1. Common Acceleration: Since the blocks are in contact and moving together under the applied force on a frictionless surface, they share a common acceleration ($a$).

• Total Mass, $M_{total} = m_A + m_B + m_C = 4 + 2 + 1 = 7 \text{ kg}$.
• Applied Force, $F = 14 \text{ N}$.
• According to Newton's Second Law: $a = \frac{F}{M_{total}} = \frac{14}{7} = 2 \text{ m/s}^2$ [NCERT Class 11, Physics Part I, Sec 4.5].

2. Contact Force ($F_{AB}$): The contact force exerted by block A on block B ($F_{AB}$) is the force responsible for accelerating the remaining mass consisting of blocks B and C.

• Mass to be accelerated by $F_{AB}$ is $(m_B + m_C)$.
• $F_{AB} = (m_B + m_C) \times a$
• $F_{AB} = (2 + 1) \times 2 = 3 \times 2 = 6 \text{ N}$.

(Alternatively, using the free body diagram of block A: $F - F_{BA} = m_A a \Rightarrow 14 - F_{AB} = 4(2) \Rightarrow F_{AB} = 6 \text{ N}$ due to Newton's Third Law where action $F_{AB}$ equals reaction $F_{BA}$ magnitude).