The centre of mass of a system of three particles of masses 1 g , 2 g and 3 g is taken as the origin of a coordinate system. The position vector of a fourth particle of mass 4 g such that the centre of mass of the four particle system lies at the point (1,2,3) is α( hati+2 hatj+3 hatk), where α is a constant. The value of α is

Question

The centre of mass of a system of three particles of masses 1 g , 2 g and 3 g is taken as the origin of a coordinate system. The position vector of a fourth particle of mass 4 g such that the centre of mass of the four particle system lies at the point (1,2,3) is α( hati+2 hatj+3 hatk), where α is a constant. The value of α is (A) (10/3) (B) (5/2) (C) (1/2) (D) (2/5).

Tardigrade Admin · Accepted Answer

Let (x, y, z) coordinates of three particles of masses 1 g , 2 g and 3 g be (x1, y1, z1),(x2, y2, z2) and (x3, y3, z3) respectively. ∴ The x -coordinate of the centre of mass is X CM =(m1 x1+m2 x2+m3 x3/m1+m2+m3) or 0=(1 x1+2 x2+3 x3/1+2+3) or x1+2 x2+3 x3=0 ...(i) Let the fourth particle of mass 4 g be placed at (x4, y4, z4) so that the centre of mass of the four particles system lies at (1,2,3) ∴ 1=(x1+2 x2+3 x4+4 x4/1+2+3+4) ....(ii) or x1+2 x2+3 x4+4 x4=10 Subtract (i) from (ii) , we get 4 x4=10 text or x4=(10/4)=(5/2) ; ∴ α=(5/2)

$\frac{10}{3}$

$\frac{5}{2}$

$\frac{1}{2}$

$\frac{2}{5}$ .

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52​.

$\frac{10}{3}$

$\frac{5}{2}$

$\frac{1}{2}$

$\frac{2}{5}$ .