To solve the problem, we will follow these steps:
### Step 1: Determine the initial position of the center of mass (CM)
Given that we have four identical particles, each with a mass of 1 kg, arranged at the corners of a square with a side length of \(2\sqrt{2}\) m, we can find the coordinates of the particles:
- Particle 1 at (0, 0)
- Particle 2 at (0, \(2\sqrt{2}\))
- Particle 3 at (\(2\sqrt{2}\), 0)
- Particle 4 at (\(2\sqrt{2}\), \(2\sqrt{2}\))
The formula for the center of mass (CM) of a system of particles is given by:
\[
\text{CM} = \left( \frac{\sum m_ix_i}{\sum m_i}, \frac{\sum m_iy_i}{\sum m_i} \right)
\]
For our case, since all masses are equal (1 kg), the total mass \(M = 4 \text{ kg}\).
Calculating the x-coordinate of the CM:
\[
x_{CM} = \frac{1 \cdot 0 + 1 \cdot 0 + 1 \cdot 2\sqrt{2} + 1 \cdot 2\sqrt{2}}{4} = \frac{4\sqrt{2}}{4} = \sqrt{2}
\]
Calculating the y-coordinate of the CM:
\[
y_{CM} = \frac{1 \cdot 0 + 1 \cdot 2\sqrt{2} + 1 \cdot 0 + 1 \cdot 2\sqrt{2}}{4} = \frac{4\sqrt{2}}{4} = \sqrt{2}
\]
Thus, the initial position of the center of mass is:
\[
\text{CM}_{initial} = (\sqrt{2}, \sqrt{2})
\]
### Step 2: Determine the new position of the center of mass after removing one particle
Let’s assume we remove Particle 1 at (0, 0). The remaining particles are now:
- Particle 2 at (0, \(2\sqrt{2}\))
- Particle 3 at (\(2\sqrt{2}\), 0)
- Particle 4 at (\(2\sqrt{2}\), \(2\sqrt{2}\))
Now, the total mass \(M = 3 \text{ kg}\).
Calculating the new x-coordinate of the CM:
\[
x_{CM}' = \frac{1 \cdot 0 + 1 \cdot 2\sqrt{2} + 1 \cdot 2\sqrt{2}}{3} = \frac{4\sqrt{2}}{3}
\]
Calculating the new y-coordinate of the CM:
\[
y_{CM}' = \frac{1 \cdot 2\sqrt{2} + 1 \cdot 0 + 1 \cdot 2\sqrt{2}}{3} = \frac{4\sqrt{2}}{3}
\]
Thus, the new position of the center of mass is:
\[
\text{CM}_{final} = \left(\frac{4\sqrt{2}}{3}, \frac{4\sqrt{2}}{3}\right)
\]
### Step 3: Calculate the shift in the center of mass
The shift in the center of mass can be calculated using the distance formula:
\[
d = \sqrt{(x_{CM}' - x_{CM})^2 + (y_{CM}' - y_{CM})^2}
\]
Substituting the values:
\[
d = \sqrt{\left(\frac{4\sqrt{2}}{3} - \sqrt{2}\right)^2 + \left(\frac{4\sqrt{2}}{3} - \sqrt{2}\right)^2}
\]
This simplifies to:
\[
d = \sqrt{2 \left(\frac{4\sqrt{2}}{3} - \sqrt{2}\right)^2}
\]
Calculating the expression inside the square root:
\[
\frac{4\sqrt{2}}{3} - \sqrt{2} = \frac{4\sqrt{2} - 3\sqrt{2}}{3} = \frac{\sqrt{2}}{3}
\]
Thus,
\[
d = \sqrt{2 \left(\frac{\sqrt{2}}{3}\right)^2} = \sqrt{2 \cdot \frac{2}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3} \text{ m}
\]
### Final Answer
The shift in the center of mass is \(\frac{2}{3} \text{ m}\).
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