Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 11 - Impulse and Momentum - Exercises and Problems - Page 291: 80


The third piece moves forward with a speed of $1.5\times 10^7~m/s$

Work Step by Step

We can find the mass $m_3$ of the third piece. $m_1+m_2+m_3 = m_0$ $m_3 = m_0-m_1-m_2$ $m_3 = (2.0\times 10^6~kg)-(5.0\times 10^5~kg)-(8.0\times 10^5~kg)$ $m_3 = 7.0\times 10^5~kg$ We can use conservation of momentum to find the velocity $v_3$ of the third piece after the explosion. $p_f=p_0$ $m_1~v_1+m_2~v_2+m_3~v_3 = m_0~v_0$ $v_3 = \frac{m_0~v_0-m_1~v_1-m_2~v_2}{m_3}$ $v_3 = \frac{(2.0\times 10^6~kg)(5.0\times 10^6~m/s)-(5.0\times 10^5~kg)(-2.0\times 10^6~m/s)-(8.0\times 10^5~kg)(1.0\times 10^6~m/s)}{7.0\times 10^5~kg}$ $v_3 = 1.5\times 10^7~m/s$ The third piece moves forward with a speed of $1.5\times 10^7~m/s$.
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.