Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 9 - Exercises and Problems - Page 163: 21


$v=.268 \ Mm/s$

Work Step by Step

We first use the equation for kinetic energy to find the speed of the alpha particle. Thus: $v = \sqrt{\frac{2K}{m}}= \sqrt{\frac{2(5.15\times10^6)}{4 \ amu}}$ We know that momentum is conserved, so it follows: $239v+4(\sqrt{\frac{2(5.15\times10^6)}{4 \ amu}})=0$ $v=.268 \ Mm/s$
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