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: 8.28

Answer

The person's speed will be 0.421 m/s

Work Step by Step

Note that the initial momentum is zero so the horizontal component of the person's momentum will be equal in magnitude to the horizontal component of the rock's momentum. Note that the person and the rock will be moving in opposite directions. $m_p~v_p = m_r~v_{r,x}$ $v_p = \frac{m_r~v_{r,x}}{m_p}$ $v_p = \frac{(3.00~kg)(12.0~m/s)~cos(35.0^{\circ})}{70.0~kg}$ $v_p = 0.421~m/s$ The person's speed will be 0.421 m/s.
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