University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 14 - Periodic Motion - Problems - Exercises - Page 465: 14.83

Answer

$T = 0.421~s$

Work Step by Step

We can use conservation of momentum to find the speed of the block just after the bullet strikes it. Let $M$ be the total mass of the block and the bullet. Let $m_b$ be the mass of the bullet. $Mv_2 = m_bv_1$ $v_2 = \frac{m_bv_1}{M}$ $v_2 = \frac{(0.00800~kg)(280~m/s)}{1.00~kg}$ $v_2 = 2.24~m/s$ We can find the force constant of the spring. $\frac{1}{2}kA^2 = \frac{1}{2}Mv_{max}^2$ $k = \frac{Mv_{max}^2}{A^2}$ $k = \frac{(1.00~kg)(2.24~m/s)^2}{(0.150~m)^2}$ $k = 223~N/m$ We can find the period. $T = 2\pi~\sqrt{\frac{M}{k}}$ $T = 2\pi~\sqrt{\frac{1.00~kg}{223~N/m}}$ $T = 0.421~s$
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