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(C) Let the horizontal block be $A$ and the hanging block be $B$. Both have mass $m = 4 \ kg$.Initially,$v_A = 4 \ m/s$ and $v_B = 2 \ m/s$.The string

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$4$

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(C) Let the horizontal block be $A$ and the hanging block be $B$. Both have mass $m = 4 \ kg$.Initially,$v_A = 4 \ m/s$ and $v_B = 2 \ m/s$.The string becomes slack because $v_A > v_B$. The block $B$ moves under gravity with acceleration $g = 10 \ m/s^2$.Let $t$ be the time when the string becomes taut again. In this time,the displacement of $B$ relative to $A$ must be zero.$s_B = v_B t + \frac{1}{2} g t^2 = 2t + 5t^2$$s_A = v_A t = 4t$Setting $s_A = s_B$: $4t = 2t + 5t^2 \implies 5t^2 = 2t \implies t = 0.4 \ s$.At $t = 0.4 \ s$,the velocity of $A$ is $v_A = 4 \ m/s$ and the velocity of $B$ is $v_B' = v_B + gt = 2 + 10(0.4) = 6 \ m/s$.When the string becomes taut,both blocks must have the same velocity $v$ along the string. Since the string is inextensible,the impulse $J$ acts on both blocks to equalize their velocities.For block $A$: $J = m(v - v_A) = 4(v - 4)$For block $B$: $-J = m(v - v_B') = 4(v - 6)$Adding the two equations: $0 = 4(2v - 10) \implies v = 5 \ m/s$.Substituting $v$ back into the impulse equation: $J = 4(5 - 4) = 4 \ N-s$.

Two blocks of same mass $4 \ kg$ are placed as shown in the diagram. The initial velocities of the blocks are $4 \ m/s$ (horizontal) and $2 \ m/s$ (downward). The string is initially taut. Find the impulse on the $4 \ kg$ block when the string again becomes taut (in $N-s$).

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