Essential University Physics: Volume 1 (4th Edition)

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

Chapter 9 - Exercises and Problems - Page 173: 81


$v_1=v/6$ $v_2=5v/6$

Work Step by Step

We know from conservation of momentum that: $mv = mv_{1f}+mv_{2f}$ $v = v_{1f}+v_{2f}$ $v -v_{1f}= v_{2f}$ 5/18 of the initial kinetic energy is lost, so: $(1-5/18)(\frac{1}{2}mv^2)=\frac{1}{2}mv_{1f}^2+\frac{1}{2}mv_{2f}^2$ $(1-5/18)(\frac{1}{2}mv^2)=\frac{1}{2}mv_{1f}^2+\frac{1}{2}m(v -v_{1f})^2$ $(1-5/18)(v^2)=v_{1f}^2+(v -v_{1f})^2$ $-\frac{5v^2}{18}=2v_{1f}^2-2vv_{1f}$ To find the ratio, we set v equal to one: $-\frac{5}{18}=2v_{1f}^2-2v_{1f}$ $v_{1f}=v/6$ This means: $v_2=5v/6$
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