(i) Resolve forces for particle A:

The normal reaction force is given by:

\(R = mg \cos 30^{\circ}\)

The frictional force is:

\(F = 0.2 \times mg \cos 30^{\circ}\)

For particle B, the tension is:

\(T = 4g\)

For particle A, resolve parallel to the plane:

\(T = mg \sin 30^{\circ} + F\)

Substitute for T and F:

\(40 = mg \sin 30^{\circ} + 0.2 \times mg \cos 30^{\circ}\)

Solve for m:

\(m = \frac{40}{5 + \sqrt{3}} = 5.94\)

\((ii) For m = 3, resolve forces and use energy principles:\)

Resolve forces for particle A:

\(R = 3g \cos 30^{\circ}\)

\(F = 0.2 \times 3g \cos 30^{\circ}\)

Apply Newton's Second Law:

\(4g - T = 4a\)

\(T - 3g \sin 30^{\circ} - F = 3a\)

Solve for acceleration a:

\(a = \frac{25}{3 \sqrt{3}} - 7\)

Use the equation:

\(v^2 = u^2 + 2as\)

\(Since initial velocity u = 0, solve for s:\)

\(s = \frac{0.5 \times 2.83}{2} = 0.710 \text{ m}\)

First, calculate the normal reaction \(R\) on the 5 kg particle:

\(R = 5g \cos \alpha = 4g\)

The frictional force \(F\) is:

\(F = 0.5 \times 4g = 2g\)

Apply Newton's second law to the 5 kg particle on the slope:

\(T - 2g - 5g \sin \alpha = 5a\)

Since \(\sin \alpha = \frac{3}{5}\), substitute to get:

\(T - 5g = 5a\)

For the 10 kg particle hanging vertically:

\(10g - T = 10a\)

Combine the equations:

\(10g - 5g = 15a\)

Solve for \(a\):

\(a = \frac{g}{3} = 3.33 \text{ m/s}^2\)

Substitute \(a\) back to find \(T\):

\(T = 10g - 10 \left( \frac{g}{3} \right) = \frac{20g}{3} = 66.7 \text{ N}\)

Using Newton's second law for particle P:

\(10 - 0.5g - T = 0.5a\)

For particle Q:

\(T - 0.3g = 0.3a\)

For the system of both particles:

\(10 - 0.5g - 0.3g = 0.8a\)

Solving for acceleration \(a\):

\(10 - 0.8g = 0.8a\)

\(a = \frac{10 - 0.8 \times 9.8}{0.8} = 2.5 \text{ m/s}^2\)

Substitute \(a = 2.5\) into the equation for Q:

\(T - 0.3 \times 9.8 = 0.3 \times 2.5\)

\(T = 0.3 \times 9.8 + 0.75 = 3.75 \text{ N}\)

(i) For block B in equilibrium, the tension in S₂ must balance the weight of both blocks. Therefore, the tension in S₂ is given by:

\(T_{S_2} = (3 + 2)g = 5g = 50 \text{ N}\)

For block A, the tension in S₁ must balance the weight of block A alone. Therefore, the tension in S₁ is:

\(T_{S_1} = 3g = 30 \text{ N}\)

(ii) After S₂ breaks, apply Newton's second law to each block. For block A:

\(3g - T - 1.6 = 3a\)

For block B:

\(2g + T - 4 = 2a\)

Adding these equations:

\((3g + 2g) - (1.6 + 4) = (3 + 2)a\)

\(5g - 5.6 = 5a\)

\(a = \frac{5g - 5.6}{5} = 8.88 \text{ m/s}^2\)

Substitute \(a\) back into one of the equations to find \(T\):

\(3g - T - 1.6 = 3 \times 8.88\)

\(T = 3g - 1.6 - 26.64\)

\(T = 1.76 \text{ N}\)

(i) In equilibrium, the forces on each particle must balance. For particle A, the tension in S₁ must balance the weight of A and the tension in S₂. Therefore, we have:

\(T_1 = W_A + T_2\)

For particle B, the tension in S₂ must balance the weight of B:

\(T_2 = W_B\)

Given that the weight of each particle is \(W = mg = 0.2 \times 9.8 = 1.96 \text{ N}\), we find:

\(T_2 = 1.96 \approx 2 \text{ N}\)

Substituting back, we find:

\(T_1 = 1.96 + 1.96 = 3.92 \approx 4 \text{ N}\)

(ii) When S₁ is cut, both particles fall with acceleration. Using Newton's second law for particle A:

\(T_2 + 2 - 0.4 = 0.2a\)

For particle B:

\(2 - T_2 - 0.2 = 0.2a\)

Solving these equations simultaneously, we find:

\(a = 8.5 \text{ m/s}^2\)

\(T_2 = 0.1 \text{ N}\)