A 5.0 kg trolley moving east at 4.0 m/s meets a 3.0 kg trolley moving west at 6.0 m/s.

Step 1. Momentum. $5.0\ \text{kg} \times 4.0\ \text{m/s} + 3.0\ \text{kg} \times (-6.0\ \text{m/s}) = 20\ \text{kg m/s} - 18\ \text{kg m/s} = 2.0\ \text{kg m/s}$.

Step 2. Velocity. $v = \dfrac{2.0\ \text{kg m/s}}{8.0\ \text{kg}} = 0.25\ \text{m/s}$, east.

Step 3. Energy. Before: $\tfrac12 \times 5.0\ \text{kg} \times 16\ \text{m}^2/\text{s}^2 + \tfrac12 \times 3.0\ \text{kg} \times 36\ \text{m}^2/\text{s}^2 = 40\ \text{J} + 54\ \text{J} = 94\ \text{J}$. After: $\tfrac12 \times 8.0\ \text{kg} \times (0.25\ \text{m/s})^2 = 0.25\ \text{J}$. Lost: $93.75\ \text{J}$.