Three particles of masses 2 kg, 3 kg and 4 kg are placed at the vertices A, B and C of an equilateral triangle ABC. If A and B lie at (0,0) and (2,0) m, the co-ordinates of their centre of mass are

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(32m and 76m)

(109m and 439m)

(712m and 34m)

(712m and 712m)

answer is B.

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Detailed Solution

. xcm=m1x1+m2x2+m3x3m1+m2+m3 XCM=2(0)+3(2)+4(1)9=6+49=109 Ycm=2(0)+3(0)+4(3)9 =439 xcm,ycm=109,439

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