Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 7 - Linear Momentum - General Problems - Page 197: 85

Answer

The speed of the block of mass $m$ is $3\times \sqrt{\frac{k~D^2}{12~m}}$ The speed of the block of mass $3m$ is $\sqrt{\frac{k~D^2}{12~m}}$

Work Step by Step

Since the spring is compressed a distance $D$, the energy in the system is $\frac{1}{2}k~D^2$ which is stored as elastic potential energy in the spring. Let $v_A$ be the speed of the block of mass $m$. Let $v_B$ be the speed of the block of mass $3m$. We can use conservation of momentum to set up an equation. $m~v_A = 3m~v_B$ $v_A = 3~v_B$ We can use conservation of energy to find the speeds of the two blocks. $KE = EPE$ $\frac{1}{2}m~v_A^2 + \frac{1}{2}(3m)~v_B^2 =\frac{1}{2}k~D^2$ $m~(3v_B)^2 + 3m~v_B^2 =k~D^2$ $v_B^2 = \frac{k~D^2}{12~m}$ $v_B = \sqrt{\frac{k~D^2}{12~m}}$ We can use $v_B$ to find $v_A$. $v_A = 3~v_B = 3\times \sqrt{\frac{k~D^2}{12~m}}$ The speed of the block of mass $m$ is $3\times \sqrt{\frac{k~D^2}{12~m}}$ The speed of the block of mass $3m$ is $\sqrt{\frac{k~D^2}{12~m}}$
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