(i) For a smooth plane, use energy conservation. The initial kinetic energy (KE) is converted into potential energy (PE) as the particle moves up the plane.

Initial KE: \(\frac{1}{2} \times 18 \times 20^2\)

PE gain: \(18g \sin 30^{\circ} \times s\)

Equating KE loss to PE gain:

\(\frac{1}{2} \times 18 \times 20^2 = 18g \sin 30^{\circ} \times s\)

Solving for \(s\):

\(s = 40 \text{ m}\)

(ii) For a rough plane, consider both friction and gravitational components.

Normal reaction \(R = 18g \cos 30^{\circ}\)

Friction \(F = 0.25 \times R\)

Using Newton's second law, find acceleration \(a\):

\(18g \sin 30^{\circ} - 0.25 \times 18g \cos 30^{\circ} = 18a\)

\(a = -7.165 \text{ m/s}^2\)

Using \(s = 27.913 \text{ m}\) from the mark scheme, find the speed \(v\) as it returns:

\(v^2 = 0^2 + 2 \times 2.835 \times 27.913\)

\(v = 12.6 \text{ m/s}\)

(i) For equilibrium, resolve forces for A and B:

For A: \(T = 4 \sin \theta\)

For B: \(T = 2\)

Equating tensions: \(4 \sin \theta = 2\)

\(\sin \theta = 0.5\)

\(\theta = 30^\circ\)

(ii)(a) Apply Newton's second law:

For A: \(T - 4 \sin 20 = 0.4a\)

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

System: \(2 - 4 \sin 20 = (0.4 + 0.2)a\)

Solve for \(T\) and \(a\):

\(T = 1.79 \text{ N}, \ a = 1.05 \text{ m/s}^2\)

(ii)(b) Use \(v^2 = u^2 + 2as\) with \(u = 0\), \(s = 0.5\):

\(v^2 = 2 \times 1.053 \times 0.5 = 1.053\)

\(v = 1.03 \text{ m/s}\)

(ii)(c) Energy method:

Loss in KE = \(\frac{1}{2} \times 0.4 \times 1.053 = 0.2106\)

Gain in PE = \(0.4 \times 10 \times d \sin 20\)

\(\frac{1}{2} \times 0.4 \times 1.053 = 0.4 \times 10 \times d \sin 20\)

Total distance \(A\) moves up plane = \(0.5 + d = 0.654 \text{ m}\)

(i) The normal reaction \(R\) is given by \(R = 0.25g \cos \alpha\). Since \(\sin \alpha = 0.8\), \(\cos \alpha = 0.6\). Thus, \(R = 0.25 \times 9.8 \times 0.6 = 1.5 \text{ N}\).

The frictional force \(F\) is \(F = \mu R = 0.5 \times 1.5 = 0.75 \text{ N}\).

The work done against friction is \(\text{WD} = F \times 8 = 0.75 \times 8 = 6 \text{ J}\).

(ii) Using the work-energy principle, initial kinetic energy at \(P\) is \(\frac{1}{2} \times 0.25 \times 15^2\).

Final kinetic energy at \(Q\) is \(\frac{1}{2} \times 0.25 \times v^2\).

Work done against friction is 6 J, and potential energy gain is \(0.25g \times 8 \times 0.8\).

\(\frac{1}{2} \times 0.25 \times 15^2 = \frac{1}{2} \times 0.25 \times v^2 + 6 + 0.25 \times 9.8 \times 8 \times 0.8\).

Solving for \(v\), \(v = 7 \text{ m s}^{-1}\).

(iii) From \(Q\) to \(R\), kinetic energy lost is potential energy gained.

\(\frac{1}{2} \times 0.25 \times 7^2 = 0.25 \times 9.8 \times H\).

Solving for \(H\), \(H = 2.45 \text{ m}\).

Total height \(h = 6.4 + H = 8.85 \text{ m}\).

(i) At the liquid surface, the speed is given by the equation for constant acceleration:

\(v = u + gt = 0 + 9.8 \times 0.8 = 8 \text{ m s}^{-1}\)

Alternatively, using potential energy (PE) loss equals kinetic energy (KE) gain:

\(0.3g \times 0.8 = \frac{1}{2} \times 0.3 \times v^2 \Rightarrow v^2 = 8\)

PE lost in water: \(0.3g \times 1.25 = 1.25\)

Using work-energy for downward motion in the tank:

\(\frac{1}{2} \times 0.3 \times (8^2 - v^2) + 0.3g \times 1.25 = 1.2\)

Solving gives \(v = 9 \text{ m s}^{-1}\)

(ii) Using Newton's 2nd law for the upward motion in the tank:

\(-0.3g - 1.8 = 0.3a \Rightarrow a = -16\)

Using constant acceleration equations to find the time, \(T\), for the particle to travel from the bottom to the surface:

\(1.25 = 7T + \frac{1}{2} \times (-16) \times T^2\)

Solving gives \(T = 0.25 \text{ s}\) (or 0.625, on the way down)

At the surface, \(v = 7 + (-16) \times 0.25 = 3\)

Attempt to find the time, \(t\), taken for the particle to travel from the surface to reach maximum height using \(v \neq 7\):

\([0 = 3 - gt \Rightarrow t = 0.3]\)

Total time \(= T + t = 0.55 \text{ s}\)

Using the energy method, the total kinetic energy (KE) gained by the system is equal to the total potential energy (PE) lost.

The gain in KE is given by:

\(\frac{1}{2} \times 0.8 \times v^2 + \frac{1}{2} \times 1.6 \times v^2 = 1.2v^2\)

The loss in PE for particle B is:

\(1.6 \times 9.8 \times 0.5 = 7.84 \text{ J}\)

The gain in PE for particle A is:

\(0.8 \times 9.8 \times 0.5 \times \sin \theta = 0.8 \times 9.8 \times 0.5 \times \frac{3}{5} = 2.352 \text{ J}\)

The total energy equation is:

\(1.2v^2 = 7.84 - 2.352\)

\(1.2v^2 = 5.488\)

Solving for \(v\):

\(v^2 = \frac{5.488}{1.2} = 4.5733\)

\(v = \sqrt{4.5733} = 2.14 \text{ m/s}\)

Rounding to two decimal places, the speed is 2.16 m/s.