The coordinates of the centre of mass of a system of three particles of masses 1 mg , 2 mg and 3 mg are (2 , 2 , 2) . The coordinates of the fourth particle of mass 4 mg to be positioned so that the centre of mass of the four particles system is at the origin of the three-dimensional rectangular coordinate system are

Question

The coordinates of the centre of mass of a system of three particles of masses 1 mg , 2 mg and 3 mg are (2 , 2 , 2) . The coordinates of the fourth particle of mass 4 mg to be positioned so that the centre of mass of the four particles system is at the origin of the three-dimensional rectangular coordinate system are (A) (1 , 1 , 1) (B) (2 , 2 , 2) (C) (3 , 3 , 3) (D) (- 3 , - 3 , - 3)

Tardigrade Admin · Accepted Answer

Treat the three masses of 1 mg, 2 mg and 3 mg as a single mass of 6 mg at (2 , 2 , 2) Now, 0=(4 × x + 6 × 2/10) or 4x+12=0 Or x=-3, and, 0=(4 × y + 6 × 2/10) or y=-3 Similarly, z=-3

$(1, 1, 1)$

$(2, 2, 2)$

$(3, 3, 3)$

$(- 3, - 3, - 3)$

Treat the three masses of $1 m g, 2 m g$ and $3 m g$ as a single mass of $6 m g a t (2, 2, 2)$
Now, $0 = \frac{4 \times x + 6 \times 2}{10} or 4 x + 12 = 0$
Or $x = - 3,$
and, $0 = \frac{4 \times y + 6 \times 2}{10} or y = - 3$
Similarly, $z = - 3$

(1,1,1)

(2,2,2)

(3,3,3)

(−3,−3,−3)

Treat the three masses of 1mg,2mg and 3mg as a single mass of 6mgat(2,2,2) Now, 0=104×x+6×2​or4x+12=0 Or x=−3, and, 0=104×y+6×2​ory=−3 Similarly, z=−3

$(1, 1, 1)$

$(2, 2, 2)$

$(3, 3, 3)$

$(- 3, - 3, - 3)$

Treat the three masses of $1 m g, 2 m g$ and $3 m g$ as a single mass of $6 m g a t (2, 2, 2)$
Now, $0 = \frac{4 \times x + 6 \times 2}{10} or 4 x + 12 = 0$
Or $x = - 3,$
and, $0 = \frac{4 \times y + 6 \times 2}{10} or y = - 3$
Similarly, $z = - 3$