College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Review & Synthesis: Chapters 6-8 - Review Exercises - Page 324: 25


The velocity after the collision is $~2.06~m/s$

Work Step by Step

By conservation of momentum, the final momentum of the system is equal to the initial momentum. We can find the east component $p_x$ of the momentum: $p_x = (2.00~kg)(2.70~m/s) = 5.40~kg~m/s$ We can find the south component $p_y$ of the momentum: $p_y = (1.50~kg)(3.20~m/s) = 4.80~kg~m/s$ We can find the magnitude of the momentum: $p = \sqrt{p_x^2+p_y^2} = \sqrt{(5.40~kg~m/s)^2+(4.80~kg~m/s)^2} = 7.225~kg~m/s$ We can find the final velocity $v_f$ after the collision: $m_f~v_f = p$ $v_f = \frac{p}{m_f}$ $v_f = \frac{7.225~kg~m/s}{2.00~kg+1.50~kg}$ $v_f = 2.06~m/s$ The velocity after the collision is $~2.06~m/s$.
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