(a) Apply the conservation of momentum. The initial momentum of the system is given by the momentum of particle A since B is at rest:

\(6 \times 2.5 = 2.5v + 5v\)

Solving for \(v\):

\(15 = 7.5v\)

\(v = 2 \text{ m s}^{-1}\)

(b) Calculate the kinetic energy before and after the collision. The kinetic energy before the collision is:

\(\text{KE(before)} = \frac{1}{2} \times 2.5 \times 6^2 = 45 \text{ J}\)

The kinetic energy after the collision is:

\(\text{KE(after)} = \frac{1}{2} \times 7.5 \times 2^2 = 15 \text{ J}\)

The loss of kinetic energy is:

\(\text{Loss of KE} = 45 - 15 = 30 \text{ J}\)