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

\(4 \times 10 + [0] = 4 \times 0.5v + 2v\)

Solving for \(v\):

\(40 = 2v + 2v\)

\(40 = 4v\)

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

(b) For the collision between B and C, using conservation of momentum:

\(2 \times 10 + [0] = 2v + 3v\)

Solving for \(v\):

\(20 = 5v\)

\(v = 4 \text{ m/s}\)

Since \(v_A > v_B\), another collision occurs.

(c) For the coalescence of A and B, using conservation of momentum:

\(4 \times 5 + 2 \times 4 = 4v + 2v\)

Solving for \(v\):

\(34 = 6v\)

\(v = \frac{14}{3} \text{ m/s}\)

Initial kinetic energy:

\(KE_{\text{initial}} = \frac{1}{2} \times 4 \times 10^2 = 200 \text{ J}\)

Final kinetic energy:

\(KE_{\text{final}} = \frac{1}{2} \times 6 \times \left( \frac{14}{3} \right)^2 + \frac{1}{2} \times 1 \times 12^2\)

\(KE_{\text{final}} = \frac{412}{3} \text{ J}\)

Loss of kinetic energy:

\(200 - \frac{412}{3} = \frac{188}{3} \text{ J}\)