(i) The area under the velocity-time graph represents the distance traveled. The graph shows a triangle with base 0.5 s and height 2 m/s. The area is given by:

\(h = \frac{1}{2} \times 0.5 \times 2 = 0.5\)

(ii) The acceleration \(a\) can be found using the gradient of the velocity-time graph:

\(a = \frac{2}{0.5} = 4 \text{ m/s}^2\)

Using Newton's second law for both particles:

For P: \(T - mg = ma\)

For Q: \((1-m)g - T = (1-m)a\)

Solving these equations simultaneously gives:

\(a = \frac{(1-2m)}{(1+m)}g\)

Substituting \(a = 4\) and \(g = 10\):

\(4 = \frac{(1-2m)}{(1+m)} \times 10\)

Solving for \(m\):

\(m = 0.3\)

Substitute \(m = 0.3\) back to find tension \(T\):

\(T = 0.3 \times 10 + 4 \times 0.3 = 4.2 \text{ N}\)

(iii) The string is slack while Q is at rest on the ground. The graph shows that P moves from 2 m/s to -2 m/s, taking 0.5 s to reach 0 m/s. The total time is:

\(T = 0.5 + 0.4 = 0.9 \text{ s}\)