In the figure shown,both blocks are released | vedclass

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(A) Let the mass of block $A$ be $m_A = 4 \ kg$ and block $B$ be $m_B = 1 \ kg$. The initial vertical separation between their bottom edges is $6 \ m$

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$1$

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(A) Let the mass of block $A$ be $m_A = 4 \ kg$ and block $B$ be $m_B = 1 \ kg$. The initial vertical separation between their bottom edges is $6 \ m$ (based on the geometry of the system).From the Free Body Diagram $(FBD)$ of the blocks:For block $A$: $m_A g - T = m_A a \implies 4g - T = 4a \quad \dots(1)$For block $B$: $T - m_B g = m_B a \implies T - 1g = 1a \quad \dots(2)$Adding equations $(1)$ and $(2)$: $3g = 5a \implies a = \frac{3g}{5} = \frac{3 	imes 10}{5} = 6 \ m/s^2$.The acceleration of block $A$ relative to block $B$ is $a_{rel} = a - (-a) = 2a = 2 	imes 6 = 12 \ m/s^2$.To cross each other,the relative displacement required is $s_{rel} = 6 \ m$.Using the equation of motion $s = ut + \frac{1}{2}at^2$ with $u = 0$:$6 = 0 + \frac{1}{2} 	imes 12 	imes t^2$$6 = 6t^2 \implies t^2 = 1 \implies t = 1 \ s$.

In the figure shown,both blocks are released from rest. Find the time taken for them to cross each other. (Assume $g = 10 \ m/s^2$)

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