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(C) Let the initial velocity of block $A$ be $v$. Since the collisions are perfectly inelastic,the blocks stick together after each collision.$1$. Col

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$4$

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(C) Let the initial velocity of block $A$ be $v$. Since the collisions are perfectly inelastic,the blocks stick together after each collision.$1$. Collision between $A$ and $B$:By conservation of linear momentum: $mv = (m + m)v_1 = 2mv_1 \implies v_1 = v/2$.$2$. Collision between the combined mass $(A+B)$ and $C$:By conservation of linear momentum: $(2m)v_1 = (2m + M)v_f$.Substituting $v_1 = v/2$: $(2m)(v/2) = (2m + M)v_f \implies mv = (2m + M)v_f \implies v_f = \frac{mv}{2m + M}$.$3$. Kinetic Energy Analysis:Initial kinetic energy $K_i = \frac{1}{2}mv^2$.Final kinetic energy $K_f = \frac{1}{2}(2m + M)v_f^2 = \frac{1}{2}(2m + M) \left( \frac{mv}{2m + M} \right)^2 = \frac{m^2v^2}{2(2m + M)}$.Given that $5/6^{th}$ of the initial kinetic energy is lost,the final kinetic energy is $1/6^{th}$ of the initial kinetic energy:$K_f = \frac{1}{6} K_i \implies \frac{m^2v^2}{2(2m + M)} = \frac{1}{6} \left( \frac{1}{2}mv^2 \right)$.Simplifying: $\frac{m}{2m + M} = \frac{1}{6} \implies 6m = 2m + M \implies M = 4m \implies M/m = 4$.

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