Answer
$0.98c$
Work Step by Step
We can find the required speed as follows:
$E=\frac{m_{\circ}c^2}{\sqrt{1-v^2/c^2}}=bm_{\circ}c^2$
This simplifies to:
$v=c\sqrt{1-\frac{1}{b^2}}$
$\implies v=c\sqrt{1-\frac{1}{(5.5)^2}}$
$v=0.98c$
Physics Technology Update (4th Edition)
by Walker, James S.
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: 62
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