Rutgers Physics 382 Mechanics II (Spring’19/Gershtein)

Final Exam – May 10, 2019

This is a closed book/notes exam. A one-sided 8.5x11 sheet with only formulae is allowed. Please attach the sheet to your solutions. Calculators are not needed. Exam duration – 3 hours. 1. (5 pts) Spaceship A moves with speed 0.2c (as viewed from some inertial frame) and is chased by spaceship B with speed of 0.5c in the same frame.

a) what is the speed of spaceship B as observed from spaceship A? b) what is the speed of spaceship A as observed from spaceship B?

2. (5 pts) Differential cross section of a scattering process is given by !"!! = !

! ∙ (cos!)!. Calculate the total cross section. 3. (5 pts) An infinitely long string is tied to the wall at x=0 and occupies negative x values. It is kept under tension of T and has linear density µ. At x=-∞ the string is vibrated, inducing the wave propagating along the x axis with amplitude A.

a) Determine the amplitude, frequency, and phase of the reflected from the wall wave.

b) Determine the amplitude, frequency, and the wave number of the standing wave resulting from the interference of the incident and reflected waves

4. (7 pts) A rubber cube (with known BM and SM) is placed into a trough running along the x-axis. The trough dimensions are such that the cube snugly fits (see figure). Pressure of σΟ is applied uniformly to the top of the cube. Assuming there is no friction between the cube and the trough, the stress and strain tensors should be diagonal, i.e.

Σ =!! 0 00 !! 00 0 −!!

and E =!! 0 00 !! 00 0 !!

.

a) Find the five unknowns: !! , !!, !!, !!, and !!

(Recall the generalized Hooke’s law !E = 9 ⋅BM ⋅

!Σ− (3⋅BM − 2 ⋅SM ) ⋅ tr

!Σ⋅!1

18 ⋅BM ⋅SM ).

(HINT: two out of the five are easily reasoned to be zeroes, start with figuring out which ones)

b) What is the change in volume of the cube (ΔV/V)?

± . .

/

5. (6 pts) A beam of light has frequency !! and is travelling along the x-axis in some inertial system. It strikes a mirror that also moves along the x-axis with speed !. Find the frequency of the reflected light 6. (6 pts) Mass !! can move without friction on the table. It is attached by a light string to a mass !! as shown in the figure. a) write the Hamiltonian of the system and the Hamilton equations of motion (but do not solve them) b) find the solution to the Hamilton equation of motion where M2 is at rest 7. (6 pts) A particle of known mass m has energy E0 in the lab frame. It decays into two photons. The opening angle between the photons in the lab frame is Θ. What are the energies of the two photons?

MIRROR

mini ¥.I

.imr

7'

w,

=o

!!

!!"

m, E0

γ1

γ2

Θ

② B o .se A 0.2C④ : I > → I →

B A at resta) ④ i I > Is I ④in ,owith speed are

0.5 - 0.2 d

OB ' = C = oc= Iz

b) ④

Efrat

Ei moves,

seed oie

0.2 - 0.5

Yt " ' oE- -I?ge= - Is

② drFor

= A? cosy

Foot . f A'

wig dr = A' f? coiydy § nine do =

= A? If . 2 = Lo AZ

③ %Incoming wave :µ

F f , = Has ( Kx - wt ) K - - E. wifereflected wave fr - - B cos

Ckxtwtt8)

so full wave

f- = A cos Ckx - wt ) + Bus ( Kx-i wt t S )Boundary condition : f ( o , t ) = O

Acn ( - wt ) t B cos ( wt -1

s ) - - O

a) A = B,

S - - r

Reflected wave fr = A cos ( kxtwt + is )same amplitude , frequency , phase = 5

③b) Full wave

f- = Has ( Kx - wt ) t Acosckxtwttr ) =

= A ( cos @x - wt ) - cos ( kxxwt ) ) =

= A ( cowswt tniukxiiuwt - eoktiiukxiiuwt )

f = LA sinkx . sin wt

twice the amplitudeSame frequency and wave number .

④pi

" '• ⇐ .

E-I

Foot!!I it¥

.it#..e:.:.:d/Ey - - O → can not expand ! Tx = O → no pressure along xE=÷sm . ( 9 BME

- Gbm - zoom ) Cry - r . ) . I )Ex - - ipfzmsm( 3 BM - ISM ) fry - r . )

O = igbmgm ( 9 BM . Ty - 13 BM - Ism ) @y - ro )).

I

0 = igbmgm ( 9 BM . rg - 13 BM - Ism )@ y- ro ))

{Ez = ÷j ( g BM ( - r . ) - ( 3 BM

- 25M )@y

- ro ))

D= Ty ( 9 BM - 3 BM -125M ) t To ( 3 BM- 25M )

Ty = - of .3BM-2sM_ .6 BM t ISM

Ty - To = - ro (3134-252 GBM

6 But 25M+ ") = - To Fsm

3 BM - 25M

E×

= jam ( 313M - 25M ) . fro ) . 9613172sci-Fi Ism

3 BM - 25M

Ex = I . -25M 6 BM -125M

Ez - - Ein ( 9 BM - GBM -zsmj.fm?z-sm)=3BMt4SM= - Fm ( I - 337.7%1 : - Esm Fsn

Ez = - Ijm . 3BMt4SM6 BM t 2 SM

tr ¥ = ÷m 6¥ m ( 3dm- 2 SM -3/BM - 4S m) = -ro.GS/2S/M.(6BMt2sy

✓ = asTo . FBI6 BM +2 S M

⑤

④ : mins'

imw

,= ?

Boost into the frame of the mirror :

④ :untieminora

w't

4- rector of incident tphekeu :

in K : ( WE , WE , 0,0 ) Lorentztranfovnatren :

ink : ( WE , WE,

o,

o ) Ie =rtEtpE )-

- II. rate )

WE , WE'

zits

4- rector of reflected photon:

ink : ( Ee , - WI,

o,

o )Lorentz tranfanabren :

ink :(I , - E, o , o ) Is rtE-ipf-EH.IM. p

I - - WE t.fpr.ci - H

W , = WotfHP

⑥ it me

;→••L deg . of freedom : Z , y

cheese z = o so that if•Mz 2=0 then ✓ =D .

2- 4 Iif ) U= mzgz= Tz = 'zmzI2 IF, = Im , ( E 'tI - - I fuz-im.IE 't

'

zu ,Z' if

'- mzgz

pz=f¥t= Chinna)E Eh = PI( Mithra)Z

Pg-

- f÷= mizzi go ! PIhe ,2z4

Tl - - zYuI -121¥ + mga

a).

fe=zY÷⇒+aPm÷ + megaZ :

ftp..az . . PI F-ftp..it .Mit Mzon OH

of= - P'z== Mag - mPg, Ty

= - P'y = d .

b) pi - - o ⇒ Put,

-

- mzg ⇒ mI¥ = mzgm

, zig ! mzg

if - - Eft Fz

⑦Pe = I t Pz

M'

= L Er,

Era ( I - cost )

Er,

+ Era = Eo,

so

2

ur

Er,

( E . - Er , ) = -2 ( I - an O )

m2

Er ? - Eo Er , + Eog =D

← got . egu for Er ,

Er,

-

- Ef ± FEET-424 - ero )