# C spaceship 3 finds that epsilon explodes before

This preview shows pages 57–64. Sign up to view the full content.

(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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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.
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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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.
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 = = = × × = ×
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 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern