Centre of mass of three particles of masses 1 kg, 2 kg and 3 kg lies at the point (1,2,3) and centre of mass of another system of particles 3 kg and 2 kg lies at the point (-1,3,-2). Where should we put a particle of mass 5 kg so that the centre of mass of entire system lies at the centre of mass of first system?

Question

Centre of mass of three particles of masses 1 kg, 2 kg and 3 kg lies at the point (1,2,3) and centre of mass of another system of particles 3 kg and 2 kg lies at the point (-1,3,-2). Where should we put a particle of mass 5 kg so that the centre of mass of entire system lies at the centre of mass of first system? (A) (0,0,0) (B) (1,3,2) (C) (-1,2,3) (D) (3,1,8)

Tardigrade Admin · Accepted Answer

According to the definition of centre of mass, we can imagine one particle of mass

(1 + 2 + 3) k g

at

(1, 2, 3)

; another particle of mass

(2 + 3) k g

at

(- 1, 3, - 2)

.
Let the third particle of mass 5 kg put at

(x_{3}, y_{3}, z_{3})

i.e.

m_{1} = 6 k g, (x_{1}, y_{1}, z_{1}) = (1, 2, 3)

m_{2} = 5 k g, (x_{2}, y_{2}, z_{2}) = (- 1, 3, - 2)

m_{3} = 5 k g, (x_{3}, y_{3}, z_{3}) = ?

Given,

(X_{CM}, Y_{CM}, Z_{CM}) = (1, 2, 3)

X_{CM} = \frac{m _{1} x _{1} + m _{2} x _{2} + m _{3} x _{3}}{m _{1} + m _{2} + m _{3}}

or

1 = \frac{6 \times 1 + 5 \times ( - 1 ) + 5 x _{3}}{6 + 5 + 5}

5 x_{3} = 16 - 1 = 15

or

x_{3} = 3

Similarly,

y_{3} = 1

{ and

z_{3} = 8

$(0, 0, 0)$

$(1, 3, 2)$

$(- 1, 2, 3)$

$(3, 1, 8)$

(0,0,0)

(1,3,2)

(−1,2,3)

(3,1,8)

$(0, 0, 0)$

$(1, 3, 2)$

$(- 1, 2, 3)$

$(3, 1, 8)$