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 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 $1\,g, 2\,g , 3\,g$ and $4\,g$ 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, $\frac{4 \alpha}{10} = 1$
$\Rightarrow \alpha = \frac{5}{2}$
$\Rightarrow y_{CM} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3 + m_4 y_4}{m_1 + m_2 + m_3 + m_4}$
$= \frac{8 \alpha}{10} = 2$
$\Rightarrow \alpha =\frac{5}{2}$
$\Rightarrow z_{CM} = \frac{m_1 z_1 + m_2 z_2 + m_3 z_3 + m_4 z_4}{m_1 + m_2 + m_3 + m_4}$
$=\frac{12 \alpha}{10} = 3$
$\Rightarrow \alpha = 5/2$