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: 8.21

Answer

(a) The speed of puck A before the collision was 0.790 m/s. (b) The change in kinetic energy during the collision is -0.0023 J.

Work Step by Step

(a) $m_Av_{A1} = m_Av_{A2}+m_Bv_{B2}$ $v_{A1} = \frac{m_Av_{A2}+m_Bv_{B2}}{m_A}$ $v_{A1} = \frac{(0.250~kg)(-0.120~m/s)+(0.350~kg)(0.650~m/s)}{0.250~kg}$ $v_{A1} = 0.790~m/s$ The speed of puck A before the collision was 0.790 m/s. (b) $K_1 = \frac{1}{2}m_Av_{A1}^2$ $K_1 = \frac{1}{2}(0.250~kg)(0.790~m/s)^2$ $K_1 = 0.0780~J$ $K_2 = \frac{1}{2}m_Av_{A2}^2+ \frac{1}{2}m_Bv_{B2}^2$ $K_2 = \frac{1}{2}(0.250~kg)(0.120~m/s)^2+ \frac{1}{2}(0.350~kg)(0.650~m/s)^2$ $K_2 = 0.0757~J$ We can find the change in kinetic energy during the collision. $\Delta K = K_2-K_1$ $\Delta K = (0.0757~J)-(0.0780~J)$ $\Delta K = -0.0023~J$ The change in kinetic energy during the collision is -0.0023 J.
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