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(D) Initial momentum of the system (block $C$) is $P_i = mv_0$.After the elastic collision between $C$ (mass $m$) and $A$ (mass $m$),since they have e

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$\frac{2}{3}m\left(\frac{v_0}{x_0}\right)^2$

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(D) Initial momentum of the system (block $C$) is $P_i = mv_0$.After the elastic collision between $C$ (mass $m$) and $A$ (mass $m$),since they have equal masses,they exchange their velocities. Thus,$C$ comes to rest and $A$ starts moving with velocity $v_0$.Now,the system consists of blocks $A$ and $B$ connected by a spring,with $A$ moving at $v_0$ and $B$ at rest.Let $v$ be the common velocity of $A$ and $B$ when the spring compression is $x_0$. By the law of conservation of linear momentum:$mv_0 = (m + 2m)v$$mv_0 = 3mv$$v = \frac{v_0}{3}$By the law of conservation of energy,the initial kinetic energy of $A$ equals the sum of the kinetic energy of the system $(A+B)$ and the potential energy of the spring:$\frac{1}{2}mv_0^2 = \frac{1}{2}(3m)v^2 + \frac{1}{2}kx_0^2$Substitute $v = \frac{v_0}{3}$:$\frac{1}{2}mv_0^2 = \frac{1}{2}(3m)\left(\frac{v_0}{3}\right)^2 + \frac{1}{2}kx_0^2$$\frac{1}{2}mv_0^2 = \frac{1}{2}(3m)\frac{v_0^2}{9} + \frac{1}{2}kx_0^2$$\frac{1}{2}mv_0^2 = \frac{1}{6}mv_0^2 + \frac{1}{2}kx_0^2$$\frac{1}{2}kx_0^2 = \frac{1}{2}mv_0^2 - \frac{1}{6}mv_0^2 = \frac{1}{3}mv_0^2$$k = \frac{2}{3}m\left(\frac{v_0}{x_0}\right)^2$

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