(i) For a smooth surface, apply Newton's second law to the system:

For A: \(T = 0.8a\)

For B: \(2 - T = 0.2a\)

Combine: \(0.2g = (0.2 + 0.8)a\)

Solve for \(a\): \(a = 2 \, \text{ms}^{-2}\)

Use the equation \(s = \frac{1}{2} a t^2\) to find \(t\):

\(2.5 = \frac{1}{2} \times 2 \times t^2\)

\(t = 1.58 \text{ s}\)

(ii) For a rough surface, the normal force \(N = 8\) and friction \(F = 0.1 \times N = 0.8\).

Apply Newton's second law:

For A: \(T - 0.8 = 0.8a\)

For B: \(2 - T = 0.2a\)

Combine: \(0.2g - 0.8 = (0.2 + 0.8)a\)

Solve for \(a\): \(a = 1.2 \, \text{ms}^{-2}\)

Use the equation \(v^2 = u^2 + 2as\) to find \(v\):

\(v^2 = 0 + 2 \times 1.2 \times 2.5\)

\(v = \sqrt{6} = 2.45 \text{ ms}^{-1}\)

(i) After B reaches the floor, A continues at constant speed until it reaches the pulley (no tension and the surface is smooth). Thus A's speed when B reaches the floor is the same as A's speed (3 m s^-1) when it reaches the pulley. Until the instant when B reached the floor, A and B have the same speed and hence B reaches the floor with speed 3 m s^-1.

(ii) Using energy conservation:

Loss of potential energy (PE) of B is given by:

\(0.15gh\)

Gain of kinetic energy (KE) of the system is:

\(\frac{1}{2} (0.35 + 0.15) \times 3^2\)

Equating loss of PE to gain of KE:

\(0.15gh = \frac{1}{2} \times 0.5 \times 9\)

\(1.5h = 0.25 \times 9\)

\(h = 1.5\)

(i) Using Newton's second law for both particles, we have:

\(T - 0.2 \times 3 = 0.3a \quad \text{and} \quad 7 - T = 0.7a\)

Solving these equations gives the acceleration:

\(a = 6.4 \text{ m/s}^2\)

Using the equation \(v = u + at\) to find the speed when the string breaks:

\(v = 0 + 6.4 \times 0.25 = 1.6 \text{ m/s}\)

Using \(s = ut + \frac{1}{2}at^2\) to find the distance moved before the break:

\(s = 0 + \frac{1}{2} \times 6.4 \times (0.25)^2 = 0.2 \text{ m}\)

Using \(v^2 = u^2 + 2gs\) to find the speed when B hits the floor:

\(v^2 = 1.6^2 + 2 \times 9.8 \times (0.5 - 0.2)\)

Solving gives:

\(v = 2.93 \text{ m/s}\)

(ii) For finding the distance travelled by A after the break from \(v^2 = u^2 + 2as\):

Distance travelled after break:

\(= (0 - 1.6^2) \div (2 \times -2) = 0.64 \text{ m}\)

Total distance travelled:

\(= 0.2 + 0.64 = 0.84 \text{ m}\)

(i) For particle A on the table, applying Newton's second law:

\(T - \frac{2}{7} \times 1.26 \times g = 1.26 \times a\)

For particle B hanging, applying Newton's second law:

\(0.9 \times g - T = 0.9 \times a\)

Solving these equations simultaneously gives:

\(0.9g - \frac{2}{7} \times 1.26g = (0.9 + 1.26) \times a\)

\(a = 2.5 \text{ m s}^{-2}\)

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

\(T = \frac{2}{7} \times 1.26 \times g + 1.26 \times 2.5\)

\(T = 6.75 \text{ N}\)

(ii) Using \(v^2 = 2ah\) for particle B:

\(v^2 = 2 \times 2.5 \times 0.45\)

\(v = 1.5 \text{ m s}^{-1}\)

(iii) For particle A, using \(v^2 = v_B^2 + 2as\):

\(v^2 = 2.25 + 2 \times \left( -\frac{20}{7} \right) \times 0.03\)

\(v = 1.44 \text{ m s}^{-1}\)

(i) Applying Newton's second law to particle A:

\(0.32g - T = 0.32a\)

For particle B:

\(T = 0.48a\)

Solving these equations simultaneously:

\(0.32g = 0.32a + 0.48a\)

\(0.32g = 0.8a\)

\(a = \frac{0.32g}{0.8} = 4 \text{ m/s}^2\)

Substitute back to find tension:

\(T = 0.48 \times 4 = 1.92 \text{ N}\)

(ii) Using the equation \(s = \frac{1}{2}at^2\) for A:

\(0.98 = \frac{1}{2} \times 4 \times t^2\)

\(t^2 = \frac{0.98}{2}\)

\(t = 0.7 \text{ s}\)

(iii) For B to reach the pulley:

Velocity when A hits the floor:

\(v = at = 4 \times 0.7 = 2.8 \text{ m/s}\)

Time for B to travel remaining distance:

\(t = \frac{1.4 - 0.98}{2.8} = 0.15 \text{ s}\)

Total time:

\(0.7 + 0.15 = 0.85 \text{ s}\)