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 9 - Impulse and Momentum - Exercises and Problems - Page 241: 29


The cart bounces back up the ramp a distance of 0.50 meters.

Work Step by Step

We can find the magnitude of the cart's acceleration when it is on the ramp. $mg~sin(\theta) = ma$ $a = g~sin(\theta)$ $a = (9.80~m/s^2)~sin(30.0^{\circ})$ $a = 4.90~m/s^2$ We can find the cart's speed just before it hits the rubber block. $v^2 = v_0^2+2ad = 0+2ad$ $v = \sqrt{2ad}$ $v = \sqrt{(2)(4.90~m/s^2)(1.0~m)}$ $v = 3.13~m/s$ We can use the graph to find the magnitude of the impulse that the rubber block exerts on the cart. The impulse is equal to the area under the force versus time graph. $J = \frac{1}{2}F_{max}~t$ $J = \frac{1}{2}(200~N)(26.7\times 10^{-3}~s)$ $J = 2.67~N~s$ We can find the velocity after the cart bounces off the rubber block. $p_f = p_0+J$ $m~v_f = m~v_0+J$ $v_f = \frac{m~v_0+J}{m}$ $v_f = \frac{(0.50~kg)(3.13~m/s)-2.67~N~s}{0.50~kg}$ $v_f = -2.21~m/s$ We can find the distance $d$ the cart rolls back up the ramp. We can let $v_0 = 2.21~m/s$ for this part of the question. $d = \frac{v^2-v_0^2}{2a}$ $d = \frac{0-(2.21~m/s)^2}{(2)(-4.90~m/s^2)}$ $d = 0.50~m$ The cart bounces back up the ramp a distance of 0.50 meters.
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