Mass centers of a system of three particles of masses `1,2,3kg` is at the point (1m,2m,3m) and mass center of another group of two partices of masses `2kg` and `3kg` is at point `(-1m,3m,-2m)`. Where a `5kg` particle should be placed, so that mass center of the system of all these six particles shifts to mass center of the first system?

Options

A. `(1m,-3m,2m)`
B. `(3m,3m,2m)`
C. `(-1m,2m,3m)`
D. `(3m,1m,8m)`

Explanation

To find the position where a 5 kg particle should be placed so that the center of mass of the entire system of six particles shifts to the center of mass of the first system (1 m, 2 m, 3 m), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Masses and Their Positions:**
- First system has three particles with masses:
- \( m_1 = 1 \, \text{kg} \) at \( (x_1, y_1, z_1) = (1, 2, 3) \)
- \( m_2 = 2 \, \text{kg} \) at \( (1, 2, 3) \)
- \( m_3 = 3 \, \text{kg} \) at \( (1, 2, 3) \)
- Second system has two particles with masses:
- \( m_4 = 2 \, \text{kg} \) at \( (-1, 3, -2) \)
- \( m_5 = 3 \, \text{kg} \) at \( (-1, 3, -2) \)
- We need to find the position \( (x, y, z) \) of the 5 kg particle.

2. **Calculate the Center of Mass of the First System:**
The center of mass \( (X_{cm1}, Y_{cm1}, Z_{cm1}) \) of the first system is given as:
\[ (X_{cm1}, Y_{cm1}, Z_{cm1}) = (1, 2, 3) \]

3. **Calculate the Center of Mass of the Second System:**
The center of mass \( (X_{cm2}, Y_{cm2}, Z_{cm2}) \) of the second system can be calculated as:
\[ X_{cm2} = \frac{m_4 \cdot x_4 + m_5 \cdot x_5}{m_4 + m_5} = \frac{2 \cdot (-1) + 3 \cdot (-1)}{2 + 3} = \frac{-2 - 3}{5} = -1 \]
\[ Y_{cm2} = \frac{m_4 \cdot y_4 + m_5 \cdot y_5}{m_4 + m_5} = \frac{2 \cdot 3 + 3 \cdot 3}{2 + 3} = \frac{6 + 9}{5} = 3 \]
\[ Z_{cm2} = \frac{m_4 \cdot z_4 + m_5 \cdot z_5}{m_4 + m_5} = \frac{2 \cdot (-2) + 3 \cdot (-2)}{2 + 3} = \frac{-4 - 6}{5} = -2 \]

4. **Combine the Two Systems:**
Now, we have a total mass of:
\[ M = m_1 + m_2 + m_3 + m_4 + m_5 + m_6 = 1 + 2 + 3 + 2 + 3 + 5 = 16 \, \text{kg} \]

5. **Set Up the Center of Mass Equation:**
The center of mass of the entire system (including the 5 kg particle) should equal the center of mass of the first system:
\[ (X_{cm}, Y_{cm}, Z_{cm}) = \left( \frac{1 \cdot 1 + 2 \cdot 1 + 3 \cdot 1 + 2 \cdot (-1) + 3 \cdot (-1) + 5 \cdot x}{16}, \frac{1 \cdot 2 + 2 \cdot 2 + 3 \cdot 2 + 2 \cdot 3 + 3 \cdot 3 + 5 \cdot y}{16}, \frac{1 \cdot 3 + 2 \cdot 3 + 3 \cdot 3 + 2 \cdot (-2) + 3 \cdot (-2) + 5 \cdot z}{16} \right) \]

6. **Equate to the Center of Mass of the First System:**
Set the equations for \( X_{cm}, Y_{cm}, Z_{cm} \) equal to \( (1, 2, 3) \) and solve for \( x, y, z \).

For \( x \):
\[ \frac{6 + 5x}{16} = 1 \implies 6 + 5x = 16 \implies 5x = 10 \implies x = 2 \]

For \( y \):
\[ \frac{12 + 5y}{16} = 2 \implies 12 + 5y = 32 \implies 5y = 20 \implies y = 4 \]

For \( z \):
\[ \frac{18 + 5z}{16} = 3 \implies 18 + 5z = 48 \implies 5z = 30 \implies z = 6 \]

7. **Final Position of the 5 kg Particle:**
The 5 kg particle should be placed at the coordinates:
\[ (x, y, z) = (2, 4, 6) \]