Answer:
[tex](x_{3},y_{3})=(-1.50\ m, -1.43\ m)[/tex]
Step-by-step explanation:
Given:
Mass of the first particle [tex]m_{1} =2.0\ kg[/tex]
Mass of the second particle [tex]m_{2} =4.0\ kg[/tex]
Mass of the third particle [tex]m_{3} =\ ?[/tex]
xy coordinate of the first particle [tex](x_{1},y_{1})=(-1.20 m, 0.500 m)[/tex]
xy coordinate of the second particle [tex](x_{2},y_{2})=(0.600 m, -0.750 m)[/tex]
coordinate of the centre of mass of three-particle system [tex](x_{cm},y_{cm})=(-0.500 m, -0.700 m)[/tex]
We need to find the xy coordinate of the third particle.
Solution:
First we calculate total mass of the three-particle system.
[tex]M = m_{1}+m_{2}+m_{3}[/tex]
[tex]M = 2.0+4.0+3.0[/tex]
[tex]M = 9.0\ kg[/tex]
x coordinate of the centre of mass of the three-particle system are given as;
[tex]x_{cm}=\frac{1}{M}(m_{1}x_{1}+ m_{2}x_{2}+ m_{3}x_{3})[/tex]
Rewrite the equation for [tex]x_{3}[/tex]
[tex]m_{3}x_{3}=Mx_{cm}-m_{1}x_{1}-m_{2}x_{2}[/tex]
Part a.
[tex]x_{3}=\frac{1}{m_{3}} (Mx_{cm}-m_{1}x_{1}-m_{2}x_{2})[/tex]
Substitute all given values in above equation.
[tex]x_{3}=\frac{1}{3.0} (9.0\times (-0.5)-2.0\times (-1.20)-4.0\times (0.6))[/tex]
[tex]x_{3}=-1.50\ m[/tex]
Part b.
Similarly for [tex]y_{3}[/tex] coordinate.
[tex]y_{3}=\frac{1}{m_{3}} (My_{cm}-m_{1}y_{1}-m_{2}y_{2})[/tex]
Substitute all given values in above equation.
[tex]y_{3}=\frac{1}{3.0} (9.0\times (-0.7)-2.0\times 0.5-4.0\times (-0.75))[/tex]
[tex]y_{3}=-1.43\ m[/tex]
Therefore, xy coordinate of the third particle [tex](x_{3},y_{3})=(-1.50\ m, -1.43\ m)[/tex]
