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: 49

Answer

$-1.6m/s$

Work Step by Step

We know that $v_2=\pm\frac{m_1v_1}{\sqrt{m_2^2-(m_2^2-m_1^2)\frac{v_1^2}{c^2}}}$ We plug in the known values to obtain: $v_2=\pm\frac{(88Kg)(2.0m/s)}{\sqrt{(120kg)^2-(120Kg)^2-(88Kg)^2\frac{(2.0m/s)^2}{(3.0m/s)^2}}}$ $v_2=-1.6m/s$
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