Three particles are placed in the xy plane. A 40g particle is located at (6,3) m

Question

Three particles are placed in the xy plane. A 40g particle is located at (6,3) m, and a 50g particle is positioned at (-4, -2) m. Where must a 20g particle be placed so that the center of mass of this three-particle system is located at the origin?

Explanation / Answer

(xcm,ycm ) -> (0,0) co-ordinates of centre of mass. let (x,y) are co-ordinates of 20g mass. (40+50+20)*(xcm,ycm) = 40*(6,3) + 50*(-4,-2)+20*(x,y) , 110*(0,0) = 40*(6,3) + 50*(-4,-2)+20*(x,y) , Now compare,x-coordinates on both sides : 110*0 = 40*6+50*(-4)+20*x , then x=( 50*4-40*6)/20 = -2 m lly for y , 110*0 = 40*3+50*(-2)+20*y , then y = ( 50*2-40*3)/20 = -1 m (x,y) = (-2,-1) m

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