Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 9 - Section 9.2 - Momentum - Example: 9.5

Answer

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: $\theta = tan^{-1}(-3.05/3.3)=-43^{\circ}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.