Answer
a) $(\frac{9}{4}, 1)$
b) $(\frac{46}{17}, \frac{14}{17})$
Work Step by Step
a)
$M_{x}$ = $m(1)+m(2)+m(0)+m(1)$ = $4m$
$M_{y}$ = $m(1)+m(1)+m(4)+m(3)$ = $9m$
the total mass of the system is $4m$
so the coordinates of the center of mass are
$(\frac{M_{y}}{M},\frac{M_{x}}{M})$ = $(\frac{9m}{4m}, \frac{4m}{4m})$ = $(\frac{9}{4}, 1)$
b)
With the indicated masses of the particles,
$M_{x}$ = $3(1)+2(2)+5(0)+7(1)$ = $14$
$M_{y}$ = $3(1)+2(1)+5(4)+7(3)$ = $46$
the total mass of the system is $17$
so the coordinates of the center of mass are
$(\frac{M_{y}}{M},\frac{M_{x}}{M})$ = $(\frac{46}{17}, \frac{14}{17})$
