(a) To find the speed of P immediately before the collision, use the equation of motion:

\(v^2 = u^2 + 2as\)

where \(u = 20\) m/s, \(a = -0.4 \times g = -4\) m/s², and \(s = 12.5\) m.

Substitute the values:

\(v^2 = 20^2 + 2(-4)(12.5)\)

\(v^2 = 400 - 100 = 300\)

\(v = \sqrt{300} = 10\sqrt{3} \text{ m/s}\)

(b) Use conservation of momentum to find the speed of R after the collision:

\(6 \times 10\sqrt{3} = (6 + 2)v'\)

\(v' = \frac{60\sqrt{3}}{8} = 7.5\sqrt{3} \text{ m/s}\)

Initial kinetic energy:

\(KE_i = \frac{1}{2} \times 6 \times (10\sqrt{3})^2 = 900 \text{ J}\)

Final kinetic energy:

\(KE_f = \frac{1}{2} \times 8 \times (7.5\sqrt{3})^2 = 675 \text{ J}\)

Loss of kinetic energy:

\(\text{Loss} = 900 - 675 = 225 \text{ J}\)

(c) To find the distance travelled by R before coming to rest, use:

\(0 = (7.5\sqrt{3})^2 + 2(-4)s\)

\(0 = 168.75 - 8s\)

\(s = \frac{168.75}{8} = 21.1 \text{ m}\)