Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 9 - Impulse and Momentum - Exercises and Problems - Page 241: 26


$F_{max} = 0.96~N$

Work Step by Step

We can find the impulse $J$ that the wall exerted on the ball as: $p_0+J=p_f$ $J = p_f-p_0$ $J = m~(v_f-v_0)$ $J = (0.060~kg)[(-32~m/s)-(32~m/s)]$ $J = -3.84~N~s$ The magnitude of the impulse is equal to the area under the force versus time graph; $area = 3.84~N~s$ $\frac{1}{2}F_{max}(2~s)+F_{max}(2~s)+\frac{1}{2}F_{max}(2~s) = 3.84~N~s$ $2~F_{max}(2~s) = 3.84~N~s$ $F_{max} = \frac{3.84~N~s}{(2)(2~s)}$ $F_{max} = 0.96~N$
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