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 173: 80

Answer

The proof is below.

Work Step by Step

We know that the velocity of the balls when they hit the ground is: $Mgh=\frac{1}{2}Mv^2\\ v=4.42\sqrt{h}$ We know the following equation for elastic collisions: $v_{1f}=\frac{m_1-m_2}{m_1+m_2}v_{1i}+\frac{2m_2}{m_1+m_2}v_{2i}$ Since the large mass is much more massive and the velocities are initially equal, this becomes: $v_{1f}=v+2v=3v$ Thus, we find: $gh_f=\frac{1}{2}176.58h$ $h_f=9h$
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