Centre of mass of three particles of masses 1 kg, 2 kg and 3 kg lies at the point (1, 2, 3) and centre of mass of another system of particles 3 kg and 2 kg lies at the point (-1, -3, -2). Where should we put a particle of mass 5 kg so that the centre of mass of entire system lies at the centre of mass of first system?

Options

A. (0,0,0)
B. (1,3,2)
C. (-1,2,3)
D. None of these

Answer

A

Explanation

Center of mass of Ist system already lies at (1,2,3). Therefore center of mass of 3kg and 5kg should lie at (1,2,3).
`therefore` `(3((-hat(i)+3hat(j)-2hat(k))+5r_(5))/(3+5) = (hat(i)+2hat(j)+3hat(k))`
Solving this we get, `r_(5) = (11)/(5)hat(i) + 7/5hat(j) + 6hat(k)`
i.e., 5 kg mass should be kept at (11//5,7//5,6)`