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TRex3.CH-2,12.solutions

# TRex3.CH-2,12.solutions - 24 25 13 Chapter 2 3 Using the...

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Unformatted text preview: 24. 25. 13. Chapter 2 3. Using the vector triangle shown, the U W C V Speed rm speed of the light coming toward the mirror is Va? — v2 and the same on the return trip. Therefore the total time is distance t2 2 — . . U . Notice that smﬂ = —, so 9 = SKI—l ' c c . As in Problem 3, sinﬁ = Ulfm, so 6‘ = sin’1(v1/vg) = sin’1 [ﬁ—ﬁéﬂ = 19.9 ' and u = \/ u;i — v15 = ‘f (0.94 lcm/h)2 — (0.32 km/h) = 0.88 km/h . When the apparatus is rotated by 90°, the situation is equivalent, except that we have effec- tively interchanged £1 and £2. Interchanging £1 and 22 in Equation (2.3) leads to Equation (2.4). From the Lorentz transformations At’ = '7 [At — UAI/c2] But At' = D in this case, so solving for v we ﬁnd 1,: = cht/Ax Inserting the values At = t2 — t1 = ~a/2c and AI: = 2:2 F 591 = a, vi: M = #6/2 a We conclude that the frame K’ travels at a speed c/2 in the —\$—direction. Note that there is no motion in the transverse direction. L = Lyn/“y so clearly 1 = 2 in this case. Thus 2 = ——\/i_—___27§ and solving for '0 we ﬁnd —v c 2 The clocksI rates differ by a factor of '7 = 1/ (/1 — Eli/62. Since {3 is very small we will-nee the binomial theorem approximation 7 m 1 + (62 / 2. Then the time differenCe is Atzt—t'zt-rytzth—l) Using 7 — 1 z 62 / 2 and the fact that the time for the trip equals distance divided by speed, 2 375 _ 8 x 106 m(3’_5%'.00x1iin in At = t 2 ._ _——— ([3 m 375m/s 2 = 1.67 >< 10-8 s = 16.7 ns 27. Spacetime invariant (see Section 2.9): c213? _— A32 = c2At'2 # As”; We know Ax = 4 km, At = 0, and AI’ = 5 km. Thus ,2 _ as? — A32 _ (5000 m)2 — (4000 m)2 _ = 1.0 x10‘” s2 c2 (3.00 x 108 m/s)2 At and At] = 1.0 X 10’5 s 32. Velocity addition , it: ~— 11 u _‘______.__—— x : 1 — var/c2 with v = —0.8c and u: = 0.8c. I 0.8a — (-0.86) 2 L65 : 0.976c :___#__,____ “I 1—(—e.8c)(0.8c)/c2 1.64 * 37. Classical: t = 4205 m 21.43 x 10—5 s 0.98:: Then - N = No exp : 14.6 or about 15 muons 1/2 Relativistic: 1 43 104 t'=t/'y: ‘ X5 5 22.86x10’65 N = No exp = 2710 muons t1/2 Because of the exponential nature of the decay curve, a factor of ﬁve (shorter) in time results in many more muons surviving. L L L L 2L 4D.T=t1+t2=;+:c_+___,:__ v c 1.! , Hank sends signals at rate f, so Mary receives fl" = 2fL/o signals. L T! = fl + t} = 2—— 71; Mary sends signals at rate f, so Frank receives fT’ = 2fL/7'u signals. 1—13 1—092 f=fo 1+16:(400m) H092 :82 Hz * 54. The Doppler shift to higher wavelengths is (with A0 = 589 nm) 1+,0’ A2700 :A nm 0 lug Solving for £3 we ﬁnd 15’ = 0.171. Then (0.171) (3.00 x 103 111/3) 25 f 2 = 2.052 x106 5 m s v t:—-: G. which is 23.75 days. One problem with this analysis is that we have only computed the 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 :r-direction and the force be in the y-direction. Then dﬁ’ = F dt = 7m d1? and (it? is also in the y—direction. So we have —- d1? F: m— : ' d 7 dt “’13” 57. _, _, me = mv=—————— p 7 x/l—U2/C2 _. d" Fzﬂ The momentum is the product of two factors that contain the velocity, 50 we apply the product rule for derivatives: _. d mﬁ' F : mﬁl:\/1—v§/c2 diT/dt __ Td 1 = m|:\/1—U§/CQTLE(\/1-Ui/C§ _. _, 1 21.: 3dv : 7ma+mv 'Y E; ﬁc andv: 2 . 76. The energy needed equals the kinetic energy of the spaceship. 1 — 1) mac2 K H CE” I .5 3 n M H /"_"‘\ H | 7:1) M II 1 (J1 — 0.35 or 4.35% of 1021 J. 80. AB {mu—rnp—melc2 H [1.008665 11 — 1.007276 11 — 0.000549 11] 02 ( 82. a) E = M1358 + 53 = 1/(30 GeV)2 + (511 keV)2 2 30.0 GeV K=E—E0=30.0GeV )1 931.494 MeV c2-u b) E = \/p2c2 + 133 z 1/(30 GeV)2 + (0.938 GeV)2 = 30.015 GeV K = E — E0 = 30.015 GeV A 0.938 GeV = 29.08 GeV — 1) (104 kg] (3.00 x 108 m/s)2 = 4.35 x 1019 J ) = 0.782 MeV 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. 1)] 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 two parallel wires carrying current in the same direction attract each other. That is, the same result is achieved in the “lab” frame. 92. a) K = E+ E0 = 200Eg so K = 199193 = 199 (511 keV) = 102 MeV 26. b) '7 = 200 1 25‘ = 1/1— —2 = 0.9999875 7 v = 0.9999875c c) p : E2 _ D _ (200 x 511 keV}2 — (511 km? 6 c = 102 MeV/c CNN?! ER ‘1‘ R=Rge_lt=%att=T=3GGGs 1112 ln2 1112 . tI/g — T — ET —— E—5 (3600 s)=1550 3 ~ 26 minutes ...
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