(a) Using conservation of momentum for the collision between A and B:

Initial momentum: \(0.3 \times 2 + 0.4 \times 0 = 0.6\).

Final momentum: \(0.3 \times 0.6 + 0.4 \times v\).

Equating initial and final momentum:

\(0.6 = 0.18 + 0.4v\).

Solving for \(v\):

\(0.4v = 0.42\).

\(v = 1.05\) m/s.

(b) Using conservation of momentum for the collision between B and C:

Initial momentum: \(0.4 \times 1.05 + m \times 0 = 0.42\).

Final momentum: \((0.4 + m) \times 0.5\).

Equating initial and final momentum:

\(0.42 = (0.4 + m) \times 0.5\).

\(0.42 = 0.2 + 0.5m\).

\(0.5m = 0.22\).

\(m = \frac{11}{25}\).

(c) To find the time from when B and C collide until A collides with the combined particle:

Distance traveled by B and C together: \(0.5t\).

Distance traveled by A: \(0.6t\).

Since the total distance is 2.1 m, we have:

\(0.6t - 0.9 = 0.5t\).

\(0.1t = 0.9\).

\(t = 9\) s.

Using conservation of momentum:

\(6 \times 5 + 2 \times (-3) = 6v_A + 2v_B\)

\(30 - 6 = 6v_A + 2v_B\)

\(24 = 6v_A + 2v_B\)

Given that the difference in speeds is 2 m s^-1, we have:

\(v_B = v_A + 2\)

Substitute into the momentum equation:

\(24 = 6v_A + 2(v_A + 2)\)

\(24 = 6v_A + 2v_A + 4\)

\(24 = 8v_A + 4\)

\(20 = 8v_A\)

\(v_A = 2.5 \text{ m s}^{-1}\)

Then:

\(v_B = 2.5 + 2 = 4.5 \text{ m s}^{-1}\)

Calculate initial kinetic energy (KE):

\(\text{Initial KE} = \frac{1}{2} \times 6 \times 5^2 + \frac{1}{2} \times 2 \times (-3)^2\)

\(= 75 + 9 = 84 \text{ J}\)

Calculate final kinetic energy (KE):

\(\text{Final KE} = \frac{1}{2} \times 6 \times (2.5)^2 + \frac{1}{2} \times 2 \times (4.5)^2\)

\(= 18.75 + 20.25 = 39 \text{ J}\)

Loss of KE:

\(\text{Loss of KE} = 84 - 39 = 45 \text{ J}\)