3063 Exam1 Solutions Sp06 (1)

8/9/2019 3063 Exam1 Solutions Sp06 (1)

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PHY3063 Spring 2006 R. D. Field

Exam 1 Solutions Page 1 of 7 January 31, 2006

PHY 3063 Exam 1 Solutions

Problem 1 (25 points):A rod with is at rest in the O-frame (parallel to the

x-axis) with the left end of the rod (L) at x = 0 andright end (R) at x = Rand a cart with rest length L0

= 300 m is at rest in the O'-frame (parallel to the

x'-axis) with the right end of the cart (r) at x' = 0

and left end (l) at x' = -L0. as shown in the figure.

The O'-frame is moving to the right along the

positive x-axis of the O-frame with velocity V =

0.8c and the origins of the two frames coincide at t= t' = 0. (Note that c = 310

8m/s.) Consider the

following four events:

Event Description

A Right end of cart crosses left end of rod (Lr)B Right end of cart crosses right end of rod (Rr)

C Left end of cart crosses left end of rod (Ll)

D Left end of cart crosses right end of rod (Rl)

In the O-frame event B and event C occursimultaneously (i.e. 0== BC ttt ).Part A (10 points): Plot the four events A, B, C, and D on the space-time plots below and labelthe (ct, x) coordinates of each event in the O-frame and the (ct',x') coordinates of each event in

the O'-frame. (Express the coordinates in terms of c, , , and L0).

.

O-frame

x

ct

Path of r

Path of l

A = (0,0)

(L0/,0) = C B = (L0/,L0/)D = (2L0/,L0/)

O'-frame

x'

ct'

Path of L

Path of R

(0,0) = A

B = (L0/2,0)(L0/,- L0) = C

D = (L0(2-2)/,-L0)

Solution:

O-frame: The key to plotting the events is to first plot the paths of the right (r) and left (l) end ofthe cart in the O-frame. Both r and l follow straight (parallel) lines with /1tan = . Events Aand C lie at x = 0 and events B and D lie at x = R. Also, /RctB = and )/(/ 0 LLctC == ,

where I used /0LL = . Thus, /0LR = and )/(2 0 LctD = .O'-frame: We can use the Lorentz transformations to compute the coordinates of the four events

in the O'-frame as follows:

y

x

z

y'

z'

x'

V

OO'

Cart at rest inthe O'-frame.

Rod at rest in

O-frame

L R

l r


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0)('

0)('

==

==

AAA

AAA

ctxx

xctct

00

00

)('

)('

LL

ctxx

LLxctct

CCC

CCC

=

==

=

==

0)('

)1()('

00

2

02000

=

==

==

==

LLctxx

LLLLxctct

BBB

BBB

000

200

000

2)('

)2(22

)('

LLL

ctxx

LL

LLLxctct

DDD

DDD

=

==

==

==

Both R and L follow straight (parallel) lines with /1tan = . Events A and B lie at x' = 0 and

events C and D lie at x' = -L0. Also, /' 0LctC= , )/(' 20 LctB = , and /)2()/(/' 20

2

00 == LLLctD .Part B (5 points):How long (in microseconds) does it take for the cart to cross over the rod

according to an observer in the O-frame (i.e. what is AD ttt = )?Answer: 1.5 sSolution: From part A we see that

sssm

m

V

Lttt AD

5.1105.1

)/103(8.0

)6.0)(300(22 68

0 ==

=== ,

where I used 6.0/1 = .Part C (5 points):How long (in microseconds) does it take for the rod to cross under the cart

according to an observer in the O'-frame (i.e. what is AD ttt ''' = )?Answer: 1.7 sSolution: From part A we see that

sssm

m

V

Lttt AD 7.1107.1

)/103(8.0

)36.1)(300()2(''' 6

8

20 ==

===

Part D (5 points):What is the time difference (in microseconds) between event C and event B

according to an observer in the O-frame (i.e. what is BC ttt ''' = )? Which event occurs first?

Answer: 0.8 s, B occurred firstSolution: From part A we see that

sssm

mV

LVL

VLttt Bc

8.0108.0

)/103(8.0)8.0)(300(''' 68

2

202

00 ======


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Problem 2 (25 points):Three particles with equal rest masses of1 GeV/c

2start at

the origin at t = 0 and travel in different directions but

with equal speeds V in the O-frame. In the O-frameparticle A travels to the left along the x-axis with vx = -V,

particle B travels to the right along the x-axis with vx = V,particle C travels to the up along the y-axis with vy = V.

The O'-frame is moving to the right along the x-axis alsowith speed V relative to the O-frame (the origins of the

two frames coincide at t = t' = 0 and c = 3x108

m/s). Note

that particle B is at rest in the O'-frame.

Part A (10 points): If V = 0.75c what are the following quantities as observed in the O-frame:

A = vA/c = B = vB/c = C = vC/c =KEA(in GeV) = KEB(in GeV) = KEC(in GeV) =

EA(in GeV) = EB(in GeV) = EC(in GeV) =

cpA(in GeV) = cpB(in GeV) = cpC(in GeV) =

where v is the speed (i.e. magnitude of the velocity), KE is the relativistic kinetic energy, E isthe relativistic energy, and pis the magnitude of the relativistic momentum.

Answer:

A = vA/c = B = vB/c = C = vC/c = 0.75KEA(in GeV) = KEB(in GeV) = KEC(in GeV) = 0.512 GeV

EA(in GeV) = EB(in GeV) = EC(in GeV) = 1.512 GeV

cpA(in GeV) = cpB(in GeV) = cpC(in GeV) = 1.134 GeV

Solutions: In the O-frame all the particles have the same speed v = 0.75c with = 1.512 and = 1.134 and hence for Particle A in the O-frame

A = vA/c = 0.75, GeVcmKE AA 512.0)1(2 = , GeVcmE AA 512.1

2 = ,

and GeVcmVmccp AAA 134.12 == .

For Particle B in the O-frame we get the same

B = vB/c = 0.75, GeVcmKE BB 512.0)1(2 = , GeVcmE BB 512.1

2 = , and

GeVcmcp BB 134.12 = .

Similarly for Particle B in the O-frame we get the same

C = vC/c = 0.75, GeVcmKE CC 512.0)1(2 = , GeVcmE CC 512.1

2 = , and

GeVcmcp CC 134.12 = .

Part B (15 points): If V = 0.75c what are the following quantities as observed in the O'-frame

'A = v'A/c = 'B = v'B/c = 'C = v'C/c =KE'A(in GeV) = KE'B(in GeV) = KE'C(in GeV) =E'A(in GeV) = E'B(in GeV) = E'C(in GeV) =

cp'A(in GeV) = cp'B(in GeV) = cp'C(in GeV) =

y

x

z

y

z

x

V=0.75c

O'O

A B

C


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where v' is the speed, KE' is the relativistic kinetic energy, E' is the relativistic energy, and p' isthe magnitude of the relativistic momentum.

Answer:

'A = v'A/c = 0.96 'B = v'B/c = 0 'C = v'C/c = 0.899

KE'A(in GeV) = 2.57 GeV KE'B(in GeV) = 0 KE'C(in GeV) = 1.29 GeVE'A(in GeV) = 3.57 GeV E'B(in GeV) = 1.0 GeV E'C(in GeV) = 2.29 GeVcp'A(in GeV) = 3.43 GeV cp'B(in GeV) = 0 cp'C(in GeV) = 2.06 GeV

Solutions: In the O'-frame all three particles have different speeds. Particle B has 0' =Bv andParticle A has

222 /1

2

/)(1

)()'(

cV

V

cVv

Vvv

xA

xAxA +

=

= and hence 96.01

2''

2=

+==

c

v AA and

22

22

222

22

2

2

)1(

)1(

4)1(

)1(

'1

1'

+

=+

+=

=

A

A so that 5714.3)1()1(

)1(' 22

2

2

+=+

=

A , and

42857.3'' AA , where I uses Vv xA =)( . Particle C has

VcVv

Vvv

xC

xCxC =

=

2/)(1

)()'( and

V

cVv

vv

xC

yC

yC ==

)/)(1(

)()'(

2.

Thus,222 /11)'()'(' +=+= Vvvv yCxCC so that 8992.0/11'

2 += C and

4

2

2

222

2

222

2

222

2

1)1()1()/11(1

1

'1

1'

=

=

=

+=

+=

=

V

C .

Hence, 2857.2' 2 = C , and 0553.2'' CC , where I uses 0)( =xCv and Vv yA =)( .

For Particle A in the O'-frame we have

'A = v'A/c = 0.96, GeVcmKE AAA 57.2)1'('2 = , GeVcmE AAA 57.3''

2 = , and

GeVcmcp AAAA 43.3'''2 = .

For Particle B in the O'-frame

'B = v'B/c = 0, 0' =BKE , GeVcmE BB 0.1'2 == , 0' =Bcp .

For Particle C in the O'-frame we have

'C = v'C/c 0.899, GeVcmKE CCC 286.1)1'('2 = , GeVcmE CCC 286.2''

2 = , and

GeVcmcp CCCC 055.2'''2 = .


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Problem 3 (25 points):Laboratory Frame

(before collision)

E1 p1

p2 = 0

Laboratory Frame

(after collision)

E3 p3

Particle 1 with a rest mass of M1 = 2 GeV/c2

collides with particle 2 which is at rest with a restmass of M2 = 1 GeV/c

2. The two particles fuse together to form a new particle 3 with a rest

mass of M3 = 4 GeV/c2

as shown in the figure.

Part A (15 points): What is the energy (in GeV), magnitude of the momentum (in GeV/c), and

speed = v/c ofparticle 1 in the Lab frame?Answer:

(E1)lab

= 5.5 GeV(P1)

lab= 5.123 GeV/c

(1)lab

= 0.9315

Solution: If we define 221 )~~( pps += , then in the Lab Frame

lab

lablab

lab

lab

lab

lablab

lab

pccMEcME

pccMEppcEEpps

)()(2)(

)()()()()~~(

2

1

222

21

2

2

2

1

2

1

222

21

2

21

22

21

2

21

++=

+=++=+=rr

Where I used 2212

1

22

1 )()()( cMpcE lablab = and hence

2

2

22

2

22

11

2

)()(

cM

cMcMsElab

= .

Remembers is an invariant so it is the same in every frame. It is easy to evaluate it in the center-

of-mass frame22

3

2

3

22

3

2

3

2

21 )()()()~()~~( cMpcEppps cm

CM

cmcm ===+=r

,

where I used the fact that (p3)cm = 0 and hence

GeVGeVcM

cMcMcMElab 5.5

2

)1()2()4(

2

)()()( 222

2

2

22

2

22

1

22

31 =

=

=

GeVcMEcp lablab 123.5)()()( 2212

11 ==

9315.05.5

123.5)()(

1

11 === lab

lablab

E

cp

Part B (10 points): What is the energy (in GeV), magnitude of the momentum (in GeV/c), and

speed = v/c ofparticle 3 in the Lab frame?Answer:

(E3)lab = 6.5 GeV(P3)

lab= (P1)

lab= 5.123 GeV/c

(3)lab

= 0.79

Solution: If Momentum conservation tells us that (p1)lab

= (p3)lab

and hence

GeVGeVGeVcMcpcMcpElab 5.6)4()123.5()()()()( 222232

1

22

3

2

33 =+=+=+= ,

and


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79.05.6

123.5

)(

)()(

3

33 === lab

lablab

E

cp .

Problem 4 (25 points):

An observer at rest at the origin in the O-frame watches a comet pass by him. The comet emits

monochromatic light with a proper wavelength of0 (i.e.0 is the wavelength of light emitted bythe comet as observed in the comet rest frame). The comet travels at a constant velocity of

xVv =r

measured in the O-frame and crosses the y-axis a distance of10,000 miles from theobserver (i.e. distance of closest approach). The light emitted by the comet when it was

approaching the observer and was at distance of100 million miles away appears to be violet in

color by the observer (obs = 400 nm) and the light emitted by the comet when it was at its pointof closest approach appears to be red to the observer (obs = 800 nm).Part A (6 points): What is the speed = V/c of the comet in the rest frame of the observer O?Answer: = 0.5Solution: Define the unit vector yxr sincos += with cosrx = and sinrd= , where d isthe distance of closest approach and r is the distance to the comet (see the figure). We know that

000 )/1()cos1()1( rxr +=+=+=r

.

At the point of closest approach = 90o and 0 =closestobs . When the comet was approaching andwas 100 million mile away 180o and 0)1( =towardobs and we see that

=1

closest

obs

toward

obs and hence 5.0800

40011 ===

nm

nmclosest

obs

toward

obs

.

Part B (6 points): What is the proper wavelength 0 of the light emitted by the comet (in nm)?Answer: 692.8 nm

Solution: We see from part A that

nmnmclosestobs 8.692

155.1

8000 ==

,

where I used 155.1)5.0(1/11/1 22 == .Part C (6 points): What is the wavelength of the light (in nm) observed by the observer that was

emitted by the comet when it moves away and was at distance of100 million miles away?

Answer: 1,200 nm

Solution: When the comet was receding and was 100 million mile away 0o and0)1( +=

away

obs and we see that2

0

2

0

2 )1)(1( =+=awayobstoward

obs

V

x-axis

Closest

ApproachComet

Comet Path

Observer at rest

in the O-frame

y-axis

r

d


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Hence,

nmnm

nmtoward

obs

toward

obs 200,1400

)8.692( 220 ===

.

Part D (7 points):How far away (in miles) from the observer was the comet when it emitted the

light that is observed by the observer to be at a wavelength ofobs = 517 nm?Answer: 14,142 miles

Solution: From part A 0)cos1( +=obs and henceclosest

obsobs )cos1( += .Thus,

7075.01800

517

5.0

11

1cos =

=

=

nm

nmclosest

obs

obs

and hence 135o and

milesmilesddd

r 142,14)000,10(22)135sin(sin

====o

.

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3063 Exam1 Solutions Sp06 (1)