Question

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

• A. 10/3
• B. 5/2
• C. 1/2
• D. 2/5

Answer 1

Answer: (B)

The coordinates (x, y, z) of masses $1g,2g,3g$ and $4g$ are
$(x_1=0,y_1=0,z_1=0),(x_2=0,y_2=0,z_2=0)$
$\Rightarrow x_{CM}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3+m_4{x}_4}{m_1+m_2+m_3+m_4}$
$=\frac{4\alpha }{1+2+3+4}$
$=\frac{4\alpha }{10}$
Hence, $\frac{4\alpha }{10}=1$
$\Rightarrow \alpha =\frac{5}{2}$
$\Rightarrow y_{CM}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3+m_4{y}_4}{m_1+m_2+m_3+m_4}$
$=\frac{8\alpha }{10}=2$
$\Rightarrow \alpha =\frac{5}{2}$
$\Rightarrow z_{CM}=\frac{m_1{z}_1+m_2{z}_2+m_3{z}_3+m_4{z}_4}{m_1+m_2+m_3+m_4}$
$=\frac{12\alpha }{10}=3$
$\Rightarrow \alpha =5/2$