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 264: 8.14

Answer

$p = (-2.44~kg~m/s)~\hat{i}+(2.80~kg~m/s)~\hat{j}$

Work Step by Step

$F_x = (0.280~N/s)~t$ $J_x = \int_{0}^{2.00}F_x~dt$ $J_x = \int_{0}^{2.00}(0.280~N/s)~t~dt$ $J_x = (0.140~N/s)~t^2\vert_{0}^{2.00}$ $J_x = 0.560~kg~m/s$ $p_{x2} = p_{x1}+J_x$ $p_{x2} = -3.00~kg~m/s+0.560~kg~m/s$ $p_{x2} = -2.44~kg~m/s$ $F_y = (-0.450~N/s^2)~t^2$ $J_y = \int_{0}^{2.00}F_y~dt$ $J_y = \int_{0}^{2.00}(-0.450~N/s)~t^2~dt$ $J_y = (-0.150~N/s)~t^3\vert_{0}^{2.00}$ $J_y = -1.20~kg~m/s$ $p_{y2} = p_{y1}+J_y$ $p_{y2} = (4.00~kg~m/s)+(-1.20~kg~m/s)$ $p_{y2} = 2.80~kg~m/s$ $p = (-2.44~kg~m/s)~\hat{i}+(2.80~kg~m/s)~\hat{j}$
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