Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 7 - Impulse and Momentum - Problems - Page 194: 40


0.557 m/s

Work Step by Step

We obtain: $(Mc)(Vcox)+(Mmc)(Vmcox)= (Mc+Mmc)Vfx$ $150(.8)cos25^{\circ}+440(.5)=(150+440)Vfx$ $108.8+220=590Vfx$ $328.8/590=Vfx= 0.557 m/s$
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