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 7 - Newton's Third Law - Exercises and Problems: 28

Answer

(a) $F = 2800~N$ (b) $F = 2800~N$ (c) The forehead is not in danger of fracturing. The cheekbone is in danger of fracturing.

Work Step by Step

(a) We can find the deceleration of the ball. $a = \frac{v-v_0}{t}$ $a = \frac{0-30~m/s}{1.5\times 10^{-3}~s}$ $a = -2.0\times 10^4~m/s^2$ We can use the magnitude of the deceleration to find the magnitude of the force that stops the ball. $F = ma$ $F = (2.0\times 10^4~m/s^2)(0.140~kg)$ $F = 2800~N$ (b) By Newton's third law, the ball exerts an equal and opposite force on the head. Therefore the magnitude of the force that the ball exerts on the head is 2800 N. (c) Since the forehead can withstand a force of 6000 N, the forehead is not in danger of fracturing. Since the cheekbone can only withstand a force of 1300 N, the cheekbone is in danger of fracturing.
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