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: 12

Answer

The glider is in contact with the spring for 0.20 seconds.

Work Step by Step

We can find the impulse exerted on the glider by the spring. $p_0+J = p_f$ $J = p_f-p_0$ $J = m~(v_f-v_0)$ $J = (0.60~kg)[3.0~m/s-(-3.0~m/s)]$ $J = 3.6~N~s$ The impulse is equal to the area under the force versus time graph. We can find the time $t$ that the glider was in contact with the spring. $\frac{1}{2}F_{max}~t = J$ $t = \frac{2J}{F_{max}}$ $t = \frac{(2)(3.6~N~s)}{36~N}$ $t = 0.20~s$ The glider is in contact with the spring for 0.20 seconds.
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