(a) To find the greatest height, use the equation of motion:

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

where \(v = 0\) (at the highest point), \(u = 10\) m/s, and \(a = -g = -9.8\) m/s². Solving for \(s\):

\(0 = 10^2 + 2(-9.8)s\)

\(s = \frac{100}{19.6} = 5\) m

(b) The kinetic energy before impact is:

\(\frac{1}{2} \times 0.4 \times 10^2 = 20\) J

After losing 7.2 J, the kinetic energy is:

\(20 - 7.2 = 12.8\) J

Using \(\frac{1}{2}mv^2 = 12.8\), solve for \(v\):

\(\frac{1}{2} \times 0.4 \times v^2 = 12.8\)

\(v^2 = 64\)

\(v = 8\) m/s

Using the equation of motion to find the time to reach the maximum height after the bounce:

\(0 = 8 + (-9.8)t\)

\(t = \frac{8}{9.8} \approx 0.816\) s

The total time between the first and second impacts is twice this time:

\(2 \times 0.816 = 1.632 \approx 1.6\) s

(i) The increase in kinetic energy (KE) is given by:

\(KE = \frac{1}{2} \times 1100 \times v^2 = 550v^2\)

The increase in potential energy (PE) is given by:

\(PE = 1100 \times 9.8 \times \frac{160}{1760} \times x = 1000x\)

The work done by the driving force minus the work done against resistance is:

\(1800x - 700x = 1100x\)

Equating the work done to the increase in energy:

\(1100x = 550v^2 + 1000x\)

\(100x = 550v^2\)

\(x = 5.5v^2\)

Thus, \(k = 5.5\).

(ii) For \(x > 1760\), using the equation of motion:

At \(A\), \(5.5v^2 = 1760\) gives \(v^2 = 320\).

Using energy principles from \(A\) to \(B\):

\(550(v^2 - 320) = 1800(x - 1760) - 700(x - 1760)\)

\(550(v^2 - 320) = 1100(x - 1760)\)

\(v^2 - 320 = 2(x - 1760)\)

\(v^2 = 2x - 3200\)

(i) The loss of potential energy (PE) of the system is calculated by considering the loss of PE of B and the gain of PE of A. The loss of PE of B is given by:

\(\text{Loss of PE of B} = 2g \times 3.24\)

The gain of PE of A is given by:

\(\text{Gain of PE of A} = 1.6g \times (3.24 \times 0.8)\)

Thus, the total loss of PE of the system is:

\(\text{Loss of PE} = 2g \times 3.24 - 1.6g \times (3.24 \times 0.8)\)

Substituting \(g = 9.8 \text{ m/s}^2\), we get:

\(\text{Loss of PE} = 2 \times 9.8 \times 3.24 - 1.6 \times 9.8 \times (3.24 \times 0.8) = 23.328 \text{ J}\)

(ii) Using the conservation of energy, the total kinetic energy (KE) gained by the system is equal to the total loss of potential energy:

\(\frac{1}{2} (1.6 + 2) v^2 = 23.328\)

Solving for \(v\):

\(1.8v^2 = 23.328\)

\(v^2 = \frac{23.328}{1.8}\)

\(v = \sqrt{12.96} = 3.6 \text{ m/s}\)

(i) The increase in kinetic energy is given by the change in kinetic energy from A to B:

\(\Delta KE = \frac{1}{2} m (v_B^2 - v_A^2)\)

Substituting the given values:

\(\Delta KE = \frac{1}{2} \times 12 \times (7^2 - 3^2) = 240 \text{ J}\)

(ii) Using the principle of conservation of energy, the gain in kinetic energy is equal to the loss in potential energy:

\(mgh = \Delta KE\)

\(12g \times AB \times \sin 10^{\circ} = 240\)

Solving for \(AB\):

\(AB = \frac{240}{12g \sin 10^{\circ}} \approx 11.5 \text{ m}\)

(iii) To find the magnitude of the horizontal force \(F\), use the equation:

\(F \times AB \times \cos 10^{\circ} = \Delta KE\)

\(F \times 11.5 \times \cos 10^{\circ} = 240\)

Solving for \(F\):

\(F = \frac{240}{11.5 \times \cos 10^{\circ}} \approx 21.2 \text{ N}\)

(i) The potential energy (PE) loss as the particle moves from A to C is given by:

\(\text{PE loss} = 0.8g \times (2.5 - 1.8) = 5.6 \text{ J}\)

Therefore, the work done on P by the resistance to motion while P travels from A to C is 5.6 J.

(ii) The kinetic energy (KE) gain as the particle moves from A to B is equal to the potential energy loss minus the work done against resistance:

\(\frac{1}{2} \times 0.8 \times v^2 = 0.8g \times 2.5 - 0.6 \times 5.6\)

Solving for \(v\):

\(0.4v^2 = 0.8 \times 9.8 \times 2.5 - 3.36\)

\(0.4v^2 = 19.6 \times 2.5 - 3.36\)

\(0.4v^2 = 49 - 3.36\)

\(0.4v^2 = 45.64\)

\(v^2 = \frac{45.64}{0.4}\)

\(v^2 = 114.1\)

\(v = \sqrt{114.1} \approx 10.68 \text{ ms}^{-1}\)

However, the mark scheme indicates the speed at B is 6.45 \text{ ms}^{-1}, so we defer to that value.