1. **State the problem:** A lorry of mass 18t moving at 35m/s collides with a stationary wall of mass 2.5t. Simultaneously, a car of mass 1.9t moves in the opposite direction at 55m/s and also collides with the wall. The three bodies stick together after collision. Find the final velocity of the combined mass. 2. **Formula used:** We use the conservation of momentum principle for inelastic collisions where bodies stick together: $$m_1 v_1 + m_2 v_2 + m_3 v_3 = (m_1 + m_2 + m_3) v_f$$ where $v_f$ is the final velocity of the combined mass. 3. **Assign values and directions:** - Lorry mass $m_1 = 18$ t, velocity $v_1 = +35$ m/s (positive direction) - Wall mass $m_2 = 2.5$ t, velocity $v_2 = 0$ m/s (stationary) - Car mass $m_3 = 1.9$ t, velocity $v_3 = -55$ m/s (opposite direction, negative) 4. **Calculate total initial momentum:** $$p_{initial} = 18 \times 35 + 2.5 \times 0 + 1.9 \times (-55)$$ $$p_{initial} = 630 + 0 - 104.5 = 525.5$$ 5. **Calculate total mass:** $$m_{total} = 18 + 2.5 + 1.9 = 22.4$$ 6. **Apply conservation of momentum to find final velocity:** $$525.5 = 22.4 \times v_f$$ 7. **Isolate $v_f$ and simplify:** $$v_f = \frac{525.5}{22.4}$$ $$v_f = \frac{\cancel{525.5}}{\cancel{22.4}}$$ (showing cancellation of common factors is not applicable here as numbers are not factorable simply, so we keep fraction as is) 8. **Calculate final velocity:** $$v_f \approx 23.46 \text{ m/s}$$ **Final answer:** The final velocity of the combined mass after collision is approximately **23.46 m/s** in the original positive direction of the lorry.