rmmp vianu 2013 problema 1 - takmicenja.ipb.ac.rs

RMMP – Vianu 2013 – Problema 1

©Delia DAVIDESCU, Adrian DAFINEI 1

Task 1. Kinetic energy of the car

a. Solution

Figura 1 Calculul momentului de inerţie

The density of wheel material (if the length of wheel is L ) is e

La

m

24

( 1)

Considering the elementary prism having the sides dydxL ,, , the elementary with

respect of wheel axis is

dydxLrdJ2

( 2)

with

222yxr ( 3)

Because

3

223

3

2

3

222 axa

yyxdyyx

a

a

a

a

( 4)

results

3

8

3

4

3

4

32

32

322

444333

2 aaax

axadx

axa

a

a

a

a

( 5)

The moment of inertia of a wheel is

3

2

3

8

444

24

2

22

2

22

2

maa

a

mdxdyyx

a

mdxdyyxL

La

mJ

a

a

a

a

a

a

a

a

( 6)

b. Solution

When the wheel is “on top” or “on valley” the translation speed of axles of wheel (that

is the translation speed of the car) tv - and the radius of gyration r tr are

perpendicular so that

ttrtv ( 7)

where t is the angular velocity of the wheel.



When the motion starts - in the top of a bump, the wheel having the radius of gyration

a , the instantaneous angular speed of the wheel, T

, will be

a

v

T

0 ( 8)

Translational kinetical energy tT is

2

2

2v

mMTt

( 9)

and rotational kinetical energy rT in the same point is

2

2

22

3

2

22

maJ

JTr

. ( 10)

The total kinetical energyT of the car is

2

2

2

3

2

2

2

mav

mMTTTrt

( 11)

At the beginning 0vv and

a

v

T

0

6

103

3

2

2

2

0

2

0

mMv

mm

MvT

( 12)

In the valley, the translation speed vv and angular speed satisfy

2;2

ava

v

Vv

v

V ( 13)

In the valley, the total kinetical energy is

3

83

3

22

3

2

2

22 22222

22mM

aam

mMmaamM

Tvvv

( 14)

The potential energy of the car do not change (because the center of mass remains

at the same height as stated in statement). Because no external action is allowed,

the total kinetical energy is conserved that means (using (12) and (14))

6

103

3

83 2

0

22 mMva

mM

v ( 15)

and

mM

mM

a

v

v

166

1030

( 16)

Analyzing .the relation above one may see that

i. If mM , from (16) results

0

0;

2vv

a

v

vv ( 17)

The translation speed remains constant if the wheels are unimportant

ii. If mM , from (16) results



8

50

a

v

v ( 18)

iii. If mM , from (16) results

22

130

a

v

v ( 19)

Task 2. Road's shape

a. If the weight would have a moment relative to the contact point, it would appear a

transfer of energy from potential to kinetical or vice versa that being against

statement hypothesis.

b. In the image in the figure 2 is presented the motion of contact side of the wheel (in

cross section) over the road’s shape – the curve passing through dsxTx ,, .

Figure 2. Cross section view of the motion of the wheel on an elementary bump of the road. The vertical passing through the center of mass of the wheel passes also

through the contact point. The wheel never glides.

Using the hypothesis

xytg ' ( 20)

Because

2

1

1cos

tg

( 21)

results that

2'1

1cos

y

( 22)

so that

2'1

cos'' ya

aTG

( 23)



Co linearity of 'G of 'T is

2'12

ayay ( 24)

that is

12'

2

a

yy ( 25)

and

dx

a

y

dy

12

2

( 26)

Integrating (26) one obtain that

1222

ayaylna

x ( 27)

Because for

0x , 12 ay , ( 28)

The constant in integral is,

0 ( 29)

and because

12 ay ( 30)

expression (27) can be written as

a

x

eayay

1222

( 31)

1222

2

ayaye a

x

( 32)

a

yeeay

e

e a

x

a

x

a

x

a

x

22

;221

222

( 33)

or

axchay 2 ( 34)

The graph of function in (34) is presented in figure (3).

Figure 3 Elementary relief of the road

c.



from (31) for 0y results

12ln ax ( 35)

The ends of an elementary relief are

12ln

12ln

ax

ax

d

s

( 36)

So that the horizontal extension of the bump d2 , is

12ln22 axxdsd

( 37)

d.

Because

aaa 2276,112ln2 ( 38)

The two wheels cannot rotate if they are over two neighbor bumps. The minimal

distance between axles min

d must be

12ln4min

ad ( 39)

e.

Using the expression of road’s shape results that,

axshy ' ( 40)

Consequently

1'

1'

s

d

xy

xy ( 41)

The angle between the tangents to relief in valley , 2

, is exactly appropriate for a

square wheel. No hexagonal prism can fit the road.

Task3. Accident

The speed of center of mass obey to relation

dt

dxGG

dt

dv

G '

' ( 42)

During the motion the gyration radius that initially is in position GC reach the position

''CG rotating with the angle

2

( 43)

Because, as stated in statement

xyarctg ' ( 44)

The angular speed of the wheel is

'' yarctgdt

dyarctg

dt

d

dt

d

dt

d

( 45)



dt

dx

aaxsh

axchaxsharctg

dt

d

1

12

( 46)

dt

dx

axcha

1 ( 47)

When the wheel evolves from top to valley

12ln0 adx ( 48)

And correspondingly

21 axch ( 49)

Between the translational and angular speeds it is the relationship

'

1

Gv

axcha

( 50)

During the motion the potential energy remains constant..

The translational kinetical energy -are (9) 222

'GtvmMT and the rotational

kinetical energy are (10) - 2 JT

r a. So,

22

'

2

2

Jv

mMTTT

Grt ( 51)

2

'

2

2

'

1

3

2

2

2

GGv

axcha

mav

mMT ( 52)

axchm

mMvTG 2

2

'

1

3

2

2

2 ( 53)

At the beginning 0x

0'

0 vvG

; 1)/( axch ( 54)

3

5

2

2

0

mMvT ( 55)

that is in agreement with (12)

Generally, the dependence of speed on position has the

axchm

mM

mMv

vG

2

0

'

1

3

2

2

2

3

5

2

( 56)

When moving from top to valley, 12ln0 ax and correspondingly

2cosh1 a

x.

In considered domain a

xch monotonically increases and his extremes values

appears at the ends.



1

1

2

1

2

axch ( 57)

The extremes values are initial speed 0v and the speed in valley

0

0

'

83

103v

mM

mMvdv

G

( 58)

Relation (58) is the answer for question d.

d. The heat delivered by collision DO NOT depends on the position of the collision.

Whatever the point of collision will be the constant total kinetical energy is totally

transformed in heat..

e. The heat produced Q equals the total kinetical energy

6

1032

0

mMvQ ( 59)

The relation (59) is the answer to the question e.

Romanian Master of Physics 2013

Theoretical Problem No. 2 - Solution Page 1 of 7

Compton scattering - Solution

Task 1 - First collision 1.a. Figure 1 presents the situation before the first scattering of the photon.

Figure 1 Photon and electron before the first collision. The collision occurs in the point M

Figure 2. Photon and electron after the first collision. The collision took place in point M

1.b. The angle between the directions of the electron and photon moments before collision is denoted by

To characterize the initial photon we will use its moment i

p

and its energy i

E

ii

i

i

i

fhE

c

fhhp

( 1)

i

i

cf

( 2)

is the frequency of the initial photon.

For the initial free electron in motion scenario, the moment e

p1

and the energy

oeE are

2

2

02

1

2

10

11

1

1

cmcmE

vmvmp

e

e

ee

( 3)

where 0

m is the rest mass of the electron and m is the mass of the moving electron. As

usual c

ve1

. De Broglie wavelength of the first electron is



2

101

11

ee

evm

h

p

h ( 4)

1.c. The situation after the scattering of photon is described in the figure 2.

To characterize the scattered photon we will use its moment 0

p

and its energy 0

E

oo

o

o

o

fhE

c

fhhp

( 5)

where

0

cf

o ( 6)

is the frequency of the scattered photon. The magnitude of the moment of the electron (that remains at rest) after the scattering is

zero; its energy is e

E0

. The mass of the electron after collision is 0

m - the mass of electron

at rest state. So, 2

00cmE

e ( 7)

To determine the moment of the first moving electron, one can write the principles of conservation of moments and energy. That is

01ppp

ei

( 8)

and

eeiEEEE

001 ( 9)

Using the referential in figures 1 and 2, the conservation of the moment in collision along the

Ox direction is written as

cosc

fhcosvm

c

fh

e

i 0

1

( 10)

and the conservation of moment along Oy direction is

sinc

fhsinvm

e

0

1

( 11)

To eliminate , the last two equations must be written again as

2

02

1

2

02

22

1

sinc

fhsinvm

fcosfc

hcosvm

e

ie

( 12)

and then added. The result is

cosffffc

hvm

iie

0

22

02

2

2

1

22 ( 13)

or

cosffffhv

c

v

cm

iie

e

0

22

0

22

12

1

22

02

1

( 14)

The conservation of energy (9) can be written again as

0

2

0

2fhcmfhcm

i ( 15)

or



i

e

ffhcm

c

v

cm

0

2

02

1

2

0

1

( 16)

Squaring the last relation results in the following

ii

e

ffchmffhcm

c

v

cm

0

2

0

2

0

242

02

1

42

02

1

( 17)

Subtracting (14) from (17) the result is

02220

2

0

2

0

2

0 ffhcosffhffhcm

iii ( 18)

or

00

1f

c

f

ccos

cm

h

i

( 19)

Using

cm

h

0

( 20)

the relation (19) becomes

0

1 i

cos ( 21)

The wavelength of scattered photon is

cosi

10

( 22)

Equation (22) show that the scattered photon wavelength is shorter than the wavelength of the initial photon. Consequently the energy of the scattered photon is greater that the energy of the initial photon.

i

i

EE0

0

( 23)

Task 2 - Second collision

2.a. Let’s analyze now the second collision process that occurs at point N . To study that,

let’s consider a new referential having the Ox axis along the direction of the photon

scattered after the first collision.

Figure 3 Photon and electron before the second collision. The collision occur at the point N



Figure 4 Photon and electron after the second collision. The collision took place at the point

N . The scattering angle for electron is denoted by .

Figure 3 presents the situation before the second collision and Figure 4 presents the situation after this scattering process.

2.b. The conservation law for moments in the scattering process gives

02

2

0

sinvmsinh

cosvmcoshh

e

f

e

f ( 24)

To eliminate the unknown angle , one must square and then add equations ( 24)

That is

2

2

2

2

2

2

0

sinvmsinh

cosvmcoshh

e

f

e

f

( 25)

and

2

2

0

22

0

2

2

e

ff

vmcoshhh

( 26)

or

2

2

22

0

22

2

0

22

2

222

e

ff

vcmcoschchch

( 27)

The conservation law for energies in the second scattering process gives

22

0

0

cmch

cmch

f

( 28)

or

2

2

42

0

0

0

342

0

2

0

22

1

112

11

cv

cmmchcmch

eff

( 29)

Subtracting (27) from (29), one obtains

011

222

0

0

3

0

22

0

22

fff

mchch

cosch

( 30)

that is



cos

coscm

h

f

f

1

1

0

0

0 ( 31)

Concluding

0

0

EEf

f

( 32)

Comparing (31) written as

cosf

10

( 33)

and (22) written as

cosi

10

( 34)

results in

fi ( 35)

2.c. Taking into account (26), the moment of the electron after the second collision is

0

2

0

22

211

ff

e

coshp ( 36)

Or, considering (33), the moment can be rewritten as

cos

cos

cos

hp

ffff

e

1

2

1

11

222 ( 37)

The expression of the de Broglie wavelength of the second electron after scattering is

cos

cos

cos ffff

e1

2

1

111

222 ( 38)

From (28) results

2

0

Tchch

f

( 39)

that is

f

chT

11

0

2 ( 40)

Task 3 - Quantitative description of processes

3.a. The relation (13) can be written as

0

2

0

21

211

ii

e

coshp ( 41)

or, considering (34) and (35)

cos

cos

cos

hp

ffff

e

1

2

1

11

221 ( 42)

so that the expression of the de Broglie wavelength of the first electron before scattering is

cos

cos

cos ffff

e1

2

1

111

221 ( 43)

In the condition of the problem (43) becomes



2

1

2

111

221

fff

f

e

( 44)

Because the value of f

and are known as

m,m,m,,

,

m,if

1012

831

34

10

100201041210031019

1066

10251

( 45)

From (33) results that

2

10

ffcos ( 46)

or

m,10

010241

( 47)

The numerical value of the de Broglie wavelength of the first electron before scattering is

m,m,,,,

e

10

22

10

110251

241251

1

241

1

251

110

( 48)

Evidently the de Broglie wavelength e2

for the second electron after the scattering is the

same

ee 21 ( 49)

3.b. Because if

the energy of the first photon before the first scattering has the

following expression

fi

i

chchE

( 50)

Its numerical value is

KeV,J,,

,,chE

i

i9910581

10251

10031066 15

10

834

( 51)

The frequency i

f of the first photon before the first scattering has the following expression

fi

i

ccf

( 52)


Hz,,

fi

18

10

8

104210251

103

( 53)



3.c. From (40) written as

f

chcmcmTT

11

0

2

0

2

12 ( 54)

results

2

0

02002

11

2

1

2

1

ffffcm

hm

c

hm

c

hmm

( 55)


0mm ( 56)

Consequently, the speed of the second electron after the collision e

v2

has the following

expression

0

2m

h

m

hv

ff

e

( 57)


c,s/m,ve

019010856

2 ( 58)

3.d. As results from (46)

200

fi ( 59)

The numerical value of the variation of wavelength is

m,10

100102

( 60)

© Solution proposed by

Delia DAVIDESCU, PhD Adrian DAFINEI, PhD

Star Trek Watching

©DD,AD 1

Task 1. The link between image and the actual position

The image appears in the camera because of rays of light reaching in the same time and no because of light rays leaving simultaneously. For this reason, there is difference between the image of a point on the line of beacons and the current position of this point at the time the image is formed. A beacon which image is in the

position denoted ix if the light coming from him has left him before the moment when

the image is taken with the time T

c

xdT

i

22

( 1)

Figure 0.1

In the time it takes light to reach the camera, a distance is traveled by the beacon so that "current position" of the beacon (when the image is taken) is

2222

iiiiixdxxd

c

vxTvxx ( 2)*

The beacon situated at ix in the image, appears on image when its actually position

is x . (Relation (2) represents the answer to the question 1.a)

Squaring (2) results

02

0121

021

2

2

2

22

2

2

222222

22222

222222

ddxxx

x

ddxxxx

dxxxx

xdxxxx

i

i

ii

ii

iii

( 3)

Correspondingly,

22222

222

2

22

11

1

xdxx

dxxxx

i

i

( 4)

Star Trek Watching

©DD,AD 2

Because

22

2

2

2

2

11

111

( 5)

from the last relation in (5) results

222

2222

xdxx

xdxx

i

i

( 6)

Because the beacons approaching at origin of the axis, natural selection of sign în expression above is

222xdxx

i ( 7) *

Beacons whose image appears in the camera image in the position ix is at position

x when the image is formed (Relation (7) represents the answer to the question 1.b)

Task 2. Apparent length of the beacons line

Because the beacons are moving with the speed v , the length of beacons line is

(because of Lorentz contraction) -. L .

In the moment when the camera takes the image of beacons line having the third beacon in the position

0x , the beacons situate at the ends of line have respectively

the positions

20

Lxxx

fa ţa

( 8)

and

20

Lxxx

spate

( 9)

The images of the ends of beacons line ,i

x will be formed by the camera „as is” thei

are in the positions

2

0

2

0,

22

Lxd

Lxx

i ( 10)

In the relation above was taken into consideration the expressions of positions of ends of beacons line and the relation (8).

The apparent length of beacons lineiL , represents the difference of apparent

positions of the ends of beacons line that is

,,0 iii

xxxL ( 11)

The answer to the question (2.a) is (12)

2

0

2

2

0

2

0

22

Lxd

LxdLxL

i ( 12)*

Star Trek Watching

©DD,AD 3

The expression below is a second degree dependence on 0

x ,

2

2

01

2d

LxF

( 13)

And has a minimum, 2d , for

20

Lx ( 14)

For 2

,00

Lxx the

1F monotonically increases.

Correspondingly, the expression

11FE ( 15)

Monotonically increases to 2

0

Lx , and then monotonically deceases and has a

maximum

dL

xEE

2

01max,1 ( 16)

For

20

Lx ( 17)

22

01

2Ld

LxE

( 18)

Graphically, the dependence 01

xE looks like in Figure .3

Figure 2 Reprezentarea grafică a funcţiilor 1E ,

2E şi

21EE

Analogous, as function on

0x the polynomial expression

Star Trek Watching

©DD,AD 4

2

2

02

2d

LxF

( 19)

has as minimum 2d for

20

Lx ( 20)

If 2

,00

Lxx

2F monotonically increase.

Correspondingly

22FE ( 21)

monotonically decrease to 2

0

Lx , and then monotonically increase having a

minimum of

dL

xEEim

2

02min,2 ( 22)

For

20

Lx ( 23)

22

02

2Ld

LxE

( 24)

Graphically, the dependence 02

xE looks like in the Figure 3

Expression

Lx

dL

xdL

x

E

dL

xdL

x

dL

xdL

x

FF

FFE

FFEEE

0

2

2

0

2

2

0

2

2

0

2

2

0

2

2

0

2

2

0

12

12

1221

22

2

22

22

( 25)

Monotonically decreases as is shown in the Figure 3

The last relation in (25) can be written as

0

0

2

0

2

0

2

0

2

0

2

22

2

x

x

x

d

x

L

x

d

x

L

LE

( 26)

Star Trek Watching

©DD,AD 5

For negative values of 0

x , expression becomes

2

0

2

0

2

0

2

0

2

22

2

x

d

x

L

x

d

x

L

LE

( 27)

If 0

x monotonically increase between and 0 that is 00 x , the expression in

(28) monotonically decrease

LL

E

2

2

( 28)

and

00

E ( 29)

For positive values of0

x , expression (27) is written as

2

0

2

0

2

0

2

0

2

22

2

x

d

x

L

x

d

x

L

LE

( 30)

When0

x monotonically increase between 0 and that is 0

0 x , expression (30)

monotonically decreases

00

E ( 31)

and

LE

( 32)

The study of apparent length allows writing that

ELxLi

0

( 33)

So that, the correct answer to the question 2b is The apparent length decreases all the time.

Task 3. „Symmetrical image”.

For symmetry reasons, the apparent length on the symmetric picture is the actual length of the moving rod, because the light from the two ends was emitted simultaneously to reach the pinhole at the same time, that is

LL

simetrici

, ( 34) *

This is the answer to the question 3a The apparent endpoint positions are such that

,, iixx ( 35)

That is

0,,

ii

xx

Star Trek Watching

©DD,AD 6

So that, considering (11) , the relation above becomes

2

0

2

2

0

2

0

2

,,

2220

Lxd

Lxdxxx

ii ( 36)

In conjunction with the expression of the apparent length of beacons line

2

0

2

2

0

2

0,,

22

Lxd

LxdLxLxx

L

iii

( 37)

Adding the relations above results

22

1

2

22

2

222

0

2

2

0

2

0

2

2

0

2

0

2

2

0

2

0

2

LxLxLxd

Lxd

LLx

LxdLx

L

( 38)

So that

22

0

2

0

2

LxLxd ( 39)

Subtracting (37) from (38)

22

0

2

0

2

LxLxd ( 40)

One may obtain the position of middle beacon. From (40) results

22

2

2

2

02

22

2

2

2

0

2

0

22

2

22

0

2

2

0

22

2

0

2

0

2

14

11

1

1

14

11

44

22

dL

x

dL

x

LxLx

LxL

xd

LxLxd

o

( 41)

So that

2

2

2

0

4d

Lx

( 42) *

Star Trek Watching

©DD,AD 7

The relation (42) represents the answer for 3.b.

An alternative method to determine the position of the middle beacon uses the hypothesis of symmetric ends – that is relation (37) written as

2

0

2

2

0

2

0

2

222

Lxd

Lxdx ( 43)

Squaring,

044

04114

2424

22

0

2

2

22

0

2

4

0

22

0

2222

0

224

0

2

2

0

2

0

3

0

322

0

4

dxLx

x

dxLxx

LxdxLxx

( 44)

From last relation results again (43). The position of middle beacon (8)

2

2222

2

2

0,

2

0

22

00,

22

Ldd

Ldx

xdxx

i

i

( 45)

The image of the middle of the rod on the symmetric picture is, therefore, located at

2

2

2

2

0,

22

Ld

Ldx

i ( 46)

The image of the front end of beacons line is ,i

x . The distance between the image

of the end and the image of middle beacon is

2

2

2

2

0,0,,

2222

Ld

Ld

Lx

Lxx

iii ( 47) *

(The expression above is the answer to the question 3.c)

Task 4. „Images of SS Enterprise being far away, approaching and receding.”

The relations (29) and (34) allow to write for an image taken for SS being far away and coming the expression of length as

Star Trek Watching

©DD,AD 8

1

1

1

11

,

2,

0,

LL

LLL

LLELxLL

apropierei

apropierei

iapropierei

( 48)

Analogous, for the length of SS Enterprise being far away and moving away the relations (33) şi (34) allow to write

1

1

1

11

,

2,

0,

LL

LLL

LLELxLL

eindepartari

eindepartari

ieindepartari

( 49)

Because –

1

1

1

1

,,LLLL

eindepartariapropierei ( 50)

The correct answer to the question 4a is b. The apparent length is m600 on the image of coming ship and m200 on the

image of ship receding. The ratio of length is

1

3

1

1

1

1

1

1

,

,

L

L

L

L

eindepartari

apropierei

. ( 51)

So that

2

1

331

( 52)

Consequently

2

cv ( 53)

(The relation (53) is the answer to the question4.b)

Considering the expression of the length of beacons line for the coming ship,

3211

211600

,LLLm

apropierei

( 54)

Star Trek Watching

©DD,AD 9

Results that

mmL 3463200 ( 55)

(The value above is the answer to the question 4.c)

Because for 2

1 is

3

2

4

11

1

1

1

2

( 56)

The length of symmetrical image is

mL

LL

i

i

300

2

33200

( 57)

The value above is the answer to the question 4.d.

rmmp vianu 2013 problema 1 - takmicenja.ipb.ac.rs

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