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)

Answer 1

Answer: (D)

(3,1,8)

Answer 2

According to the definition of centre of mass, we can imagine one particle of mass (1+2+3) kg at (1,2,3) ; another particle of mass (2 + 3 ) kg at (-1, 3, -2).

Let the third particle of mass 5 kg put at ($x_{3},y_{3},z_{3}$) i.e.,

$m_{1} = 6kg,(x_{1},y_{1},z_{1}) = (1,2,3)$

$m_{2} = 5kg,(x_{2},y_{2},z_{2}) = ( - 1,3, - 2)$

$m_{3} = 5kg,(x_{3},y_{3},z_{3})$ =?

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

Using $X_{CM} = \frac{m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3}}{m_{1} + m_{2} + m_{3}}$

$1 = \frac{6 \times 1 + 5 \times ( - 1) + 5x_{3}}{6 + 5 + 5}$

$5x_{3} = 16 - 1 = 15orx_{3} = 3$

Similarly, $y_{3} = 1$and $z_{3} = 3$