Question

Three masses $m_1$ = 1, $m_2$ = 2 and $m_3$ = 3 are located on the x-axis such that their center of mass is at x = 1. Another mass $m_1$ = 4 is placed at x, and the new center of mass is at x = 3. The value of $x_0$ is.

Answer 1

Correct Answer: 6

To solve this problem, we need to determine the position $x_0$ where a new mass $m_4$ = 4 is placed, given the changes in the center of mass. Initially, the center of mass $x_{cm,\text{initial}}$ of masses $m_1$, $m_2$, and $m_3$ is given as 1. We use the center of mass formula: \[ x_{cm} = \frac{\sum (m_i \cdot x_i)}{\sum m_i} \] For the initial setup with $m_1 = 1$, $m_2 = 2$, $m_3 = 3$, and their center at $x = 1$: \[ 1 = \frac{1 \cdot x_1 + 2 \cdot x_2 + 3 \cdot x_3}{6} \] Using $x_{cm,\text{initial}} = 1$, we derive: \[ 1 \cdot x_1 + 2 \cdot x_2 + 3 \cdot x_3 = 6 \] Now, when the additional mass $m_4 = 4$ is placed at $x = x_0$, the new center of mass $x_{cm,\text{new}}$ becomes 3: \[ 3 = \frac{6 + 4 \cdot x_0}{10} \] Solving for $x_0$: \[ 3 \cdot 10 = 6 + 4 \cdot x_0 \] \[ 30 = 6 + 4 \cdot x_0 \] \[ 24 = 4 \cdot x_0 \] \[ x_0 = \frac{24}{4} \] \[ x_0 = 6 \] Thus, the position of the new mass is \( x_0 = 6 \).