Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 11 - Impulse and Momentum - Exercises and Problems - Page 289: 39


$F_{max} = 960~N$

Work Step by Step

If the ball hits the wall at 32 m/s and rebounds with the same speed, then the change in velocity is 64 m/s. We can find the impulse $J$ that the wall exerted on the ball. Note that the impulse is equal to the change in momentum. $J=\Delta p$ $J = m~\Delta v$ $J = (0.060~kg)(64~m/s)$ $J = 3.84~N~s$ The impulse is equal to the area under the force versus time graph. $area = J$ $\frac{1}{2}(F_{max})(2~ms)+(F_{max})(2~ms)+\frac{1}{2}(F_{max})(2~ms) = J$ $(2)(F_{max})(2~ms) = J$ $F_{max} = \frac{J}{(2)(2~ms)}$ $F_{max} = \frac{3.84~N~s}{(2)(2\times 10^{-3}~s)}$ $F_{max} = 960~N$
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