28 visualize at t t t 0 s the origins of the s s and

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37.28. Visualize: At t = t = t = 0 s, the origins of the S, S , and S reference frames coincide. Solve: We have 1 1 2 2 2 2 1 ( / ) 1 (0.80) 1.667. v c γ = = = (a) Using the Lorentz transformations, 8 6 ( ) 1.667 1200 m (0.80)(3 10 m/s)(2.0 10 s) x x vt γ = = × × = 1200 m ( ) 8 6 2 2 8 (0.80)(3 10 m/s)(1200 m) 1.667 2.0 10 s 2.0 s 3 10 m/s vx t t c γ μ × ′ = = × = − × (b) Using v = −0.80 c , the above equations yield x = 2800 m and 8.67 s. t μ ′′ =
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37.29. Solve: We have ( ) ( ) 1 1 2 2 2 2 1 1 0.60 1.25. v c γ = = = In the earth’s reference frame, the Lorentz transformations yield ( ) ( ) ( ) ( ) ( )( ) ( ) 10 8 10 10 8 10 2 2 8 ( ) 1.25 3.0 10 m 0.60 3.0 10 m/s 200 8.25 10 m 8.3 10 m 0.60 3.0 10 m/s 3.0 10 m 1.25 200 325 s 330 s 3.0 10 m/s x x vt v x t t c γ γ = + = × + × = × × × × = + = + = ×
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37.30. Model: S is the ground’s frame and S is the rocket’s frame. S moves with velocity v = 0.5 c relative to S. Solve: (a) We have ( ) ( ) 1 1 2 2 2 2 1 / 1 0.50 1.155. v c γ = = = Applying the Lorentz transformations to the lightning strike at x = 0 m and t = 10 μ s, ( ) ( ) ( )( ) 8 5 5 2 (1.155) 0 m 0.5 3.0 10 m/s 1 10 s 1732 m 1700 m (1.155)(1 10 s 0 s) 11.55 s 12 s x x vt vx t t c γ γ μ μ ′ = = × × = − ≈ − ′ = × = For the lightning strike at x = 30 km and t = 10 μ s, ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) 4 8 5 8 4 5 2 8 1.155 3.0 10 m 0.50 3.0 10 m/s 1 10 s 32.91 m 33 m 0.50 3.0 10 m/s 3.0 10 m 1.155 1 10 s 46.2 s 46 s 3.0 10 m/s x t μ μ ′ = × × × = × × ′ = × = − ≈ − × (b) The events in the rocket’s frame are not simultaneous. The lightning is observed to strike the pole before the tree by 46 + 12 = 58 μ s.
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37.31. Model: The rocket and the earth are inertial frames. Let the earth be frame S and the rocket be frame S . S moves with v = 0.8 c relative to S. The bullet’s velocity in reference frame S is u = 0.9 c . Solve: Using the Lorentz velocity transformation equation, ( )( ) 2 2 0.9 0.8 0.36 1 / 1 0.9 0.8 / u v c c u c u v c c c c ′ + + = = = − + + − The bullet’s speed is 0.36 c along the x -direction. Note that the velocity transformations use velocity , which can be negative, and not speed.
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37.32. Model: The proton and the earth are inertial frames. Let the earth be frame S and the proton be frame S . S moves with v = 0.9 c . The electron’s velocity in the laboratory frame is 0.9 c . Solve: Using the Lorentz velocity transformation equation, ( )( ) 2 2 0.9 0.9 0.994 1 / 1 0.9 0.9 / u v c c u c uv c c c c ′ = = = − − − The electron’s speed is 0.994 c .
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37.33. Model: The earth and the other galaxy are inertial reference frames. Let the earth be frame S and the other galaxy be frame S . S moves with v = + 0.2 c . The quasar’s speed in frame S is u = + 0.8 c .
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