The coordinates of the centre of mass of objects of mass 10, 20, 30 kg are (1,1,1) m. Where should an object of mass 40 kg be placed such that the centre of mass of this new system lies at (0,0,0)?

(a) 3/2, 3/2, 3/2

(b) -3/2, -3/2, -3/2

(c) 3/4, 3/4, 3/4

(d) -3/4, -3/4, -3/4

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This interesting question is from System of Particles in division System of Particles and Rotational Motion of Physics – Class 11

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The correct choice is (b) -3/2, -3/2, -3/2

Easy explanation: Sum of masses = 10 + 20 + 30 + 40 = 100

Since initial centre of mass is at (1,1,1), we can assume an object of 60 kg is at (1,1,1).

x-coordinate;

(60*1 + 40*x)/100 = 0

x = -3/2

y-coordinate;

(60*1 + 40*y)/100 = 0

y = -3/2

z-coordinate;

(60*1 + 40*z)/100 = 0

z = -3/2.