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 267: 8.64

Answer

(a) The impulse delivered to the ball during impact is 0.474 kg m/s. (b) The average force on the ball during impact is 237 N.

Work Step by Step

(a) We can find the speed $v_1$ after falling 2.00 meters. $v_1^2 = v_0^2+2ay = 0 + 2ay$ $v_1 = \sqrt{2ay} = \sqrt{(2)(9.80~m/s^2)(2.00~m)}$ $v_1 = 6.26~m/s$ We can find the speed $v_2$ after the rebound. $v_2^2 = v^2-2ay = 0 - 2ay$ $v_2 = \sqrt{-2ay} = \sqrt{-(2)(-9.80~m/s^2)(1.60~m)}$ $v_2 = 5.60~m/s$ $J = \Delta p = m \Delta v$ $J = (0.0400~kg)(11.86~m/s)$ $J = 0.474~kg~m/s$ The impulse delivered to the ball during impact is 0.474 kg m/s. (b) $F~t = J$ $F = \frac{J}{t} = \frac{0.474~kg~m/s}{0.00200~s}$ $F = 237~N$ The average force on the ball during impact is 237 N.
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