Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 9 - Exercises and Problems - Page 172: 69

Answer

The proof is below.

Work Step by Step

We first find the velocity that would cause the cars to slide this far: $\frac{1}{2}mv^2 = F_f \times \Delta x$ $\frac{1}{2}mv^2 = mg \mu \Delta x$ $v=\sqrt{2g \mu \Delta x}=13.1\ m/s=47.19 \ km/h$ This means that the initial momentum must have been: $=(2140+1040)(47.19)=150,062$ The slowest possible speed scenario would be if the large car possessed most of this momentum. Thus, we find: $150062 =\sqrt{(2140^2v^2+1040^2(55^2)}$ $v=64.82 \ km/h$ Thus, it is clear that at least one of the cars was speeding.
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