Question

The centre of mass of a system of three particles of masses $1 \,g , 2\, g$ and $3 \,g$ is taken as the origin of a coordinate system. The position vector of a fourth particle of mass $4\, g$ such that the centre of mass of the four particle system lies at the point (1,2,3) is $\alpha(\hat{i}+2 \hat{j}+3 \hat{k}),$ where $\alpha$ is a constant. The value of $\alpha$ is

• A. $\frac{10}{3}$
• B. $\frac{5}{2}$
• C. $\frac{1}{2}$
• D. $\frac{2}{5}$.

Answer 1

Answer: (B)

Let $(x, y, z)$ coordinates of three particles of masses
$1 g , 2 g$ and $3 g$ be $\left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right)$ and $\left(x_{3}, y_{3}, z_{3}\right)$ respectively.
$\therefore $ The $x$ -coordinate of the centre of mass is
$X_{ CM }=\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3}}{m_{1}+m_{2}+m_{3}}$
or $ 0=\frac{1 x_{1}+2 x_{2}+3 x_{3}}{1+2+3}$ or $x_{1}+2 x_{2}+3 x_{3}=0 \,\,\,\, ...(i)$
Let the fourth particle of mass $4 g$ be placed at $\left(x_{4}, y_{4}, z_{4}\right)$ so that the centre of mass of the four particles system lies at $(1,2,3)$
$\therefore 1=\frac{x_{1}+2 x_{2}+3 x_{4}+4 x_{4}}{1+2+3+4} \,\,\,\,\, ....(ii)$
or $ x_{1}+2 x_{2}+3 x_{4}+4 x_{4}=10$
Subtract $(i) $ from $(ii)$ , we get
$4 x_{4}=10 \text { or } x_{4}=\frac{10}{4}=\frac{5}{2} ; \therefore \alpha=\frac{5}{2}$