Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 16 - Waves-I - Problems - Page 473: 24d

Answer

$M_2 = 312.5~g$

Work Step by Step

We can find $M_1$: $v_1 = v_2$ $\sqrt{\frac{\tau_1}{\mu_1}} = \sqrt{\frac{\tau_2}{\mu_2}}$ $\frac{\tau_1}{\mu_1} = \frac{\tau_2}{\mu_2}$ $\frac{M_1g}{\mu_1} = \frac{M_2 g}{\mu_2}$ $\frac{M_1}{\mu_1} = \frac{M-M_1}{\mu_2}$ $M_1\mu_2 = (M-M_1)(\mu_1)$ $M_1(\mu_1+\mu_2) = M\mu_1$ $M_1 = \frac{M\mu_1}{\mu_1+\mu_2}$ $M_1 = \frac{(500~g)(3.00~g/m)}{3.00~g/m+5.00~g/m}$ $M_1 = 187.5~g$ We can find $M_2$: $M_2 = M-M_1$ $M_2 = (500~g)-(187.5~g)$ $M_2 = 312.5~g$
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