Question

A $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{kg}\mathbf{}$particle has the xycoordinates $\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{20}\mathbf{}\mathbf{m}\mathbf{,}\mathbf{}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{}\mathbf{m}\mathbf{)}$. and a $\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{kg}$particle has the xycoordinates $\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{600}\mathbf{}\mathbf{m}\mathbf{,}\mathbf{}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{750}\mathbf{}\mathbf{m}\mathbf{)}$ . Both lie on a horizontal plane. At what (a) xand (b) ycoordinates must you place a$\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{kg}$particle such that center of mass of the three-particle system has the coordinates $\mathbf{(}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{}\mathbf{m}\mathbf{,}\mathbf{}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{700}\mathbf{}\mathbf{m}\mathbf{)}$

Short answer

1. The x coordinate of third particle is $-1.50\mathrm{m}$
2. The y coordinate of third particle is$-1.43\mathrm{m}$

Step-by-step solution

Given data

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$(-0.500,-0.700)$.

Understanding the concept of center of mass

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)

(a) Determination of the x coordinates for the third particle

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

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

Thus, the x coordinate of third particle is $-1.50\mathrm{m}$.

(b) Determination of the y coordinates for the third particle

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

$$\begin{gathered} y_{\mathrm{com}}=\frac{{\mathrm{m}}_1{\mathrm{y}}_1+{\mathrm{m}}_2{\mathrm{y}}_2+{\mathrm{m}}_3{\mathrm{y}}_3+...}{{\mathrm{m}}_1+{\mathrm{m}}_2+{\mathrm{m}}_3+...} \\ -0.700\mathrm{m}=\frac{(2.00\mathrm{kg})(-0.500\mathrm{m})+(4.00\mathrm{kg})(-0.750\mathrm{m})+(3.00\mathrm{kg})\left({\mathrm{y}}_3\right)}{2.00\mathrm{kg}+4.00\mathrm{kg}+3.00\mathrm{kg}} \\ (3.00\mathrm{kg}){\mathrm{y}}_3=-6.30\mathrm{kg}-\mathrm{m}-1.00\mathrm{kg}-\mathrm{m}+3.00\mathrm{kg}-\mathrm{m} \\ {\mathrm{x}}_3=-1.43\mathrm{m} \end{gathered}$$

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