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

\(0.9g - T = 0.9a\)

For particle B:

\(T - 0.6g = 0.6a\)

Adding the two equations:

\(0.9g - 0.6g = 0.9a + 0.6a\)

\(0.3g = 1.5a\)

Solving for \(a\):

\(a = \frac{0.3g}{1.5} = 2 \text{ m/s}^2\)

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

\(T = 0.6g + 0.6 \times 2 = 0.6 \times 9.8 + 1.2 = 7.2 \text{ N}\)

(ii) After A hits the floor, B moves upwards for 0.3 s with initial velocity \(u = 3 \text{ m/s}\) (from previous motion):

Using \(v = u - gt\):

\(0 = 3 - 9.8 \times 0.3\)

Solving for \(t\):

\(t = 0.3 \text{ s}\)

Using \(v^2 = u^2 + 2ah\) to find height \(h\):

\(0 = 3^2 - 2 \times 9.8 \times h\)

\(9 = 19.6h\)

\(h = \frac{9}{19.6} = 0.459 \text{ m}\)

Total height from initial position:

\(h = 2.25 \text{ m}\)

Let the tension in the string be denoted by \(T\) and the acceleration of the system be \(a\).

For particle A (mass 0.65 kg), applying Newton's second law:

\(0.65g - T = 0.65a\)

For particle B (mass 0.35 kg), applying Newton's second law:

\(T - 0.35g = 0.35a\)

Adding these two equations to eliminate \(T\):

\(0.65g - 0.35g = 0.65a + 0.35a\)

\(0.30g = a\)

Substitute \(a = 0.30g\) back into one of the equations to find \(T\):

\(0.65g - T = 0.65(0.30g)\)

\(T = 0.65g - 0.195g\)

\(T = 0.455g\)

Using \(g = 9.8 \text{ m/s}^2\), \(T = 0.455 \times 9.8 = 4.55 \text{ N}\).

The resultant force exerted on the pulley by the string is twice the tension:

\(2T = 2 \times 4.55 = 9.1 \text{ N}\).

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

For particle A: \(T - 0.3g = 0.3a\)

For particle B: \(0.7g - T = 0.7a\)

Adding these equations gives:

\(0.7g - 0.3g = (0.7 + 0.3)a\)

\(0.4g = a\)

\(a = 4 \text{ m s}^{-2}\)

(ii) Using the equation \(v^2 = u^2 + 2as\) with \(u = 0\), \(v = 1.6 \text{ m s}^{-1}\), and \(a = -4 \text{ m s}^{-2}\):

\(0 = 1.6^2 - 2 \times 4 \times s\)

\(s = \frac{1.6^2}{2 \times 4} = 0.32 \text{ m}\)

For the total height, \(s_1 + s_2 = 0.32 + 0.128 = 0.448 \text{ m}\)

(iii) Using \(v = u + at\) with \(u = 0\), \(v = 1.6 \text{ m s}^{-1}\), and \(a = 4 \text{ m s}^{-2}\):

\(1.6 = 4t_1\)

\(t_1 = \frac{1.6}{4} = 0.4 \text{ s}\)

For the descent, \(t_2 = \frac{1.6}{10} = 0.16 \text{ s}\)

Total time \(t_1 + t_2 = 0.4 + 0.16 = 0.56 \text{ s}\)

(i) Using Newton's second law for the system, the net force is given by the difference in weights:

\(0.55g - T = 0.55a\)

\(T - 0.45g = 0.45a\)

Adding these equations, we eliminate tension \(T\):

\(0.55g - 0.45g = 0.55a + 0.45a\)

\(0.10g = 1.00a\)

Thus, the acceleration \(a\) is:

\(a = \frac{0.10g}{1.00} = 1 \text{ m/s}^2\)

(ii) (a) Using the equation of motion \(s = ut + \frac{1}{2}at^2\) with \(u = 0\), \(a = 1 \text{ m/s}^2\), and \(t = 2 \text{ s}\):

For P:

\(s_P = 5 - \frac{1}{2} \times 1 \times (2)^2 = 5 - 2 = 3 \text{ m}\)

For Q:

\(s_Q = 5 + \frac{1}{2} \times 1 \times (2)^2 = 5 + 2 = 7 \text{ m}\)

(b) The speed \(v\) of the particles after 2 s is given by:

\(v = u + at = 0 + 1 \times 2 = 2 \text{ m/s}\)

(iii) After the string breaks, both particles move under gravity. For P, using \(s = ut + \frac{1}{2}gt^2\) with \(u = 2 \text{ m/s}\), \(g = 9.8 \text{ m/s}^2\), and \(s = 3 \text{ m}\):

\(3 = 2t_P + \frac{1}{2} \times 9.8 \times t_P^2\)

Solving gives \(t_P = 0.6 \text{ s}\).

For Q, using \(s = ut + \frac{1}{2}gt^2\) with \(u = 2 \text{ m/s}\), \(s = 7 \text{ m}\):

\(7 = -2t_Q + \frac{1}{2} \times 9.8 \times t_Q^2\)

Solving gives \(t_Q = 1.4 \text{ s}\).

Thus, Q reaches the ground 0.8 s later than P.

(i) Using the equation of motion: \(s = ut + \frac{1}{2} a t^2\), where \(s = 0.36\) m, \(u = 0\), and \(t = 0.6\) s. Substituting, we get:

\(0.36 = 0 + \frac{1}{2} a (0.6)^2\)

\(0.36 = 0.18a\)

\(a = \frac{0.36}{0.18} = 2 \text{ m/s}^2\)

(ii) Applying Newton's second law to particle A: \(0.45g - T = 0.45a\)

\(0.45 \times 10 - T = 0.45 \times 2\)

\(4.5 - T = 0.9\)

\(T = 4.5 - 0.9 = 3.6 \text{ N}\)

(iii) Applying Newton's second law to particle B: \(T - mg = ma\)

\(3.6 - m \times 10 = m \times 2\)

\(3.6 = 12m\)

\(m = \frac{3.6}{12} = 0.3 \text{ kg}\)

(iv) Using the equation \(v^2 = u^2 + 2as\) for particle B when it reaches maximum height (\(v = 0\)):

\(0 = (0.6 \times 2)^2 - 2 \times 10 \times s\)

\(0 = 1.44 - 20s\)

\(s = \frac{1.44}{20} = 0.072 \text{ m}\)

Maximum height above the floor = initial height + \(s = 0.72 + 0.072 = 0.792 \text{ m}\)