Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 14 - Oscillations - Exercises and Problems: 60

Answer

$A = 0.11~m$ $T = 1.7~s$

Work Step by Step

We can use conservation of momentum to find the speed of the two blocks just after the collision. $(m_1+m_2)v_2 = m_1v_1$ $v_2 = \frac{m_1v_1}{m_1+m_2}$ $v_2 = \frac{(0.25~kg)(1.20~m/s)}{0.25~kg+0.50~kg}$ $v_2 = 0.40~m/s$ Let $M$ be the the total mass of the two blocks. We then find the amplitude as: $\frac{1}{2}kA^2 = \frac{1}{2}Mv_{max}^2$ $A = \sqrt{\frac{M}{k}}~v_{max}$ $A = \sqrt{\frac{0.75~kg}{10~N/m}}~(0.40~m/s)$ $A = 0.11~m$ We then find the period as: $T = 2\pi~\sqrt{\frac{M}{k}}$ $T = 2\pi~\sqrt{\frac{0.75~kg}{10~N/m}}$ $T = 1.7~s$
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.