Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 9 - Exercises and Problems - Page 172: 62

Answer

The proof is below.

Work Step by Step

We use our answers from the last problem. The initial kinetic energy is: $K_i=\frac{1}{2}(4m)v^2+\frac{1}{2}(m)(2v)^2=4mv^2$ The final kinetic energy is $K_f=\frac{1}{2}(4m)(1.4v)^2+\frac{1}{2}(m)(.4v)^2=4mv^2$ Thus, kinetic energy is conserved.
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