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:

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