chapter 37 solutions

# chapter 37 solutions - RELATIVITY 37 Figure 37.1 37.1...

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R ELATIVITY 37.1. I DENTIFY and S ET U P : Consider the distance A to OO and B to OO as observed by an observer on the ground (Figure 37.1). Figure 37.1 E XECUTE : Simultaneous to observer on train means light pulses from and A B O arrive at OO at the same time. To observer at O light from AO has a longer distance to travel than light from B O so O will conclude that the pulse from ( ) A AO started before the pulse at ( ). B BO To observer at O bolt A appeared to strike first. E VALUATE : Section 37.2 shows that if they are simultaneous to the observer on the ground then an observer on the train measures that the bolt at B O struck first. 37.2. (a) 2 1 γ 2.29. 1 (0.9) = = - 6 6 γ (2.29) (2.20 10 s) 5.05 10 s. t τ - - = = x = (b) 8 6 3 (0.900) (3.00 10 m s) (5.05 10 s) 1.36 10 m 1.36 km. d vt - = = = ; = 37.3. I DENTIFY and S ET U P : The problem asks for u such that 0 1 / . 2 t t ∆ = E XECUTE : 0 2 2 1 / t t u c ∆ = - gives ( 29 2 2 8 8 0 1 1 / (3.00 10 m/s) 1 2.60 10 m/s 2 u c t t � � = - ∆ = - = � � � � ; 0.867 u c = Jet planes fly at less than ten times the speed of sound, less than about 3000 m/s. Jet planes fly at much lower speeds than we calculated for u . 37.4. I DENTIFY : Time dilation occurs because the rocket is moving relative to Mars. S ET U P : The time dilation equation is 0 t t γ ∆ = ∆ , where t 0 is the proper time. E XECUTE : (a) The two time measurements are made at the same place on Mars by an observer at rest there, so the observer on Mars measures the proper time. (b) 0 2 1 (75.0 s) 435 s 1 (0.985) t t γ μ μ ∆ = ∆ = = - E VALUATE : The pulse lasts for a shorter time relative to the rocket than it does relative to the Mars observer. 37.5. (a) I DENTIFY and S ET U P : 8 7 0 2.60 10 s; 4.20 10 s. t t - - = x ∆ = ; In the lab frame the pion is created and decays at different points, so this time is not the proper time. E XECUTE : 2 2 0 0 2 2 2 says 1 1 / t u t t c t u c ∆ = - = - 2 2 8 0 7 2.60 10 s 1 1 0.998; 0.998 4.20 10 s u t u c c t - - = - = - = = E VALUATE : , u c < as it must be, but u / c is close to unity and the time dilation effects are large. (b) I DENTIFY and S ET U P : The speed in the laboratory frame is 0.998 ; u c = the time measured in this frame is , t so the distance as measured in this frame is d u t = E XECUTE : 8 7 (0.998)(2.998 10 m/s)(4.20 10 s) 126 m d - = = E VALUATE : The distance measured in the pion’s frame will be different because the time measured in the pion’s frame is different (shorter). 37-1 37

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37-2 Chapter 37 37.6. γ 1.667 = (a) 8 0 1.20 10 m 0.300 s. γ γ(0.800 ) t t c = = = (b) 7 (0.300 s) (0.800 ) 7.20 10 m. c = (c) 0 0.300 sγ 0.180 s. t = = (This is what the racer measures your clock to read at that instant.) At your origin you read the original 8 8 1.20 10 m 0.5 s. (0.800) (3 10 m s) n = Clearly the observers (you and the racer) will not agree on the order of events! 37.7. I DENTIFY and S ET U P : A clock moving with respect to an observer appears to run more slowly than a clock at rest in the observer’s frame. The clock in the spacecraft measurers the proper time 0 . t 365 days 8760 hours. t ∆ = = E XECUTE : The clock on the moving spacecraft runs slow and shows the smaller elapsed time.
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