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 1043: 51

Answer

$9.4\times 10^7Kg$

Work Step by Step

We know that $\frac{m_1v_{1i}}{\sqrt{1-\frac{v_{1i}^2}{c^2}}}=\frac{(m_1+m_2)v}{\sqrt{1-\frac{v^2}{c^2}}}$ We plug in the known values to obtain: $\frac{(8.2\times 10^6Kg)(0.5c)}{\sqrt{1-\frac{(0.50c)^2}{c^2}}}=\frac{(8.2\times 10^7Kkg)+m_2(0.26c)}{\sqrt{1-\frac{(0.260c)^2}{c^2}}}$ This simplifies to: $m_2=9.4\times 10^7Kg$
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