Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 7 - Linear Momentum - Problems - Page 193: 38

Answer

When the ball hits the floor, the impulse was $0.13~kg\cdot m/s$

Work Step by Step

We can find the speed $v_1$ just before the ball hits the floor when y = 1.5 m. $v_1^2 = 2gy$ $v_1 = \sqrt{2gy} = \sqrt{(2)(9.80~m/s^2)(1.5~m)}$ $v_1 = 5.42~m/s$ We can find the speed $v_2$ just after the ball hits the floor when y = 0.85 m. $v_2^2 = 2gy$ $v_2 = \sqrt{2gy} = \sqrt{(2)(9.80~m/s^2)(0.85~m)}$ $v_2 = 4.08~m/s$ If we let the up direction be positive, then $v_1 = -5.42 ~m/s$. We can use the change in momentum to find the impulse. $Impulse = \Delta p$ $Impulse = m \Delta v = m(v_2-v_1)$ $Impulse = (0.014~kg)(4.08~m/s-(-5.42~m/s))$ $Impulse = (0.014~kg)(9.50~m/s)$ $Impulse = 0.13~kg\cdot m/s$ When the ball hit the floor, the impulse was $0.13~kg\cdot m/s$
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