Question 9.1: A 2.00 kg particle has the xy coordinates (-1.20 m, 0.500 m)......
A 2.00 kg particle has the xy coordinates (-1.20 m, 0.500 m), and a 4.00 kg particle has the xy coordinates (0.600 m, -0.750 m). Both lie on a horizontal plane. At what (a) x and (b) y coordinates must you place a 3.00 kg particle such that the center of mass of the three-particle system has the coordinates (-0.500 m, -0.700 m)?
We use Eq. 9-5 to solve for (x_{3},y_{3}).
x_{\mathrm{com}}={\frac{1}{M}}\sum_{i=1}^{n}m_{i}x_{i}, y_{\mathrm{com}}={\frac{1}{M}}\sum_{i=1}^{n}m_{i}y_{i}, z_{\mathrm{com}}={\frac{1}{M}}\sum_{i=1}^{n}m_{i}z_{i}, (9-5)
(a) The x coordinate of the system’s center of mass is:
x_{{com}}={\frac{m_{{1}}x_{{1}}+m_{2}x_{\mathrm{2}}+m_{\mathrm{3}}x_{\mathrm{3}}}{m_{{1}}+m_{\mathrm{2}}+m_{\mathrm{3}}}}={\frac{(2.00\ \mathrm{kg})(-1.20\ \mathrm{m})+(4.00\ \mathrm{kg})(0.600\ \mathrm{m})+(3.00\ \mathrm{kg})\,x_{3}}{2.00\ \mathrm{kg}+4.00\ \mathrm{kg}+3.00\ \mathrm{kg}}}
=-0.500\ m.
Solving the equation yields {{x}}_{3} = –1.50 m.
(b) The y coordinate of the system’s center of mass is:
y_{{com}}={\frac{m_{1}y_{1}+m_{2}y_{2}+m_{3}y_{3}}{m_{1}+m_{2}+m_{3}}}=\frac{(2.00\ \mathrm{kg})(0.500\ \mathrm{m})+(4.00\ \mathrm{kg})(-0.750\ \mathrm{m})+(3.00\ \mathrm{kg})y_{3}}{2.00\ \mathrm{kg}+4.00\ \mathrm{kg}+3.00\ \mathrm{kg}}
=-0.700\ m
Solving the equation yields y_{3} = –1.43 m.