In the figures (i), (ii) and (iii) shown, the | vedclass

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(D) Let the mass of $A, B$ and $C$ be $m$. In each case, $C$ sticks to $B$, so the combined mass becomes $2m$.Case $(i)$: $C$ strikes $B$ (which is co

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none of these

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(D) Let the mass of $A, B$ and $C$ be $m$. In each case, $C$ sticks to $B$, so the combined mass becomes $2m$.Case $(i)$: $C$ strikes $B$ (which is connected to $A$ via a string over a fixed pulley). By conservation of linear momentum for the system $(B+C)$, the impulse from the string acts on $B$. However, since $A$ is also connected, the system $(A+B+C)$ moves together. The momentum $mu$ is shared by $2m$ (mass of $B+C$) and $m$ (mass of $A$). Since the string is inextensible, $v_A = v_B = v$. By conservation of momentum: $mu = (m+m+m)v \implies v = u/3$. Wait, let's re-evaluate: The impulse $J$ acts on $B$ from the string. For $(B+C)$, $mu - J = (2m)v$. For $A$, $J = mv$. Adding these, $mu = 3mv \implies v = u/3$. Case $(ii)$: $C$ strikes $B$. The spring is massless and initially uncompressed. The impulse acts only on $B+C$. $mu = (2m)v_B \implies v_B = u/2$. The spring does not transmit the impulse instantaneously.Case $(iii)$: $C$ strikes $B$. The string is inextensible. Similar to case $(i)$, the momentum $mu$ is shared by the whole system $(A+B+C)$. $mu = (3m)v \implies v = u/3$.Correct analysis: Case $(i)$: $B$ is connected to $A$ by a string. $mu = (2m+m)v_1 \implies v_1 = u/3$.Case $(ii)$: $B$ is connected to $A$ by a spring. The spring does not act instantaneously. $mu = (2m)v_2 \implies v_2 = u/2$.Case $(iii)$: $B$ is connected to $A$ by a string. $mu = (2m+m)v_3 \implies v_3 = u/3$.Ratio $v_1 : v_2 : v_3 = u/3 : u/2 : u/3 = 2 : 3 : 2$.

In the figures $(i), (ii)$ and $(iii)$ shown, the objects $A, B$ and $C$ have the same mass $m$. The string, spring, and pulley are massless. Object $C$ strikes object $B$ with velocity $u$ in each case and sticks to it. Find the ratio of the velocity of $B$ in case $(i)$ to $(ii)$ to $(iii)$.

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