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