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: 63


.9153 meters per second

Work Step by Step

We use conservation of momentum to find the velocity of each one following the collision: $350=950\Delta v_s\\ \Delta v_s=.3684 \ m/s $ And: $350=640\Delta v_b\\ \Delta v_b=.5469 \ m/s $ Since they are moving apart, we add these values to find: $\Delta v = .5469+.3684=\fbox{.9153 meters per second}$
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