f5ch38 - CHAPTER 38 RELATIVITY 1035 CHAPTER 38 Answer to...

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Unformatted text preview: CHAPTER 38 RELATIVITY 1035 CHAPTER 38 Answer to Checkpoint Questions 1. 2. 3. 4. 5. (a) same (speed of light postulate); (b) no (the start and end of the ight are spatially separated); (c) no (again, because of the spatial separation) (a) Sally's; (b) Sally's (a) negative; (b) positive; (c) negative (a) right; (b) more (a) equal; (b) less Answer to Questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. all tie (pulse speed is c) 0 0 (a) C1 ; (b) C1 (a) C1 ; (b) C1 (a) Sam; (b) neither (a) 3, 2, 1; (b) 1 and 3 tie, then 2 (a) U{V{W{N{M{G{F{A{B{C{D{E (use either L= c or 2L= c to nd the time for each lane; pick lanes with large values); (b) you; (c) you (a) negative; (b) positive (a) no; (b) yes; (c) yes (c), then (b) and (d) tie, then (a) less (a) 3, tie of 1 and 2, then 4; (b) 4, tie of 1 and 2, then 3; (c) 1, 4, 2, 3 (b), (a), (c), (d) greater than f1 1036 CHAPTER 38 RELATIVITY 14. (a) 3, then 1 and 2 tie; (b) 2, then 1 and 3 tie; (c) 2, 1, 3; (d) 2, 1, 3 Solutions to Exercises & Problems 1E (a) (b) (c) (d) (e) 2 m/in.)(1 7 = v = (1 in./y)(0:2540 :10 108 m/s y=3:15 10 s) = 3 10 c 30 18 : 10 m/mi)(1 = v = (55 mi/h)(1:609 108 m/s h=3600 s) = 8:2 10 8 : c 3:00 3 = (1200 km/h)(10m/km)(1 h=3600 s) = 1:1 10 6 : 3:00 108 m/s 3 m/km) = 3:7 10 5 : = (11:2 km/s)(10 m/s 8 3:00 10 3 = v = (3:0 1000km/s)(10 m/km) = 0:10 : c 3: 108 m/s 4 3 (a) The time it takes for an electron with a horizontal component of velocity v to travel a horizontal distance L is 20 10 m t = L = (0:992)(300 108 m/s) = 6:7 10 v : 2 10 2E s: (b) During this time it falls a vertical distance 1 y = 1 gt2 = 2 (9:8 m/s2 )(6:72 10 2 3P 10 s)2 = 2:2 10 18 m: 60 :c : v = s = 2:0 y:+ly:0 y = (600y )(1600y) = 0:75c: t 6 2: + : y CHAPTER 38 RELATIVITY 1037 Solve from = (1 2 ) 1=2 : = (1 2 )1=2 : p (a) = p 1 (1:01) 2 = 0:140: (b) = p1 (10:0) 2 = 0:9950: (c) = p1 (100) 2 = 0:999 950: (d) = 1 (1000) 2 = 0:999 999 50: Use the time dilation equation t = tp where t0 is the proper time interval, = 0, p 2 , and = v=c. Thus t = t = 1 2 and the solution for is 1= 1 0 s 5E 4E = 1 t0 t 2 : The proper time interval is measured by a clock at rest relative to the muon. That is, t0 = 2:2 s and t = 16 s. This means s = 1 2:2 s 16 s 2 = 0:99 : The muon speed is v = c = 0:99c = (0:99)(3:00 108 m/s) = 3:0 108 m/s. 6P Use Eqs. 38-6 through 38-8: t = s = s 1 v 2 t0 = v v c p 3 2 = 1:05 : 10 :m 1108(0:992) = 4:45 10 (0 992)(3 00 m/s) s 13 s: The mean life time of a pion measured in a frame xed on the Earth is t = t0 , so the distnace it can travel is 9 8 s = vt = vt0 = (0:99)(3:00 p 10 m/s)(26 10 s) = 55 m : 1 (0:99)2 7P 1038 CHAPTER 38 RELATIVITY 8P (a) Find v = c from = (1 2 ) 1=2 = t=t0 : s v = c = 1 t0 t 2 s = 1 1 y 2 c = 0:999 999 50 : 1000 y (b) Yes. In fact if you do not travel in a straight line then you would be undergoing an acceleration, meaning that you are no longer in an inertial frame of reference in which the Lorentz transformation is valid. The length L of the rod, as measured in a frame in which it is moving with speed v parallel p to its length, is related to its rest length L0 by L = L0 = , where = 1= 1 2 and = v=c. Since must be greater than 1, L is less than L0 . For this problem L0 = 1:70 m p and = 0:630, so L = (1:70 m) 1 (0:630)2 = 1:32 m. 10E 9E (a) Use Eq. 38-12 and solve for v = c: s v = c = c 1 L L0 2 s =c 1 2 1 2 = 0:866c : (b) The factor is = (1 2 ) 1=2 = [1 (0:866)2 ] 1=2 = 2:00: 11E The coordinate dierences between the two ends of the meter stick in S 0 frame is x0 = p p L0 cos 30 = (1:0 m)( 3=2) = 3 m=2 and y0 = L0 sin 30 = 0:50 m. When measured in S , length contraction gives x = x0 = and y = y0 . So the length in S is L = (x) + (y) = 2 2 p s x0 2 = ( 3 m=2)2 [1 (0:90)2 ] + (0:50 m)2 = 0:63 m : q p + (y0 )2 = q (x0 1 2 )2 + (y0 )2 p 12E L = L0 = L0 1 2 = (3:00 m) 1 (0:999 9987)2 = 0:0153 m : p p CHAPTER 38 RELATIVITY 1039 13E jLj = L0 1 L = L0 L = L0 (1 2 s p 1 2) 4 = 2(6:370 104 m) 41 1 3:0 10 m/s 3:00 108 m/s 2 3 5 = 0:064 m : (a) The rest length L0 (= 130 m) of the spaceship and its length L as measured by the p p timing station arep related by L = L0 = = L0 1 2 , where = 1= 1 2 and = v=c. Thus L = (130 m) 1 (0:740)2 = 87:4 m. (b) The time interval for the passage of the spaceship is m t = L = (0:740)(387:4 108 m/s) = 3:94 10 7 s : v :00 (a) The speed of the traveler is v = 0:99c which is the same as 0:99 ly/y. Let d be the distance traveled. Then the time for the trip, as measured in the frame of the Earth, is t = d=v = (26 ly)=(0:99 ly/y) = 26 y. (b) The signal, presumed to be a radio wave, travels with speed c and so takes 26:0 y to reach Earth. The total time elapsed, in the frame of the Earth, is 26:3 y + 26:0 y = 52:3 y. (c) The proper time interval is measured by a clock in the spaceship, so t0 = t= . Now p p = 1= 1 2 = 1= 1 (0:99)2 = 7:09. Thus t0 = (26:3 y)=(7:09) = 3:7 y. (a) 16P 15P 14E 2 1 630 m/s = 2 3:00 108 m/s = 2:21 10 12 : (b) Let jt t0 j = t0 ( 1) = = 1:00 s and solve for t0 : jLj = L0 (1 1 ) = 1 L L 0 p 0 1 1 2 1 1 2 = 1 2 2 2 2 t0 = 1 = 1 + 1 2 1 = 2 2 ) 1=2 (1 1 2 2(1:00 10 6 s)(1 = [(630 m/s)=(3:00 d=86400 s) = 5:25 d : 108 m/s)]2 1040 CHAPTER 38 RELATIVITY (a) In principle, yes. If the person moves fast enough, then from the time dilation argumnet his travel time measurd from the Earth is much longer than a normal lifetime. Alternatively, from the length contraction argument the distance he needs to cover (measured with respect to his spaceship) is much less than 23; 000 ly. Either way, we conclude that it is possible for him to reach the center of the galaxy in a normal lifetime. (b) Solve v from = (1 2 ) 1=2 = t=t0 23; 000 y=30 y: s 17P v = c = c 1 30 y 23000 y 2 = 0:999 999 15c : The proper time is not measured by clocks in either frame S or frame S 0 since a single clock at rest in either frame cannot be present at the origin and at the event. The full Lorentz transformation must be used: 18E x0 = (x vt) t0 = (t x=c) ; where = v=c = 0:950 and = 1= 1 2 = 1= 1 (0:950)2 = 3:2026. Thus p p x0 = (3:2026) 100 103 m (0:950)(3:00 108 m/s)(200 10 6 s = 1:38 105 m = 138 km and t0 = (3:2026) 200 10 6 s (0:950)(100 103 m) = 3:74 10 4 s = 374 s : 3:00 108 m/s 19E (a) The coordinates in the S 0 frame are: x x0 = (x vt) = p vt 2 1 8 :400)(3:00 108 m/s)(2:50 s) = 0 ; = 3:00 10 m (0p 1 (0:4)2 y0 = y = 0; z0 = z = 0, and (0:400)(3:00 103 m) = 2:29 s : 0 = t vx = p 1 t 2:50 s c2 3:00 108 m/s 1 (0:400)2 CHAPTER 38 RELATIVITY 1041 (b) Let v ! v to obtain 8 8 x p x0 = p + vt 2 = 3:00 10 m + (0:400)(3:00 2 10 m/s)(2:50 s) = 6:55 108 m ; 1 (0:400) 1 y0 = y = 0; z0 = z = 0, and (0:400)(3:00 108 m) = 3:16 s : 0 = t + vx = p 1 2:50 s + 3:00 108 m/s t c2 1 (0:400)2 For event 1 and for event 2 20E t01 = t1 + vx1 = (0) = 0 ; 2 c Therefore the dierence in time order is possible. 21E vx2 = p 1 60)(3: 3 4:0 10 6 s (0:3:00 0108 10 m) c2 m/s 1 (0:60)2 = 2:5 10 6 s = 2:5 s : Since x2 =t2 = 3:0103 m=(4:010 6 s) = 7:5108 m/s > c, the two events are independent. t02 = t2 (a) Take the ashbulbs to be at rest in frame S and let frame S 0 be the rest frame of the second observer. Clocks in neither frame measure the proper time interval between the ashes, so the full Lorentz transformation must be used. Let tb be the time and xb be the coordinate of the blue ash, as measured in frame S . Then the time of the blue ash, as measured in frame S 0 , is 0 = tb xb ; tb c where = v=c = 0:250 and = 1= 1 2 = 1= 1 (0:250)2 = 1:0328. Similarly, let tr be the time and xr be the coordinate of the red ash, as measured in frame S . Then the time of the red ash, as measured in frame S 0 , is p p t0r = tr xr : c Now subtract the rst Lorentz transformation equation from the second. Recognize that tb = tr since the ashes are simultaneous in S . Let x = xr xb = 30:0 km and let t0 = t0r t0b . Then 3 t0 = cx = (1:0328)(0:250)(30 10 m) = 2:58 10 5 s : 3:00 108 m/s 1042 CHAPTER 38 RELATIVITY (b) Since t0 is negative, t0b is greater than t0r . The red bulb ashes rst in S 0 . 22E From Eq. 1 in Table 38-2, x0 = x= vt0 . Then from Eq. 2 t0 = t= vx0 =c2 . Combine these two equations to obtain 0 0 = x v t vx : x c2 Solve for x0 : x0 = (x vt). Here we used Similarly we can get from Eq. 2 to Eq. 20 . 23P 1 = p 1 v2 =c2 : This is Eq. 10 . (a) The Lorentz factor is =p 1 = 1:25 : 1 (0:600)2 1 2 (b) In the unprimed frame the time for the clock to travel from the origin to x = 180 m is m t = x = (0:600)(3180 108 m/s) = 1:00 10 6 s : v :00 The proper time interval between the two events (clock at origin and clock at x = 180 m) is measured by the clock itself. The reading on the clock at the beginning of the interval is zero, so the reading at the end is t 10 6 t0 = = 1:00 1:25 s = 8:00 10 7 s : 24P =p 1 (a) Let the separation between the two events in S 0 be x0 = (x vt) = 0 ; then S 0 frame must move toward S frame, along their common axis (the x axis), at a speed of v = x = (5:00 10 (720 m)c 108 m/s) = 0:480c : 6 s)(3:00 t (b) Since t0 = t vx = [1 (0:48)2 ] 1=2 5:00 s (0:48)(720 m) c2 3:00 108 m/s = 4:39 s > 0 ; CHAPTER 38 RELATIVITY 1043 the big ash would occur rst. (c) t0 = 4:39 s, as calculated in (b). Let x0 = (x vt) = 0 to get the speed v of the S 0 frame: v = x=t. Now v < c so x=t < c, and t > x = 3:00 720 m m/s = 2:40 s ; c 108 i.e., the time interval in S frame cannot be less than 2:40 s. 26E 25P There are two possible solutions, depending on the relative directions of the velocity of the particle and the velocity of frame S , both as measured in S 0 . First suppose both are in the positive x direction. Then v = 0:40c and u = 0:60c. According to the velocity transformation equation (Eq. 38-23) 0 : v0 + v = 1 + uv0u 2 = 1 + (0::40c)( 0060cc)=c2 = 0:263c : =c 40c :60 Notice that in the equation u is the velocity of S 0 relative to S . Since S is moving with speed 0:60c in the positive x direction according to S 0 , S 0 is moving with the same speed but in the negative x direction according to S . Now suppose the velocity of frame S is in the negative x direction when viewed from S 0 . Then u = +0:60c and 60c v = 1 + 0:40cc+ 0::60c)=c2 = 0:81c : (0:40 )(0 27E (a) Use Eq. 38-23: 62c v0 + v = 1 + uv0u 2 = 1 + 0:47cc+ 0::62c)=c2 = 0:84c ; =c (0:47 )(0 in the direction of increasing x (since v > 0). The classical theory predicts that v = 0:47c + 0:62c = 1:1c > c. (b) Now v0 = 0:47c so : 62c v0 + v = 1 + uv0u 2 = 1 + ( 0047cc+ 0::62c)=c2 = 0:21c ; =c :47 )(0 still in the direction of increasing x. The classical prediction is v = 0:62c 0:47c = 0:15c: 1044 CHAPTER 38 RELATIVITY Let the reference frame be S in which in particle approaching the South Pole is at rest, and the frame that is xed on Earth be S 0 . Then u = 0:60c and v0 = 0:80c. The relative speed is now the speed of the other particle is S : 28E v0 + 60c v = 1 + uv0u 2 = 1 + 0:80cc+ 0::60c)=c2 = 0:95c : =c (0:80 )(0 (a) Let frame S 0 be attached to us and frame S be attached to Galaxy A. Take the positive axis to be in the direction of motion of Galaxy A, as seen by us. In S 0 our velocity is v0 = 0 and the velocity of Galaxy A is 0:35c in the positive x direction. This means u = 0:35c. Our velocity, as observed from Galaxy A, is 29E v0 + v = 1 + uv0u 2 = u = 0:35c : =c The negative sign indicates motion in the negative x direction. (b) In frame S 0 the velocity of Galaxy B is v0 = 0:35c, so in S it is 0 : v = 1v++ u0 = 1 + ( 0::35c)( 0035cc)=c2 = 0:62c : uv 0 35c :35 The negative sign again indicates motion in the negative x direction. Denote the reference frame xed on the earth as S and that xed on Q1 be S 0 . Then the velocity of Q2 measured on Q1 is u 0 800c v0 = 1 v vu=c2 = 1 0:400cc)(0::800c)=c2 = 0:588c ; (0:400 where the minus sign indicated that Q2 is moving away from Q1 (i.e., towards the Earth). Calculate the speed of the micrometeorite relative to the spaceship. Let S 0 be the reference frame for which the data is given and attach frame S to the spaceship. Suppose the micrometeorite is going in the positive x direction and the spaceship is going in the negative x direction, both as viewed from S 0 . Then in Eq. 38-23, v0 = 0:82c and u = 0:82c. Notice that u in the equation is the velocity of S 0 relative to S . Thus the velocity of the micrometeorite in the frame of the spaceship is 82c v0 + v = 1 + uv0u 2 = 1 + 0:82cc+ 0::82c)=c2 = 0:9806c : =c (0:82 )(0 31P 30P CHAPTER 38 RELATIVITY 1045 The time for the micrometeorite to pass the spaceship is 350 t = L = (0:9806)(3:00m 108 m/s) = 1:2 10 6 s : v (a) vr = 2v = 2(17; 000 mi/h) = 34; 000 mi/h: (b) The correct formula for vr is vr = 2v=(1 + v2 =c2 ) so the fractional error is 1 1 1 1 + v2 =c2 = 1 1 + [(17; 000 mi/h)=(6:17 108 mi/h)]2 = 6:4 10 10 32P : The speed of the spaceship after the rst increment is v1 = 0:5c. After the second one, it becomes v0 : : v2 = 1 + v+vv1 2 = 10+50c:+ 0)50c2 = 0:80c ; 0 =c (0 50c 2 =c 1 33P and after the third one v0 50c v3 = 1 + v+vv2 2 = 1 + 0:50cc+ 0::80c)=c2 = 0:929c : 0 2 =c (0:50 )(0 Continuing with this process, you can get v4 = 0:976c; v5 = 0:992c; v6 = 0:997c and v7 = 0:999c. Thus seven increments are needed. (a) In the messenger's rest system (called Sm ), the velocity of the armada is 0 95c v0 = 1 v vvvm 2 = 1 0:80cc)(0::95c)=c2 = 0:625c : (0:80 m =c The length of the armada as measured in Sm is p L L1 = 0 = (1:0 ly) 1 ( 0:625)2 = 0:781 ly : 34P v 0 Thus the length of the trip is 0 t0 = jL0 j = 00::781 ly = 1:25 y : v 625c 1046 CHAPTER 38 RELATIVITY (b) In the armada's rest frame (called Sa ), the velocity of the messenger is 0 80c v0 = 1 v vvva 2 = 1 0:95cc)(0::80c)=c2 = 0:625c : (0:95 a =c Thus the length of the trip is 0 ly t0 = L0 = 01::625c = 1:6 y : 0 v (c) Measured in system S , the length of the armada is L = L0 = 1:0 ly 1 (0:80)2 = 0:60 ly ; p so the length of the trip is 0 t = v L v = 0:95c:60 ly:80c = 4:0 y : 0 m a 35E The spaceship is moving away from the Earth, so the frequency received is given by Eq. 3825: s f = f0 1 + ; 1 r where f0 is the frequency in the frame of the spaceship, = v=c, and v is the speed of the spaceship relative to the Earth. Thus f = (100 MHz) 1 + 0::9000 = 22:9 MHz : 1 0 9000 36E (a) Use the classical Doppler shift equation in Chapter 18: f 0 = fv=(v + vs ); or 0 = (v + vs )=v: Solve for vs : 0 vs = v 1 = c 3 1 = 2c > c : (b) Now use Eq. 38-25 to nd vs = s c : 2 2 1 1 0 2 1 s = 1 + (f=f0 )2 = 1 + (0 =)2 = 1 + (1=3)2 = 0:8 ; (f=f0 ) ( =) (1=3) CHAPTER 38 RELATIVITY 1047 so vs = s c = 0:8c: 37E Use the transverse Doppler shift formula, Eq. 38-27: f = f0 1 2 a, or p 1 2 : Solve for 0 : p 1 = 0 1 0 = 0 p 1 2 1 1 = (589:00 mm) p 1 1 (0:100)2 1 = +2:97 nm : 38P The spaceship is moving away from the Earth, so the frequency received is given by Eq. 3825: s f = f0 1 + ; 1 where f0 is the frequency in the frame of the spaceship, = v=c, and v is the speed of the spaceship relative to the Earth. The frequency f and wavelength are related by f = c, so if 0 is the wavelength of the light as seen on the spaceship and is the wavelength detected on Earth, then s = 0 1 + = (450 nm) 1 + 0:20 = 550 nm : 1 1 0:20 r This is in the yellow-green portion of the visible spectrum. 39P (a) This is a time dilation problem. The answer is (see Eqs. 38-6 to 38-8) s = 0 = p 0 1 2 s : (b) Use Eq. 38-25 and note that fR = R 1 and f0 = 0 1 : R = f1 = f0 1 + 1 R 40E s 1 = 0 1+ = c+v : 1 0 c v r Use W = K = me c2 me0 c2 = me0 c2 ( 1) and me0 c2 = 0:511 MeV for an electron. 1048 CHAPTER 38 RELATIVITY (a) W = me0 c2 (b) (c) p 1 2 1 1 = (0:511 MeV) p 1 1 (0:50)2 1 = 0:079 MeV : W = (0:511 MeV) p 1 1 (0:990)2 W = (0:511 MeV) p 1 1 (0:9990)2 1 = 3:11 MeV : 1 = 10:9 MeV : 41E (a) (b) 2(6:37 106 v = 2Re c = (1:00 s)(3:00 10m)c = 0:134c : 8 m/s) Tc K = (me me0 )c2 = me0 c2 ( 1) = (0:511 MeV) (c) The classical value of K is p 1 1 (0:134)2 1 = 4:65 keV : 1 1 1 Kc = 2 me0 v2 = 2 me0 c2 2 = 2 (0:511 MeV)(0:134)2 = 4:59 keV: The percent error is then jK Kc j = 4:65 keV 4:59 keV = 1:3% : K 4:65 keV 42E Solve and from K = me0 c2 ( 1) = me0 c2 [(1 2 ) 1=2 1] to obtain s = 1 and = K=me0 c2 + 1. K +1 me0 c2 2 CHAPTER 38 RELATIVITY 1049 (a) Now K = 1:00 keV so s = 1 1:00 keV + 1 0:511 MeV 2 = 0:0625 and = 1:00 keV=0:511 MeV + 1 = 1:00196 : (b) Now K = 1:00 MeV so s = 1 1:00 MeV + 1 0:511 MeV 2 = 0:941 and = 1:00 MeV=0:511 MeV + 1 = 2:96 : (c) Now K = 1:00 GeV so s = 1 1:00 GeV + 1 0:511 MeV 2 = 0:999 999 87 and = 1:00 GeV=0:511 MeV + 1 = 1960 : Use the formulas found in 42E. p (a) me0 c2 = 0:511 MeV so = 10:0 MeV=0:511 MeV + 1 = 20:6 and = 1 2 = p 1 (20:6) 2 = 0:9988: p (b) Now m0 c2 = 937 MeV so = 10:0 MeV=938 MeV + 1 = 1:01 and = 1 2 = p 1 (1:01) 2 = 0:145: (c) Now m0 c2 =p :00 u)(932 MeV/u) = 3:73 103 MeV so = 10:0 MeV=3:73 103 MeV = (4 p 1:0027 and = 1 2 = 1 (1:0027) 2 = 0:073. 44E 43E Use the two expressions for the total energy: E = me0 c2 + K and E = me0 c2 , where me0 is p the rest mass of an electron, K is the kinetic energy,p = 1= 1 2 . Thus me0 c2 +K = and me0 c2 and = (me0 c2 + K )=me0 c2 . This means 1 2 = (me0 c2 )=(me0 c2 + K ) and s = 1 Now me0 c2 = 0:511 MeV so s me0 c2 me0 c2 + K 2 : = 1 0:511 MeV 0:511 MeV + 100 MeV 2 = 0:999 987 : 1050 CHAPTER 38 RELATIVITY The speed of the electron is 0:999 987c or 99:9987% the speed of light. 45E (a) For a proton (b) For an electron E = mp c2 = mp0 c2 = p 938 MeV 2 = 6:65 GeV ; 1 (0:990) K = E mp0 c2 = 6:65 GeV 938 MeV = 5:71 GeV : p p = mp v = mp0 v (938 MeV)(0:990)=c = 6:59 GeV=c ; 1 (0:990)2 E = me c2 = me0 c2 = p0:511 MeV 2 = 3:62 MeV ; 1 (0:990) K = E me0 c2 = 3:625 MeV 0:511 MeV = 3:11 MeV ; p p = me v = me0 v (0:511 MeV)(0:990)=c = 3:59 MeV=c : 1 (0:990)2 46E It doesn't mater how the energy is generated, just how much the total amount is. Since the rest energy E0 and the mass m of the quasar are related by E0 = mc2 , the rate P of energy radiation and the rate of mass loss are related by P = dE0 =dt = (dm=dt)c2 . Thus 1041 W dm = P = 24 dt c2 (3:00 108 m/s)2 = 1:11 10 kg/s : Since a solar mass is 2:0 1030 kg and a year is 3:156 107 s, dm = (1:11 1024 kg/s) 3:156 107 s/y = 18 smu/y : dt 2:0 1030 kg/smu 47E E = (2:2 1012 KW h)(103 W/kW)(3600 s/h) = 88 kg : m = c2 (3:00 108 m/s) 48E Use the work-energy theorem: W = K = ( f 1)me0 c2 ( i 1)me0 c2 = ( f i )me0 c2 : CHAPTER 38 RELATIVITY 1051 (a) W= q 1 f 1 2 p 1 i 1 2 me0 c2 1 p = 1:0 keV : 1 (0:18)2 = (0:511 MeV) p 1 1 (0:19)2 (b) W = (0:511 MeV) p 1 2 p 1 2 = 1:1 MeV : 1 (0:99) 1 (0:98) Note that 1:1 MeV 1:0 keV, meaning that it gets more and more dicult to accelerate the electron as its speed approaches c. 49P (a) Let K = ( 1)m0 c2 = 2m0 c2 , or = 3. Then v = c = 1 2 c = 1 3 2 c = 0:943c. p (b) Now E = mc2 = m0 c2 = 2m0 c2 so = 2. Thus v = 1 2 2 = 0:866c: p p Since the total momentum of the two particles is zero in S 0 , it must be that the velocities of these two particles is equal in magnitude and opposite in direction in S 0 . Let the velocity of S 0 be u relative to S , then the particle which is at rest in S must have a velocity of 0 v1 = u in S 0 , while the velocity of the other particle is 50P u 0 v2 = 1 v2uv u 2 = 1 c=2 2c2 : uc= 2 =c 0 0 Let v2 = v1 = u to obtain c=2 u = u : 1 u=2c Solve for u: u = 0:27c. (a) Classically, let eV = K = 1 me0 c2 , we get 2 MeV = 256 kV : e V = m20 c = 0:511e e 2 2 51P 1052 CHAPTER 38 RELATIVITY (b) Now K = eV = ( 1)me0 c2 so v = c = c 1 12 = c 1 r s eV + 1 me0 c2 2 s =c 1 1 +1 2 2 = 0:746c : (a) In general, the momentum of a particle with mass m and speed v is given by p = mv, where is the Lorentz factor. Since the particle of the problem has momentum mc, mc = mv or 1 = , where = v=c. Now nd an expression for in terms ofp. Since p p 2 ) = ( 2 2 . Thus = 2, 2 = 1 2 = 1= 1 1)= 1= and 1 = 2 1. p (1= The solution is = 2 = 1:41. (b) The speed parameter is 2 p = 1 = 2 1 = 0:707 : 2 52P p p The speed of the particle is v = c = 0:707c. p (c) The kinetic energy is K = ( 1)mc2 = ( 2 1)mc2 = 0:414 mc2 . 53P Let E = p2 c2 + (mc2 )2 = 3mc2 and solve for p: p = 8mc. p p 54P (a) The photon, which always moves at the speed of light. (b) For the electron, e = Ke =me0 c2 + 1 = 0:40=0:511 + 1 = 1:78. For the proton, p = Kp =mp0 c2 + 1 = 10=938 + 1 = 1:011. Since e > p we have ve > vp , so the proton is moving the slowest. p 2 (c) and (d) For the photon pph = Eph =c = 2:0 eV=c. For the electron pe = qEe m20 c4 =c e p 2 c4 =c 2 2 MeV=c = 0:75 MeV=c. For the proton p = 2 Ep mp0 = (0:40 + 0:51) (0:511) p p = (938 + 10)2 (938)2 MeV=c = 137 MeV=c: So the photon has the greatest momentum while the photon has the least. 55P The energy equivalent of one tablet is mc2 = (32010 6 kg)(3:00108 m/s)2 = 2:881013 J. This provides the same energy as (2:88 1013 J)=(1:30 108 J/gal) = 2:21 105 gal of gasoline. The distance the car can go is d = (2:21 105 gal)(30:0 mi/gal) = 6:65 106 mi. CHAPTER 38 RELATIVITY 1053 56P (a) From E = K + mc2 = p2 c2 + (mc2 )2 we solve for m: ) m = (pc2Kc2K : 2 2 p (b) As u=c ! 0 we have p = mv mv mc, so p E = K + mc = mc 1 + mc 2 2 s 2 1 p mc 1 + 2 mc 2 2 : Here we used (1 + x)n 1 + nx for jxj 1. Thus 1 p K ' mc2 1 + 2 mc 2 2 mc2 = 2pm ; as expected from classical mechanics. (c) [(121 MeV=c)c]2 (55:0 MeV)2 = 106 MeV=c2 m = 207m : m= e 2(55:0 MeV)c2 0:511 MeV=c2 e (The particle is a muon.) 57P The distance traveled by the pion in the frame of the Earth is d = vt, where v is the speed of the pion and t is the pion lifetime, both as measured in that frame. The proper time interval t0 is measured in the rest frame of the pion, so t = t0 . We must calculate the speed of the pion and the Lorentz factor . Since the total energy of the pion is given by E = mc2 , E 105 MeV = mc2 = 1:35 :6 MeV = 967:1 : 139 Since = 1= 1 2 , 2 2 = 1 = (967:1) 1 = 0:999 999 5 : 967:1 p p p The speed of the pion is extremely close to the speed of light and we may approximate as 1. Thus v = 3:00 108 m/s. 1054 CHAPTER 38 RELATIVITY The pion lifetime as measured in the frame of the Earth is t = (967:1)(35:0 ns) = 3:385 104 ns = 3:385 10 5 s. The distance traveled is d = (3:00 108 m/s)(3:385 10 5 s) = 1:02104 m = 10:2 km. The altitude at which the pion decays is 120 km 10:2 km = 110 km. (a) Use = =0 = (1 2 ) 1=2 to nd v = c: 1 = 1 = 1 2 58P s s 0 2 s = 1 2:20 s 6:90 s 2 = 0:948 ; so v = 0:948 c. (b) (c) 6 90 s K = ( 1)m0 c = 2::20 s 1 (106 MeV) = 226 MeV : 2 p = E 2 (m0 c2 )2 =c = (K + m0 c2 )2 (m0 c2 )2 =c p = (226 + 106)2 (106)2 MeV=c = 314 MeV=c : p p 59P (a) E = mc2 = (3:0 kg)(0:10%)(3:00 108 m/s)2 = 2:7 1014 J. (b) The mass of TNT is 14 mTNT = (2:7 103:4J)(0:227 kg/mol) = 1:8 107 kg : 106 J (c) The fraction of mass converted in the TNT case is mTNT = (3:0 kg)(0:10%) = 1:6 10 9 ; mTNT 1:8 107 kg So the fraction in question is (0:10%)=1:6 10 9 = 6:0 106 . 60P (a) Let me be the rest mass of an electron. From K = ( 1)me0 c2 we get s = 1 K +1 me0 c2 2 s = 1 10:0 MeV + 1 0:511 MeV 2 = 0:9988 ; CHAPTER 38 RELATIVITY 1055 so classically 31 00 8 r = me0 v = (9:11 10 :60 kg)(0:9988)(3::20 T)10 m/s) = 7:76 10 4 m : qB (1 10 19 C)(2 (b) Using relativistic formula, we have 4 7 p r = me0 v = me0 v=qB = p:76 10 m2 = 1:60 10 2 m : qB 1 (0:9988) 1 2 (c) The true period is 2r = 2(1:60 10 2 m) = 3:35 10 T = v (0:9988)(3:00 108 m/s) The result is not speed-independent. In fact 10 s: me0 v p p = 2me0 2 ; T = 2r = 2v v qB 1 2 qB 1 which is a function of = v=c. 61P The radius r of the path is given in 60P as r= so p qB 1 2 p mv ; p 19 C)(1:00 T)(6:28 m) 2 2 m = qBr v1 = 2(1:60 10 (0:710)(3:00 108 m/s) 1 (0:710) = 6:64 10 27 kg : Since 1:00 u = 1:66 10 27 u, m = 4:00 u. The nuclear particle contains 4 nucleons. Since there must be 2 protons to provide the charge 2e, the nuclear particle is a helium nucleus with 2 protons and 2 neutrons. Now = K=mc2 + 1 and s 62P = 1 K +1 mc2 2 s = 1 10 GeV + 1 938 MeV 2 = 0:9963 ; 1056 CHAPTER 38 RELATIVITY so the radius is (see 60P) 27 8 1)(1 67 r = mv = (10 GeV=938 MeV +60 :10 10 kg)(0:9963)(3:00 10 m/s) 19 C)(55 10 6 T) qB (1: = 6:6 105 m = 660 km : Now = K=me0 c2 + 1 = 2:50 MeV=0:511 MeV + 1 = 5:892 and = 1 so from r = me0 v=qB we get 63P p 2 = 0:9855, 8 :11 10 31 kg)(0: B = me0 v = (5:892)(9(1:60 10 19 C)(39855)(3:00m) 10 m/s) = 0:330 T : qr :0 10 2 64P (a) (b) (c) 500 GeV = mKc2 + 1 = 938 MeV + 1 = 534 : p0 = 1 2 = 1 (534) 2 = 0:999 998 25 : 27 8 p B = mqr0 pv = (534)(1:67 10 :60 kg)(0:99999825)(3:00 10 m/s) = 2:23 T : (1 10 19 C)(750 m) p p signal 1: signal 2: signal 3: 1014 m; signal 4: 1014 m; signal 5: 66 65 sent at 6:0 h, reply received at 400 h, reported at 11:8 h, distance = 2:10 1014 m; sent at 12:0 h, reply received at 800 h, reported at 23:6 h, distance = 4:191014 m; sent at 18:0 h, reply received at 1200 h, reported at 35:5 h, distance = 6:29 sent at 24:0 h, reply received at 1600 h, reported at 47:3 h, distance = 8:38 sent at 30:0 h, reply received at 2000 h, reported at 59:1 h, distance = 1:051015 m (a) From Eq. 38-9 (length contraction) the length L0c of the car according to Garageman is L0c = Lc = Lc 1 2 = (30:5 m) 1 (0:9980)2 = 1:93 m : p p CHAPTER 38 RELATIVITY 1057 (b) Since the xg axis is xed to the garage xg2 = Lg = 6:00 m. As for tg2 , note from Fig. 38-25(b) that, at tg = tg1 = 0 the coordinate of the front bumper of the limo in the xg frame is L0c , meaning that the front of the limo is still a distance Lg L0c from the back door of the garage. Since the limo travels at a speed v, the time it takes for the front of the limo to reach the back door of the garage is given by 0 6:00 1 93 tg = tg2 tg1 = Lg v Lc = 0:9980(3m :108m = 1:36 10 8 s : :00 m/s) Thus tg2 = tg1 + tg = 0 + 1:36 10 8 s = 1:36 10 8 s. (c) The limo is inside the garage between times tg1 and tg2 , so the time duration is tg2 tg1 = 1:36 10 8 s. (d) Again from Eq. 38-9 (length contraction) the length L0g of the garage according to Carman is L0g = Lg = Lg 1 2 = (6:00 m) 1 (0:9980)2 = 0:379 m : p p (e) Again, since the xc axis is xed to the limo xc2 = Lc = 30:5 m. Now calculate tc2 . From the two diagrams in part (h) below we know that, at tc = tc2 (when event 2 takes place) the distance between the rear bumper of the limo and the back door of the garage is given by Lc L0g . Since the garage travels at a speed v, the front door of the garage will reach the rear bumper of the limo a time tc later, where tc satis es L L0 30:5 m 0 tc = tc1 tc2 = c v g = 0:9980(3:00 :379 m = 1:01 10 7 s : 108 m/s) Thus tc2 = tc1 tc = 0 1:01 10 7 s = 1:01 10 7 s. (f) From Carman's point of view the answer is obviously no. (g) Event 2 occurs rst according to Carman, since tc2 < tc1 . (h) Lc L'g xc L'g Lc xc event 2 event 1 (i) Both Carman and Garageman are correct in their respective reference frame. 1058 CHAPTER 38 RELATIVITY (a) The separation between the two bursts is vt. Project this length onto the direction perpendicular to the light rays headed to Earth to obtain Dapp = vt sin . (b) Firstly, burst 1 is emitted a time t ahead of burst 2. Secondly, burst 1 has to travel an extra distance L than burst 2 before reaching the Earth, where L = vt cos (see Fig. 3826), which requires an additional time t0 = L=c. Thus the apparant time is given by 67 Tapp = t t0 = t (c) vt cos = t 1 c v cos : c :980) sin 30:0 Vapp = Dapp = 1 (v=c) sin c = 1 (0(0:980) cos 30:0 c = 3:24 c : Tapp (v=c) cos ...
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This note was uploaded on 01/22/2012 for the course PHYS 2101 taught by Professor Grouptest during the Fall '07 term at LSU.

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