Question

The center 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 center 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

The Correct Option is B

The coordinates $(x, y, z)$ of masses $1\, g,\, 2\, g,\, 3\, g$ and $4\, g$ are
$\left(x_{1}=0, y_{1}=0, z_{1}=0\right)\left(x_{2}=0, y_{2}=0, z_{2}=0\right)$
$\left(x_{3}=0, y_{3}=0, z_{3}=0\right)\left(x_{4}=\alpha, y_{4}=2 \alpha, z_{4}=3 \alpha\right)$
$x_{ CM }=\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3}+x_{4} x_{4}}{m_{1}+m_{2}+m_{3}+m_{4}}$
$x_{ CM }= \frac{4 \alpha}{1+2+3 +4}=\frac{4 \alpha}{10}$
$1=\frac{4 \alpha}{10}$
$\Rightarrow \alpha=\frac{5}{2}$
$y_{ CM }= \frac{m_{1} y_{1}+m_{2} y_{2}+m_{3} y_{3}+m_{4} y_{4}}{m_{1}+m_{2}+m_{3}+m_{4}}$
$2= \frac{4 \times 2 \alpha}{10}$
$\alpha =\frac{20}{8}=\frac{5}{2}$
$z_{ CM } =\frac{m_{1} z_{1}+m_{2} z_{2}+m_{3} z_{3}+m_{4} z_{4}}{m_{1}+m_{2}+m_{3}+m_{4}}$
$3 =\frac{4 \times 3 \alpha}{10}$
$\alpha =\frac{5}{2}$