(i) Using the equation of motion:

\(v = u + at\)

where \(v = 3.5\) m/s, \(u = 0\), and \(t = 2\) s, we find:

\(3.5 = 0 + 2a\)

Solving for \(a\), we get \(a = 1.75\) m/s².

The acceleration \(a = g \sin \alpha\), where \(g = 9.8\) m/s². Thus:

\(1.75 = 9.8 \sin \alpha\)

Solving for \(\sin \alpha\), we get \(\alpha = 10.1^{\circ}\) or \(0.176\) radians.

(ii) For particle P, the distance \(s_P\) is given by:

\(s_P = \frac{1}{2} a (2^2) + a(2)t + \frac{1}{2} at^2\)

For particle Q, the distance \(s_Q\) is:

\(s_Q = \frac{1}{2} a t^2\)

The separation is given by:

\(s_P - s_Q = 4.9\)

Substituting the expressions for \(s_P\) and \(s_Q\):

\(\frac{1}{2} \times 1.75 \times 4 + 1.75 \times 2 \times t + \frac{1}{2} \times 1.75 \times t^2 - \frac{1}{2} \times 1.75 \times t^2 = 4.9\)

Simplifying, we get:

\(2 \times 1.75 + 2 \times 1.75t = 4.9\)

Solving for \(t\), we find \(t = 0.4\) s.