(a) For particle Q, apply Newton's second law along the plane:

\(-2mg \sin α - F = 2ma\)

Where \(F = 0.6 \times 2mg \cos α\).

Substitute \(\sin α = 0.8\) and \(\cos α = 0.6\):

\(-2m(9.8)(0.8) - 0.6 \times 2m(9.8)(0.6) = 2ma\)

\(-15.68m - 7.056m = 2ma\)

\(-22.736m = 2ma\)

\(a = -11.368 \approx -11.6 \text{ m/s}^2\)

(b) For particle P, apply Newton's second law:

\(mg \sin α - 0.6mg \cos α = ma\)

\(9.8 \times 0.8 - 0.6 \times 9.8 \times 0.6 = a\)

\(7.84 - 3.528 = a\)

\(a = 4.312 \text{ m/s}^2\)

Using constant acceleration equations, find the time \(T_1\) for Q to come to rest:

\(10 - 11.6T_1 = 0\)

\(T_1 = \frac{25}{29} \approx 0.862 \text{ s}\)

Calculate the distance traveled by P and Q:

\(s_P = \frac{1}{2} \times 4.4 \times (0.862)^2 = 1.635 \text{ m}\)

\(s_Q = 10 \times 0.862 - \frac{1}{2} \times 11.6 \times (0.862)^2 = 4.31 \text{ m}\)

Find the extra distance needed for collision:

\(d = 6.4 - (s_P + s_Q) = 0.455 \text{ m}\)

Find the time \(T_2\) for P to travel this extra distance:

\(d = 4.4T_2\)

\(T_2 = \frac{0.455}{4.4} \approx 0.103 \text{ s}\)

Total time before collision:

\(T = T_1 + T_2 = 0.862 + 0.103 = 0.965 \text{ s}\)

(c) Using conservation of momentum for the collision:

\(m \times 4.4 \times 0.965 + 2m \times 0 = (m + 2m)v\)

\(4.4m \times 0.965 = 3mv\)

\(v = \frac{4.4 \times 0.965}{3} \approx 1.79 \text{ m/s}\)