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# solution_#_3 - Chapter 2 Special Theory of Relativity 26 a...

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Unformatted text preview: Chapter 2 Special Theory of Relativity * 26. a) L’ = L/’y = L./1— 0%? = 3.58 x 104 km 1/1— 0.942 = 1.22 x 104 km b) Earth’s frame: 3.58 X 107 m = = ———————-— = . 7 t M" (0.94) (3.00 x 108 m/s) 0 12 5 Golf ball’s frame: t’ = t/y = 0.127 s 1/1 — 0.942 = 0.0433 s 27. Spacetime invariant (see Section 2.9): czAt2 ~‘— A002 = czAt’2 — Ax’ 2 We know A0: = 4 km, At = 0, and Ax’ = 5 km. Thus I2 _ 2 2 _ 2 At'2 = Act: 2 Ax = (5000 m) (4000 1211) = 1.0 x 10_10 82 C (3.00 x 108 m/s) and At, = 1.0 x 10—5 s * 31. Start from the formula for velocity addition, Equation (2.23a): u’z + v u:c = —— 1 + 'uuQE/c2 1 a) u _ O.7c+0.8c —l£E—096c a” ” 1 + (0.7c)(0.8c)/c2 _ 1.56 _ ' b) u _ —0.7c+ 0.8c ~ 0.1c __ O 23c ” ‘ 1 + (—0.7c)(0.8c)/c2 _ 0.44 _ ' 32. Velocity addition 11' '_ um —— v ' - m —. 1 —— Hum/C2 with v = —0.8c and u3 = 0.8a. 0.8c — (—0.8c) 1.6c I = ———-—————'-————- = —-——- = _ 7 “z 1 — (—0.8c)(0.8c)/c2 1.64 0 9 66 * 37. Classical: 4205 . In t: = .43 10*5 0.9a: 1 x S Then 1 2 h N = No exp [ﬂ] = 14.6 or about 15 muons 1/2 Relativistic: 5 t, =t/7 = 1_43><_510_s = 2.86 x 10—6 s N = No exp [M] = 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. 8 41. 32 =\$2+y2+22_c2t2 Using the Lorentz transformation .92 : 72(ml+vtl)2+yl2+zl2_c272(t/+U\$I/c2)2 \$1272 (l—vz/c2)+y’2+z’2—c2t'272 (1;,02/c2) \$I2+yl2+zl2_c2tl2=sl2 ll N; 44. In order for two events to be simultaneous in K’, the-two events must lie along the 93’ axis, or along a line parallel to the :c’ axis. The slope of the as’ axis is ﬂ = v/c, so v/c = slope = %. Solving for v, we ﬁnd 1) = c2At/Az. Since the slope of the :1," axis must be less than one, we see that An: > cAt, so 32 = sz —- c2At2 > 0 is required. 57. ﬁ—vmi— m” \/1 —112/02 s dp‘ F: — dt The momentum is the product of two factors that contain the velocity, so we apply the product rule for derivatives: 13 mi WW dt x/l—vZ/c2 || l__| H a. | '21 e \ to E: \ ('3 {0 e1 Ecle l—l I l—l e m \ n [O V L—l H ‘2 3 Q + 3 61 /T\ [0| H V + 27) 3d'u 02 7 dt '02 = 7ma+73mfi<c—2> 2 2 'u v _. = 73ma[1—c—2+Z§]—73ma 58. From the previous problem F = 73m a. We have a‘ = 1019 m/s2 and m = 1.67 x 10—27 kg. a) 7 = ———1—— = ——1——— = 1.00005 x/l—vQ/c2 \/1—0.012 F = (1.00005)3 (1.67 x 10-27 kg) (1019 m/s2) =1.67 x 10—8 N b) As in (a) 7 = 1.005 and F = 1.70 x 10—8 N c) As in (a) 'y = 2.294 and F = 2.02 x 10‘7 N d) As in (a) 'y = 7.0888 and F = 5.95 X 10‘6 N * 60. The initial momentum is 1 = mv = —————m 0.5c = 0.57735 mc P0 ”Y ,——~1 —O.52 ( ) a) p/po = 1.01 vmv .01 = —— 1 0.57735 mc 75 = (1.01) (057735 c) = .58312 0 Substituting for 7 and solving for v, 1 1 4/2 0 0 = —— —— = .5 v l(.58312c)2+c2] 4c 14 b) Similarly c) Similarly ...
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solution_#_3 - Chapter 2 Special Theory of Relativity 26 a...

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