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 170: 17

Answer

$v=.268 \ Mm/s$

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: $239v+4(\sqrt{\frac{2(5.15\times10^6)}{4 \ amu}})=0$ $v=.268 \ Mm/s$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.