C spaceship 3 finds that epsilon explodes before

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(c) Spaceship 3 finds that Epsilon explodes before Delta. (d) No. The explosions are far enough apart that Delta can have no causal influence on Epsilon. Thus there’s no difficulty if Epsilon explodes before Delta in some reference frames.
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37.57. Model: Let S be the earth’s reference frame and S be the reference frame of one rocket. S moves relative to S with v = 0.75 c . The speed of the second rocket in the frame S is u = + 0.75 c . Visualize: Solve: Using the Lorentz velocity transformation equation, ( ) ( )( ) 2 2 0.75 0.75 0.96 1 1 0.75 0.75 c c u v u c uv c c c c − − ′ = = = Assess: In Newtonian mechanics, the Galilean transformation of velocity will give u = 0.75 c ( 0.75 c ) = 1.50 c . This is not permissible according to the theory of relativity.
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37.58. Model: The earth’s frame is S and the rocket’s frame is S . S (the rocket) moves relative to S (the earth) with velocity v , which we want to know. We are given 0.90 u c = and 0.95 . u c ′ = Visualize: Solve: We want to solve for v in the Lorentz velocity transformation equation: 2 2 1 / 1 / u v u u u v uv c uu c ′ = = 2 2 0.90 0.95 0.34 1 / 1 (0.90 )(0.95 )/ u u c c v c uu c c c c = = = − Assess: We expected the rocket to be moving to the left.
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37.59. Model: Use the relativistic expression for kinetic energy in Equation 37.44: p 0 ( 1) K E γ = . The electric potential energy of the electron is transformed into its kinetic energy. Visualize: Use the conservation of energy equation f i f i 0 J. U U k K + = Solve: f i f f ( )( ) 0 J 0 J e V V K K e V + = = Δ ( ) 2 31 8 2 1 2 1 0 99 p 0 p 6 f 19 1 (9 1 10 kg)(3 0 10 m/s) ( 1) ( 1) 3 1 10 V 1 6 10 C E mc K V e e e γ γ − . . × . × Δ = = = = = . × . × Assess: Three million volts is easily obtainable.
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37.60. Model: Use the relativistic expression for the kinetic energy in Equation 37.44. The electric potential energy of the proton is transformed into its kinetic energy. Solve: The conservation of energy equation U f U i + K f K i = 0 J is ( e )( V f V i ) + K f – 0 J = 0 J K f = e Δ V ( γ p – 1) mc 2 = e Δ V ( )( ) ( )( ) 19 6 p 2 2 2 2 27 8 1.6 10 C 50 10 V 1 1 1 1.05323 1 / 1.67 10 kg 3.0 10 m/s e V mc v c γ × × Δ = + = + = = × × 1 – v 2 / c 2 = 0.90148 v = 0.314 c
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37.61. Model: Let S be the ground’s reference frame and S the muon’s reference frame. S travels with a speed of v relative to S. Solve: (a) The half-life of a muon at rest is 1.5 μ s. That is, the half-life in the muon’s rest frame S is 1.5 μ s. So, Δ t = Δ τ = 1.5 μ s. The half-life of 7.5 μ s, when muons have been accelerated to very high speed, means that Δ t = 7.5 μ s. Thus ( ) 2 2 2 2 2 1.5 s 7.5 s 1 0.20 1 1 t v c v c v c τ μ μ Δ Δ = = = = v = 0.98 c (b) The muon’s total energy is ( ) ( )( ) 2 2 2 31 8 11 p 2 2 1 1 207 9.11 10 kg 3.0 10 m/s 8.5 10 J 0.20 1 E mc mc v c γ = = = × × = ×
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37.62. Model: Let S be the sun’s reference frame and S be the rocket’s reference frame. S moves with speed v = 0.8 c relative to S. The flare’s speed in the frame S is u = 0.9 c .
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