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(C) Let the acceleration of block $C$ relative to the wedge $D$ be $a$. Since the strings are inextensible,the acceleration of blocks $A$ and $B$ rela

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$2$

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(C) Let the acceleration of block $C$ relative to the wedge $D$ be $a$. Since the strings are inextensible,the acceleration of blocks $A$ and $B$ relative to the wedge $D$ will also be $a$ in the horizontal direction.For block $C$ (mass $m_C = 2 \, kg$): The downward force is $m_C g = 20 \, N$. There are two strings pulling it up with tension $T$. The equation of motion is: $20 - 2T = m_C a = 2a$  ---$(1)$For block $A$ (mass $m_A = 1 \, kg$): The only horizontal force is tension $T$. The equation of motion is: $T = m_A a = 1a$  ---$(2)$Substituting $T = a$ from $(2)$ into $(1)$: $20 - 2a = 2a \Rightarrow 4a = 20 \Rightarrow a = 5 \, m/s^2$.Using the kinematic equation $v^2 = u^2 + 2as$ with $u = 0$,$a = 5 \, m/s^2$,and $s = x_0 = 0.1 \, m$:$v^2 = 0 + 2 	imes 5 	imes 0.1 = 1 \Rightarrow v = 1 \, m/s$.Thus,the velocity of $A$ relative to $D$ is $v_A = 1 \, m/s$ (towards the center) and the velocity of $B$ relative to $D$ is $v_B = 1 \, m/s$ (towards the center).The velocity of $A$ with respect to $B$ is $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$. Since they move in opposite directions towards each other,the magnitude is $v_{AB} = v_A + v_B = 1 + 1 = 2 \, m/s$.

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