1. Solution
Part 1a
a.
( )
( ) smVmevaccv
smVmevretv
/610044.2/22
0
/610956.1/22
0
×=+=
×=−=
(0.5 pts)
( )2/accx, Ttaccvtretvretx −== (0.5 pts)
T
retvaccv
Taccv
accxretx 61.11
)(2buncht =
−
=→= (0.3 pts)
.210272.2 mbunchtretvb −×== (0.2 pts)
b. The phase difference:
∆ϕ = ± π2)( n
T
tbunch
− =±0.61 π2× =±220°. (1.0 pts)
OR
∆ϕ = ±140°
Part 1b
A
LL
N
M
n=ρ
(0.3 pts)
where nL is the number of molecules per cubic meter in the liquid phase
Average distance between the molecules of water in the liquid phase:
3
1
3
1
)()(
AL
LL
N
M
nd
ρ
==
−
(0.2 pts)
PaV=nRT,
where n is the number of moles (0.6 pts)
M
RT
N
Mn
M
RT
M
RT
V
nM
P
A
V
Va === ρ
where nV is the number of molecules per cubic meter in the vapor phase. (0.9 pts)
3
1
3
1
)()(
Aa
VV
NP
RT
nd ==
−
(0.2 pts)
12)( 3
1
==
MP
RT
d
d
a
L
L
V ρ
(0.3 pts)

2. Part 1c
a. (0.5 pts.)
b. Vi >> Vf (0.2 pts)
c. ( )RC/T
if e1VV −
−= (0.2 pts)
If
Vi >>Vf , T/RC<<1, ( )RCTe RCT
/1/
−≈−
then
T = (Vf / Vi) RC (0.2 pts)
d. R (0.2 pts)
e. SG and R (0.2 pts)
f. Correct circuit (0.4 pts)
'
0V (0.3 pts)
fi VV −′ with the correct polarity (0.3 pts)
Total (1.0 pts)
+
R
C
-
SG
Vi
(Vı
-Vf
)
+
-
V’0
T
t
Vf
Vi
V0

3. Part 1d
As the beam passes through a hole of diameter D the resulting uncertainty in the y-
component of the momentum;
Dyp

≈∆ (0.6 pts)
and the corresponding velocity component;
MDyv

≈∆ (0.4 pts)
Diameter of the beam grows larger than the diameter of the hole by an amount
∆D= ∆vy.t ,
where t is the time of travel. (0.2 pts)
If the oven temperature is T, a typical atom leaves the hole with kinetic energy
kTMvKE
2
3
2
1 2
== (0.4 pts)
M
kT
v
3
= (0.2 pts)
Beam travels the horizontal distance L at speed v in time
v
L
t = , so (0.2 pts)
MkTD
L
M
kT
MD
L
MDv
L
vtD y
33

==≈∆=∆
(0.4 pts)
Hence the new diameter after a distance L will be;
Dnew=D +
MkTD
L
3

(0.1 pts)

4. Part 2a
The total energy radiated per second = 4πR2
σT4
, where σ is the Stephan-Boltzmann
constant. The energy incident on a unit area on earth per second is;
24
424
π
σπ TR
P = yielding, 
2/1
4/ 



= TPR σ (1) (0.8 pts)
The energy of a photon is hf=hc/λ. The equivalent mass of a photon is h/cλ. Conservation
of photon energy:
λλλ
hc
c
h
R
Gmhc
=⋅−
0
0
0
(0.8 pts)
yielding
( )
λ
λλ
∆
∆+
= 2
00
c
Gm
R (2)
and (2) yields,
( )
( )

λλ
σλ
∆+
∆
=
0
2/142
0
/
G
TPc
m (3) (0.2 pts)
The stars are rotating around the center of mass with equal angular speeds:
ω = (2π/2τ) = π/τ (4) (0.2 pts)
The equilibrium conditions for the stars are;
( )
2
2
2
102
21
0
ωω Mrrm
rr
GMm
==
+
(5) (0.8 pts)
with
2
1
θ∆
=r ,
2
2
φ∆
= r (6) (0.4 pts)
Substituting (3), (4) and (6) into (5) yields
( ) ( )( )
2/1
2
0
2/
2/1
4/28














∆+∆∆+∆




∆
=
φθλλτπφ
σλ TPc
 . (0.8 pts)
Part 2b
Conservation of angular momentum for the ordinary star;
0
2
00
2
ωω rmmr = (7) (0.6 pts.)
Conservation of angular momentum for dm:
dmrdmr ff
22
ωω = (8) (0.6 pts)
where ωf is the angular velocity of the ring. Equilibrium in the original state yields,
2/1
3
0
0










=
r
GM
ω (9) (0.8 pts)
and (7), (8) and (9) give,
2/1
0
2
00








=
r
GM
mr
rm
ω ,
2/1
0
2
00








=
r
GM
fmr
rm
fω (10) (0.4 pts)
Conservation of energy for dm;

5. fr
dm22
2
1
r
dm222
02
1 GM
ffrdm
GM
rvdm −=−



 + ωω (11) (1.2 pts)
Substituting (10);
0
11
2
11
222
0
2
02
0 =








−−








−+
ff
rr
GM
rrm
GMrm
v (12)
Since r0>> rf , if r> r0 , r-1
and r-2
terms can be neglected. Hence,










−








+= 11
2/1
2
2
00
2
0
2
0 GMm
rmGM
rf
ν
ν
. (0.8 pts)
To show that r>r0 change in the linear momentum of the ordinary star in its reference
frame:
dt
dm
v
dt
dv
mmr
r
GMm gasr
0
2
2
−=−+− ω (13) (0.8 pts)
and (13) implies the existence of an outward force initially and hence r starts growing.
Using (7) one can write
3
2
0
4
0
2
02
mr
rm
mr
ω
ω = .
Hence, r
forcelCentrifuga
forcenalGravitatio 2
mα . (0.4 pts)
where m is definitely decreasing. If r starts decreasing at some time also, this ratio starts
decreasing, which is a contradiction.
So r>r0. (0.4 pts)

6. Part 3a
The net force on a charged particle must be zero in the steady state
BxvqEqF +==0
∧
=−= yvBBxvE (0.4 pts)
vBwHV =
ρρρ
vBLh
w
vBwLh
Lh
w
HV
R
HV
I ====
, direction: -
∧
y (0.6 pts)
BxIF = =
ρ
LhwvB2
, direction: (-
∧
−=
∧∧
xzxy )
Force is in the -x direction (0.8 pts)
This creates a back pressure Pb
ρρ
LvB
hw
LhwvB
Pb
22
== (0.6 pts)
Fnet=(P-Pb)hw, (0.6 pts)
v=αFnet (0.4 pts)
v=α(P-Pb)hw= α
ραρ P
LBvv
v
P
vLvB
P
2
0
0
0
2
)( −=−
0
2
0
)1( v
P
LBv
v =+
ρ
1
2
010
−










+=
ρP
LBv
vv
LBvP
P
vv 2
0
0
+
=
ρ
ρ
(0.6 pts)
x
z y
B
P v
L

7. Part 3b
From conservation of energy:
ρ
whLBv
IVPower H
22
0
==∆
(1.0 pts)
or,
to recover v0 the pump must supply an additional pressure ∆P=Pb
00 hwvPPhwvPower b=∆=∆
ρ
whLBv 22
0
=
Part 3c
1.
n
c
u =
cn
v
v
n
c
c
v
n
c
v
n
c
u
+
+
=
+
+
=
1
2
1
' (0.5 pts)
For small v (v<<c);
neglect the terms containing
2
2
c
v
in the expansion of
1)1( −+
cn
v
)
2
1
1()1)((
1
1
)('
n
v
n
c
cn
v
v
n
c
cn
v
v
n
c
u −+≈−+≈
+
+=
)
2
1
1('
n
vuuu −≈−=∆ (0.5 pts)
Tf∆=∆ πφ 2 , )12(
22u
u
T, −≈
∆
=∆= n
c
Lv
L
u
L
T (0.5 pts)
v=v0 so that, 0)12(
2
2 vn
c
L
f −=∆ πφ (0.5 pts)
2. 0
2
2
)1(2 vn
c
L
f −=∆ πφ
a phase of π/36 results in (0.4 pts)
fnL
c
v
)1(72 2
2
0
−
= (0.2 pts)
smx
xxx
x
v /102.3
25)156.2(1072
109 14
1
16
0 =
−
= −
which is not physical.
(0.4 pts)

8. 3. For v=20 m/s, f≈4x1014
Hz. But for this value of f, skin depth is about 25
nm. This means that amplitude of the signal reaching the end of the
tube is practically zero. Therefore mercury should be replaced with
water. (0.6 pts)
On the other hand if water is used instead of mercury, at 25 Hz δ≈3x105
m. Signal reaches to the end but v≈6x1014
m/s, is still nonphysical.
Therefore frequency should be readjusted. (0.6 pts)
For v=20 m/s electromagnetic wave of f≈8x1014
Hz has a skin depth of
about δ≈5.6 cm in water and the emerging wave is out of phase by π/36
with respect to the incident wave. (The amplitude of the wave reaching
to the end of the section is about 17% of the incident amplitude). (0.6
pts)
Therefore mercury should be replaced with water and frequency should
be adjusted to f≈8x1014
Hz. The correct choice is (iii) (0.2 pts)