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)

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_10= 6 \,kg, (x_1, y_1, z_1) = (1,2, 3) $
$m_2 = 5 \,kg,(x_2, y_2, z_2) = ( - 1,3, - 2)$
$m_3 = 5\,kg,(x_3, y_3, z_3) = $?
Given, $\left(X_{CM}, Y_{CM}, Z_{CM}\right) = \left( 1, 2, 3\right) $
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\times1+5\times\left(-1\right)+5x_{3}}{6+5+5} $
$ 5x_{3} = 16 -1 $
$ = 15$ or $x_{3} =3 $
Similarly, $y_3 = 1$ and $z_3 = 8 $