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)?

Step-by-step explanation:

Given:

Mass of the first particle

Mass of the second particle

Mass of the third particle

xy coordinate of the first particle

xy coordinate of the second particle

coordinate of the centre of mass of three-particle system

We need to find the xy coordinate of the third particle.

Solution:

First we calculate total mass of the three-particle system.

x coordinate of the centre of mass of the three-particle system are given as;

Rewrite the equation for

Part a.

Substitute all given values in above equation.

Part b.

Similarly for coordinate.

Substitute all given values in above equation.

Therefore, xy coordinate of the third particle

x = 25/2304

step-by-step explanation:

step-by-step explanation:

no, her answer is not correct.   we square the lengths of all sides and then determine whether the pythagorean theorem holds here:

9² + 15² = 306.   this is not equal to the square of the longest side ( that square being 395.   answer c is correct.