CollegePhysics StudentSolutionsManual...

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College Physics Student Solutions Manual Chapter 28 201 CHAPTER 28: SPECIAL RELATIVITY 28.2 SIMULTANEITY AND TIME DILATION 1. (a) What is γ if c v 250 . 0 = ? (b) If c v 500 . 0 = ? Solution (a) Using the definition of γ , where c v 250 . 0 = : 0328 . 1 ) 250 . 0 ( 1 1 2 / 1 2 2 2 / 1 2 2 = = = c c c v γ (b) Again using the definition of γ , now where c v 500 . 0 = : 15 . 1 1547 . 1 ) 500 . 0 ( 1 2 / 1 2 2 = = = c c γ Note that γ is unitless, and the results are reported to three digits to show the difference from 1 in each case. 6. A neutron lives 900 s when at rest relative to an observer. How fast is the neutron moving relative to an observer who measures its life span to be 2065 s? Solution Using 0 t t Δ = Δ γ , where 2 / 1 2 2 1 = c v γ , we see that: . 1 2944 . 2 s 900 s 2065 2 / 1 2 2 0 = = = Δ Δ = c v t t γ Squaring the equation gives 2 2 2 1 2 2 2 2 v c c c v c = = γ Crossmultiplying gives 2 2 2 2 γ c v c = , and solving for speed finally gives:

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CHAPTER  28:  SPECIAL  RELATIVITY  28.2  SIMULTANEITY  AND  TIME  DILATION  

1.   (a)  What  is  γ  if   cv 250.0= ?  (b)  If   cv 500.0= ?  

Solution   (a)  Using  the  definition  of  γ ,  where   cv 250.0= :  

0328.1)250.0(112/1

2

22/1

2

2

=⎥⎦

⎤⎢⎣

⎡−=⎟⎟

⎞⎜⎜⎝

⎛−=

−−

cc

cv

γ  

(b)  Again  using  the  definition  of  γ ,  now  where   cv 500.0= :  

15.11547.1)500.0(12/1

2

2

==⎥⎦

⎤⎢⎣

⎡−=

cc

γ  

Note  that  γ  is  unitless,  and  the  results  are  reported  to  three  digits  to  show  the  difference  from  1  in  each  case.  

6.   A  neutron  lives  900  s  when  at  rest  relative  to  an  observer.  How  fast  is  the  neutron  moving  relative  to  an  observer  who  measures  its  life  span  to  be  2065  s?  

Solution  Using   0tt Δ=Δ γ ,  where  

2/1

2

2

1−

⎟⎟⎠

⎞⎜⎜⎝

⎛−=cv

γ ,  we  see  that:  

.12944.2s 900s 2065

2/1

2

2

0

⎟⎟⎠

⎞⎜⎜⎝

⎛−===

Δ

Δ=

cv

tt

γ  

Squaring  the  equation  gives  22

21

2

222

vcc

cvc

−=⎟⎟

⎞⎜⎜⎝

⎛ −=

γ  

Cross-­‐multiplying  gives   2

222

γcvc =− ,  and  solving  for  speed  finally  gives:  

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ccccccv 900.090003.0)2944.2(

11112/1

2

2/1

22

22 ==⎥

⎤⎢⎣

⎡−=⎟⎟

⎞⎜⎜⎝

⎛−=−=γγ

 

11.   Unreasonable   Results   (a)   Find   the   value   of  γ   for   the   following   situation.   An   Earth-­‐bound   observer  measures   23.9   h   to   have   passed  while   signals   from   a   high-­‐velocity  space   probe   indicate   that   h 0.24 have   passed   on   board.   (b)   What   is   unreasonable  about  this  result?  (c)  Which  assumptions  are  unreasonable  or  inconsistent?  

Solution   (a)  Using  the  equation   ,0tt Δ=Δ γ we  can  solve  for  γ :    

996.09958.0h 24.0h 23.9

0

===Δ

Δ=tt

γ  

(b)  γ  cannot  be  less  than  1.  

(c)  The  earthbound  observer  must  measure  a  longer  time  in  the  observer's  rest  frame  than  the  ship  bound  observer.  The  assumption  that  time  is  longer  in  the  moving  ship  is  unreasonable.  

28.3  LENGTH  CONTRACTION  

14.   (a)  How  far  does  the  muon  in  Example  28.1  travel  according  to  the  Earth-­‐bound  observer?  (b)  How  far  does  it  travel  as  viewed  by  an  observer  moving  with  it?  Base  your  calculation  on  its  velocity  relative  to  the  Earth  and  the  time  it  lives  (proper  time).  (c)  Verify  that  these  two  distances  are  related  through  length  contraction   3.20=γ .  

Solution   Using  the  values  given  in  Example  28.1:  

(a)  

km 1.39m 101.386s) 1087.4)(m/s 10998.2)(950.0(

s) 1087.4)(950.0(368

60

=×=××=

×=Δ=−

cctvL

 

(b)   km 0.433m 104.329)s 1052.1)(m/s 10998.2)(950.0( 2680 =×=××=Δ= −ctvL  

(c)   km 0.433m 104.333.20

m 101.386 23

0 =×=×

==γL

L  

 

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28.4  RELATIVISTIC  ADDITION  OF  VELOCITIES  

22.   If  a  spaceship  is  approaching  the  Earth  at   c100.0  and  a  message  capsule  is  sent  toward  it  at   c100.0  relative  to  the  Earth,  what  is  the  speed  of  the  capsule  relative  to  the  ship?  

Solution   Using  the  equation  )/(1 2cuv

uvuʹ′+

ʹ′+= ,  we  can  add  the  relativistic  velocities:  

[ ] cccc

cccuvuvu 198.0

/)100.0)(100.0(1100.0100.0

)/(1 22 =+

+=

ʹ′+

ʹ′+=  

28.   When  a  missile  is  shot  from  one  spaceship  towards  another,  it  leaves  the  first  at  c950.0  and  approaches  the  other  at   c750.0 .  What  is  the  relative  velocity  of  the  two  

ships?  

Solution   We  are  given:   cu 750.0=  and   cu 950.0=ʹ′ .  We  want  to  find   v ,  starting  with  the  

equation  )/(1 2cuv

uvuʹ′+

ʹ′+= .  First  multiply  both  sides  by  the  denominator:  

uvcuuvu ʹ′+=ʹ′

+ 2  ,  then  solving  for   v  gives:  

[ ] cccccc

cuuuuv 696.0

1/)950.0)(750.0(750.0950.0

1)/( 22 −=−

−=

−ʹ′−ʹ′

=  

The  velocity  v  is  the  speed  measured  by  the  second  spaceship,  so  the  minus  sign  indicates  the  ships  are  moving  apart  from  each  other  (v  is  in  the  opposite  direction  as  u ).  

34.   (a)  All  but  the  closest  galaxies  are  receding  from  our  own  Milky  Way  Galaxy.  If  a  galaxy  ly 100.12 9×  away  is  receding  from  us  at   c900.0 ,  at  what  velocity  relative  to  us  must  we  

send  an  exploratory  probe  to  approach  the  other  galaxy  at   c990.0 ,  as  measured  from  that  galaxy?  (b)  How  long  will  it  take  the  probe  to  reach  the  other  galaxy  as  measured  from  the  Earth?  You  may  assume  that  the  velocity  of  the  other  galaxy  remains  constant.  (c)  How  long  will  it  then  take  for  a  radio  signal  to  be  beamed  back?  (All  of  this  is  possible  in  principle,  but  not  practical.)  

Solution   Note  that  all  answers  to  this  problem  are  reported  to  5  significant  figures,  to  distinguish  the  results.  

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(a)  We  are  given   cv 900.0−=  and   cu 990.0= .  Starting  with  the  equation  

)/(1 2cuvuvuʹ′+

ʹ′+= ,  we  now  want  to  solve  for  uʹ′ .  First  multiply  both  sides  by  the  

denominator:   uvcuvuu ʹ′+=ʹ′+ 2 ;  then  solving  for  the  probe’s  speed  gives:  

cccc

cccuvvuu 99947.0

]/)900.0)(990.0[(1)900.0(990.0

)/(1 22 =−−

−−=

−=ʹ′  

(b)  When  the  probe  reaches  the  other  galaxy,  it  will  have  traveled  a  distance  (as  seen  on  Earth)  of   0 vtxd += because  the  galaxy  is  moving  away  from  us.  As  seen  from  

Earth,   cu 9995.0=ʹ′ ,  so   .0

uvtx

udt

ʹ′+

=ʹ′

=    

Now,   vtxtu +=ʹ′ 0  and   y 101.2064ly 1y) (1

0.9000.99947ly 1012.0 11

90 ×=×

−×

=−ʹ′

=c

ccvuxt  

(c)  The  radio  signal  travels  at  the  speed  of  light,  so  the  return  time   tʹ′  is  given  by  

,0

cvtx

cdt

+==ʹ′  assuming  the  signal  is  transmitted  as  soon  as  the  probe  reaches  

the  other  galaxy.  Using  the  numbers,  we  can  determine  the  time:  

y 101.2058y) 10.2064(0.900c)(1ly 101.20 111110

×=×+×

=ʹ′c

t  

28.5  RELATIVISTIC  MOMENTUM  

39.   What  is  the  velocity  of  an  electron  that  has  a  momentum  of   m/skg 1004.3 21 ⋅× − ?  Note  that  you  must  calculate  the  velocity  to  at  least  four  digits  to  see  the  difference  from  c .  

Solution  Beginning  with  the  equation  

( )[ ] 21221 cu

mumup−

== γ  we  can  solve  for  the  speed  u .  

First  cross-­‐multiply  and  square  both  sides,  giving   22

2

2

2

1 upm

cu

=−  

Then,  solving  for   2u gives   ( ) ( ) ( )222

2

2222

//1/1

cpmp

cpmu

+=

+=  

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Finally,  taking  the  square  root  gives  ( )222 / cpmpu

+= .  

Taking  the  values  for  the  mass  of  the  electron  and  the  speed  of  light  to  five  significant  figures  gives:  

( ) ( ) ( )[ ]m/s 102.988

kg 102.9979/m/skg 103.34kg 109.1094

m/skg 103.04

8

1/222821231

21

×=

⎭⎬⎫

⎩⎨⎧ ×⋅×+×

⋅×=

−−

u  

28.6  RELATIVISTIC  ENERGY  

48.   (a)  Using  data  from  Table  7.1,  calculate  the  mass  converted  to  energy  by  the  fission  of  1.00  kg  of  uranium.  (b)  What  is  the  ratio  of  mass  destroyed  to  the  original  mass,  

mm /Δ ?  

Solution   (a)  From  Table  7.1,  the  energy  released  from  the  nuclear  fission  of  1.00  kg  of  uranium  is   J 108.0 13×=ΔE .  So,   2

released0 mcEE Δ==Δ ,  we  get  

( )( )

g .890kg 108.89m/s 103.00

J 100.8 428

13

2 =×=×

×=

Δ=Δ⇒ −

cEm  

(b)  To  calculate  the  ratio,  simply  divide  by  the  original  mass:  

( )( )

10.98 108.89kg 1.00

kg 108.89 442

4−−

×=×=×

mm

 

52.   A  π -­‐meson  is  a  particle  that  decays  into  a  muon  and  a  massless  particle.  The  π -­‐meson  has  a  rest  mass  energy  of  139.6  MeV,  and  the  muon  has  a  rest  mass  energy  of  105.7  MeV.  Suppose  the  π -­‐meson  is  at  rest  and  all  of  the  missing  mass  goes  into  the  muon’s  kinetic  energy.  How  fast  will  the  muon  move?  

Solution   Using  the  equation   2relKE mcΔ= ,  we  can  determine  the  kinetic  energy  of  the  muon  

by  determining  the  missing  mass:   222rel )1()(KE cmcmmmc µµπ γ −=−=Δ=  

Solving  for  γ  will  give  us  a  way  of  calculating  the  speed  of  the  muon.  From  the  

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equation  above,  we  see  that:  

.32072.1MeV 105.7MeV 139.61 ===

−−=+

−=

µ

π

µ

µµπ

µ

µπγmm

mmmm

mmm

 

Now,  use   ( )[ ] 21221 −

−= cvγ  or   ,11 22

2

γ=−

cv

 so  

cccv 0.6532(1.32072)

1111 22 =−=−=γ

 

58.   Find  the  kinetic  energy  in  MeV  of  a  neutron  with  a  measured  life  span  of  2065  s,  given  its  rest  energy  is  939.6  MeV,  and  rest  life  span  is  900s.  

Solution   From  Exercise  28.6,  we  know  that  γ =  2.2944,  so  we  can  determine  the  kinetic  energy  of  the  neutron:   ( )( ) .MeV 1216MeV 939.612944.2)1(KE 2

rel =−=−= mcγ  

64.   (a)  Calculate  the  energy  released  by  the  destruction  of  1.00  kg  of  mass.  (b)  How  many  kilograms  could  be  lifted  to  a  10.0  km  height  by  this  amount  of  energy?  

Solution   (a)  Using  the  equation   2mcE = ,  we  can  calculate  the  rest  mass  energy  of  a  1.00  kg  mass.  This  rest  mass  energy  is  the  energy  released  by  the  destruction  of  that  amount  of  mass:

( )( ) J 108.99J 108.988m/s 102.998kg 1.00 161682released ×=×=×== mcE    

(b)  Using  the  equation   mgh=PE ,  we  can  determine  how  much  mass  can  be  raised  to  

a  height  of  10.0  km:   ( )( ) kg 109.17m 1010.0m/s 9.80

J 108.99PE 1132

16

×=×

×==

ghm