Four particles A, B, C and D with masses $$m_A = m$$, $$m_B = 2m$$, $$m_C = 3m$$ and $$m_D = 4m$$ are at the corners of a square. They have accelerations of equal magnitude with directions as shown. The acceleration of the centre of mass of the particles is:

Problem Analysis

• Masses of the particles: $$m_A = m$$, $$m_B = 2m$$, $$m_C = 3m$$, $$m_D = 4m$$
• Total mass of the system ($$M$$):
• Accelerations ($$\vec{a}$$): Each particle has an acceleration of magnitude $$a$$. Based on standard square corner configurations in these types of problems, their vector directions are:
  • o $$\vec{a}_A = -a\hat{i}$$
• o $$\vec{a}_B = a\hat{j}$$
• o $$\vec{a}_C = a\hat{i}$$
• o $$\vec{a}_D = -a\hat{j}$$
• Along $$\hat{i}$$ (x-axis): $$-ma + 3ma = 2ma$$
• Along $$\hat{j}$$ (y-axis): $$2ma - 4ma = -2ma$$
• Correct Option: (3) $$\frac{a}{5}(\hat{i} - \hat{j})$$

$$M = m_A + m_B + m_C + m_D = m + 2m + 3m + 4m = 10m$$

Step-by-Step Solution

Step 1: Formula for Acceleration of Centre of Mass ($$\vec{a}_c$$)

$$\vec{a}_c = \frac{m_A\vec{a}_A + m_B\vec{a}_B + m_C\vec{a}_C + m_D\vec{a}_D}{M}$$

Step 2: Substitute the Vector Values

Substitute the masses and acceleration vectors into the formula:

$$\vec{a}_c = \frac{m(-a\hat{i}) + 2m(a\hat{j}) + 3m(a\hat{i}) + 4m(-a\hat{j})}{10m}$$

$$\vec{a}_c = \frac{-ma\hat{i} + 2ma\hat{j} + 3ma\hat{i} - 4ma\hat{j}}{10m}$$

Step 3: Simplify Component-wise

Combine the terms:

$$\vec{a}_c = \frac{2ma\hat{i} - 2ma\hat{j}}{10m}$$

$$\vec{a}_c = \frac{2ma(\hat{i} - \hat{j})}{10m} = \frac{a}{5}(\hat{i} - \hat{j})$$

Conclusion

option (B)