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(C) $1$. Collision between $A$ and $B$: Since both blocks have the same mass $m$ and the collision is elastic,they interchange their velocities. Initi

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$0$ (at rest)

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(C) $1$. Collision between $A$ and $B$: Since both blocks have the same mass $m$ and the collision is elastic,they interchange their velocities. Initially,$v_A = v$ and $v_B = 0$. After the collision,$v_A = 0$ and $v_B = v$.$2$. Collision between $B$ and $C$: Block $B$ (mass $m$,velocity $v$) collides with block $C$ (mass $4m$,velocity $0$). The final velocity of $B$ after this elastic collision is given by the formula $v_{B}' = \left(\frac{m_B - m_C}{m_B + m_C}\right) v_B + \left(\frac{2m_C}{m_B + m_C}\right) v_C$. Substituting the values: $v_{B}' = \left(\frac{m - 4m}{m + 4m}\right) v + 0 = -\frac{3}{5} v = -0.6v$. The negative sign indicates that block $B$ moves to the left with velocity $0.6v$.$3$. Subsequent motion: Block $A$ is now at rest and block $B$ is moving to the left with velocity $0.6v$. Since block $B$ is moving away from $A$,no further collisions occur between $A$ and $B$. Thus,the final velocity of block $A$ is $0$.

Three blocks are initially placed as shown in the figure. Block $A$ has mass $m$ and initial velocity $v$ to the right. Block $B$ with mass $m$ and block $C$ with mass $4m$ are both initially at rest. Neglect friction. All collisions are elastic. The final velocity of block $A$ is

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