Solution Manual for Modern Physics for Scientists and Engineers, 4th Edition

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Solution Manual for Modern Physics for Scientists and Engineers, 4th Edition gives you all the tools you need to solve your textbook problems effectively.

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Table of Contents
CHAPTER 2 – Special Theory of Relativity 1
CHAPTER 3 – The Experimental Basis of Quantum Physics 28
CHAPTER 4 – Structure of the Atom 46
CHAPTER 5 – Wave Properties of Matter and Quantum Mechanics I 62
CHAPTER 6 – Quantum Mechanics II 78
CHAPTER 7 – The Hydrogen Atom 100
CHAPTER 8 – Atomic Physics 114
CHAPTER 9 – Statistical Physics 123
CHAPTER 10 – Molecules, Laser, and Solids 142
CHAPTER 11 – Semiconductor Theory and Devices 160
CHAPTER 12 – The Atomic Nucleus 169
CHAPTER 13 – Nuclear Interactions and Applications 187
CHAPTER 14 – Particle Physics 201
CHAPTER 15 – General Relativity 212
CHAPTER 16 – Cosmology and Modern Astrophysics – The Beginning and the End 221

Chapter 2 Special Theory of Relativity 1
Chapter 2
1. For a particle Newton’s second law says2 2 2
2 2 2
ˆˆ ˆd x d y d z
F ma m i j k
dt dt dt
 
    
  .
Take the second derivative of each of the expressions in Equation (2.1):2 2 2 2 2 2
2 2 2 2 2 2
d x d x d d y d d z
dt dt dt dt dt
z
t
y
d
   
 
. Substitution into the previous equation gives2 2 2
2 2 2
ˆˆ ˆ .
d x d d
F ma m i j k
y z F
dt dt dt
        
 

2. From Equation (2.1)ˆˆ ˆdx dy dz
p m i j k
dt dt dt
 
     .
In a Galilean transformationdx dx dy dy dz dz
v
dt dt dt dt dt dt
  
    .
Substitution into Equation (2.1) givesˆˆ ˆdx dy dz
p m v i j k p
dt dt dt
           .
However, becauseˆˆ ˆdx dy dz
p m i j k
dt dt dt
         the same form is clearly retained, given
the velocity transformationdx dx v
dt dt
   .
3. Using the vector triangle shown, the speed of light coming toward the mirror is2 2
c v
and the same on the return trip. Therefore the total time is2
2 2 2
2distance
speed
t c v
   .
Notice thatsin v
c
  , so1
sin v
c
   
  
  .

2 Chapter 2 Special Theory of Relativity
4. As in Problem 3,1 2sin /v v
  , so1 1
1 2
0.350 m/s
1.25
sin ( / ) sin 1
m/ .
s 6 3v v
    
    
  and2 2 2 2
2 1 (1.25 /s) (0.35m m/s) 1.20 m/sv v v    
.
5. When the apparatus is rotated by 90°, the situation is equivalent, except that we have
effectively interchanged1 and2 . Interchanging1 and2 in Equation (2.3) leads to
Equation (2.4).
6. Let n = the number of fringes shifted; thend
n


 . Because d c t t     , we have   2
1 2
2
c t t v
n c


   
 
. Solving for v and noting that1 +2 = 22 m,    9
8
1 2
0.005 589 10 m
3.00 10 m/s 3.47 km/s.
22 m
n
v c
 

   

7. Letting2
1 1 1
  (where/v c
  ) the text equation (not currently numbered) for1t
becomes 
2
1 1
1 2
2 1 2
1 1
t c c




 
 
which is identical to2t when1 2 so0t  as required.
8. Since the Lorentz transformations depend on c (and the fact that c is the same constant
for all inertial frames), different values of c would necessarily lead two observers to
different conclusions about the order or positions of two spacetime events, in violation of
postulate 1.
9. Let an observer in K send a light signal along the + x-axis with speed c. According to the
Galilean transformations, an observer inK measures the speed of the signal to bedx dx v c v
dt dt
    
. Therefore the speed of light cannot be constant under the Galilean
transformations.
10. From the Principle of Relativity, we know the correct transformation must be of the form
(assumingy y andz z ):;x ax bt x ax bt     
.
The spherical wave front equations (2.9a) and (2.9b) give us:;( ) ( )ct ac b t ct ac b t    
.
Solve the second wave front equation fort and substitute into the first:

Chapter 2 Special Theory of Relativity 3( )( )ac b ac b t
ct c
  
   
or2 2 2 2
( )( ) .c ac b ac b a c b    
Now v is the speed of the origin of thex -axis. We can find that speed by setting0x 
which gives0 ax bt  , or/ /v x t b a  , or equivalently b = av. Substituting this into
the equation above for2
c yields 2 2 2 2 2 2 2 2
c a c a v a c v    . Solving for a:2 2
1
1 .
/v
a c


 
This expression, along with b = av, can be substituted into the original expressions for x
andx to obtain:   ;x x vt x x vt

     
which in turn can be solved for t andt to complete the transformation.
11. Whenv c we find2
1 1
  , so:2
1
x ct
x x ct x vt
 


     

;2
/ /
1
t x c
t t x c t
 


    

;2
1
x ct
x x ct x vt
 

         

;2
/ /
1
t x c
t t x c t
 

       

.
12. (a) First we convert to SI units: 95 km/h = 26.39 m/s, so 8 8
/ (26.39 m/s) / 3.00 10 m/s 8.8 10v c
 
    
(b) 8 7
/ (240m/s) / 3.00 10 m/s 8.0 10v c
 
    
(c) sound2.3 2.3 330 m/sv v   so   8 6
/ 2.3 330 m/s / 3.00 10 m/s 2.5 10v c
 
     
(d) Converting to SI units, 27,000 km/h = 7500 m/s, so 8 5
/ (7500 m/s) / 3.00 10 m/s 2.5 10v c
 
    
(e) (25 cm)/(2 ns) =8
1.25 10 m/s so 8 8
/ (1.25 10 m/s) / 3.00 10 m/s 0.42v c
     
(f)   14 22 8
1 10 m / 0.35 10 s 2.857 10 m/s 
    , so 8 8
/ (2.857 10 m/s) / 3.00 10 m/s 0.95v c
     

4 Chapter 2 Special Theory of Relativity
13. From the Lorentz transformations2
/t t v x c
        . But0t  in this case, so
solving for v we find2 /v c t x   . Inserting the values2 1 / 2t t t a c     and2 1x x x a   
, we find 2 / 2 / 2
c a c
v c
a

   . We conclude that the frameK travels
at a speed c/2 in thex -direction. Note that there is no motion in the transverse
direction.
14. Try setting 0x x v t
      . Thus0 / 2x v t a va c      . Solving for v we find2v c 
, which is impossible. There is no such frameK .
15. For the smaller values of β we use the binomial expansion  1/22 2
1 1 / 2
 


    .
(a)2 15
1 / 2 1 3.87 10
  
    
(b)2 13
1 / 2 1 3.2 10
  
    
(c)2 12
1 / 2 1 3.1 10
  
    
(d)2 10
1 / 2 1 3.1 10
  
    
(e)   
1/2 1/22 2
1 1 0.42 1.10
   
    
(f)   
1/2 1/22 2
1 1 0.95 3.20
   
    
16. There is no motion in the transverse direction, so3.5y z  m.2 2
1 1 5 / 3
1 1 0.8
 
  
     5 2m 0.8 0 10 / 3 m
3
x x vt c
           2 2 95
/ 0 0.8 2 m / 8.9 10 s
3
t t vx c c c
 
      
17. (a) 2 2 22 2 2
8
8
3 m (5 m) (10 m) 3.86 10 s
3.00 10 m/s
x y z
t c

  
   

(b) With0.8
  we find5 / 3
  . Then5y y   m,10z z   m,   8 -85 3m 2.40 10 m/s (3.86 10 s) 10.4 m
3
x x vt
                   22 8 8 85
/ 3.86 10 s 2.40 10 m/s 3 m / 3.00 10 m/s 51.0 ns
3
t t vx c
            

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Chapter 2 Special Theory of Relativity 5
(c)     
2 2 22 2 2
8
9
10.4 m 5 m 10 m 2.994 10 m/s
51.0 10 s
x y z
t 
        
  which equals c
to within rounding errors.
18. At the point of reflection the light has traveled a distance1 1L v t c t    . On the return
trip it travels2 2L v t c t    . Then the total time is1 2 2 2 2 2
2 2 /
1 /
Lc L c
t t t c v v c
      
  .
But from time dilation we know (with0proper time 2 /t L c   ) that0
2 2
2 /
1 /
L c
t t v c
     
. Comparing these two results fort we get0
2 2 2
2
2 /2 /
1 /1
L cL c
v v c
c
 
which reduces to2 2 0
0 1 / L
L L v c

   . This is Equation (2.21).
19. (a) With a contraction of 1%,2 2
0/ 0.99 1 /L L v c   . Thus2 2
1 (0.99) 0.9801.
  
Solving for , we find0.14
  or0.14 .v c
(b) The time for the trip in the Earth-based frame is6
1
8
5.00 10 m 1.19 10 s
0.14 3.00 10 m/s
d
t v

    
 
. With the relativistic factor1.01
 
(corresponding to a 1% shortening of the ship’s length), the elapsed time on the rocket
ship is 1% less than the Earth-based time, or a difference of  1 3
0.01 1.2 10 s = 1.2 10 s. 
 
20. The round-trip distance is d = 40 ly. Assume the same constant speedv c
 for the
entire round trip. In the rocket’s reference frame the distance is only2
1 .d d
   Then
in the rocket’s frame of reference2
240ly 1distance 1 .
time 40 y 40
d
v c
y
 
 
    
Rearranging2
1
v
c

   . Solving for we find0.5
  , or0.5 0.71v c c  .
To find the elapsed time t on Earth, we know40t  y, so2
1 40y 56.6 y.
1
t t
 
  

21. In the muon’s frame0 2.2 s.μT  In the lab frame the time is longer; see Equation (2.19):0T T
 
. In the lab the distance traveled is0 09.5cm vT v T c T
     , sincev c
 .

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6 Chapter 2 Special Theory of Relativity
Therefore 2
0
9.5 cm 1
cT



 , so 
 
2
9.5 cm 1
2.2μs
v
c c



  . Now all quantities are
known except β. Solving for β we find4
1.4 10
 
  or4
1.4 10v c
  .
22. Converting the speed to m/s we find 25,000 mi/h = 11,176 m/s. From tables the distance
is8 m3.84 10 . In the earth’s frame of reference the time is the distance divided by
speed, or8
3.84 10 m 34,359 s
11,176 m/s
d
t v

   . In the astronauts’ frame the time elapsed is2
/ 1t t t

   
. The time difference is2 2
1 1 1t t t t t t

         
  .
Evaluating numerically2
5
8
11,176 m/s
34,359 s 1 1 2.4 10 s.
3.00 10 m/s
t 
         
   
23.0T T
  , so we know that2 2
1
5 / 3 1 /v c
    . Solving for v we find4 / 5.v c
24.0 /L L
 so clearly2
  in this case. Thus2 2
1
2 1 /v c
  and solving for v we find3 .
2
c
v 
25. The clocks’ rates differ by a factor of2 2
1/ 1 /v c
   . Because is very small we will
use the binomial theorem approximation2
1 / 2
   . Then the time difference is 1t t t t t t
       
. Using2
1 / 2
   and the fact that the time for the trip
equals distance divided by speed,  8
2
6
2
375 m/s
3.00 10 m/s8 10 m
/ 2 375 m/s 2
t t
 
 
     8
1.67 10 s 16.7 ns.t 
   
26. (a) 2 2 4 2 4
/ 1 / 3.58 10 km 1 0.94 1.22 10 kmL L L v c
        
(b) Earth’s frame:  
7
8
3.58 10 m
/ 0.127 s
0.94 3.00 10 m/s
t L v 
  

Golf ball’s frame:2
/ 0.127 s 1 0.94 0.0433 st t
    

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Chapter 2 Special Theory of Relativity 7
27. Spacetime invariant (see Section 2.9):2 2 2 2 2 2
c t x c t x        . We know4x  km,0t 
, and5x  km. Thus   
 
2 22 2
2 10 2
22 8
5000 m 4000 m 1.0 10 s
3.00 10 m/s
x x
t c
  
    

and5
1.0 10t 
    s.
28. (a) Converting v = 120 km/h = 33.3 m/s. Now with c = 100 m/s, we have/ 0.333v c
  
and2 2
1 1 1.061
1 1 0.333
 
  
  . We conclude that the moving
person ages 6.1% slower.
(b)/ (1 m) /(1.061) 0.942 m.L L
   
29. Converting v = 300 km/h = 83.3 m/s. Now with c =100 m/s, we have/ 0.833v c
  
and2 2
1 1 1.81.
1 1 0.833
 
  
  So the length is0 / 40 /1.81 22.1L L
   m.
30. Let subscript 1 refer to firing and subscript 2 to striking the target. Therefore we can see
that1 1x  m,2 121x  m, and1 3t  ns.2 1
distance 120 m
3 ns + 3 ns + 408 ns = 411 ns.
speed 0.98
t t c
   
To find the four primed quantities we can use the Lorentz transformations with the
known values of1x ,2x ,1t , and2t . Note that with0.8v c ,2 2
1 / 5 / 3v c
    . 
 
 
 
2
1 1 1
2
2 2 2
1 1 1
2 2 2
/ 0.56 ns
/ 147 ns
0.47 m
37.3 m
t t vx c
t t vx c
x x vt
x x vt




   
  
  
 



31. Start from the formula for velocity addition, Equation (2.23a):2
1 /
x
x
x
u v
u vu c
 
  .
(a)2
0.62 0.84 1.46 0.96
1 (0.62 )(0.84 ) / 1.52
x
c c c
u c
c c c

  

(b)2
0.62 0.84 0.22 0.46
1 ( 0.62 )(0.84 ) / 0.48
x
c c c
u c
c c c
 
  
 
32. Velocity addition, Equation (2.24):2
1 /
x
x
x
u v
u vu c

   with0.8v c  and0.8 .xu c2
0.8 ( 0.8 ) 1.6 0.976
1 ( 0.8 )(0.8 ) / 1.64
x
c c c
u c
c c c
 
   
 

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8 Chapter 2 Special Theory of Relativity
33. Conversion: 110 km/h = 30.556 m/s and 140 km/h = 38.889 m/s. Let30.556xu  m/s
and38.889v   m/s. Our premise is that100c  m/s. Then by velocity addition, 
    22
30.556 m/s 38.889 m/s 62.1 m/s.
1 / 1 38.889 m/s 30.556 m/s / 100 m/s
x
x
x
u v
u vu c
 
   
  
By symmetry
each observer sees the other one traveling at the same speed.
34. From Example 2.5 we have1 /
1 /
c nv c
u n v nc
 
    . For light traveling in opposite directions1 / 1 /
1 / 1 /
c nv c nv c
u n v nc v nc
  
     
. Because/v c is very small, use the binomial expansion:     
11 / 1 / 1 / 1 / 1 / 1 / /
1 /
nv c nv c v nc nv c v nc nv c v nc
v nc
         

, where we
have dropped terms of order2 2
/ .v c Similarly1 / 1 / / .
1 /
nv c nv c v nc
v nc
   
 Thus       22
1 / / 1 / / 1 1/ 2 1 1/ .
c v
u nv c v nc nv c v nc n v n
n n
            
Evaluating
numerically we find2
1
2(5 m/s) 1 4.35 m/s.
1.33
u  
    
 
35. Clearly the speed of B is just0.60c . To find the speed of C use0.60xu c and0.60v c 
:2 2
0.60 ( 0.60 ) 0.88 .
1 / 1 ( 0.60 )(0.60 ) /
x
x
x
u v c c
u c
vu c c c c
  
   
  
36. We can ignore the 400 km, which is small compared with the Earth-to-moon distance8
3.84 10
m. The rotation rate is1 2
2 rad 100 s 2 10
 

    rad/s. Then the speed
across the moon’s surface is  2 8 11
2 10 rad/s 3.84 10 m 2.41 10 m/s.v R
      
37. Classical:54205 m 1.43 10 s
0.98
t c

   . Then 
0
1/2
ln 2
exp 14.6
t
N N t
 
  
 
or about 15 muons.
Relativistic:5
61.43 10 s
/ 2.86 10 s
5
t t



     so 
0
1/ 2
ln 2
exp 2710 muons.
t
N N t
 
  
 
Because of the exponential nature of the decay
curve, a factor of five (shorter) in time results in many more muons surviving.
38. The circumference of the fixed point’s rotational path isE2 cos(39 )R
  , whereER 
Earth’s radius = 6378 km. Thus the circumference of the path is 31,143 km. The

Loading page 10...

Chapter 2 Special Theory of Relativity 9
rotational speed of that point is 31,143 km / 24 h 1298 km/h 360. m s5 /v    . The
observatory clock runs slow by a factor of2 13
2
1 1 / 2 1 7.22 10
1




     
 . In
41.2 h the observatory clock is slow by  13 11
41.2 h 7.22 10 2.9746 10 
   h = 107 ns.
In 48.6 h it is slow by  13 11
48.6 h 7.22 10 3.5089 10 
   h = 126 ns. The Eastward-
moving clock has a ground speed of 31,143 km/41.2 h = 755.9 km/h = 210.0 m/s and thus
has a net speed of 210.0 m/s + 360.5 m/s = 570.5 m/s. For this clock2 12
2
1 1 / 2 1 1.81 10
1




     

and in 41.2 hours it runs slow by  12 11
41.2 h 1.81 10 7.4572 10 
  
h = 268 ns. The Westward-moving clock has a
ground speed of 31,143 km/48.6 h = 640.8 km/h = 178.0 m/s and thus has a net speed of
360.5 m/s − 178.0 m/s = 182.5 m/s. For this clock2 13
2
1 1 / 2 1 1.85 10
1




     

and in 48.6 hours it runs slow by  13 12
48.6 h 1.85 10 8.991 10 
  
h = 32 ns. So our prediction is that the Eastward-
moving clock is off by 107 ns – 269 ns162  ns, while the Westward-moving clock is
off by 126 ns − 32 ns = 94 ns. These results are correct for special relativity but do not
reconcile with those in the table in the text, because general relativistic effects are of the
same order of magnitude.
39. The derivations of Equations (2.31) and (2.32) in the beginning of Section 2.10 will
suffice. Mary receives signals at a ratef  for1t and a ratef  for2t . Frank receives
signals at a ratef  for1t and a ratef  for2t .
40.1 2
2L L L L L
T t t v c v c v
       ; Frank sends signals at rate f, so Mary receives2 /f T f L v
signals.1 2
2L
T t t v

    
; Mary sends signals at rate f , so Frank receives2 /fT f L v
  signals.
41.2 2 2 2 2 2
s x y z c t    ; Using the Lorentz transformation   
2 2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2 2 2 2
2 2 2 22 2
( ) ( / )
1 / 1 /
s vt y z c t v c
v c y z c t v c
y
x x
x z c
x
st




        
       

 

      

Loading page 11...

10 Chapter 2 Special Theory of Relativity
42. For a timelike interval2 0s  so2 2 2
x c t   . We will prove by contradiction. Suppose
that there is a frameK is which the two events were simultaneous, so that0t  . Then
by the spacetime invariant22 2 2 2 2 2
x c t c tx x         . But because2 2 2
x c t   ,
this implies2 0x  which is impossible becausex is real.
43. As in Problem 42, we know that for a spacelike interval2 0s  so2 2 2
x c t   . Then in
a frameK in which the two events occur in the same place,0x  and2 2 2 2 2 2 2 2
x c t x c t c t           
. But because2 2 2
x c t   we have2 2 0c t  ,
which is impossible becauset  is real.
44. In order for two events to be simultaneous inK , the two events must lie along thex
axis, or along a line parallel to thex axis. The slope of thex axis is/v c
  , so/ slope = c t
v c x

 
. Solving for v, we find2 /v c t x   . Since the slope of thex axis
must be less than one, we see thatx c t   so2 2 2 2 0s x c t     is required.
45. parts (a) and (b) To find the equation of the line
use the Lorentz transformation. With0t  we
have 2
0 /t t vx c
    or, rearranging,/ct vx c c
 
. Thus the graph of ct vs. x is a
straight line with a slope .
(c) Now witht constant, the Lorentz transformation
gives 2
/t t vx c
   . Again we solve for ct :/ constantct x ct x
     
. This line is
parallel to the0t  line we found earlier but
shifted by the constant.
(d) Here both thex andct axes are shifted
from their normal (x, ct) orientation and they
are not perpendicular.

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Chapter 2 Special Theory of Relativity 11
46. The diagram is shown here. Note that there is only one worldline for light, and it bisects
both the x, ct axes and the,x ct  axes. Thex andct axes are not perpendicular. This can
be seen as a result of the Lorentz transformations, since0x  defines thect axis and0t 
defines thex axis.
47. The diagram shows that the events A and B that occur at the same time in K occur at
different times inK .
48. The Doppler shift gives0
1
1

  

  . With numerical values0 650
  nm and540
 
nm, solving this equation for gives0.183
  . The astronaut’s speed is7
5.50 10v c
  
m/s. In addition to a red light violation, the astronaut gets a speeding
ticket.
49. According to the fixed source (K) the signal and receiver move at speeds c and v,
respectively, in opposite directions, so their relative speed isc v . The time interval
between receipt of signals is0/ ( ) 1/t c v f
    . By time dilation( )
t
t c v

 

    .
Using0/c v
  and2 2
1
1 /v c
   we find22 2
0 0
11 /
( ) (1 )
c v c
t f c v f



  
  and0
02
(1 )1 1
11
f
f f
t




 
   
 
.

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12 Chapter 2 Special Theory of Relativity
50. For a fixed source and moving receiver, the length of the wave train is.cT vT Since n
waves are emitted during time T,cT vT
n
 
 and the frequency/f c
 iscn
f cT vT
  .
As in the text00n Tf  and0 / .T T
  Therefore2
00
0
1/ 1
1 1
fcf T
f f
cT vT





 
  
   .
51. 0
1 1 0.95
1400 kHz 224 kHz
1 1 0.95
f f


 
  
 
52. The Doppler shift function0
1
1
f f



  
is the rate at which #1 and #2 receive signals from each other and the rate at which #2
and #3 receive signals from each other. But for signals between #1 and #3 the rate is0
1 1
1 1
f f f




 
  
 
.
53. The Doppler shift function0
1
1
f f



  
is the rate at which #1 and #2 receive signals
from each other and the rate at which #2 and #3
receive signals from each other. As for #1 and
#3 we will assume that these plumbing vans are
non-relativistic v c . Otherwise it would be
necessary to use the velocity addition law and apply
the transverse Doppler shift. From the figure we see
that0 2 1
1
( )
f t t t
    . Now0 01/f t and0
2 1
2 cos2 vtx
t t c c

  
. With an angle of45 ,cos(45) 1/ 2 and 
0 0
0 0
1
1/ (2 cos ) / 1 2 cos / 1 2 /
f f
f f v cf v c v c


   
  
.
54. The Doppler shift to higher wavelengths is (with0 589
  nm)0
1
700 nm 1


 

   .
Solving for we find0.171
  . Then  8
6
2
0.171 3.00 10 m/s 1.75 10 s
29.4 m/s
v
t a

   
which is 20.25 days. One problem with this analysis is that we have only computed the

Loading page 14...

Chapter 2 Special Theory of Relativity 13
time as measured by Earth. We are not prepared to handle the non-inertial frame of the
spaceship.
55. Let the instantaneous momentum be in the x-direction and the force be in the y-direction.
Thendp F dt mdv
  anddv is also in the y-direction. So we havedv
F m ma
dt
   .
56. The magnitude of the centripetal force is2
v
ma m r
  for circular motion. For a charged
particleF qvB , so2
v
qvB m r
 or, rearrangingqBr mv p
  . Therefore.
p
r qB

When the speed increases the momentum increases, and thus for a given value of B the
radius must increase.
57.2 2
1 /
mv
p mv v c
   anddp
F dt
 . The momentum is the product of two factors that
contain the velocity, so we apply the product rule for derivatives:2 2
2 2 2 2
3
2
2
3
2
2 2
3
2 2
3
1 /
/ 1
1 / 1 /
1 2
2
1
d mv
F m dt v c
dv dt d
m v dtv c v c
v dv
ma mv c dt
v
ma ma c
v v
ma c c
ma


 


 
  
 
  
   
    
  
     
  
 
   
 
 
   
 

58. From the preceding problem3
F ma
 . We have19
10a  m/s2 and27
1.67 10m 
  kg.
(a)    
2 2 2
3 27 19 2 8
1 1 1.00005
1 / 1 0.01
1.00005 1.67 10 kg 10 m/s 1.67 10 N
v c
F

 
  
 
   
(b) As in (a)1.005
  and8
1.70 10F 
  N

Loading page 15...

14 Chapter 2 Special Theory of Relativity
(c) As in (a)2.294
  and7
2.02 10F 
  N
(d) As in (a)7.0888
  and6
5.95 10F 
  N
59.p mv
 with2 2 2
1 1 2.5516
1 / 1 0.92v c
   
  ;   
16
25
8
10 kg·m / s 1.42 10 kg
2.5516 0.92 3.00 10 m/s
p
m v



   

60. The initial momentum is   0 2
1 0.5 0.57735
1 0.5
p mv m c mc
  
 .
(a)0/ 1.01p p   
1.01 0.57735
1.01 0.57735 0.58312
mv
mc
v c c



 
Substituting for and solving for v, 
1/ 2
2 2
1 1 0.504 .
.58312
v c
cc

 
   
  
(b) Similarly 
1/2
2 2
1 1 0.536
.63509
v c
cc

 
   
  
(c) Similarly 
1/2
2 2
1 1 0.756
1.1547
v c
cc

 
   
  
61. 6.3 GeV protons have3
6.3 10K   MeV and0 7238E K E   MeV. Then2 2
0 7177 MeV/
E E
p c
c

 
. Converting to SI units13
18
8
1.60 10 J
7177 MeV/ 3.83 10 kg·m/s
MeV 3.00 10 m/s
c
p c

   
    
  
From Problem 56 we have  
18
19
3.83 10 kg m/s 1.57 T
1.60 10 C 15.2 m
p
B qr


 
  
 .
62. Initially Mary throws her ball with velocity (primes showing the measurements are in
Mary’s frame):0.0x yM Mu u u    After the elastic collision, the signs on the above
expressions are reversed, so the change in momentum as measured by Mary is0 0 0
2 2 2 2 2 2
0 0 0
2
1 / 1 / 1 /
M
mu mu mu
p u c u c u c

   
  
.

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