91Ó°ÊÓ

Mass of particle 1, ${\mathrm{m}}_1=2.00\mathrm{kg}$

Mass of particle 2, ${\mathrm{m}}_2=4.00\mathrm{kg}$

Mass of particle 3, ${\mathrm{m}}_3=3.00\mathrm{kg}$

Coordinates of particle 1 are $(-1.20,0.500)$.

Coordinates of particle 2 are $(0.600,-0.750)$.

Coordinates of center of mass are$(-0.500,-0.700)$.

For a system of particles, the whole mass of the system is concentrated at the center of mass of the system. Center of mass is the point where external forces are seems to be applied for the motion of the system as a whole.

The expression for the x coordinates of the center of mass is given as:

$x_{com}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3+...}{m_1+m_2+m_3+...}$ … (i)

Here, ${\mathrm{m}}_1,{\mathrm{m}}_2,{\mathrm{m}}_3,....$are the masses of the particles, and $x_1,x_2,x_3,...$are their respective x coordinates.

The expression for the y coordinates of the center of mass is given as:

$y_{\mathrm{com}}=\frac{{\mathrm{m}}_1{\mathrm{y}}_1+{\mathrm{m}}_2{\mathrm{y}}_2+{\mathrm{m}}_3{\mathrm{y}}_3+...}{{\mathrm{m}}_1+{\mathrm{m}}_2+{\mathrm{m}}_3+...}$ … (ii)

Using equation (i), the x coordinate for particle 3 is calculated as:

$$\begin{gathered} x_{\mathrm{com}}=\frac{{\mathrm{m}}_1{\mathrm{x}}_1+{\mathrm{m}}_2{\mathrm{x}}_2+{\mathrm{m}}_3{\mathrm{x}}_3+...}{{\mathrm{m}}_1+{\mathrm{m}}_2+{\mathrm{m}}_3+...} \\ -0.500\mathrm{m}=\frac{(2.00\mathrm{kg})(-1.20\mathrm{m})+(4.00\mathrm{kg})(0.600\mathrm{m})+(3.00\mathrm{kg})\left({\mathrm{x}}_3\right)}{2.00\mathrm{kg}+4.00\mathrm{kg}+3.00\mathrm{kg}} \\ (3.00\mathrm{kg}){\mathrm{x}}_3=--4.50\mathrm{kg}-\mathrm{m}+2.40\mathrm{kg}-\mathrm{m} \\ {\mathrm{x}}_3=-1.50\mathrm{m} \end{gathered}$$

Using equation (ii), the y coordinate for particle 3 is calculated as:

Thus, the y coordinate of third particle is$-1.43m$.