Mon.
Tues.
Wed.
Lab
Fri.
6.14-.17 Electric and Rest Energy
RE 6.e
EP6, HW6: Pr’s 58, 59, 91, 99(a-c), 105(a-c)
7.1-.4 Macroscopic Energy Quiz 6 RE 7.a bring laptop, smartphone, pad,…
L6 Work and Energy
laptop
7.5-.9 Energy Transfer
RE 7.b
 Example
vE .i
Gravitational System: Earth and Sun
mE

vE . f
rE  S . f
rE  S .i
System= Earth + Sun
Active environment = none
E  Wsystemext  0
Not changing
EE , S  Erest . E  Erest .S  K E  K S  U E , S  0
EE , S  K E  U E , S  0
K E . f  U E , S . f  K E .i  U E , S .i
1
2
2
E E. f
m v
mE ms 1
mE ms
2
G
 2 mE v E . f  G
rESf
rESi
mS
Which of the following graphs of U vs r represents the
gravitational potential energy, U = –GMm/r?
Different Initial Speeds / kinetic Energies, Different Paths
(orbit noncircular, with energy vs position.py)
In which graph does the cyan line correctly represent
the sum of kinetic energy plus potential energy?
Energy
Conceptual Understanding from Energy Diagrams
Ex. Nuclear Potential
Limits / Turning
Points
|r|
K=0
K=0
K+U = Const
Energy
Conceptual Understanding from Energy Diagrams
Ex. Nuclear Potential
Limits / Turning
Points
r1
K=0
|r|
K=0
K+U = Const
U(r1)
K(r1)+U(r1)=Const
K(r1)=Const - U(r1)
Energy
Conceptual Understanding from Energy Diagrams
Ex. Nuclear Potential
Limits / Turning
Points
r1
K=0
|r|
K=0
K+U = Const
U(r1)
K(r1)
K(r1)+U(r1)=Const
K(r1)=Const - U(r1)
Energy
Conceptual Understanding from Energy Diagrams
Ex. Nuclear Potential
Limits / Turning
Points
r1 r2
K=0
U(r1)
K(r1)
|r|
K+U = Const
K=0
K(r1)
U(r2)
K(r1)+U(r1)=Const = K(r2)+U(r2)
K(r1)=Const - U(r1)
04_potential_energy_well.py
The system is a comet and a star. In which case(s) will the
comet escape from the star and never return?
1) A
2) B
3) C
4) A,B
5) B,C
6) A,B,C
Force as negative gradient (3-D slope) of
Potential Energy
small change in potential
dU1, 2


  F12  dr12  F12. x dx  F12. y dy  F12. z dz 
Say only moves in the x direction, then
dU1, 2
dU1, 2   F12. x dx so 
U1, 2
dx
 F12. x
Similarly, if only moves in the y direction, then
dU1, 2   F12. y dy
so

dU1, 2
dy
 F12. y
x̂
or, if only moves in the z direction, then
repulsive
attractive
dx
 F12. x
repulsive

dU1, 2
dU1, 2   F12. z dz
so

dU1, 2
Moving in all directions,
dz
 F12. z

U1, 2 dU1, 2 dU1, 2
F12  F12. x , F12. y , F12. z  
,
,
x12 dy12 dz12
x̂
Newton’sElectric
Universal
LawLaw
of
Coulomb’s
Force
m
q
Gravitation

22
mq1
F21

r21

F12

1



F21 4Go
q1qm22
m
 2 rˆ21
r21
N m
G 46o.67
 910
10
C 22
kg
11
9
1
22

r21
r̂21 
r21
Gravitational
Electric Potential Energy
U1, 2.electric
U1, 2
r12
like charges
m
q1qm22
 4Go
r12
1
opposite charges
r12
U1, 2.electric
r̂21
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?
System= electron + proton
ri  10
10
rf   U e, p.elect 
m
U e, p ,elct 
U e, p ,elct
4 o
U e, p.elect 


 
2
4  8.8510 12 C 2  
Nm 


1
U e, p ,elct  2.3 10
18
4 o
re p
Active environment = none
Comparison:
Electric vs. Gravitational
 e2
r12
0
2

J
Or in eV’s (divide by electron charge)
 2.3 10
1
qe q p
1

1
e 


4 o
 ri rf 


2
19
1.6 10 C 

10 10 m

2
 e2

e
 41 o

ri
rf
18
4 o
re p
U e, p.electric
1
1
J 1.6101e19 C  14eV
∞

2

e
1
4 o
U e, p.elect
r12

me m p
U e, p. grav
G
r12
2
U e, p.elect
e
 41oG
U e, p. grav
me m p
1.6 10 C 
9 10 kg 1.7 10
9 109
6.7 10
11 Nm 2
kg 2
19
Nm 2
C2
31
U e, p.elect
U e, p. grav
2
 27
 5.6 1039
kg

Return to Rest Energy and Mass
-e
-e
Pair (electron and positron) Annihilation
E  me c  me c
2
2
ri  
initial


final
e e  
Electron and positron
ri  
U e, p
re p
Two photons (light pulses)
E  2me c 2  2 E
0.511MeV / c c
2
2
 E
0.511MeV  E
+e
Return to Rest Energy and Mass
Neutron Decay
-e
initial
E  mn c
ri  
2
final
n0  p   e  e
Proton, electron, and neutrino
neutron
Finally infinitely far apart
Nearly massless
E  mn c 2  me c 2  m p c 2  m c 2  K e  K p  K  U e, p  U e,  U , p
E  mn c 2  me c 2  m p c 2  K e  K p  K
K

2
2
2


K

K

m
c

m
c

m
c
e
p

n
e
p
 939.6MeV  0.511MeV  938.3MeV   0.79MeV
rf  
U e, p

p+-en e
Mass as Energy and Energy as Mass
Box o’ decaying Neutrons
re p E  Erest  mbox c 2

 m
n
 me  m p  m c 2  K e  K p  K  U e, p
all . particles
Viewed from outside
Peaking inside
Box’s mass includes internal kinetic and potential energies

Efree=2mc2
Mfree=Efree/c2=2m
Energy
Return to Rest Energy and Mass
O2 bonding
r
K+U=0
rfree
Efree=2mc2
Mfree=Efree/c2=2m
Energy
Return to Rest Energy and Mass
O2 bonding
rbound
U
K+U
K
rfree
r
Return to Rest Energy and Mass
O2 bonding
K+U
Efree=2mc2
Mfree=Efree/c2=2m
Ebound=2mc2+ (K+ U)
Mbound=Ebound/c2 =2m + (K+U)/c2
Note: would have shed excess energy
by emitting photon / light pulse
Energy
Energy / Mass difference
M  M free  M bound  E f  Eb  / c 2



M  2mc 2  2mc 2  K  U bound / c 2
M  K  U bound / c 2  5eV / c 2  9 10 36 kg
Noticeable?
rbound
U
K+U
K
M
 3  10 10 not really
2 mo
rfree
r
Return to Rest Energy and Mass
Nuclear Binding: Iron nucleus
If an iron nucleus were disintegrated, how much K + U energy would be
consumed /produced?
initial
final
Useful info
26
Fe56
 26 p   30n
Iron nucleus
Noticeable?
mc 2
 0.009  1%
mFe c 2
M Fe.nuc  52107 MeV / c 2
mn  939.9 MeV / c 2
Protons and neutrons
Ei  E f
Er . Fe   Er  K  
all . particles
m p  938.3MeV / c 2
U
all . pairs



mFe c  26  m p c  30  mn c    K   U 
yes
 all . particles all . pairs 


2
2
2
mFe c  26  m p c  30  mn c    K   U 
 all . particles all . pairs 



52107 MeV  26  939.9MeV   30  938.3MeV     K   U 
 all . particles all . pairs 


 482MeV    K   U 
2
2

2

all . particles
all . pairs
Rest and Electric-Potential and Kinetic
A U-235 nucleus is struck by a slow-moving neutron, so that the merge and become U236, 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?
Mon.
Tues.
Wed.
Lab
Fri.
6.14-.17 Electric and Rest Energy
RE 6.e
EP6, HW6: Pr’s 58, 59, 91, 99(a-c), 105(a-c)
7.1-.4 Macroscopic Energy Quiz 6 RE 7.a bring laptop, smartphone, pad,…
L6 Work and Energy
laptop
7.5-.9 Energy Transfer
RE 7.b