagrams
Ex. Nucleus - Proton Potential
E = Wext=0
(K + U + Erest) = 0
Kf+Uf = Ki + Ui = Constant
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Which of the following graphs of U vs r represents the
gravitational potential energy, U = –GMm/r?
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vi = 0
r
mMGU s
r
(Separation)
A)
Energy Diagrams
Ex. Gravitational, Bound
(K + U)A
Energ
y
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vi = 0
r
mMGU s
r
(Separation)
A)
Rstar
B) rmax rmin
(K + U)A
(K + U)B
Energ
y
Energy Diagrams
Ex. Gravitational, Bound
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vi = 0
r
mMGU s
r
(Separation)
Energ
y
A)
Rstar
B) rmax rmin
r C)
(K + U)A
(K + U)B
(K + U)C
Energy Diagrams
Ex. Gravitational, Bound
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Energy Diagrams
Ex. Gravitational, Un-Bound / Escape
r
mMGU
p
r
(Separation)
K+
U
rplanet
A
B
C
D
Consider an rocket launching from a planet’s surface, which of the following
represent un-bound systems (so the rocket could get away and never fall back to
the planet)?
1. A
2. B
3. A & B
4. C
5. D
6. C & D
7. A,B,C, & D
Special Case: K + U = 0 and Escape Speed
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In which graph does the cyan line correctly represent
the sum of kinetic energy plus potential energy?
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1) A 2) B 3) C 4) A,B 5) B,C 6) A,B,C
The system is a comet and a star. In which case(s) will the
comet escape from the star and never return?
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Gravitational Potential Energy
Newton’s Universal Law of
Gravitation
12r
1m
2m
12F
21F
122
12
21
rr
mmG ˆ12F
22
1110676kg
mNG
.
12r̂
12r̂12
12
r
r
1q
2q
2
29
41 109
C
mNo
122
12
214
1
rr
qqo
ˆ
Coulomb’s Electric Force Law
21
212,1
r
mmGU
21
21
41
2,1
r
qqU
o
Electric Potential Energy
21r
electricU .2,1
opposite charges
21r
electricU .2,1
like charges
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Example: Ionize Hydrogen. In a hydrogen atom the electron averages around
10-10 m from the proton. When a hydrogen atom is ionized, the electron is stripped
away. What is the change in electric potential energy when such an atom is
ionized?
per
electricpeU .,
mri
1010 fr
pe
pe
electper
qqU
o
41
.,
System= electron + proton
Active environment = none
21
2
41
.,
r
eU
oelectpe
if
elctper
e
r
eU
oo
2
41
2
41
,,
fi
e
rro
114
2
∞
0
m
CU
Nm
Celctpe 10
219
1085.84
1,,
10
106.1
2
212
JU elctpe
18
,, 103.2
Or in eV’s (divide by electron charge)
C
eJ 19106.1
118103.2
eV14
Comparison:
Electric vs. Gravitational
21
21
2
41
.,
.,
r
mmG
r
e
U
U
pegravpe
electpeo
pe
G
gravpe
electpe
mm
e
U
U
o
2
41
.,
.,
39
.,
.,106.5
gravpe
electpe
U
U
kgkg
C
kg
Nm
C
Nm
273111
2199
107.1109107.6
106.1109
2
2
2
2
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Return to Rest Energy and Mass
Pair (electron and positron) Annihilation
-e
+e
per peU ,
ir
ir
-e
+e
22 cmcmE ee
eeinitial final
Electron and positron Two photons (light pulses)
EcmE e 22 2
EccMeV 22/511.0
EMeV 511.0
Application Note:
Positron Electron Tomography (PET)
Scans
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Return to Rest Energy and Mass Neutron Decay
per peU ,
fr
ir
-e p+
2cmE n
initial final
neutron Proton, electron, and neutrino
pepepepen UUUKKKcmcmcmcmE ,,,
2222
eepn 0
n e
KKKcmcmcmE pepen 222
Finally infinitely far apart Nearly massless
222 cmcmcmKKK penpe
MeVMeVMeV 3.938511.06.939 MeV79.0
Mass as Energy and Energy as Mass
Box o’ decaying Neutrons
Viewed from outside
2cmEE boxrest
Peaking inside
particlesall
pepepe UKKKcmmm.
,
2
Box’s mass includes internal kinetic and potential energies
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What is Mass
Quantification of…
Gravitational ‘charge’
Inertia
Internal Energy
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r
Energ
y
Efree=2mc2
Mfree=Efree/c2=2m
rfree
K+U=0
Return to Rest Energy and Mass O2 bonding
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rbound r
Energ
y
U
rfree
K
K+U
Efree=2mc2
Mfree=Efree/c2=2m
Return to Rest Energy and Mass O2 bonding
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rbound r
Energ
y
U
rfree
K
K+U
Ebound=2mc2+ (K+ U)
Mbound=Ebound/c2 =2m + (K+U)/c2
K+U
Return to Rest Energy and Mass
Note: would have shed excess energy
by emitting photon / light pulse
O2 bonding
2/ cEEMMM bfboundfree
Energy / Mass difference
kgceVcUKM bound
3622 109/5/
222 /22 cUKmcmcM bound
Noticeable?
101032
om
M
Efree=2mc2
Mfree=Efree/c2=2m
not really
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Return to Rest Energy and Mass
Nuclear Binding: Iron nucleus
initial final
Iron nucleus Protons and neutrons
pairsallparticlesall
npFe UKcmcmcm..
222 3026
npFe 302626
56
fi EE
pairsallparticlesall
rFer UKEE..
.%1009.0
2
2
cm
mc
Fe
pairsallparticlesall
npFe UKcmcmcm..
222 3026
2
2
2
.
/3.938
/9.939
/52107
cMeVm
cMeVm
cMeVM
p
n
nucFe
If an iron nucleus were disintegrated, how much K + U energy would be
consumed /produced?
Useful info
pairsallparticlesall
UKMeVMeVMeV..
3.938309.9392652107
pairsallparticlesall
UKMeV..
482
Noticeable?
yes
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Rest and Electric-Potential and Kinetic A U-235 nucleus is struck by a slow-moving neutron, so that the merge and become U-
236, with mass MU-236 This nucleus is unstable to falling apart – fission. One way it
could do so is to first slosh into something of a dumbbell shape, now most of the into two
symmetric nuclei, Pd-118, with mass MPd-118 , each has ½ the original number of protons,
i.e., qPd = 46e. Having fallen apart, the two palladium nuclei no longer experience a Strong
interaction holding them together, just the Electric repulsion of each other’s protons.
Subsequently, they accelerate away.
a) What’s the final speed of one of the Pd atoms, when they have sped far, far apart?
b) What is the distance between the Pd atoms just after fission?
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Fri. 6.11, 14-17 Visualizing Electric and Rest Energy RE 6.d,e
Mon.
Tues.
Things Engineers and Physicists Do
EP6, HW6: Ch 6 Pr’s
58, 59, 91, 99(a-c),
105(a-c)