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

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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    v 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 H ydrogen 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 1 69 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

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    Chapter 2 Special Theory of Relativity 1 Chapter 2 1. For a particle Newton’s second law says 2 2 2 2 2 2 ˆ ˆ ˆ d 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 gives 2 2 2 2 2 2 ˆ ˆ ˆ . d x d d F ma m i j k y z F dt dt dt                2. From Equation (2.1) ˆ ˆ ˆ dx dy dz p m i j k dt dt dt          . In a Galilean transformation dx dx dy dy dz dz v dt dt dt dt dt dt        . Substitution into Equation (2.1) gives ˆ ˆ ˆ dx dy dz p m v i j k p dt dt dt                . However, because ˆ ˆ ˆ dx dy dz p m i j k dt dt dt              the same form is clearly retained, given the velocity transformation dx dx v dt dt    . 3. Using the vector triangle shown, the speed of light coming toward the mirror is 2 2 c v  and the same on the return trip. Therefore the total time is 2 2 2 2 2 distance speed t c v    . Notice that sin v c   , so 1 sin v c          .

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    2 Chapter 2 Special Theory of Relativity 4. As in Problem 3, 1 2 sin / v v   , so 1 1 1 2 0.350 m/s 1.25 sin ( / ) sin 1 m/ . s 6 3 v v              and 2 2 2 2 2 1 (1.25 /s) (0.35 m m/s) 1.20 m/s v v v      . 5. When the apparatus is rotated by 90°, the situation is equivalent, except that we have effectively interchanged 1 and 2 . Interchanging 1 and 2 in Equation (2.3) leads to Equation (2.4). 6. Let n = the number of fringes shifted; then d n    . Because   d c t t       , we have     2 1 2 2 c t t v n c          . Solving for v and noting that 1 + 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. Letting 2 1 1 1    (where / v c   ) the text equation (not currently numbered) for 1 t becomes   2 1 1 1 2 2 1 2 1 1 t c c         which is identical to 2 t when 1 2  so 0 t   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 in K  measures the speed of the signal to be dx 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 (assuming y y   and z 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 for t  and substitute into the first:

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    Chapter 2 Special Theory of Relativity 3 ( )( ) ac b ac b t ct c          or 2 2 2 2 ( )( ) . c ac b ac b a c b      Now v is the speed of the origin of the x  -axis. We can find that speed by setting 0 x   which gives 0 ax bt   , or / / v x t b a   , or equivalently b = av . Substituting this into the equation above for 2 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 and x  to obtain:     ; x x vt x x vt          which in turn can be solved for t and t  to complete the transformation. 11. When v c  we find 2 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 10 v c        (b)   8 7 / (240m/s) / 3.00 10 m/s 8.0 10 v c        (c)   sound 2.3 2.3 330 m/s v v    so     8 6 / 2.3 330 m/s / 3.00 10 m/s 2.5 10 v 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 10 v 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 0.42 v 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 0.95 v c      

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    4 Chapter 2 Special Theory of Relativity 13. From the Lorentz transformations 2 / t t v x c            . But 0 t    in this case, so solving for v we find 2 / v c t x    . Inserting the values 2 1 / 2 t t t a c      and 2 1 x x x a     , we find   2 / 2 / 2 c a c v c a     . We conclude that the frame K  travels at a speed c /2 in the x  -direction. Note that there is no motion in the transverse direction. 14. Try setting   0 x x v t         . Thus 0 / 2 x v t a va c       . Solving for v we find 2 v c   , which is impossible. There is no such frame K  . 15. For the smaller values of β we use the binomial expansion   1/2 2 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 1/2 2 2 1 1 0.42 1.10          (f)     1/2 1/2 2 2 1 1 0.95 3.20          16. There is no motion in the transverse direction, so 3.5 y 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 9 5 / 0 0.8 2 m / 8.9 10 s 3 t t vx c c c           17. (a)   2 2 2 2 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) With 0.8   we find 5 / 3   . Then 5 y y    m, 10 z z    m,     8 -8 5 3m 2.40 10 m/s (3.86 10 s) 10.4 m 3 x x vt                         2 2 8 8 8 5 / 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 2 2 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 distance 1 1 L v t c t     . On the return trip it travels 2 2 L v t c t     . Then the total time is 1 2 2 2 2 2 2 2 / 1 / Lc L c t t t c v v c          . But from time dilation we know (with 0 proper time 2 / t L c     ) that 0 2 2 2 / 1 / L c t t v c        . Comparing these two results for t  we get 0 2 2 2 2 2 / 2 / 1 / 1 L c L c v v c c    which reduces to 2 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    . Thus 2 2 1 (0.99) 0.9801.     Solving for  , we find 0.14   or 0.14 . v c  (b) The time for the trip in the Earth-based frame is 6 1 8 5.00 10 m 1.19 10 s 0.14 3.00 10 m/s d t v          . With the relativistic factor 1.01   (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 speed v c   for the entire round trip. In the rocket ’ s reference frame the distance is only 2 1 . d d     Then in the rocket ’ s frame of reference 2 2 40ly 1 distance 1 . time 40 y 40 d v c y          Rearranging 2 1 v c      . Solving for  we find 0.5   , or 0.5 0.71 v c c   . To find the elapsed time t on Earth, we know 40 t   y, so 2 1 40y 56.6 y. 1 t t        21. In the muon’ s frame 0 2.2 s. μ T  In the lab frame the time is longer; see Equation (2.19): 0 T T    . In the lab the distance traveled is 0 0 9.5cm vT v T c T        , since v 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 2.2 μs v c c      . Now all quantities are known except β. Solving for β we find 4 1.4 10     or 4 1.4 10 v c    . 22. Converting the speed to m/s we find 25,000 mi/h = 11,176 m/s. From tables the distance is 8 m 3.84 10  . In the earth ’ s frame of reference the time is the distance divided by speed, or 8 3.84 10 m 34,359 s 11,176 m/s d t v     . In the astronauts’ frame the time elapsed is 2 / 1 t t t       . The time difference is 2 2 1 1 1 t t t t t t                 . Evaluating numerically 2 5 8 11,176 m/s 34,359 s 1 1 2.4 10 s. 3.00 10 m/s t                       23. 0 T T    , so we know that 2 2 1 5 / 3 1 / v c     . Solving for v we find 4 / 5. v c  24. 0 / L L   so clearly 2   in this case. Thus 2 2 1 2 1 / v c   and solving for v we find 3 . 2 c v  25. The clocks’ rates differ by a factor of 2 2 1/ 1 / v c    . Because  is very small we will use the binomial theorem approximation 2 1 / 2     . Then the time difference is   1 t t t t t t           . Using 2 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/s 8 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 km L 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 s t 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 know 4 x   km, 0 t   , and 5 x    km. Thus       2 2 2 2 2 10 2 2 2 8 5000 m 4000 m 1.0 10 s 3.00 10 m/s x x t c              and 5 1.0 10 t      s. 28. (a) Converting v = 120 km/h = 33.3 m/s. Now with c = 100 m/s, we have / 0.333 v c    and 2 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 / 0.833 v c    and 2 2 1 1 1.81. 1 1 0.833        So the length is 0 / 40 /1.81 22.1 L L     m. 30. Let subscript 1 refer to firing and subscript 2 to striking the target. Therefore we can see that 1 1 x  m, 2 121 x  m, and 1 3 t  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 of 1 x , 2 x , 1 t , and 2 t . Note that with 0.8 v c  , 2 2 1 / 5 / 3 v 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     with 0.8 v c   and 0.8 . x u 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. Let 30.556 x u  m/s and 38.889 v   m/s. Our premise is that 100 c  m/s. Then by velocity addition,        2 2 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 have 1 / 1 / c nv c u n v nc          . For light traveling in opposite directions 1 / 1 / 1 / 1 / c nv c nv c u n v nc v nc              . Because / v c is very small, use the binomial expansion:       1 1 / 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 order 2 2 / . v c Similarly 1 / 1 / / . 1 / nv c nv c v nc v nc      Thus         2 2 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 find 2 1 2(5 m/s) 1 4.35 m/s. 1.33 u           35. Clearly the speed of B is just 0.60 c . To find the speed of C use 0.60 x u c  and 0.60 v 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 distance 8 3.84 10  m. The rotation rate is 1 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: 5 4205 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 6 1.43 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 is E 2 cos(39 ) R   , where E R  Earth ’s radius = 6378 km. Thus the circumference of the path is 31,143 km. The

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    Chapter 2 Special Theory of Relativity 9 rotational speed of that point is   31,143 km / 24 h 1298 km/h 360. m s 5 / v    . The observatory clock runs slow by a factor of 2 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 clock 2 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 clock 2 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 ns 162   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 rate f  for 1 t  and a rate f  for 2 t  . Frank receives signals at a rate f  for 1 t and a rate f  for 2 t . 40. 1 2 2 L L L L L T t t v c v c v        ; Frank sends signals at rate f , so Mary receives 2 / f T f L v  signals. 1 2 2 L T t t v        ; Mary sends signals at rate f , so Frank receives 2 / 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 2 2 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 s t                                    

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    10 Chapter 2 Special Theory of Relativity 42. For a timelike interval 2 0 s   so 2 2 2 x c t    . We will prove by contradiction. Suppose that there is a frame K  is which the two events were simultaneous, so that 0 t    . Then by the spacetime invariant 2 2 2 2 2 2 2 x c t c t x x             . But because 2 2 2 x c t    , this implies 2 0 x    which is impossible because x   is real. 43. As in Problem 42, we know that for a spacelike interval 2 0 s   so 2 2 2 x c t    . Then in a frame K  in which the two events occur in the same place, 0 x    and 2 2 2 2 2 2 2 2 x c t x c t c t              . But because 2 2 2 x c t    we have 2 2 0 c t    , which is impossible because t   is real. 44. In order for two events to be simultaneous in K  , the two events must lie along the x  axis, or along a line parallel to the x  axis. The slope of the x  axis is / v c   , so / slope = c t v c x    . Solving for v , we find 2 / v c t x    . Since the slope of the x  axis must be less than one, we see that x c t    so 2 2 2 2 0 s x c t      is required. 45. parts (a) and (b) To find the equation of the line use the Lorentz transformation. With 0 t   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 with t  constant, the Lorentz transformation gives   2 / t t vx c     . Again we solve for ct : / constant ct x ct x         . This line is parallel to the 0 t   line we found earlier but shifted by the constant. (d) Here both the x  and ct  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. The x  and ct  axes are not perpendicular. This can be seen as a result of the Lorentz transformations, since 0 x   defines the ct  axis and 0 t   defines the x  axis. 47. The diagram shows that the events A and B that occur at the same time in K occur at different times in K  . 48. The Doppler shift gives 0 1 1        . With numerical values 0 650   nm and 540   nm, solving this equation for  gives 0.183   . The astronaut ’ s speed is 7 5.50 10 v 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 is c v  . The time interval between receipt of signals is 0 / ( ) 1/ t c v f      . By time dilation ( ) t t c v          . Using 0 / c v   and 2 2 1 1 / v c    we find 2 2 2 0 0 1 1 / ( ) (1 ) c v c t f c v f           and 0 0 2 (1 ) 1 1 1 1 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   is cn f cT vT   . As in the text 0 0 n T f   and 0 / . T T    Therefore 2 0 0 0 1 / 1 1 1 f cf 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 function 0 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 is 0 1 1 1 1 f f f             . 53. The Doppler shift function 0 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 that 0 2 1 1 ( ) f t t t     . Now 0 0 1/ f t  and 0 2 1 2 cos 2 vt x t t c c     . With an angle of 45  , 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 (with 0 589   nm) 0 1 700 nm 1         . Solving for  we find 0.171   . 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

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    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. Then dp F dt mdv    and dv is also in the y -direction. So we have dv F m ma dt     . 56. The magnitude of the centripetal force is 2 v ma m r    for circular motion. For a charged particle F qvB  , so 2 v qvB m r   or, rearranging qBr 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     and dp 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 dt v c v c v dv ma mv c dt v ma ma c v v ma c c ma                                                                58. From the preceding problem 3 F ma   . We have 19 10 a  m/s 2 and 27 1.67 10 m    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   and 8 1.70 10 F    N

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    14 Chapter 2 Special Theory of Relativity (c) As in (a) 2.294   and 7 2.02 10 F    N (d) As in (a) 7.0888   and 6 5.95 10 F    N 59. p mv   with 2 2 2 1 1 2.5516 1 / 1 0.92 v 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 / 1.01 p 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 c c             (b) Similarly   1/2 2 2 1 1 0.536 .63509 v c c c             (c) Similarly   1/2 2 2 1 1 0.756 1.1547 v c c c             61. 6.3 GeV protons have 3 6.3 10 K   MeV and 0 7238 E K E    MeV. Then 2 2 0 7177 MeV/ E E p c c    . Converting to SI units 13 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 . 0 x y M M u u u      After the elastic collision, the signs on the above expressions are reversed, so the change in momentum as measured by Mary is 0 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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