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: 82

Answer

$8.04 \ s$

Work Step by Step

We first find the speed of the small block when it hits the large one: $v = \sqrt{2(9.81)(.25)}=2.21\ m/s$ We find the speed of each block following the collision: $v_{1f}=\frac{m_1-m_2}{m_1+m_2}v_{1i}+\frac{2m_2}{m_1+m_2}v_{2i}$ $v_{1f}=\frac{.2-.8}{1}(2.21)=-1.32\ m/s$ Using conservation of momentum, we find: $v_{2f}=.884\ m/s$ Thus, we see that it will take $\frac{1.4}{1.32}=1.06 \ s$ for the small block to return to the ramp. Then, to go up and down the incline, we find: $0=1.32t+\frac{1}{2}sin30gt^2 \\ t=.54 \ s$ When it returns, it will be going 1.32 meters per second, and the other block will be $((1.06+.54)(.884))+1.4=2.81$ meters away. Thus, we find that the time will be: $t=\frac{2.81}{.436}+1.6=8.04 \ s$
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