Question

Centre of mass of three particles of masses $10\text{Kg}$, $20\text{Kg}$ and $30\text{Kg}$ respectively is at $(0,0,0)$. Where should a particle of mass $40\text{Kg}$ be placed so that the combined centre of mass will be at $(3,3,3)$

• A. $(0,0,0)$
• B. $(7.5,7.5,7.5)$
• C. $(1,2,3)$
• D. $(4,4,4)$

Answer 1

The correct option is B $(7.5,7.5,7.5)$

We know that,

$x_{com}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}$

$0=10x_1+20x_2+30x_3.........\left(1\right)$

Here, $(x_1,x_2,x_3)$ are the $x$ - coordinates of particles of masses $10\text{Kg}$, $20\text{Kg}$ and $30\text{Kg}$ respectively.

Let the $x$ - coordinate of the particle of mass $40\text{Kg}$ be $x_4$.

So, the new $x$ - coordinate of centre of mass of the four particle system can be written as,

$x_{com}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3+m_4{x}_4}{m_1+m_2+m_3+m_4}$

$3=\frac{10x_1+20x_2+30x_3+40x_4}{10+20+30+40}$

$\Rightarrow 300=10x_1+20x_2+30x_3+40x_4......\left(2\right)$

Substituting equation $\left(1\right)$ in equation $\left(2\right)$ we get,

$300=0+40x_4$

$\Rightarrow x_4=\frac{300}{40}=7.5$

Similarly, $y_4=z_4=7.5$

So, the coordinates of the particle of mass $40\text{Kg}$ is $(7.5,7.5,7.5)$

Hence, option $\left(B\right)$ is the correct answer.