Question

A 3.06 kg particle has the xy coordinates (-1.55 m, 0.0352 m), and a 3.56 kg particle has the xy coordinates (0.286 m, -0.427 m). Both lie on a horizontal plane. At what (a) x and (b) y coordinates must you place a 3.21 kg particle such that the center of mass of the three-particle system has the coordinates (-0.130 m, -0.605 m)?

Answer 1

The coordinates at which a 3.21 kg particle must be placed in order to have the center of mass of the three-particle system at a given location is ( -0.130, -0.605 )

Step-by-step explanation:

To find the coordinates at which a 3.21 kg particle must be placed in order for the center of mass of the three-particle system to have the given coordinates, we can use the equation for the center of mass:

xcm = (m1x1 + m2x2 + m3x3) / (m1 + m2 + m3)

ycm = (m1y1 + m2y2 + m3y3) / (m1 + m2 + m3)

Substituting the given values, and solving for x3 and y3 will give us the coordinates at which the particle must be placed.

x3 = -0.130 m

y3 = -0.605 m