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

\(2.4g - T = 2.4a\)

For particle B:

\(T - 1.2g = 1.2a\)

Adding these equations:

\(2.4g - 1.2g = (2.4 + 1.2)a\)

\(1.2g = 3.6a\)

\(a = \frac{1.2g}{3.6} = \frac{10}{3}\) m/s²

Substituting \(a\) back into one of the equations:

\(T = 1.2g + 1.2 \times \frac{10}{3}\)

\(T = 12 + 4 = 16\) N

(b) Using energy conservation for particle B:

Initial potential energy = \(1.2g \times 1.5\)

Final potential energy = \(1.2g \times h\)

Using \(v^2 = u^2 + 2as\) with \(v^2 = 2 \times \frac{10}{3} \times 2.1\)

\(v^2 = 14\)

Using \(v^2 = u^2 + 2gh\) with \(u = 0\):

\(0 = 14 - 2g \times s\)

\(s = 4.3\) m

Greatest height = initial height + additional height = \(1.5 + 2.1 + 0.7 = 4.3\) m

(i) To find the speed of A at the instant B reaches the ground, we use Newton's Second Law and energy conservation.

Using Newton's Second Law for the system:

\(T - 0.35g = 0.35a\)

\(0.45g - T = 0.45a\)

Solving these equations gives:

\(0.45g - 0.35g = (0.35 + 0.45)a\)

\(0.1g = 0.8a\)

\(a = \frac{0.1g}{0.8} = 1.25 \text{ m/s}^2\)

Using the equation of motion:

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

where initial velocity \(u = 0\), \(a = 1.25 \text{ m/s}^2\), and \(s = 0.64 \text{ m}\):

\(v^2 = 2 \times 1.25 \times 0.64 = 1.6\)

\(v = \sqrt{1.6} = 1.26 \text{ m/s}\)

(ii) To find the total distance travelled by A after B reaches the ground, we consider the motion of A under gravity alone.

Using the equation of motion:

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

where \(v = 0\), \(u = 1.26 \text{ m/s}\), and \(g = 9.8 \text{ m/s}^2\):

\(0 = 1.26^2 - 2 \times 9.8 \times s\)

\(s = \frac{1.26^2}{2 \times 9.8} = 0.08 \text{ m}\)

The total distance travelled by A is twice this distance (as it moves up and then down):

\(\text{Total distance} = 2 \times 0.08 = 0.16 \text{ m}\)

(i) Apply Newton's second law to the particles. For the 1.2 kg particle:

\(12 - T = 1.2a\)

For the 0.8 kg particle:

\(T - 8 = 0.8a\)

Solving these equations simultaneously gives:

\(a = 2 ext{ m/s}^2, \, T = 9.6 ext{ N}\)

(ii) First, find the time taken for the 1.2 kg particle to reach the ground:

\(0.64 = \frac{1}{2} \times 2 \times t_1^2\)

Solving gives:

\(t_1 = 0.8 ext{ s}\)

The speed of the 0.8 kg particle when the 1.2 kg particle hits the ground is:

\(v = 2 \times 0.8 = 1.6 ext{ m/s}\)

Next, find the time and distance for the 0.8 kg particle to come to rest:

\(0 = 1.6 - 10t_2\)

Solving gives:

\(t_2 = 0.16 ext{ s}\)

The distance travelled during this time is:

\(s_2 = \frac{1}{2} \times 1.6 \times 0.16 = 0.128 ext{ m}\)

Finally, find the distance travelled in the remaining time:

\(t_3 = 1 - 0.8 - 0.16 = 0.04 ext{ s}\)

\(s_3 = \frac{1}{2} \times 10 \times 0.04^2 = 0.008 ext{ m}\)

Total distance travelled:

\(0.64 + 0.128 + 0.008 = 0.776 ext{ m}\)

(i) (a) Using Newton's Second Law for the system, we have:

\(1.3g - T = 1.3a\)

\(T - 0.7g = 0.7a\)

Adding these equations gives:

\(1.3g - 0.7g = (1.3 + 0.7)a\)

\(0.6g = 2a\)

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

Substituting back to find tension:

\(1.3g - T = 1.3 \times 3\)

\(T = 1.3g - 3.9\)

\(T = 9.1 \text{ N}\)

(b) Using the equation of motion:

\(s = \frac{1}{2} a t^2\)

\(2 = \frac{1}{2} \times 3 \times t^2\)

\(t^2 = \frac{4}{3}\)

\(t = \sqrt{\frac{4}{3}} \approx 1.15 \text{ seconds}\)

(ii) Using energy conservation or equations of motion:

At the point when the 1.3 kg particle reaches the plane, the speed \(v\) is given by:

\(v^2 = 2 \times 3 \times 2\)

\(v = \sqrt{12}\)

The 0.7 kg particle continues upwards:

\(0 = 12 - 2gs\)

\(s = \frac{12}{2g}\)

Greatest height above the plane is:

\(2 + \frac{12}{2g} = 4.6 \text{ m}\)

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

\(0.35g - T = 0.35a\)

For particle B:

\(T - 0.15g = 0.15a\)

Adding the two equations:

\((0.35 - 0.15)g = (0.35 + 0.15)a\)

Solving for acceleration \(a\):

\(0.2g = 0.5a\)

\(a = \frac{0.2g}{0.5} = 4\, \text{m/s}^2\)

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

\(T = 0.15g + 0.15 \times 4 = 2.1\, \text{N}\)

(ii) Using the equation \(v_1^2 = 0 + 2a \times 1.6\):

\(v_1^2 = 8 \times 1.6 = 12.8\)

Using the formula for greatest height \(H\):

\(H = 1.6 + \frac{0 - v_1^2}{-2g}\)

\(H = 1.6 + \frac{-12.8}{-20} = 2.24\, \text{m}\)