University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 8 - Momentum, Impulse, and Collision - Problems - Exercises - Page 266: 8.40

Answer

(a) The falcon changed the raven's direction of motion by an angle of $\theta = 48.0^{\circ}$. (b) The raven's speed after the collision is 13.5 m/s.

Work Step by Step

(a) We can find the falcon's change in momentum. $\Delta p = m~\Delta v$ $\Delta p = (0.600~kg)(25.0~m/s)$ $\Delta p = 15.0~kg~m/s$ This change in momentum will be equal to the raven's change in momentum directed at a right angle to its original direction of motion. We can find the raven's initial momentum. $p = m~v = (1.50~kg)(9.0~m/s)$ $p = 13.5~kg~m/s$ We can find the angle $\theta$ between the raven's original direction and its new direction. $tan(\theta) = \frac{15.0~kg~m/s}{13.5~kg~m/s}$ $\theta = arctan(\frac{15.0}{13.5})$ $\theta = 48.0^{\circ}$ The falcon changed the raven's direction of motion by an angle of $\theta = 48.0^{\circ}$. (b) We can find the magnitude of the raven's momentum after the collision. $p = \sqrt{(15.0~kg~m/s)^2+(13.5~kg~m/s)^2}$ $p = 20.18~kg~m/s$ We can find the raven's speed. $m~v = p$ $v = \frac{p}{m} = \frac{20.18~kg~m/s}{1.50~kg}$ $v = 13.5~m/s$ The raven's speed after the collision is 13.5 m/s.
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