Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 9 - Linear Momentum and Collisions - Problems and Conceptual Exercises - Page 291: 34

Answer

$3.71cm$

Work Step by Step

We can find the required compression done by putty-block system as follows: $P_i=m_1v_{1i}+m_2v_{2i}=0+m_2v_{2i}=m_2v_{2i}$ and $P_f=(m_1+m_2)v_f$ According to law of conservation of momentum $m_2v_{2i}=(m_1+m_2)v_f$ This can be rearranged as: $v_f=\frac{m_2v_{2i}}{m_1+m_2}$ We plug in the known values to obtain: $v_f=\frac{(0.05Kg)(2.3m/s)}{0.43Kg+0.05Kg}=0.2396m/s$ At equilibrium $\frac{1}{2}Kx^2=\frac{1}{2}mv_f^2$ This can be rearranged as: $x^2=\frac{(m_1+m_2)v_f^2}{K}$ $\implies x=\sqrt{\frac{(m_1+m_2)v_f^2}{K}}$ We plug in the known values to obtain: $x=\sqrt{\frac{(0.43Kg+0.05Kg)(0.2396m/s)^2}{20.0N/m}}$ $x=0.03712m$ $x=3.71cm$
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