To solve the problem of finding the shift in the center of mass when one of the four identical particles at the corners of a square has its mass doubled, we can follow these steps:
### Step-by-Step Solution:
1. **Initial Configuration**:
- We have four identical particles, each of mass \( m \), placed at the corners of a square with side length \( a \).
- The coordinates of the particles are:
- \( P_1(0, 0) \)
- \( P_2(a, 0) \)
- \( P_3(0, a) \)
- \( P_4(a, a) \)
2. **Calculate Initial Center of Mass (CM)**:
- The formula for the center of mass for a system of particles is given by:
\[
R_{CM} = \frac{\sum m_i r_i}{\sum m_i}
\]
- For the initial configuration:
\[
R_{CM} = \frac{m(0, 0) + m(a, 0) + m(0, a) + m(a, a)}{4m}
\]
- This simplifies to:
\[
R_{CM} = \frac{(0 + a + 0 + a, 0 + 0 + a + a)}{4} = \frac{(2a, 2a)}{4} = \left(\frac{a}{2}, \frac{a}{2}\right)
\]
3. **New Configuration After Doubling One Mass**:
- Let’s say we double the mass of the particle at \( P_1 \). Now the masses are:
- \( P_1(0, 0) \) with mass \( 2m \)
- \( P_2(a, 0) \) with mass \( m \)
- \( P_3(0, a) \) with mass \( m \)
- \( P_4(a, a) \) with mass \( m \)
4. **Calculate New Center of Mass**:
- The new center of mass is calculated as:
\[
R'_{CM} = \frac{2m(0, 0) + m(a, 0) + m(0, a) + m(a, a)}{2m + m + m + m} = \frac{(0 + a + 0 + a, 0 + 0 + a + a)}{5m}
\]
- This simplifies to:
\[
R'_{CM} = \frac{(2a, 2a)}{5} = \left(\frac{2a}{5}, \frac{2a}{5}\right)
\]
5. **Calculate the Shift in Center of Mass**:
- The initial center of mass was at \( \left(\frac{a}{2}, \frac{a}{2}\right) \) and the new center of mass is at \( \left(\frac{2a}{5}, \frac{2a}{5}\right) \).
- The shift \( d \) in the center of mass is given by:
\[
d = \sqrt{\left(\frac{2a}{5} - \frac{a}{2}\right)^2 + \left(\frac{2a}{5} - \frac{a}{2}\right)^2}
\]
- Simplifying the x-coordinates:
\[
\frac{2a}{5} - \frac{a}{2} = \frac{4a - 5a}{10} = -\frac{a}{10}
\]
- Therefore:
\[
d = \sqrt{2 \left(-\frac{a}{10}\right)^2} = \sqrt{2 \cdot \frac{a^2}{100}} = \sqrt{\frac{2a^2}{100}} = \frac{a\sqrt{2}}{10}
\]
6. **Final Result**:
- The shift in the center of mass is:
\[
d = \frac{a\sqrt{2}}{10}
\]
