Answer
$2v$
Work Step by Step
We are given that $m_1=4m_2$ and
$K.E_2=K.E_1$
$\implies \frac{1}{2}m_1v^2=\frac{1}{2}m_2(v^{\prime})^2$
$\implies \frac{1}{2}4m_2v^2=\frac{1}{2}m_2(v^{\prime})^2$
This simplifies to:
$v^{\prime}=2v$
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 7 - Work and Kinetic Energy - Problems and Conceptual Exercises - Page 213: 57
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