Question

The co-ordinates of centre of mass of particles
of mass 10, 20 and 30 gm are (1, 1, 1) cm. The
position co-ordinates of mass 40 gm which
when added to the system, the position of
combined centre of mass be at (0, 0, 0) are,
1) (3/2, 3/2, 3/2) 2) (-3/2, -3/2, -3/2)
3) (3/4, 3/4, 3/4) 4) (-3/4, -3/4, -3/4)​

Answer 1

Answer:

$= (\frac{-3}{2},\frac{-3}{2},\frac{-3}{2})$ option 2.

Explanation:

We can consider the three masses (10,20,30) are located at the center of mass as a single point mass. Therefore 60 gm mass is now located at point (1,1,1).

Let the position of the 40 gm mass be  (x,y,z) & the position of the combined center of mass is given (0,0,0)

60(1,1,1) + 40(x,y,z) = 100(0,0,0)

therefore 60 + 40x = 0

or 40x = -60

$\therefore x = \frac{-60}{40}$

$x=\frac{-3}{2}$

similarly $y = \frac{-3}{2} , z = \frac{-3}{2}$

position of the 40 gm mass $= (\frac{-3}{2},\frac{-3}{2},\frac{-3}{2})$