# Physics exercises_solution: Chapter 37

Chia sẻ: Thanh An | Ngày: | Loại File: PDF | Số trang:23

0
79
lượt xem
4

## Physics exercises_solution: Chapter 37

Mô tả tài liệu

Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 37

Chủ đề:

Bình luận(0)

Lưu

## Nội dung Text: Physics exercises_solution: Chapter 37

1. 37.1: If O  sees simultaneous flashes then O will see the A(A ) flash first since O would believe that the A  flash must have traveled longer to reach O  , and hence started first. 1 37.2: a) γ   2.29. 1  (0.9) 2 t  γ  (2.29) (2.20  10 6 s)  5.05  10 6 s. b) d  vt  (0.900) (3.00  10 8 m s) (5.05  10 6 s)  1.36  10 3 m  1.36 km. u2 37.3: 1  u 2 c 2  (1  u 2 c 2 )1 2  1   2c 2 u2 (250 m s) 2 (4 hrs)  (3600)  (t  t 0 )  (1  1  u 2 c 2 )(t )  2 t  2c 2(3.00  10 8 m s) 2  (t  t 0 )  5.00  10 9 s. The clock on the plane shows the shorter elapsed time. 1 37.4: γ  4.79. 1  (0.978) 2 γt  (4.79) (82.4  10 6 s)  3.95  10 -4 s  0.395 ms. 2 t 0 u 2  t  37.5: a) t   1 2   0  1 u 2 c2 c  t  2 2  t   2.6   u  c 1  0   c 1    t   42   u  0.998c. b) x  ut  (0.998) (3.00  10 8 m s) (4.2  10 7 s)  126 m. t 1.20  10 8 m 37.6: γ  1.667 a) t 0    0.300 s. γ γ(0.800c) b) (0.300 s) (0.800c)  7.20  10 7 m. c) t 0  0.300 s γ  0.180 s. (This is what the racer measures your clock to read at 1.20  10 8 m that instant.) At your origin you read the original  0.5 s. (0.800) (3  10 8 m s) Clearly the observes (you and the racer) will not agree on the order of events!
2. 2  4.80  10 6 m s  37.7: t 0  1  u c t  1   2 2  3.00  10 8 m s  (1 yr)    4  (t  t 0 )  (1.28  10 ) yr  1.12 hrs. The least time elapses on the rocket’s clock because it had to be in two inertial frames whereas the earth was only in one. 37.8: a) The frame in which the source (the searchlight) is stationary is the spacecraft’s frame, so 12.0 ms is the proper time. b) To three figures, u  c. Solving Eq. (37.7) for u c in terms of γ, u 1  1  (1 γ) 2  1  2 . c 2γ Using 1 γ  t 0  t  12.0 ms 190 ms gives u c  0.998. 1 1 37.9: γ   6.00 1  (u c) 2 1  (0.9860) 2 l 0 55 km a) l    9.17 km. γ 6.00 b) In muon’s frame: d  ut  (0.9860c)(2.20  10 6 s)  0.651 km. d 0.651 %   0.071  7.1%. h 9.17 c) In earth’s frame: t  t 0 γ  (2.2  10 6 s)(6.00)  1.32  10 5 s  d   ut  (0.9860c)(1.32  10 5 s)  3.90 km d  3.90 km %   7.1%. h 55.0 km 4.50  10 4 m 37.10: a) t   1.51  10  4 s. 0.99540c 1 b) γ   10.44 1  (0.9954) 2 h 45 km h    4.31 km. γ 10.44 h t c)  1.44  10 5 s, and  1.44  10 5 s; so the results agree but the particle’s 0.99540c γ lifetime is dilated in the frame of the earth.
3. 37.11: a) l 0  3600 m u2 (4.00  10 7 m s) 2  l  l0 1  2  l0 (3600 m) 1  c (3.00  108 m s) 2  (3600 m)(0.991)  3568 m. l 3600 m b) t 0  0   9.00  10 5 s. u 4.00  10 7 m s l 3568 m c) t    8.92  10 5 s. u 4.00  10 m s 7 37.12: γ  1 0.3048 , so u  c 1  (1 γ) 2  0.952c  2.86  10 8 m s . u2 l 37.13: l  l 0 1  2  l0  c 1 u2 c2 74.0 m  l0   92.5 m. 2  0.600c  1    c  37.14: Multiplying the last equation of (37.21) by u and adding to the first to eliminate t gives  u2  1 x   ut   x 1  2   x,  c     u and multiplying the first by and adding to the last to eliminate x gives c2 u  u2  1 t   2 x   γt 1  2   t ,  c  γ c   so x  γ( x  ut ) and t  γ(t   ux c 2 ), which is indeed the same as Eq. (37.21) with the primed coordinates replacing the unprimed, and a change of sign of u. v  u 0.400c  0.600c 37.15: a) v    0.806c 1  uv  c 2 1  (0.400) (0.600) v  u 0.900c  0.600c b) v    0.974c 1  uv  c 2 1  (0.900)(0.600) v  u 0.990c  0.600c c) v    0.997c. 1  uv  c 2 1  (0.990)(0.600)
4. 37.16: γ  1.667( γ  5 3 if u  (4 5)c). a) In Mavis’s frame the event “light on” has space-time coordinates x   0 and t   5.00 s, so from the result of Exercise 37.14 or  ux  Example 37.7, x  γ( x   ut ) and t  γ  t   2   x  γut   2.00  109 m, t  γt    c  8.33 s. b) The 5.00-s interval in Mavis’s frame is the proper time t 0 in Eq. (37.6), so t  γt 0  8.33 s, as in part (a). c) (8.33 s) (0.800c)  2.00  10 9 m, which is the distance x found in part (a). x  ut 37.17: Eq. (37.18): x   Eq. (37.19): x   ut   x 1  u 2 c 2 1 u c 2 2 x Equate: ( x  ut ) γ  ut   γ  xγ x x1   t     u  tγ  uγ   tγ  u  γ  γ         1  u c   1 2 1 1  u2 c2  γ  1  u c   2    γu 2 c 2 γ 1  u c  1  u c  1  u c  2 2 2 xuγ t  ux c 2  t   tγ   . c2 1  (u c) 2 37.18: Starting from Eq. (37.22), vu v  1  uv c 2 v (1  uv c 2 )  v  u v   u  v  v uv c 2  v(1  uv  c 2 ) from which Eq. (37.23) follows. This is the same as switching the primed and unprimed coordinates and changing the sign of u . 37.19: Let the unprimed frame be Tatooine and let the primed frame be the pursuit ship. We want the velocity v  of the cruiser knowing the velocity of the primed frame u and the velocity of the cruiser v in the unprimed frame (Tatooine). v u 0.600c  0.800c v    0.385c uv 1  (0.600) (0.800) 1 2 c  the cruiser is moving toward the pursuit ship at 0.385c.
5. 37.20: In the frame of one of the particles, u and v are both 0.9520c but with opposite sign.  v  (u )  0.9520c  0.9520c v    0.9988c. 1  (u ) (v) c 2 1  (0.9520)(0.9520) Thus, one particle moves at a speed 0.9988c toward the other in the other particle’s frame. v  u  0.950c  0.650c 37.21: v    0.784c. uv  1  (0.950)(0.650) 1 2 c v  u 0.700c  0.400c 37.22: a) In Eq.(39-24), u  0.400c, v  0.700c  v   1  uv c 2 1  (0.700) (0.400)  0.859c. x 8.00  10 9 m b)   31.0 s. v 0.859c v u uvv  37.23: v    v  2  v  u 1  uv c 2 c  vv   v  v  u 1  2   v  v   u   c  (1  vv  c 2 ) 0.360c  0.920c u  0.837c (1  (0.360)(0.920))  moving opposite the rocket, i.e., away from Arrakis. 37.24: Solving Eq. (37.25) for u c , (see solution to Exercise 37.25) u 1  ( f f0 )2  , c 1 ( f f0 )2 and so (a) if f f 0  0.98, u c   0.0202, the source and observer are moving away from each other. b) if f f 0  4, u c   0.882, they are moving toward each other. cu 37.25: a) f  f 0  (c  u ) f 2  (c  u ) f 02 cu c( f  f 02 ) c(( f f 0 ) 2  1) c((λ 0 λ) 2  1) 2 u   f 02  f 2 ( f f0 )2  1 ((λ 0 λ) 2  1) ((675 575) 2  1) u  c  0.159c  4.77  10 7 m s  4.77  10 4 km s  1.72  10 8 km h. ((675 575)  1) 2 b) (1.72  10 8 km h  90 km h) ( $1.00) $172 million dollars!
6. 37.26: Using u  0.600c  3 5c in Eq. (37.25) gives 1  3 5 25 f  f0  f 0  f 0 2. 1  3 5 85 dp d   2 mv  2 ( 2va c ) 1 1 2  mv  ma 37.27: a) F    dt dt  1  v 2 c 2  1 v2 c2 (1  v 2 c 2 ) 3 / 2    (1  v 2 c 2 )  v 2 c 2   ma  (1  v 2 c 2 ) 3 2   ma (1  v c )  2 2 32   32 F  v2   a  1  2  . m c    dp b) If the force is perpendicular to velocity then denominator is constant  F   dt m dv dt F a 1  v2 c2 . 1 v c 2 2 m 37.28: The force is found from Eq. (37.32) or Eq. (37.33). (a) Indistinguishable from F  ma  0.145 N. b) γ 3 ma  1.75 N. c) γ 3 ma  51.7 N. d) γma  0.145 N, 0.333 N, 1.03 N. mv 37.29: a) p   2mv 1 v2 c2 1 v2  1  2 1 v2 c2   1 2 4 c 3 2 3  v2  c v c  0.866c. 4 2 1 v b) F  γ 3 ma  2ma  γ 3  2  γ  (2)1 / 3 so 2  22 / 3  v c 1 2 c  1  2 2 / 3  0.608. 37.30: a) γ  1.01, so v c   0.140 and v  4.21  10 8 m s. b) The relativistic expression is always larger in magnitude than the non-relativistic expression.
7. mc 2 37.31: a) K   mc 2  mc 2 1 v c 2 2 1 1 v2 3   2   1 2  v   0.866c. 1 v2 c2 4 c 4 1 1 v2 35 b) K  5mc  2 6  1 2  v  c  0.986c. 1 v c 2 2 36 c 36 37.32: E  2mc 2  2(1.67  10 27 kg)(3.00  108 m s) 2  3.01  10 10 J  1.88  10 9 eV. 1 2 3 mv 4 37.33: K  ( γ  1)mc 2  mv   2 8 c2 1 3 v 4 0.02 2 if K  K 0  1.02 mv 2   v 2 8 c2 2 150 2 4  v  c2  v  c  0.163c  4.89  10 7 m s. 4 150 37.34: a) W  K  ( γ f  1)mc 2  (4.07  10 3 ) mc 2 . b) ( γ f  γ i ) mc 2  4.79mc 2 . c) The result of part (b) is far larger than that of part (a). 37.35: a) Your total energy E increases because your potential energy increases; E  mgy E  (m)c 2 so m   E c 2  mg (y ) c 2  m m  ( gy ) c 2  (9.80 m s 2 )(30 m) (2.998  10 8 m s) 2  3.3  10 13 % This increase is much, much too small to be noticed. b) E  U  1 kx 2  1 (2.00  10 2 N m)(0.060 m) 2  36.0 J 2 2 m  (E ) c 2  4.0  10 16 kg Energy increases so mass increases. The mass increase is much, much too small to be noticed.
8. 37.36: a) E0  m0 c 2 2 E  mc 2  2m0 c 2 m0  m  2 m0   2m 0 1 v2 / c2 1 v2 v2 3  1 2  2   v  c 3 4 4 c c 4 v  0.866 c  2.60  10 8 m s m0 b) 10 m0 c 2  mc 2  c2 1 v c 2 2 v2 1 v2 99 1 2   2  c 100 c 100 99 vc  0.995 c  2.98  10 8 m s 100 37.37: a) E  mc 2  K , so E  4.00mc 2 means K  3.00mc 2  4.50  10 10 J b) E 2  (mc 2 ) 2  ( pc) 2 ; E  4.00mc 2 , so 15.0(mc 2 ) 2  ( pc) 2 p  15mc  1.94  10 18 kg  m s c) E  mc 2 1 v2 c2 E  4.00mc 2 gives 1  v 2 c 2  1 16 and v  15 16c  0.968c 37.38: The work that must be done is the kinetic energy of the proton.    1  a) K  (   1)m0 c   2  1 m0 c 2  1  v2 2   c     1   (1.67  10 27 kg)(3.00  108 m s) 2   1     1  0.1 c c 2    1   (1.50  10 10 J)  1  1  0.01   7.56  10 13 J b) K  (1.50  10 10 J)  1  1   1 (0.5)2  11  2.32  10 J c) K  (1.50  10 10 J)  1  1   1 (0.9) 2  10  1.94  10 J
9. 37.39: (m  6.64  10 27 kg, p  2.10  10 18 kg  m s) a) E  (mc 2 ) 2  ( pc) 2  8.68  10 10 J. b) K  E  mc 2  8.68  10 10  (6.64  10 27 kg)c 2  2.70  10 10 J. K 2.70  10 10 J c)   0.452. mc 2 (6.64  10 27 kg)c 2 12   p 2  37.40: E  (m c  p c ) 2 4 2 2 12  mc 1   2    mc       1 p2  p2 1  mc 2 1   2m c   mc 2  2 2   mc 2  mv 2   2m 2 the sum of the rest mass energy and the classical kinetic energy. 1 37.41: a) v  8  10 7 m s  γ   1.0376 1  v2 c2 1 m  mp K 0  mv 2  5.34  10 12 J 2 K K  ( γ  1)mc 2  5.65  10 12 J   1.06. K0 b) v  2.85  108 m s  γ  3.203 1 K 0  mv 2  6.78  10 11 J 2 K  (γ  1)mc 2  3.31  10 10 J K K 0  4.88. 37.42: (5.52  10 27 kg)(3.00  10 8 m s) 2  4.97  10 10 J  3105 MeV. 37.43: a) K  qV  eV  1  K  mc 2   1  4.025mc 2  3.295  10 13 J  2.06 MeV  1 v2 c2    V  K e  2.06  10 V 6 b) From part (a), K  3.30  10 13 J  2.06 MeV
10. 37.44: a) According to Eq. 37.38 and conservation of mass-energy m 9.75 2Mc 2  mc 2   2Mc 2    1   1  1.292. 2M 2(16.7) Note that since   12 2 , we have that 1v c v 1 1  1 2  1  0.6331. c  (1.292) 2 b) According to Eq. 37.36, the kinetic energy of each proton is  1.00 MeV  K  (  1) Mc 2  (1.292  1)(1.67  10  27 kg)(3.00  108 m s) 2   1.60  10 13 J      274 MeV. c) The rest energy of  0 is mc 2  (9.75  10 28 kg)(3.00  108 m s) 2  1.00 MeV 1.601013 J  548 MeV. d) The kinetic energy lost by the protons is the energy that produces the  0 , 548 MeV  2(274 MeV). 37.45: a) E  0.420 MeV  4.20  10 5 eV. (9.11  10 31 kg)(3.00  10 8 m s) 2 b) E  K  mc  4.20  10 eV  2 5 1.6  10 19 J eV  4.20  10 5 eV  5.11  10 5 eV  9.32  10 5 eV. 2 mc 2  mc 2  c) E   v  c 1   E   1 v c 2 2   2  5.11  10 5 eV   c 1   9.32  10 5 eV   0.836c  2.51  10 m s .  8   1 2 2K 2(4.20  105 eV)(1.6  10 19 J eV) d) Nonrel: K  mv  v   2 m 9.11  10 31 kg  3.84  108 m s . 37.46: a) The fraction of the initial mass that becomes energy is (4.0015 u) 1  6.382  10 3 , and so the energy released per kilogram is 2(2.0136 u) (6.382  10 3 )(1.00 kg)(3.00  10 8 m s) 2  5.74  1014 J. 1.0  1019 J b)  1.7  10 4 kg. 5.74  10 J kg 14
11. 37.47: a) E  mc 2 , m  E c 2  (3.8  10 26 J) (2.998  10 8 m s) 2  4.2  10 9 kg 1 kg is equivalent to 2.2 lbs, so m  4.6  10 6 tons b) The current mass of the sun is 1.99  10 30 kg, so it would take it (1.99  10 30 kg) (4.2  10 9 kg s)  4.7  10 20 s  1.5  1013 y to use up all its mass. 37.48: a) Using the classical work-energy theorem we obtain m(v 2  v 2 0 ) (2.00  10 12 kg)[(0.920)(3.00  10 8 m s)]2 x    1.81 m. 2F 2(4.20  10 4 N) b) Using the relativistic work-energy theorem for a constant force (Eq. 37.35) we obtain (   1)mc 2 x  . F For the given speed,   1 2  2.55, thus 1 0.920 (2.55  1)(2.00  10 12 kg)(3.00  10 8 m s) 2 x   6.65 m. (4.20  10 4 N) c) According to Eq. 37.30, 3 3 3 F  v2 2 (4.20  10 4 N)  v 2  2  v2 2 a  1  2   1  2   (2.10  1016 m s 2 )1  2  , m c    (2.00  10 12 kg)  c     c    2 which yields, i) a  2.07  10 m s ( β  0.100) 16 2 ii) a  1.46  1016 m s ( β  0.462) 2 iii) a  0.126  1016 m s ( β  0.920). d 1200 m 37.49: a) d  ct  t    4.00  10 6 s (v is very close to c). c 3  10 m s 8 2 2 t u2 u  t  But 0  1  2     1   0  t c c  t  2  t   (1  )  1   0  2  t  12 12   t 0  2    2.6  10 8  2     1  1      1  1     6   2.11  10 5.   t       4.00  10      t   4.00  10 6 b) E  mc 2    t  mc   2.6  10 8  139.6 MeV  2    0    E  2.15  10 MeV. 4
12. 1 37.50: One dimension of the cube appears contracted by a factor of , so the volume in γ S  is a 3 γ  a 3 1  (u c) 2 . 37.51: Need a  b  l 0  a, l  b l b b     1  u 2 c2 l0 a 1.40b 2 2 b  1   u  c 1    c 1    0.700c. a  1.40   2.10  108 m s . 37.52: The change in the astronaut’s biological age is t 0 in Eq. (37.6), and t is the distance to the star as measured from earth, divided by the speed. Combining, the astronaut’s biological age is 42.2 yr c 42.2 yr 19 yr   19 yr   24.7 yr. γu γ(0.9910) 1 v γ2  1 v 99 37.53: a) E  mc 2 and   10    2  c 1  (v c ) 2 c γ c 100  0.995.   v 2 2  b) ( pc)  m v γ c , E  m c    γ  1 2 2 2 2 2 2 2 4  c     E  ( pc) 2 2 1 1     0.01  1%. E 2 2 v 2 1  (10 (0.995)) 2 1    c
13. 37.54: a) Note that the initial velocity is parallel to the x-axis. Thus, according to Eqn. 37.30, 3 Fx  v  (3.00  10 12 N) 2 2 3 ax  1  2    c   27 2 (1  0.900 )  1.49  1014 m s . 2 2 m  (1.67  10 kg) Now note that the initial velocity is perpendicular to the y-axis. Thus, according to Eqn. 37.33, 1 Fy  v 2  2 (5.00  10 12 N) 1 ay  1  2    c   27 2 (1  0.900 )  1.31  1015 m s . 2 2 m  (1.67  10 kg) b) The angle between the force and acceleration is given by Fx a x  Fy a y ( 3.001012 N)( 1.491014 m s 2 )  ( 5.001012 N)(1.311015 m s 2 ) cos θ  Fa   ( 3.001012 N)2  ( 5.001012 N) 2 ( 1.491014 m s 2 ) 2  (1.311015 m s 2 ) 2 θ  24.5. 37.55: a) K  20  1012 eV  3.204  10 6 J  1  1 K  mc 2   1, so  2.131  10 4  1 v 2 c 2  1 v c 2 2   v2 1 v 2 (c  v)(c  v) 2(c  v) 1 2  ;1 2   since c  v  2c c (2.131  10 4 ) 2 c c2 c v  (1  )c so 1  v 2 c 2  2 and   1.1  10 9 m m b) mrel    (2.1  10 4 )m 1 v c 2 2 2 37.56: a) (8.00 kg)(1.00  10 4 )(3.00  10 8 m s) 2  7.20  1013 J. b) (E t )  (7.20  1013 J) (4.00  10 6 s)  1.80  1019 W. E (7.20  1013 J) c) M   2  7.35  10 9 kg. gh (9.80 m s )(1.00  10 m) 3 37.57: Heat in Q  mL f  (4.00 kg)(3.34  10 5 J kg )  1.34  10 6 J Q (1.34  10 6 J)  m    1.49  10 11 kg. c 2 (3.00  10 m s) 8 2 p ( E c) E 37.58: a) v    , where the atom and the photon have the same m m mc E magnitude of momentum, E c . b) v    c, so E  mc 2 . mc
14. c c 37.59: Speed in glass v    1.97  10 8 m s n 1.52 1   1.326 1 v2 c2  K  (   1)mc 2  (0.326)(0.511 MeV)  0.167 MeV  1.67  10 5 eV. 1 2 37.60: a) 80.0 m s is non-relativistic, and K  mv  186 J. 2 b) (   1)mc 2  1.31  1015 J. c) In Eq. (37.23), v   2.20  10 8 m s , u  1.80  10 8 m s , and so v  7.14  10 7 m s . 20.0 m d)  13.6 m.  20.0 m e)  9.09  10 8 s. 2.20  10 m s 8 t 13.6 m f) t    6.18  10 8 s, or t    6.18  10 8 s.  2.20  10 8 m s 37.61: x  2  c 2 t  2   ( x  ut ) 2  2  c 2  2 t  ux c 2  2  x  ut  c(t  ux c 2 )  u 1  x1    x(u  c)  t (u  c)  x  ct  c c  x 2  c 2t 2 . 37.62: a) From Eq. (37.37), 1 3 v4 3 (3.00  10 4 m s) 4 K  mv 2  m 2  (90.0 kg)  304 J. 2 8 c 8 (3.00  10 8 m s) 2 2 (3 8) mv 4 c 2 3  v  b) 2     7.50  10  9. (1 2)mv 4c
15. dv F 37.63: a  (1  v 2 c 2 ) 3 2 dt m v dv F t F    dt  t 0 (1  (v 2 2 32 c )) m 0 m vc v c dx x v F  c c   t 0 (1  x ) 2 32 1  x2 0 1  (v c ) 2 m   v  2   Ft  2  Ft  2 2  v2    1         v 2  Ft    m  c   m   mc    Ft m v . So as t  , v  c. 1  ( Ft mc) 2 37.64: Setting x  0 in Eq. (37.21), the first equation becomes x    ut and the last, upon multiplication by c, becomes ct   ct. Squaring and subtracting gives c 2 t  2  x  2  γ 2 (c 2 t 2  u 2 t 2 )  c 2 t 2 , or x   c t  2  t 2  4.53  10 8 m.
16. 37.65: a) Want t   t 2  t1   x2  x1 x   x1  ( x1  ut1 ) γ  x2  ( x2  ut 2 ) γ  u   t 2  t1 t  ( x  x1 )  And t   γ t 2  t1  u 2   . Since u  x t ,   c2   ux   x 2  x 2 t   γ t  2   γ t  2   t   t 1   x  c 2  t 2  2 2 c   c t  t c    x There’s no physical solution for t   x  ct. c ux ux b) Simultaneously  t   0  t1  21  t 2  22 c c u c t 2  t  2 x  u  . c t Also 1 1  t 2  x   x 2  x1    (x  ut )   x  c 2   1  (u c) 2 1  c 2 t 2 x 2  x   c 2 t 2  x 1   x  x 2  c 2 t 2 . x 2 1 1 c) Part (b): t  (x) 2  (x ) 2  (5.00 m) 2  (2.50 m) 2  1.44  10 8 s. c c 37.66: a) (100 s)(0.600)(3.00  10 8 m s)  1.80  1010 m. b) In Sebulbas frame, the relative speed of the tachyons and the ship is 3.40c, and so the time t 2  100 s  1.80  1010 m  118 s. At t 2 Sebulba measures that Watto is a distance from him of 3.4c (118 s)(0.600)(3.00  10 8 m s)  2.12  1010 m. c) From Eq. (37.23), with v   4.00c and u  0.600c, v  2.43c, with the plus sign indicating a direction in the same direction as Watto’s motion (that is, away from Sebulba). d) As the result of part (c) suggests, Sebulba would see the tachyons moving toward Watto and hence t 3 is the time they would have left Sebulba in order to reach Watto at the distance found in part 2.12  1010 m (b), or 118 s   89 s, and so Sebulba receives Watto’s message before 2.43c even sending it! Tachyons seem to violate causality.
17. 37.67: Longer wavelength (redshift) implies recession. (The emitting atoms are moving (λ λ) 2  1 away.) Using the result of Ex. 37.26: u  c 0 (λ 0 λ)  1  (656.3 953.4) 2  1  u  c   0.3570c  1.071  10 m s 8  (656.3 953.4)  1 2 37.68: The baseball had better be moving non-relativistically, so the Doppler shift formula (Eq. (37.25)) becomes f  f 0 (1  (u c)). In the baseball’s frame, this is the frequency with which the radar waves strike the baseball, and the baseball reradiates at f. But in the coach’s frame, the reflected waves are Doppler shifted again, so the detected frequency is f (1  (u c))  f 0 (1  (u c)) 2  f 0 (1  2(u c)), so f  2 f 0 (u c) and the f fractional frequency shift is  2(u c). In this case, f0 f (2.86  10 7 ) u c (3.00  108 m) 2 f0 2  42.9 m s  154 km h  92.5 mi h. 37.69: a) Since the two triangles are similar: H  A  mc 2   E. b) O  H 2  A 2  E 2  (mc 2 ) 2  pc. c) K  E  mc 2 . The kinetic energy can be obtained by the difference between the hypoteneuse and adjacent edge lengths. 37.70: a) As in the hint, both the sender and the receiver measure the same distance. However, in our frame, the ship has moved between emission of successive wavefronts, and we can use the time T  1 f as the proper time, with the result that f  f 0  f 0 . 12 cu  1  0.758  b) Toward: f  f 0  345 MHz   930 MHz cu  1  0.758  f  f 0  930 MHz  345 MHz  585 MHz. Away: 12 cu  1  0.758  f  f0  345 MHz   128 MHz cu  1  0.758  f  f 0  217 MHz. c) γf 0  1.53 f 0  528 MHz, f  f 0  183 MHz. The shift is still bigger than f 0 , but not as large as the approaching frequency.
18. 37.71: The crux of this problem is the question of simultaneity. To be “in the barn at one time” for the runner is different than for a stationary observer in the barn. The diagram below, at left, shows the rod fitting into the barn at time t  0 , according to the stationary observer. The diagram below, at right, is in the runner’s frame of reference. The front of the rod enters the barn at time t1 and leaves the back of the barn at time t 2 . However, the back of the rod does not enter the front of the barn until the later time t 3 .
19. 37.72: In Eq. (37.23), u  V , v   (c n), and so ( c n)  V (c n )  V v  . cV 1  (V nc) 1 2 nc For V non-relativistic, this is v  ((c n)  V )(1  (V nc))  (c n)  V  (V n 2 )  (V 2 nc) c  1    1  2  V n  n   1  so k  1  2 . For water, n  1.333 and k  0.437.  n  dv 37.73: a) a   dt    (dt  udx c 2 ) dt  dv vu u dv   dv (1  uv c ) (1  uv c ) c 2 2 2 2 dv 1 v u u   2 2  2  . dv 1  uv c 2 (1  uv c )  c   1 (v  u ) u c 2   1 u2 c2   dv   dv   1  uv c 2 (1  uv c 2 ) 2    dv (1  uv c 2 ) 2       (1  u c ) 2 2 dv (1  uv c 2 ) 2 dv (1  u 2 c 2 ) 1 a   . . γdt  udx c 2 dt (1  uv c ) γ(1  uv c 2 ) 2 2  a (1  u 2 c 2 ) 3 2 (1  uv c 2 ) 3 . b) Changing frames from S   S just involves changing 3  uv   a  a , v   v   a  a (1  u 2 c 2 ) 3 2 1  2  .  c 
20. 37.74: a) The speed v is measured relative to the rocket, and so for the rocket and its occupant, v   0. The acceleration as seen in the rocket is given to be a   g , and so the acceleration as measured on the earth is 32 du  u2  a  g 1  2  .  c  dt   b) With v1  0 when t  0 , 1 du dt  g (1  u 2 c 2 ) 3 2 t1 1 v1 du 0 dt  g 0 (1  u 2 c 2 ) 3 2 v1 t1  . g 1  v12 c 2 c) dt   dt  dt / 1  u 2 c 2 , so the relation in part (b) between dt and du, expressed in terms of dt  and du, is 1 du 1 du dt   dt   . 1  u 2 c 2 g (1  u c ) g (1  u 2 c 2 ) 2 2 2 32 Integrating as above (perhaps using the substitution z  u c ) gives c v   t1  arctanh  1 . g c For those who wish to avoid inverse hyperbolic functions, the above integral may be done by the method of partial fractions; du 1  du du  gdt      , (1  u c)(1  u c) 2 1  u c 1  uc   which integrates to c  c  v1   t1  1n  . 2 g  c  v1     d) Solving the expression from part (c) for v1 in terms of t1 , (v1 c)  tanh( gt1 c), so that  1  (v1 c) 2  1 cosh ( gt1 c), using the appropriate indentities for hyperbolic functions. Using this in the expression found in part (b),  c tanh( gt1 c) c t1    sinh( gt1 c),  g 1 cosh( gt1 c) g which may be rearranged slightly as gt1  gt    sinh  1 . c  c   If hyperbolic functions are not used, v1 in terms of t1 is found to be  gt c v1 e gt1 c  e 1 1  c e gt1 c  e  gt1 c