(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:

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

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