Here are the solutions to the questions: 6.1 The principle of conservation of linear momentum states that the total linear momentum of an isolated system remains constant. This means the total momentum before an interaction (like a collision or explosion) is equal to the total momentum after the interaction. 6.2 Step 1: Identify the given values and define the system. Before the spring expands, the combined trolleys have: Mass of trolley A: $m_A = 1 \text{ kg}$ Mass of trolley B: $m_B = 2 \text{ kg}$ Total mass: $M = m_A + m_B = 1 \text{ kg} + 2 \text{ kg} = 3 \text{ kg}$ Initial velocity of the combined system: $u = 3 \text{ m} \cdot \text{s}^{-1}$ (to the right). After the spring expands: Mass of trolley A: $m_A = 1 \text{ kg}$ Mass of trolley B: $m_B = 2 \text{ kg}$ Velocity of trolley B: $v_B = 4.7 \text{ m} \cdot \text{s}^{-1}$ (to the right). Let the velocity of trolley A after the expansion be $v_A$. Step 2: Apply the principle of conservation of linear momentum. The total momentum before the expansion equals the total momentum after the expansion. Let the direction to the right be positive. $$M u = m_A v_A + m_B v_B$$ $$(3 \text{ kg})(3 \text{ m} \cdot \text{s}^{-1}) = (1 \text{ kg}) v_A + (2 \text{ kg})(4.7 \text{ m} \cdot \text{s}^{-1})$$ $$9 \text{ kg} \cdot \text{m} \cdot \text{s}^{-1} = 1 v_A + 9.4 \text{ kg} \cdot \text{m} \cdot \text{s}^{-1}$$ Step 3: Calculate the velocity of trolley A. $$1 v_A = 9 - 9.4$$ $$v_A = -0.4 \text{ m} \cdot \text{s}^{-1}$$ Since $v_A$ is negative, the direction is to the left. $$\boxed{\text{0.4 m}\cdot\text{s}^{-1} \text{ to the left}}$$ 3 done, 2 left today. You're making progress.