Question

The coordinates of centre of mass of particles of mass $10,20,30gm$ are $(1,1,1)cm$. The position of coordinates of mass $40gm$ which when added to the system, the position of combined centre of mass be at $(0,0,0)$ are

• A. $(3/2,3/2,3/2)$
• B. $(-3/2,-3/2,-3/2)$
• C. $(3/4,3/4,3/4)$
• D. $(-3/4,-3/4,-3/4)$

Answer 1

The correct option is B $(-3/2,-3/2,-3/2)$

Let initially the centre of masses 10g, 20g and 30g are placed at a distance $x_1,x_{2}and{x}_3$.Then x coordinate of centre of mass is given by

$\frac{10x_1+20x_2+30x_3}{10+20+30}=1$

$10x_1+20x_2+30x_3=\mathrm{60......}\left(1\right)$

When 40g mass is added the new centre of mass becomes:

$10x_1+20x_2+30x_3+40x_4=\mathrm{0.........}\left(2\right)$

$60+40x_4=0$

$x_4=\frac{-60}{40}$

$x_4=-\frac{3}{2}$

New coordinates of the combined system is $(-\frac{3}{2},-\frac{3}{2},-\frac{3}{2})$