1. **Problem statement:** A mini bus of mass 2500 kg traveling at 40 m/s collides head-on with a lorry of mass 3500 kg moving at 20 m/s. After collision, the lorry moves at 1.2 m/s in its initial direction. Find the speed of the mini bus after collision. 2. **Formula used:** Conservation of linear momentum states that total momentum before collision equals total momentum after collision. $$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$$ where $m_1, m_2$ are masses, $u_1, u_2$ initial velocities, and $v_1, v_2$ final velocities. 3. **Assign values:** - $m_1 = 2500$ kg (mini bus), $u_1 = 40$ m/s - $m_2 = 3500$ kg (lorry), $u_2 = -20$ m/s (opposite direction) - $v_2 = 1.2$ m/s (lorry final velocity) - $v_1 = ?$ (mini bus final velocity) 4. **Apply conservation of momentum:** $$2500 \times 40 + 3500 \times (-20) = 2500 \times v_1 + 3500 \times 1.2$$ Calculate left side: $$100000 - 70000 = 2500 v_1 + 4200$$ Simplify: $$30000 = 2500 v_1 + 4200$$ 5. **Solve for $v_1$:** $$2500 v_1 = 30000 - 4200 = 25800$$ $$v_1 = \frac{25800}{2500} = 10.32 \text{ m/s}$$ **Answer:** The mini bus moves at 10.32 m/s after collision in its initial direction.