国际物理林匹克竞赛试题IPhO2001-THEO-SLN

Part 1a

v

v202e/mV1.956106m/s

a.

retvv26acc02e/mV

2.04410m/sxretvret

t,

x

acc

v

acctT/2v

x

accTretx

acc tbunch

2(vacc

v.61T

ret

)

11b

v

rettbunch

2.27210

2

m.

b.The phase difference: =

(tbunchn)2

T

=0.61

2

=220. OR

= 140

Part 1b

ML

nL

N

A

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:

1

dL(nL)

3

(

M

1

)3

L

N

A

PaV=nRT,

where n is the number of moles

PnMRTRTnVMRTa

V

M

V

M

N

A

M

where nV is the number of molecules per cubic meter in the vapor phase.

1

1

d(n3

RTVV)(

P)

3

aNA

dV(

RT1

L

)3

12

dL

PaM

(0.5 pts)

(0.5 pts) (0.3 pts)

(0.2 pts)

(1.0 pts)

(0.3 pts)

(0.2 pts)

(0.6 pts)

(0.9 pts)

(0.2 pts)

(0.3 pts)

Part 1c

a.

V0

Vi

Vf

t

T

b.Vi >> Vf(0.2 pts)

c.

V

T/RC

f

Vi1

eIf

Vi >>Vf,

T/RC<<1, e

T/RC

then

T = (Vf / Vi) RC

d.R

e.SG and R f.

Correct circuit

V’0

(V?-Vf)

-

+

R

+

SG

C

Vi

-

V'

0Vi

V

f

with the correct polarity

Total

(0.5 pts.)

(0.2 pts)

1T/RC

(0.2 pts) (0.2 pts) (0.2 pts) (0.4 pts)

(0.3 pts) (0.3 pts) (1.0 pts)

Part 1d

As the beam passes through a hole of diameter D the resulting uncertainty in the y-

component of the momentum;

py

D

and the corresponding velocity component;

vy

MD

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.

If the oven temperature is T, a typical atom leaves the hole with kinetic energy

KE

1Mv

2

3kT

2

2

v

3kTMBeam travels the horizontal distance L at speed v in time

t

Lv, so

D

tvLLLy

vMD

MD

3kTD

3MkT

M

Hence the new diameter after a distance L will be;

DLnew=D +

D

3MkT

(0.6 pts)

(0.4 pts)

(0.2 pts)

(0.4 pts) (0.2 pts)

(0.2 pts) (0.4 pts)

(0.1 pts)

Part 2a

The total energy radiated per second = 4RT, where 2

4

is the Stephan-Boltzmann

constant. The energy incident on a unit area on earth per second is;

24

2

P

4RT yielding,

RP/T

4

1/ (1) 42The energy of a photon is hf=hc/. The equivalent mass of a photon is h/c

of photon energy:

hc

Gm0

hhc

0

Rc

0

yielding

R

Gm

0

02

(2)

c

and (2) yields,

2

2

mc

P/T

41/0

G

(3) 0

The stars are rotating around the center of mass with equal angular speeds:

= (2/2) = / (4)

The equilibrium conditions for the stars are;

GMm02

r0r1

Mr

22

(5) 1

r2

m2

with

r1

Substituting (3), (4) and (6) into (5) yields

2

, r2

2

(6) 241/2

1/2

8cP/T

. /

2

2

0

Part 2b

Conservation of angular momentum for the ordinary star; mr

2

m2

0r00

(7)

Conservation of angular momentum for dm:

r2

dm

r

2f

f

dm (8)

where f is the angular velocity of the ring. Equilibrium in the original state yields,

1/2GM

0

(9) r30

and (7), (8) and (9) give,

(0.8 pts) . Conservation

(0.8 pts)

(0.2 pts)

(0.2 pts)

(0.8 pts)

(0.4 pts)

(0.8 pts)

(0.6 pts.) (0.6 pts)

(0.8 pts)

mr002mr

1/2

GMr0

m

,

f

r002mr

f12

1/2

GMr0

(10) (0.4 pts)

Conservation of energy for dm;

12dm

2v0

r2

2

GMdm

r

dmr

2f

f2

GMdmrf

(11) (1.2 pts)

Substituting (10);

2v0

m0r0GM

m

2

2

1r

2

1r

2f

2GM

1r

1rf

0 (12)

Since r0>> rf , if r> r

GM

f

20

0 , r

-1

and r

20r0

-2

terms can be neglected. Hence,

1/2

r1

m

202

1. (0.8 pts)

GMm

To show that r>r0 change in the linear momentum of the ordinary frame:

GMmr

2

star in its reference

mr

2

m

dvrdt

dmv0

gas

dt

(13) (0.8 pts)

and (13) implies the existence of Using (7) one can write

mr

2

an outward force initially and hence r starts growing.

m0r0

mr

243

20

.

2

Hence,

GravitatioCentrifuga

nalforcelforce

m

r. (0.4 pts) ratio starts

where m is definitely decreasing. If r starts decreasing at some time also, thisdecreasing, which is a contradiction. So r>r0.

(0.4 pts)

Part 3a

L

P

v

B

The net force on a charged particle must be zero in the steady state

F0

qEqvxBEvxBvBy

VHvBwVIHVHvBwLh

vBLh

R

w

w

, direction: -yLh

2

F

IxB=

vB

Lhw

, direction: (-yxzx)

Force is in the -x direction This creates a back pressure P

b

PvB2

Lhw

vB2

L

b

hw

Fnet=(P-Pb)hw, v=Fnet (0.4 pts) 2

v=(P-Pb)hw=

(P

vB2

L

)

v0vv0BLP

v0

P

v2

v(10BL)

vP

0

v

B2

1v

v0

L

01

Pv

vP0

P

v2

0BL

z

yx

(0.4 pts)

(0.6 pts)

(0.8 pts)

(0.6 pts) (0.6 pts)

(0.6 pts)

Part 3b

From conservation of energy:

Power

Vv20

B2

whL

HI

or,

to recover

v0thepump must supply an additional pressure

P=Pb

2

2

Power

Phwv

PwhL

0

bhwv

v0B0

Part 3c

c

c1.

u

cvvu'

nn

n

1

c

v

1

vnc2cn

For small v (v<v2 in the expansion of

(1

v)

1

c2cn

u'(cv)1(cv)(1vn

)cv(111

vn

cn

n

)

n2

cn

u

u'u

v(1

1)

n22fT, T

L,TuLLv(n2

1)

uu2

c2

v=v0so that,

2f

Ln

21)v

c2(02.

2fL(n

2

c2

1)v0

a phase of π/36 results in

c2v0

72L(n

2

1)f9x10

16

v14

0

m/swhich is not physical. 72x10

1

3.2x10

x(2.56

1)x25

(1.0 pts)

(0.5 pts)

(0.5 pts)

(0.5 pts)

(0.5 pts)

(0.4 pts) (0.2 pts)

(0.4 pts)

3.

For v=20 m/s, f4x10 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 3x10

14

m. Signal reaches to the end but v6x10m/s, is still nonphysical. Therefore frequency should be readjusted. (0.6 pts) For v=20 m/s electromagnetic wave of f8x10Hz has a skin depth ofabout 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 14

5

14

to the end of the section is about 17% of the incident amplitude). Therefore mercury should be replaced with water and frequency should

be adjusted to f8x1014

Hz. The correct choice is (iii)

(0.6 pts) (0.2 pts)