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# ssm_ch37 - Chapter 37 11(a The rest length L0(= 130 m of...

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Chapter37 11 (a) The rest length L 0 (= 130 m) of the spaceship and its length L as measured by the timing station are related by L = L 0 = L 0 p 1 ¡ ¯ 2 , where ° = 1 = p 1 ¡ ¯ 2 and ¯ = v=c . Thus L = (130 m) p 1 ¡ (0 : 740) 2 = 87 : 4 m. (b) The time interval for the passage of the spaceship is ¢ t = L v = 87 : 4 m (0 : 740)(2 : 9979 £ 10 8 m = s) = 3 : 94 £ 10 ¡ 7 s : 19 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: x 0 = ° [ x ¡ vt ] t 0 = ° [ t ¡ ¯x=c ] ; where ¯ = v=c = 0 : 950 and ° = 1 = p 1 ¡ ¯ 2 = 1 = p 1 ¡ (0 : 950) 2 = 3 : 2026. Thus x 0 = (3 : 2026) £ 100 £ 10 3 m ¡ (0 : 950)(2 : 9979 £ 10 8 m = s)(200 £ 10 ¡ 6 s ¤ = 1 : 38 £ 10 5 m = 138 km and t 0 = (3 : 2026) 200 £ 10 ¡ 6 s ¡ (0 : 950)(100 £ 10 3 m) 2 : 9979 £ 10 8 m = s ¸ = ¡ 3 : 74 £ 10 ¡ 4 s = ¡ 374 ¹ s : 21 (a) The Lorentz factor is ° = 1 p 1 ¡ ¯ 2 = 1 p 1 ¡ (0 : 600) 2 = 1 : 25 : (b) In the unprimed frame, the time for the clock to travel from the origin to x = 180 m is t = x v = 180 m (0 : 600)(2 : 9979 £ 10 8 m = s) = 1 : 00 £ 10 ¡ 6 s : The proper time interval between the two events (at the origin and 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 0 = t ° = 1 : 00 £ 10 ¡ 6 s 1 : 25 = 8 : 00 £ 10 ¡ 7 s : 238

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