Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 6 - Exercises and Problems: 57

Answer

Please see the work below.

Work Step by Step

We know that $KE=(\frac{1}{2})mv^2$ This simplifies to $v=\sqrt{\frac{2KE}{m}}$ Hence the required ratio is $\frac{v_1}{v_2}=(\sqrt{\frac{KE_1}{KE_2}})(\sqrt{\frac{m_2}{m_1}})=(\sqrt{1}{1})(\sqrt{\frac{1}{4}})=0.5$
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