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 - Section 9.2 - Momentum - Example - Page 158: 9.5


4.49 Mm/s at 43 degrees below the horizontal.

Work Step by Step

Using conservation of momentum, we find the x and y components of the velocity to be as follows: $v_{px}=\frac{(5)(1.6)-4cos(33^{\circ})(1.4)}{1}=3.3$ $v_{py}=\frac{(1.4)(4)(sin33^{\circ})}{1}=-3.05$ We find the speed to be: $v=\sqrt{3.3^3+(-3.05)^2}=4.49 \ Mm/s$ We find the angle: $\theta = tan^{-1}(-3.05/3.3)=-43^{\circ}$
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