Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 9 - Exercises and Problems - Page 165: 73


18.6 percent

Work Step by Step

Using conservation of momentum, we find: $v=v_{1f}cos(37^{\circ})+2v_{2f}cos\theta$ $v=.799v_{1f}+2v_{2f}cos\theta$ We also know: $v_{1f}sin(37^{\circ})=v_{2f}sin\theta$ $v_{1f}=-1.66v_{2f}sin\theta$ Combining these equations gives: $v=-1.32v_{2f}sin\theta+2v_{2f}cos\theta$ We know from conservation of kinetic energy that: $\frac{1}{2}v^2 =\frac{1}{2}v_{1f}^2+v_{2f}^2$ Combining these equations, we find the value of the ratio of the kinetic energies: $\frac{K_{2f}}{K_{i}}=(\frac{v_{1i}}{1.11v_{2f}})^2=.813 $ $1-.814=\fbox{18.6 percent}$
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