Modern Physics for Scientists and Engineers 4th Edition Thornton Solutions Manual

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Chapter 2 Special Theory of Relativity 1

1 Chapter 2 Special Theory of Relativity

F ma d2 m 2 2 2 F ma md x i d y jd z k F. mdx mdx 2 2 k . ˆ ˆ ˆ 2 2

Modern Physics for Scientists and Engineers 4th Edition Thornton

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Chapter 2

x d2 y d2 z 1. For a particle Newton’s second law says iˆ jˆ ˆ

dt dt dt

Take the second derivative of each of the expressions in Equation (2.1): d2 x d2 x d2 y d2 y d2 z d2_z

. Substitution into the previous equation gives dt2 dt2 dt2 ˆ dt2 dt2 ˆ dt2 ˆ dt2 dt2 dt2 2. From Equation (2.1) iˆ dy jˆ dzk ˆ . dt dt dt In a Galilean transformation dx dxv dy dy dz dz. dt dt dt dt dt d t Substitution into Equation (2.1) gives v i dy j dz k p.

dt dt dt

However, because p m dxiˆ dy jˆ dz k ˆ the same form is clearly retained, given dt dt dt

the velocity transformationdx dxv. dt dt

3. Using the vector triangle shown, the speed of light coming toward the mirror is c2

v2

and the same on the return trip. Therefore the total time ist2 distance

speed 2

. c2_v2

Notice that sin v, so sin1 v

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c c

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1 2 1 2 1 2 1 2 3.0 1 2 2 1 1 1 2 1 1 1 1 2 1 2 1 2 1

4. As in Problem 3, sin v /v, so sin1

(v / v ) sin 10.350 m/s 16.3

and 1.25 m/s

2 2 2 2

v v2 v1 (1.25 m/s) (0.35 m/s) 1.20 m/s .

5. When the apparatus is rotated by 90°, the situation is equivalent, except that we have effectively interchanged and . Interchanging and in Equation (2.3) leads to Equation (2.4).

6. Letn= the number of fringes shifted; then n d. Becaused c t t , we have c t t v2

n . Solving forvand noting that + = 22 m,

v c n c2 0 108 m/s 0.005589 10 m 3.47 km/s. 9 22 m 7. Letting (where t 1 becomes

v / c ) the text equation (not currently numbered) for

t 2 _c1 2

c 12

which is identical tot 2when so t 0 as required.

8. Since the Lorentz transformations depend onc(and the fact thatcis the same constant for all inertial frames), different values ofcwould 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 speedc. According to the Galilean transformations, an observer in K measures the speed of the signal to be

dx dx

v c v. Therefore the speed of light cannot be constant under the Galilean dt dt

transformations.

10. From the Principle of Relativity, we know the correct transformation must be of the form (assuming y y and z z ):

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x ax bt ; x ax bt.

The spherical wave front equations (2.9a) and (2.9b) give us: ct (acb)t ; ct (ac b)t.

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ct (ac b)(ac b)t orc2 (ac b)(ac b) a2 c2 b2 . c

Nowvis the speed of the srcin of the x-axis. We can find that speed by setting x 0 which gives 0 ax bt, orv x/t b/a, or equivalentlyb=av. Substituting this into the equation above forc2

a 1 . 1v2 /c 2 yields c2 a2 c2 a2 v2 a2 c2 v2 . Solving fora:

This expression, along withb = av, can be substituted into the srcinal expressions for x and xto obtain:

x x vt ; x x vt

which in turn can be solved fortand t to complete the transformation. 11. Whenv cwe find 1 2 1, so: x xct 1 2 x ct x vt ; t t x / c t x / c t ; 1 2 x x ct x ct x vt ; 12 t tx/c t x /c t . 12

12. (a) First we convert to SI units: 95 km/h = 26.39 m/s, so v / c (26.39 m/s) /3.00 108 m/s 8.8108 (b) v/c (240 m/s) /3.00 108 m/s8.0 107 (c)v2.3v_sound 2.3330 m/sso v/c2.3 330 m/s/ 3.00 108 m/s 2.5 106 (d) Converting to SI units, 27,000 km/h = 7500 m/s, so v/c (7500 m/s) / 3.00 108 m/s2.5105 (e) (25 cm)/(2 ns) = 1.25108 m/sso v/c (1.25108 m/s) / 3.00 108 m/s 0.42 (f) 11014 m / 0.35 1022 s 2.857 108 m/s , so

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v / c (2.857 108

m/s) /3.00 108

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13. From the Lorentz transformationst t vx/c2

. But t 0in this case, so solving forvwe find v c2

t / x . Inserting the valuest t c2_a_{/ 2c}

2 t 1 a/ 2cand

x x2 x1 a, we findv c/ 2 . We conclude that the frame K travels

a

at a speedc/2 in the x-direction. Note that there is no motion in the transverse direction.

14. Try setting x 0 x vt. Thus 0 x vt a va/ 2c. Solving forvwe find v2c, which is impossible. There is no such frame K .

15. For the smaller values of β we use the binomial expansion 1 21/ 2

12 / 2 . (a) (b) (c) (d) (e) (f) 1 2 / 2 1 3.87 1015 12 / 2 1 3.2 1013 1 2 / 2 1 3.11012 12 / 2 1 3.11010 1 21/ 2 1 0.4221/ 21.10 1 21/ 2 1 0.9521/ 2 3.20

16. There is no motion in the transverse direction, so

1 1 _{5 / 3} y z 3.5 m. 1 2 1 0.82 x x vt 52m 0.8c010 / 3 m 3 t t vx /c2 5 0 0.8c2 m/c2 8.9 109 s 3 x2 y2 z 2 _{3 m}2_{(5 m)}2 _{(10 m)}2 17. (a) t 3.86 108 s c 3.00 108 m/s

(b) With 0.8 we find 5 / 3. Then y y5 m, z z 10m,

x x vt 53 m 2.40108

m/s(3.8610-8

s) 10.4 m 3

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t t vx/c2 5

3.86 108s 2.40 108m/s 3 m/ 3.00 108

m/s 2 51.0 ns 3

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(c) x 2_y2_z2 t 10.4 m25 m210 m2 2.994 108_{m/s which equals}_c 51.0 109 s to within rounding errors.

18. At the point of reflection the light has traveled a distance L vt 1 ct 1. On the return

2 Lc 2 L/c trip it travels L vt 2 ct 2. Then the total time ist t 1t 2

2 2 2 2

.

c v 1v /c But from time dilation we know (witht proper time 2 L0/c) that

t t 2 L0/c

. Comparing these two results fort we get 2 L/c 2 L0/c

2 2 2 1v2 /c2 1v c2 1v /c which reduces to L L0 1v2/c2 L0 . This is Equation (2.21).

19. (a) With a contraction of 1%, L / L0 0.99 1v 2

/c2

. Thus 1 2

(0.99)2

0.9801. Solving for , we find 0.14 orv 0.14c.

(b) The time for the trip in the Earth-based frame is d 5.00 106m

t 1.19 101

s .With the relativistic factor 1.01

v 0.14 3.00 108m/s

(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

0.011.2 101

s = 1.2 103

s.

20. The round-trip distance isd= 40 ly. Assume the same constant speedv c for the entire round trip. In the rocket’s reference frame the distance is onlyd d 12

. Then

in the rocket’s frame of referencev distance d 40ly 1

2

c 1 2_.

time 40 y 40 y Rearranging v

c

1 2. Solving for we find 0.5 , orv 0.5c 0.71c.

To find the elapsed timeton Earth, we know t 40 y, so t t 1 40y

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56.6 y. 21. In the muon’s f rameT 0 2.2 μs. In the lab frame the time is longer; see Equation (2.19):

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9.5 cm 12

v 9.5 cm 12

Therefore , so

cT 0

. Now all quantities are c c2.2μs

known exce pt β. Solving for β we find 1.4 104

orv1.4 104

c.

22. Converting the speed to m/s we find 25,000 mi/h = 11,176 m/s. From tables the distance is 3.84 108

m . In the earth’s frame of reference the time is the distance divided by d 3.84 108_m

speed, ort 34, 359 s . In the astronauts’ frame the time elapsed is v 11,176 m/s

t t/ t 1 2

. The time difference is t t t tt 1 2 t 1 12 .

2 Evaluating numericallyt 34, 359 s 1 1 11,176 m/s 2.4 10 5 s. 3.00 108 m/s 23.T T 0, so we know that 5 / 3 1

. Solving forvwe findv 4c/ 5. 1v2

/c2

24. L L0/ so clearly 2 in this case. Thus 2

v 3c. 2 1 1v 2 /c

2 and solving forvwe find

25. The clocks’ rates differ by a factor of 1 / 1 v2

/c2

. Because is very small we will use the binomial theorem approximation 12

/ 2 . Then the time difference is t t t t t t 1. Using 1 2

/ 2 and the fact that the time for the trip equals distance divided by speed,

2 375 m/s 6 8 t t2 / 2 8 10 m3.00 10 m/s 375 m/s 2 t1.67 108 s 16.7 ns. 26. (a) L L/ L 1v2_/_c2 _3.58104_km _{1 0.94}2 _{1.22 10}4_km (b) Earth’s frame:

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3 . 5 8 107 m t L/v 0.127 s 0.94 3.00 108 m/s

Golf ball’s f rame: t t / 0.127 s 1 0.942

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x

27. Spacetime invariant (see Section 2.9): c2

t2 x2 c2 t 2 x2 . We know x 4 km, 2 2 2 2 t 0 , and x 5km. Thust 2 xx 5000 m 4000 m 1.0 1010 s2 and t 1.0 105 s. c2 3.00 108 m/s2

28. (a) Convertingv = 120 km/h = 33.3 m/s. Now withc= 100 m/s, we have v / c 0.333 and 1 1 1.061. We conclude that the moving person ages 6.1% slower.

1 2

1 0.3332

(b) L L/ (1 m) /(1.061) 0.942 m.

29. Convertingv= 300 km/h = 83.3 m/s. Now withc=100 m/s, we have v / c 0.833

and 1 1 1.81. So the length is L L0 / 40 / 1.81 22.1 m.

1 2

1 0.8332

30. Let subscript 1 refer to firing and subscript 2 to striking the target. Therefore we can see that x1 1 m, x2 121m, andt 1 3 ns.

t ₂ t ₁ distance_{3 ns +}120 m _{3 ns + 408 ns = 411 ns.} speed 0.98c

To find the four primed quantities we can use the Lorentz transformations with the known values of

t t x1,

vx

x2,t 1, andt 2. Note that withv0.8c,

/c2 0.56 ns 1v 2 /c2 _{5 / 3 .} 1 1 1 t t vx /c2 147 ns 2 2 2 x₁ x₁vt ₁ 0.47 m x₂ x₂vt ₂ 37.3 m

31. Start from the formula for velocity addition, Equation (2.23a):

u u xv _. x

1vu /c

2

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x x 1 (0.62c)(0.84c) /c2 1.52 (b)u 0.62c0.84c 0.22c 0.46c x 1 (0.62c)(0.84c) /c2 0.48

32. Velocity addition, Equation (2.24):u u xv _{with v 0.8 c and}_u _0.8c. x

1vu /c2 x

u x 0.8c(0.8c) 1.6c 0.976c 1 (0.8c)(0.8c) /c2

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x

33. Conversion: 110 km/h = 30.556 m/s and 140 km/h = 38.889 m/s. Letu x 30.556

andv 38.889 m/s. Our premise is thatc100 m/s. Then by velocity addition, m/s u v 30.556 m/s 38.889 m/s u x _{62.1 m/s. By symmetry} x 1vu /c2 1 38.889 m/s30.556 m/s / 100 m/s2 each observer sees the other one traveling at the same speed.

34. From Example 2.5 we haveu c1nv/c. For light traveling in opposite directions n1v/nc

u c1nv/c1nv/c. Becausev/cis very small, use the binomial expansion: n1v/nc 1v/nc

1nv/c

1nv/c1v/nc1 1nv/c1v/nc1nv/c v/nc, where we 1v/nc

have dropped terms of order v2

/c2 . Similarly1nv/c1nv/c v/nc. Thus 1v/nc u c1 nv/c v/nc 1 nv/c v/nc 2v11 / n _2v_{11/ n}2 . Evaluating n n numerically we find u 2(5 m/s) 1 1 4.35 m/s. 1.332

35. Clearly the speed of B is just 0.60c . To find the speed of C useu_x 0.60cand v 0.60 c : u u xv 0.60c(0.60c)

0.88c.

x

1vu /c2

1 (0.60c)(0.60c) /c2

36. We can ignore the 400 km, which is small compared with the Earth-to-moon distance 3.84 108

m. The rotation rate is 2 rad 100 s1

2 102

rad/s. Then the speed across the moon’s surface isv R 2 102

rad/s3.84 108 m 2.411011 m/s. 37. Classical: t 4205 m 1.43105 s . Then N N _expln 2t _14.6 0.98c or about 15 muons. 1.43105 s Relativistic:t t/ 2.86 106 s 5

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E 0 t 1/ 2 _so ln 2t N N 0exp t 1/ 2

2710 muons. 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 2 R cos(39) , where RE

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rotational speed of that point isv31,143 km/ 24 h 1298 km/h 360.5 m/s . The observatory clock runs slow by a factor of 1 1 2_{/ 2 1 7.22 10}13

. In 1 2

41.2 h the observatory clock is slow by 41.2 h 7.22 1013

2.9746 1011

h = 107 ns. In 48.6 h it is slow by 48.6 h 7.22 1013

3.5089 1011

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

1 1 2_{/ 2 1 1.8110}12_{and in 41.2 hours it runs slow by}

1 2

41.2 h 1.811012

7.45721011

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

1 1 2

/ 2 1 1.85 1013

1 2 and in 48.6 hours it runs slow by

48.6 h 1.851013

8.9911012

h = 32 ns. So our prediction is that the Eastward-moving clock is off by 107 ns – 269 ns 162 ns, while the Westward-Eastward-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

signals at a rate f fort 1and a rate ffort 2. t

1 and a rate ffort 2. Frank receives

40.1 T t t2 L L L L

v c v c f T 2 f L / v signals.

2 L

; Frank sends signals at rate f , so Mary receives v

T t t 2 L; Mary sends signals at rate f, so Frank receives f T 2 f L/ v signals.

1 2 v 41. s2 x2 y2 z 2 c2 t 2

; Using the Lorentz transformation s2 2 ( x vt )2 y2 z 2 c22 (t vx /c2 )2 x22 1v2 /c2 y2 z 2 c2 t 22 1v2 /c2

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x2

y2_z2

c2

t 2

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ned, copied or duplicated, or posted to a ned, copied or duplicated, or posted to a

or in part. or in part.

42. For a timelike interval s2

0 so x2

c2

t2

. We will prove by contradiction. Suppose that there is a frame K is which the two events were simultaneous, so that t 0. Then by the spacetime invariant x2 c2 t2 x2 c2 t 2 x2 . But because x2 c2 t2 , this implies x2

0 which is impossible because xis real. 43. As in Problem 42, we know that for a spacelike interval s2

0 so x2

c2

t2

. Then in a frame K in which the two events occur in the same place, x 0and

x2 _c2_t2_x2 _c2_t2 _c2_t2_{. But because x}2 _c2_t2_{we have c}2_t2 ₀_{, 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 xaxis. The slope of the xaxis is v / c , so

v/cslope =ct. Solving forv, we find v c2

t / x . Since the slope of the xaxis x

must be less than one, we see that x ct so s2

x2

c2

t2

0 is required.

45. parts (a) and (b) To find the equation of the line use the Lorentz transformation. With t 0we havet 0 t vx/c2

or, rearranging, ct vx / c c . Thus the graph ofctvs. xis a straight line with a slope .

(c) Now with t constant, the Lorentz transformation givest t vx/c2

. Again we solve forct: ct x ct / xconstant . This line is parallel to the t 0 line we found earlier but shifted by the constant.

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from their normal ( x,ct ) orientation and they are not perpendicular.

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0

46. The diagram is shown here. Note that there is only one worldline for light, and it bisects both the x,ctaxes and the x, ct axes. The xand ct axes are not perpendicular. This can

be seen as a result of the Lorentz transformations, since t 0defines the xaxis.

x 0defines the ct axis and

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 values0 650 nm and

540 nm, solving this equation for gives 0.183 . The astronaut’s speed is v c 5.50 107

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 speedscandv, respectively, in opposite directions, so their relative speed is c v . The time interval between receipt of signals ist / (c v) 1 / f . By time dilationt t .

1 c 1v2_/c2 ₁2

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Using c/v0 and 1v2 /c2 we find t _f 0(c v) f 0(1 ) and f 1 f 0(1 ) _f 1 . t 12 0 1

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0

50. For a fixed source and moving receiver, the length of the wave train iscT vT . Sincen waves are emitted during timeT , cT vTand the frequency f

n cf T/ f c / is f 1 2 cn . cT vT 1 As in the textn f 0T 0 andT 0T/ . Therefore f

0 _f . 0 51. f f 0 1 1 1400 kHz 1 0.95 1 0.95 224 kHz cT vT 1 1

52. The Doppler shift function f f 0

1 1

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 f f 1 ₁ f01.

1

1 53. The Doppler shift function f f 0

1 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-relativisticv c. Otherwise it would be necessary to use the velocity addition law and apply the transverse Doppler shift. From the figure we see

that f 1 . Now

t ₀(t ₂t ₁₎ f 0 1/t 0 and

t ₂t ₁ 2 x 2vt 0cos _{. With an angle of 45}_{, cos(45) 1/ 2}

c c and

f 1 f 0 f 0 _.

1/ f 0(2vcos ) /cf 0 1 2vcos /c 1 2v/c

1 54. The Doppler shift to higher wavelengths is (with0 _{589 nm) 700 nm} 0 _.

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v 0.1713.00 108m/s Solving for we find 0.171. Thent 1.75 106

s a 29.4 m/s2

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F dt md v v direction. v F m ase. v mv F v F m d v/dt vd a mv 1 2 a 3 mav 2 a 1v a

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-Thend andd is also in the y-direction. So we have d

dt v2

m .

56. The magnitude of the centripetal force is ma m r v2

for circular motion. For a charged particle F qvB , soqvB m

r r p.

qB

or, rearranging qBr mv p. Therefore

When the speed increases the momentum increases, and thus for a given value of Bthe radius must incre

57. m

1v2

/c2 and

d

dt . The momentum is the product of two factors that

contain the velocity, so we apply the product rule for derivatives: dt m 1v2 /c2 d m 1 1v2 /c2 _dt 1v2 /c2 m 2v 3dv 2 c2 _dt m 3_m c2 v2 c 2 c 2

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a

3

58. From the preceding problem F 3

ma. We have a 1019 m/s2and m 1.67 1027 kg. 1 1 1.00005 (a) 1v2 /c2 1 0.012 F 1.000053 1.67 1027 kg 1019 m/s2 1.67 108 N (b) As in (a) 1.005 and F 1.70 108 N

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(c) As in (a) 2.294and F 2.02 107 N (d) As in (a) 7.0888 and F 5.95106 N 59. p mvwith 1 1 2.5516 ; 1v2 /c2 16 1 0.922 m p v 10 kg·m / s 1.42 1025 kg 2.55160.923.00 108 m/s

60. The initial momentum is p0 mv

1 1 0.52 m0.5c 0.57735mc. (a) p/ p0 1.01 1.01 mv 0.57735mc v1.010.57735c 0.58312c 1 1

Substituting for and solving forv,v

1/ 2 0.504c. .58312c2 c2 1 1 (b) Similarlyv 1/ 2 0.536c .63509c2 c2 1/ 2 1 1 (c) Similarlyv 0.756c 1.1547c2 c2

61. 6.3 GeV protons have K 6.3103

MeV and E K E 0 7238 MeV. Then

E2

p 0

7177 MeV/c. Converting to SI units c 1.60 1013 J c p7177 MeV/c 3.831018 kg·m/s MeV 3.00 108 m/s p 3.831018 kg m/s

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0

From Problem 56 we have B 1.57 T .

qr 1.60 1019

C15.2 m

62. Initially Mary throws her ball with velocity (primes showing the measurements are in Mary’s frame): u 0

x u yu . After the elastic collision, the signs on the above expressions are reversed, so the change in momentum as measured by Mary is

p M mu0 mu0 2mu0 . 1 u2 /c2 _{1 u}2 /c2 _{1 u}2 /c2 0 0 0

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F x F y 0 0 0 0 F F M

Now for Frank ’s ball, we knowu 0 andu

x y u0 . The velocity transformations give for Frank’s ball as measured by Mary:u

x v u y u0 1v

2

/c2

. To find for Frank ’s ball, note thatu F 2_u 2 _v2

u2 1 v2 /c2 . Then 1 1 1 . Using 1 2_u_/ 2 ₁_v2_/_c2_u2₁_v2_/c2_/c2 ₁_u2_/_c2₁_v2_/_c2 F c 0 0

p mu along with the reversal of velocities in an elastic collision, we find p m u 1v2 /c2 mu 1v2 /c2 2 mu 1v2 /c2 F 0 0 0 2mu 1v2 /c2 _2mu 0 0 1u2 /c2 1v2 /c2 1u2 /c2

Finally p p p 2mu02mu0 _{0 as required.}

1u2

/c2

63. To prove by contradiction, suppose that K 1mv2

. Then 2 K E E 0 mc 2 _mc2 1mc2 ₁_mv2 . This implies 1v2 / 2c2 , or 2 1v2 / 2c2

, which is clearly false.

64. The source of the energy is the internal energy associated with the change of state, commonly called that latent heat of fusion L

f. Letm be the mass equivalent of 2 grams

and M be the mass of ice required. m E c2 L M f c2 Rearranging 2 _{0.002 kg 3.00 10}8 _m/s2 M mc 5.39 108 kg L 334 103 _J/kg 65. In general K 1mc2

, so 1 K . For 9 GeV electrons: mc2

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1 9000 MeV1.76 104

0.511 MeV Then from the definition of we have

1 1 2 1 1 1 1.6 109 . Thus 1.76 1042 v 11.6 109 c 0.9999999984c.

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For 3.1 GeV positrons: 13100 MeV 6068 and 0.511 MeV Thusv11.4 108 c 0.999999986c. 1 1 1 1.4 108 . 60682

66. Note that the proton’s mass is 938 MeV/c2. In general K 1mc2

, so 1 K . mc2

Then from the definition of we have 1 1

. For the first section

2 K 0.750

MeV, and 10.750 MeV1.00080 with

938 MeV 1 1 2 1 1 0.040 . 1.000802 Thusv= 0.04cat the end of the first stage. For the other stages the computations are similar, and we tabulate the results:

K (GeV) 0.400 1.43 0.71 8 9.53 0.994 150 160.9 0.99998 1000 1067 0.9999996 67. (a) 511 keV/c 2 0.020c p mu 10.22 keV/c; 1 0.0202 511 keV/c2 (c2 ) E mc2 511.102 keV; 1 0.022

K E E 0 511.102 keV 511.00 keV 102eV

The results for (b) and (c) follow with similar computations and are tabulated:

p(keV/c) E(keV) K(keV) 0.20 104.3 521.5 10.5 0.90 1055 1172 661 1 3 3c 68. E 2 E 0 E 0 so 2 . Then 1 2 2

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and v .

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0

69. For a constant force, work = change in kinetic energy = Fd mc2

/ 4, because 2 _{80 kg 3.00 10}8 _m/s2 25%1 / 4 . d mc 2.25 1017m = 23.8 ly. 4 F 4 8 N 70. E K E 0 2 E 0 E 0 3 E 0 E 0 so 3 . Then 1 1 2 ₁ 1 32 2 2 0.943. Thusv = 0.943c. 3 71. (a) E K E 0 0.1 E 0 E 0 1.1 E 0 E 0, so 1.1 . Then 1 1 2 1 1 0.417 andv= 0.417c. 1.12 (b) As in (a) 2andv 3c/ 2. (c) As in (a) 11andv= 0.996c. 2 72. K E E E 0 _E _{so K E} E 0 _{. Rearranging 1} 2 E 0 _which 1 2 1 2 _{K E} 0 can be written as 2 2 1 E 0 K E 0 or 2 E 1 0 . K E ₀ 73. Using E mc2

along with p mvwe see that E/mc2

p/mv. Solving forv/cwe

find v / c pc / E .

74. The speed is the same for protons, electrons, or any particle. K 1mc2 1.01 1mv2 0.505mc2 2 so 1 1 1 0.505 2 . 2 2

Rearranging and solving for , we find

1 0.114 orv= 0.114c.

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75. Converting 0.1 ounce = 2.835103 kg. E mc2 2.835103 kg 3.00 108 m/s2 2.551014 J .

Eating 10 ounces results in a factor of 100 greater mass-energy increase, or 2.551016

J. This is a small increase compared with your srcinal mass-energy, but it will tend to increase your weight; depending on how they are prepared, peanuts generally contain about 100 kcal of food energy per ounce.

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1 2

V

76. The energy needed equals the kinetic energy of the spaceship. 1 K 1mc2 1 mc2 1 2 1 104 kg 3.00 108 m/s 4.35 1019 J 1 0.32 or 4.35% of 10 21 J.

77. Up to Equation (2.57) the derivation in the text is complete. Then using the integration by parts formula, xdy xy ydxand noting that in this case x uand y u, we

have ud ( u) u2 udu. Thus u K m 0 u d( u) mu 2 mu du u mu2_m _du 1u2 /c2

Using integral tables or simple substitution: u K mu2_mc2 1u2 /c2 0 mu2_mc2 ₁_u2 /c2 mc2 mc2 _mc2 1u2 /c2 mc2 1u2 /c2 mc2_mc2 _mc2_{( 1)} 78. Converting 0.11 cal ·g1 ∙ C 1 = 460 J ·kg1 ∙ C1 c

(specific heat at constant volume). From thermodynamics the energy Eused to change the temperature byTismcVT. Thus

E 1000 kg 460 J/(kg C 0.5C 2.30 105

J

5

and m E 2.30 10 J 2.56 1012

kg . The source of this energy is the internal c

2

3.00 108m/s

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2 0 0 2 79. E b 2m p2mn mHe c 2 2 1.007276 u 2(1.008665 u) 4.001505 uc2931.494 MeV 28.3 MeV c2 ·u 80. E mn m p me c 1.008665 u 1.007276 u 0.000549 uc2931.494 MeV 0.782 MeV c2 ·u 81. E K E 0 1 TeV 938 MeV 1 TeV ;

E2_E2 ₁_{TeV 938MeV}2 938 MeV2 p 0 1.000938 TeV/c c c E E 0 1.000938 TeV_{1067 ;} 2 1 1 1 8.78107 E 0 0.000938 TeV 18.78107 1 4.39107 ; and v c 0.999999561c 82. (a) E p2 c2

E 2 _40GeV2_{511 keV}2 _{40.0 GeV}

K E E 0 40.0 GeV

(b) E p2_c2_E2 _{40 GeV}2_{0.938 GeV}2 _40.011GeV

K E E 0 40.011 GeV 0.938 GeV 39.07 GeV

83. E K E 0 200 MeV 106 MeV 306 MeV

E2_E2 _{306 MeV}2_{106 MeV}2 p 0 287.05 MeV/c c c E 306 MeV 2.887

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E 0 106 MeV

1 1₂0.938 so v 0.938c

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n

84. (a) The mass-energy imbalance occurs because the helium-3 (3

He) nucleus is more tightly bound than the two separate deuterium nuclei (2

H) . (Masses from Appendix 8.) E [2m(2 H)] [m m(3 He)]c2 2(2.014102 u) (1.008665 u 3.016029 u)c2931.494 MeV 3.27 MeV c2 u (b) The initial rest energy is 2m(2

H) = 2(2.014102 u)931.494 MeV= 3752 MeV. c2

u Thus the answer in (a) is about 0.09% of the initial rest energy.

85. (a) The mass-energy imbalance occurs because the helium-4 (4

He) is more tightly bound than the deuterium (2

H) and tritium nuclei (3

H) . E [m(2 H)+m(3 H)] [m m(4 He)]c2 (2.014102 u 3.016029 u) (1.008665 u 4.002603 u)c2 931.494 MeV c2 u 17.6 MeV

(b) The initial rest energy is m(2

H)+m(3 H) c2 (5.030131 u)c2 931.494 MeV = c2 u 4686 MeV. Thus the answer in (a) is about 0.37% of the initial rest energy.

86. (a) In the inertial frame moving with the negative charges in wire 1, the negative charges in wire 2 are stationary, but the positive charges are moving. The density of the positive charges in wire 2 is thus greater than the density of negative charges, and there is a net attraction between the wires.

(b) By the same reasoning as in (a), note that the positive charges in wire 2 will be stationary and have a normal density, but the negative charges are moving and have an increased density, causing a net attraction between the wires.

(c) There are two facts to be considered. First, (a) and (b) are consistent with the physical result being independent of inertial frame. Second, we know from classical physics that

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two parallel wires carrying current in the same direction attract each other. That is, the same result is achieved in the “lab” frame.

As in the solution to Problem 21 we have v d 1

2

wheredis the length of the c ct

87.

particle track and t the particle’s lifetime in its rest frame. In this problem t 8.2 1011

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2

andd= 24 mm. Solving the above equation we find 0.698. Then

E E 0 1672 MeV _{2330 MeV} 1 2 1 0.6982 88. dp d d v 1/ 2 F mv mv 1 2 dt dt dt c m dv mv3 2v dv dt 2 c2 dt mdv1 v 22 m dv1 v 2 /c2 dt c2 dt 1 v2_/ c2 m dv 1 m 3 dv m dv 1 dt 1 v2_/ c2 _dt dt 1 v2 /c23/ 2

89. (a) The numbernreceived by Frank at fis half the number sent by Mary at that rate, or fL / v . The detected time of turnaround is

n fL/v L 1 L1 L L

t .

f 1 / 1 v 1 v v c

(b) Similarly, the number nreceived by Mary at fis

n fT f 1 L f L 1 . Her turnaround time isT / 2 L/ v.

2 1 v v

(c) For Frank, the timet 2for the remainder of the trip ist 2=T− t 1= L/v− L/c.

Number of signals = f t ₂ f 1 L/v L/c fL.

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1 v fL fL 2 fL

. v v v Mary’s age = Total number received 2 L.

f v

(d) For Mary,t 2 T t 1 L/ v.

Number of signals = f t 2 f

1 L fL_{1 .}

Total number received =

1 v v fL

1 1 2 fL.

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Frank ’s age = Total number received 2 L.

f v

90. In the fixed frame, the distance is Δ x= 8 ly and the elapsed time is Δt= 10 y, so the interval is s2 x2 c2 t2 64 ly2 100 ly2 = 36 ly2

. In the moving frame, Mary’s clock is at rest, so x 0 , and the time interval ist 6 y. Thus the interval is

s2_x2 _c2_t2

0 ly2

36 ly2

= 36 ly2

. The results are the same, as they should be, because the spacetime interval is the same for all inertial frames.

91. (a) From Table 2.1 number fL1 1 0.8 55.9 56 52y1 _{4.3 ly} v L L 4.3 ly 4.3 ly 0.8c fL 2 (b)t 1 9.68 y ; num ber f t 1 1 167.7 168 v c 0.8c c v (c) Frank: f t f L1 2 168 so the total is 168 + 168 = 336. 2 v Mary: number 2 f L/v559 (d) Frank:T 2 L/v10.75 y Mary: T 2 L/v6.45 y

(e) From part (c), 559 weeks = 10.75 years and 336 weeks = 6.46 years, which agrees with part (d).

92. Notice that the radar is shifted twice, once upon receipt by the speeding car and again upon reemission (reflection) of the beam. For this double shift, the received frequency fis

f f 0 1 1 1 1 1 1

For speeds much less thanc, a Taylor series approximation gives an excellent result: f 1

1 11 1 2 . f 0 1

Converting 80 mph to 35.8 m/s gives β = 1.19 × 10−7, so the received frequency is approximately

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f f 01 2 10.4999975 GHz , which is 2.5 kHz less than the srcinal frequency.

93. For this transformationv= 0.8c(so γ = 5/3),u_x 0 andu y 0.8c. Applying the

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c2 x y 0 x u u xv 0 0.8c _0.8c_; _u yu 0.8c 0.48c 1vu x c2 1 0 1vu x 5_{1 0} 3 The speed isu u2 u2

0.93c, safely under the speed of light. 94. (a) (b) K E E 0 250 E 0. Thus 250 K 249 E 0 249 511keV127.2 MeV 1 1 0.999992 ₂ v0.999992c E2 E2 _{250 511 keV}2 511 keV2 (c) p 0 128 MeV/c c c 95. (a) For the proton: p mu 1 938 MeV /c2_0.9c_{1940 MeV /c}

. 1 0.92

For the electron: E p2

c2

E 2 _{1940 MeV}2_{0.511 MeV}2 _{1940 MeV .}

E E 0 _{1940 MeV} 0.511 MeV 3797 1 1 2 1 1 37972 1 6.94 10 8 1 3.97 108 v1 3.97 108c 1 (b) For the proton: K 1 E ₀

1 0.92 1 938 MeV1214 MeV .

For the electron: K E 0 1214 MeV 0.511 MeV _{2377 .}

1 1

2

E 0

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23772 0 . 5 1 1 M e V 1 1 . 7 7 1 0 7 1 8 . 8 5 1 0 8 v1 8.85108 c

96. In the frame of the decaying K 0

meson, the pi mesons must recoil with equal speeds in opposite directions in order to conserve momentum. In that reference frame the available kinetic energy is 498 MeV 2 140 MeV 218 MeV . The pi mesons share this equally,

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2

0 0

so each one has a kinetic energy of 109 MeV in that frame. The speed of each pi meson can be found:

K E 0 109 MeV 140 MeV _{1.779 so u} ₁1

c 0.827c .

E 0 140 MeV

The greatest and least speeds in the lab frame are obtained when the pi mesons are released in the forward and backward directions. Then by the velocity addition laws:

vmax vmin 0.9c0.827c 0.990c 1 0.90.827 0.9c0.827c 0.285c 1 0.90.827

97. (a) The round-trip distance is L0 8.60 ly . Assume the same constant speedv c for

the entire trip. In the rocket’s frame, the distance is only L L1

L 12

. Mary will age in the rocket’s reference frame a total of 22 y, and in that frame

distance L 8.60 ly 12

v 0.39c

time 22y 22y 0.392

1 2

. Therefore, v 0.39 12

. c

Solving for we find

1.00 0.39

2 , or 0.36 so Mary’s speed isv 0.36c .

(b) To find the elapsed time on Earth, we know that T 0 22y , so

T T 0

1

22y 23.6 y. Frank will be 53.6 years old when Mary returns at the 1 2

age of 52 y.

98. Withv 0.995c then 0.995and 10.01. The distance out to the star is L0 5.98 ly . In the rocket’s reference frame, the distance is only

L L1 5.98 ly

0.597 ly.

0

10.01

(a) The time out to the star in Mary’s frame istime distance 0.597 ly 0.6 y. The speed 0.995c

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time for the return journey will be the same. When you include the 3 years she spends at the star, her total journey will take 4.2 y.

(b) To find the elapsed time on Earth for the outbound journey, we know thatT 0 0.6y so

T T 0 (10.01)0.6 y 6.006 y. The return journey will take an equal time. The 3 years the

spaceship orbits the star will be equivalent for both observers. Therefore Frank will measure a total elapsed time of 15.012 y, which makes him 10.8 years older than her.

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0 0

99. (a) The Earth to moon distance is 3.84 108

m. The rotation rate is 2 rad 0.030 s1

1.885101

rad/s. Then the speed across the moon’s surface is v R1.885101

rad/s3.82 108

m 7.24 107

m/s .

(b) The required speed can be found using c R 2 f R which requires the frequency c 3.00 108

m/s

to be f 0.124 Hz .

2 R _{2 3.84 10}8

m/s 100. With the data given, 3.28106

, which is very small. We will use the binomial approximation theorem. From Equation (2.21), we know that

1 1 2 2

L L₀ . 1 1 / 2 .

(a) The percentage of length contraction would be: % change L0 L L 0 100% L L1 L 0 100% 1 1 2 100% 1 (1 2 / 2) 100% 100% 5.37 1010 2

(b) The clocks’ rates differ by a factor 1 / 1 2

. The clock on the SR-71 measures the proper time and Equation 2.19 tells us thatT T 0

2

so the time difference is t T T 0 T 0T 0 T 0( 1) . Using 1 / 2 and the fact that the time for

the trip in the SR-71 equals the distance divided by the speed

2 983 m/s 6 8 t t2 / 2 3.2 10 m 3.00 10 m/s 983 m/s 2 1.75 108 s 17.5 ns

101. As the spaceship is approaching the observer, we will make use of Equation (2.32), 1

f 1

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This equation indicates that the Doppler shifted frequency will be larger than the frequency measured in a frame where the observer is at rest with respect to the source. Since c f , this means the Doppler shifted wavelengths will be lower. As given in the problem, we see that the difference in wavelengths for an observer at rest with respect to

the source is 0 0.5974nm . We want to find a speed so that the Doppler shifted

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2 1 c c f f c 1 2 f 2 f 1 f 1 f 2 1 1 f _0,1 f _0,2 c 1 1 c 1 f 0,1 f 0,2 1 f f 1 f f 0,1 0,2 1 0,1 0,2 1 c c 1 0,2 0,1 1 1 f _0,2 f _0,1 1 0 1 2 2

We can complete the algebra to show that v 2.47 107

m/s .

0

0.0825 so that

2 2

102. As we know that the quasars are moving away at high speeds, we make use of Equation (2.33) and the equation c f . Using a prime to indicate the Doppler shifted frequency

1 (or wavelength), Equation (2.33) indicates that frequency is given by f

1 f 0,or f ₀ 1 so f 1 0 c/ f z ₁ 1 0 0 c/ f 0

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f 0

1 1 1

f 1

Therefore z12 1 . We can complete the algebra to show that 1

z121 z121

z121

and thusv c.

z121 For the values of zgiven, v 0.787c for

z1.9 and v 0.944c for z 4.9 . 103. (a) In the frame of the decaying K 0

meson, the pi mesons must recoil with equal momenta in opposite directions in order to conserve momentum. In that reference frame the available kinetic energy is 498 MeV 2 135 MeV 228 MeV .

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4

(b) The pi mesons share this equally, so each one has a kinetic energy of 114 MeV in that frame. The energy of each pi meson is E K E 0 114 MeV 135 MeV 249 MeV .

The momentum of each pi meson can be found: E2

E2 _{249 MeV}2_{135 MeV}2

p 0 _{209.2 MeV/c c} _c

104. (a) For each second, the energy used is 3.9 × 1026J. The mass used in each second is

26 m E c2 3.9 10 J 4.310 9 kg 3.0 108 m/s2

(b) The time to use this mass is

30 t msun efficiency 2.0 10 kg 0.007 3.31018 s 1.0 1011 y. m/t 4.3109 kg/s

That is 20 times longer than the expected lifetime of the sun. 105. a) The planet’s orbital speed isv p

2r

1.30 104m/s . If the system’s center of mass T

is at rest, conservation of momentum gives the star ’s speed: v

m_pv_p 1.90 1027

m pv p msvs, so s 301.30 10 m/s = 12.4 m/s

1.99 10 ms

(b) The redshifted wavelength is 0

1

1 550 nm + 2.3 10

5

nm

Similarly, the blueshifted wavelength is 0

1

1 550 nm 2.3 10

5

nm

These small differences can be detected using modern spectroscopic tools.

106. To a good approximation, the shift is the same in each direction, so the redshift for the approaching light is half the difference, or 0.0045 nm. Using this in the equation

0 1

1 leads to a speed parameter β = 6.9 × 10

−6

, orv= β c= 2070 m/s. The speed

is equal to the circumference 2π Rdivided by the period, so the period is

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2 6.96 108

m

T 2.11106

s = 24.5 days.

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Modern Physics for Scientists and Engineers 4th Edition Thornton

Solutions

Manual

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