Physics: Principles with Applications (7th Edition)

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

Chapter 6 - Work and Energy - Problems - Page 165: 24

Answer

The original speeds of the two cars were 5.7 m/s and 11.4 m/s.

Work Step by Step

Let the masses of the two cars be m and 2m. Let the original speeds of the cars be $v_1$ and $v_2$ respectively. $\frac{1}{2}(2m)v_1^2 = \frac{1}{2}\times\frac{1}{2}mv_2^2$ $4v_1^2 = v_2^2$ $2v_1 = v_2$ When both cars increase their speed by 8 m/s: $\frac{1}{2}(2m)(v_1+8)^2 = \frac{1}{2}m(v_2+8)^2$ $\frac{1}{2}(2m)(v_1+8)^2 = \frac{1}{2}m(2v_1+8)^2$ $2v_1^2 + 32v_1+128 = 4v_1^2+32v_1+64$ $v_1^2 = 32$ $v_1 = \sqrt{32}$ $v_1 = 4\sqrt{2} = 5.7 ~m/s$ $v_2 = 2v_1 = 11.4 ~m/s$ The original speeds of the two cars were 5.7 m/s and 11.4 m/s.
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