If the centre of mass of three particles of masses of `1kg, 2kg, 3kg` is at `(2,2,2)`, then where should a fourth particle of mass `4kg` be placed so that the combined centre of mass may be at `(0,0,0).`

Options

A. `vec(r )=(-1,-1,-1)`
B. `vec(r )=(-2,-2,-2)`
C. `vec(r )=(-3,-3,-3)`
D. `vec(r )=(-4,-4,-4)`

Answer

C

Explanation

We know `L=|omega`
Also `E_(k)=(1)/(2)|omega^(2)" or "|=(2E_(k))/(omega^(2))`
`therefore " "L=(2E_(k))/(omega^(2))omega=(2E_(k))/(omega),L'=(2((E_(k))/(2)))/(2omega)=(L)/(4)`