Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 29 - Relativity - Problems and Conceptual Exercises - Page 1042: 44

Answer

(a) $0.50c$ (b) more than (c) $0.73c$

Work Step by Step

(a) We know that $v_{13}=\frac{v_{23}-v_{21}}{1-\frac{v_{23}v_{21}}{c^2}}$ We plug in the known values to obtain: $v_{13}=\frac{0.80c-0.50c}{1-\frac{(0.80c)(0.50c)}{c^2}}$ $\implies v_{13}=0.50c$ (b) It is given that the speed of ship A relative to Earth $(v_{23})$ increases by $0.10c$ and therefore, the relative speed between ship A and ship B will be more than $0.10c$. (c) We know that $v_{21}=\frac{v_{23}-v_{13}}{1-\frac{v_{23}v_{13}}{c^2}}$ We plug in the known values to obtain: $v_{21}=\frac{0.90c-0.50c}{1-\frac{(0.90c)(0.50c)}{c^2}}=0.73c$
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