Question

A 1.00 kg particle has the xy coordinates (-1.20 m, 0.500 m) and a 4.50 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 4.00 kg particle such that the center of mass of the three-particle system has the coordinates (-0.500 m, -0.700 m)

Answer 1

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

Step-by-step explanation:

The center of mass of a system of particles (
`\vec r_(cm)`), measured in meters, is defined by this weighted average:

`\vec r_(cm) = (\Sigma_(i=1)^(n)\,m_(i)\cdot \vec r_(i))/(\Sigma_(i=1)^(n)\,m_(i))` (1)

Where:

`m_(i)` - Mass of the i-th particle, measured in kilograms.

`\vec r_(i)` - Location of the i-th particle with respect to origin, measured in meters.

If we know that
`\vec r_(cm) = (-0.500\,m,-0.700\,m)`,
`m_(1) = 1\,kg`,
`\vec r_(1) = (-1.20\,m, 0.500\,m)`,
`m_(2) = 4.50\,kg`,
`\vec r_(2) = (0.600\,m, -0.750\,m)` and
`m_(3) = 4\,kg`, then the coordinates of the third particle are:

`(-0.500\,m, -0.700\,m) = ((1\,kg)\cdot (-1.20\,m,0.500\,m)+(4.50\,kg)\cdot (0.600\,m,-0.750\,m)+(4\,kg)\cdot \vec r_(3))/(1\,kg+4.50\,kg+4\,kg)`

`(-4.75\,kg\cdot m, -6.65\,kg\cdot m) = (-1.20\,kg\cdot m, 0.500\,kg\cdot m) + (2.7\,kg\cdot m, -3.375\,kg\cdot m) +(4\cdot x_(3),4\cdot y_(3))`

`(4\cdot x_(3), 4\cdot y_(3)) = (-6.25\,kg\cdot m,-3.775\,kg\cdot m)`

`(x_(3),y_(3)) = (-1.562\,m,-0.944\,m)`

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.