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

For particle P: \(7g - T = 7a\)

For particle Q: \(T - 3g = 3a\)

Adding these equations gives: \(7g - 3g = 10a\)

Solving for acceleration, \(a = \frac{4g}{10} = 4 \text{ m/s}^2\)

Using the equation of motion for speed of P: \(v^2 = u^2 + 2as\)

Where initial speed \(u = 0\), \(a = 4 \text{ m/s}^2\), and \(s = 0.4 \text{ m}\):

\(v^2 = 0 + 2 \times 4 \times 0.4\)

\(v^2 = 3.2\)

\(v = \sqrt{3.2} = 1.79 \text{ m/s}\)

(ii) To determine if Q comes to rest, use:

\(0 = 3.2 + 2 \times (-g) \times s\)

Solving for \(s\):

\(s = 0.16 \text{ m}\)

Since \(0.16 + 0.4 = 0.56 \text{ m}\), particle Q does not come to rest before it reaches the pulley.