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 $4 g$ 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. 44837
• B. 44683
• C. 44563
• D. 44597

Answer 1

Answer: (B)

44683

Answer 2

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 \hspace15mm 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} \hspace25mm = \frac{4 \alpha}{1 + 2 + 3 + 4} \hspace25mm = \frac{4 \alpha}{10} Hence, \hspace25mm \frac{4 \alpha}{10} = 1 \Rightarrow \hspace25mm \alpha = \frac{5}{2} \Rightarrow \hspace15mm 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 \hspace15mm \alpha =\frac{5}{2} \Rightarrow \hspace15mm 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} \hspace25mm =\frac{12 \alpha}{10} = 3 \Rightarrow \, \, \alpha = 5/2