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 $\left(x_3, y_3, z_3\right)$ i.e.
$m_1=6\, kg,\left(x_1, y_1, z_1\right)=(1,2,3)$
$m_2=5\, kg,\left(x_2, y_2, z_2\right)=(-1,3,-2)$
$m_3=5\, kg,\left(x_3, y_3, z_3\right)=?$
Given, $\left(X_{ CM }, Y_{ CM }, Z_{ CM }\right)=(1,2,3)$
$X_{CM}=\frac{m_1 x_1+m_2 x_2+m_3 x_3}{m_1+m_2+m_3}$
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 $z_3=8$