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: 38

Answer

(a) The racket is moving at a speed of 6.4 m/s after the collision. (b) The average force exerted by the racket on the ball is 360 N The racket's force is 610 times greater than the gravitational force on the ball.

Work Step by Step

(a) We first find the impulse $J$ that the racket exerted on the ball; $p_0+J=p_f$ $J = p_f-p_0$ $J = m~(v_f-v_0)$ $J = (0.060~kg)[(40~m/s)-(-20~m/s)]$ $J = 3.6~N~s$ The ball also exerted the same magnitude of impulse on the racket in the opposite direction. Using this, we can find the speed of the racket after the collision: $p_f = p_0+J$ $m~v_f = m~v_0 +J$ $v_f = \frac{m~v_0 +J}{m}$ $v_f = \frac{(1.0~kg)(10~m/s)-3.6~N~s}{1.0~kg}$ $v_f = 6.4~m/s$ The racket is moving at a speed of 6.4 m/s after the collision. (b) We can use the impulse to find the average force exerted by the racket on the ball. $F~t = J$ $F = \frac{J}{t}$ $F = \frac{3.6~N~s}{10\times 10^{-3}~s}$ $F = 360~N$ The average force exerted by the racket on the ball is 360 N. We then find the gravitational force on the ball: $F_g = m~g$ $F_g = (0.060~kg)(9.80~m/s^2)$ $F_g = 0.59~N$ We can find the ratio of the racket's force and the gravitational force on the ball. $\frac{F}{F_g} = \frac{360~N}{0.59~N} = 610$ The racket's force is thus 610 times greater than the gravitational force on the ball.
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