(a) To find the speed of Q after the collision, we use the conservation of momentum. Before the collision, the velocity of P is 5 m s^-1 (calculated using energy conservation on the incline: \(v^2 = 0 + 2 \times 2.5 \times a\) where \(a = g \sin 30^{\circ} = 5\)).

Using conservation of momentum:

\(0.3 \times 5 + 0 = 0.3 \times 2 + 0.2 \times w\)

Solving for \(w\):

\(1.5 = 0.6 + 0.2w\)

\(0.2w = 0.9\)

\(w = 4.5\) m s^-1

(b) For the rough plane, after the collision, P and Q coalesce and move with speed 1.2 m s^-1. Using conservation of momentum:

\(0.3 \times z + 0 = 0.5 \times 1.2\)

Solving for \(z\):

\(0.3z = 0.6\)

\(z = 2\) m s^-1

To find the coefficient of friction \(\mu\), consider the work-energy principle:

Friction force on P after reaching the horizontal plane \(F = \mu \times 0.3g\).

Using energy conservation:

\(\mu \times 0.3g \times 1.5 = \frac{1}{2} \times 0.3 \times 5^2 - \frac{1}{2} \times 0.3 \times 2^2\)

Solving for \(\mu\):

\(\mu \times 0.3g \times 1.5 = 3.75 - 0.6\)

\(\mu \times 0.3g \times 1.5 = 3.15\)

\(\mu = \frac{3.15}{0.3 \times 9.8 \times 1.5}\)

\(\mu = 0.7\)