Here are the solutions to the questions: 7.1 An action-reaction pair of forces involved in the launching of the rocket is: • The force exerted by the rocket on the expelled gases (downwards). • The force exerted by the expelled gases on the rocket (upwards). $$\boxed{\text{Force of rocket on gases (downwards) and force of gases on rocket (upwards)}}$$ 7.2 The principle of conservation of momentum states that in an isolated system (where no external forces act), the total momentum before a collision or explosion is equal to the total momentum after the collision or explosion. 7.3 Step 1: Identify the given values and define the system. Before the explosion, the combined rocket (sections A and B) has: Total mass $M = m_A + m_B = 3 \text{ kg} + 2 \text{ kg} = 5 \text{ kg}$. Initial velocity $u = 5 \text{ m} \cdot \text{s}^{-1}$ (upwards). After the explosion: Mass of section A: $m_A = 3 \text{ kg}$. Mass of section B: $m_B = 2 \text{ kg}$. Velocity of section B: $v_B = 8 \text{ m} \cdot \text{s}^{-1}$ (upwards). Let the velocity of section A after the explosion be $v_A$. Step 2: Apply the principle of conservation of momentum. The total momentum before the explosion equals the total momentum after the explosion. Let upwards be the positive direction. $$M u = m_A v_A + m_B v_B$$ $$(5 \text{ kg})(5 \text{ m} \cdot \text{s}^{-1}) = (3 \text{ kg}) v_A + (2 \text{ kg})(8 \text{ m} \cdot \text{s}^{-1})$$ $$25 \text{ kg} \cdot \text{m} \cdot \text{s}^{-1} = 3 v_A + 16 \text{ kg} \cdot \text{m} \cdot \text{s}^{-1}$$ Step 3: Calculate the velocity of section A. $$3 v_A = 25 - 16$$ $$3 v_A = 9$$ $$v_A = \frac{9}{3}$$ $$v_A = 3 \text{ m} \cdot \text{s}^{-1}$$ Since $v_A$ is positive, the direction is upwards. $$\boxed{\text{3 m}\cdot\text{s}^{-1} \text{ upwards}}$$ That's 2 down. 3 left today — send the next one.