Essential University Physics: Volume 1 (4th Edition)

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

Chapter 9 - Exercises and Problems - Page 171: 28


7.8 percent of the energy is lost.

Work Step by Step

Since the gold has an initial velocity of 0, the equation for the speed of the gold after the collision simplifies to: $v_{2f}=\frac{2m_1}{m_1+m_2}v_{1i}$ $v_{2f}=\frac{2(4)}{4+197}v_{1i}$ $v_{2f}=.0398v_{1i}$ We square this value since it is squared in the kinetic energy equation: $.0398^2=.00158$ We know that the second particle is 49.25 times more massive, so it follows: $=.00158\times49.25=.078$ 7.8 percent of the energy is lost.
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