(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.