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 collisi

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 $v_A$.

The total momentum before the expansion equals the total momentum after the expansion. Let the direction to the right be positive.

$$M u = m_A v_A + m_B v_B$$ $$(3 \text{ kg})(3 \text{ m} \cdot \text{s}^{-1}) = (1 \text{ kg}) v_A + (2 \text{ kg})(4.7 \text{ m} \cdot \text{s}^{-1})$$ $$9 \text{ kg} \cdot \text{m} \cdot \text{s}^{-1} = 1 v_A + 9.4 \text{ kg} \cdot \text{m} \cdot \text{s}^{-1}$$

$$1 v_A = 9 - 9.4$$ $$v_A = -0.4 \text{ m} \cdot \text{s}^{-1}$$

Since $v_A$ is negative, the direction is to the left.

$$\boxed{\text{0.4 m}\cdot\text{s}^{-1} \text{ to the left}}$$