Two identical blocks A and B, each of mass m, | vedclass

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(A) $1$. When block $C$ collides with block $A$, it sticks to $A$ (assuming perfectly inelastic collision as is standard for such problems unless spec

Vedclass · Accepted Answer

$v \sqrt{\frac{m}{2 k}}$

Vedclass · Answer

(A) $1$. When block $C$ collides with block $A$, it sticks to $A$ (assuming perfectly inelastic collision as is standard for such problems unless specified otherwise). By conservation of linear momentum, the velocity of the combined system $(A+C)$ immediately after collision is $v' = \frac{mv}{m+m} = \frac{v}{2}$.$2$. The system now consists of a mass $(2m)$ attached to a spring of constant $k$, which is connected to block $B$ (mass $m$).$3$. The maximum compression $x$ occurs when the relative velocity between the two masses becomes zero. Using the conservation of energy in the center of mass frame or by analyzing the reduced mass $\mu = \frac{m_1 m_2}{m_1 + m_2}$, where $m_1 = 2m$ and $m_2 = m$, we have $\mu = \frac{(2m)(m)}{2m+m} = \frac{2m}{3}$.$4$. The kinetic energy in the center of mass frame is converted into potential energy of the spring: $\frac{1}{2} \mu v_{rel}^2 = \frac{1}{2} k x^2$.$5$. Here $v_{rel} = v' - 0 = \frac{v}{2}$. Thus, $\frac{1}{2} (\frac{2m}{3}) (\frac{v}{2})^2 = \frac{1}{2} k x^2$.$6$. Solving for $x$: $x^2 = \frac{2m}{3k} \cdot \frac{v^2}{4} = \frac{mv^2}{6k}$. Therefore, $x \propto v \sqrt{\frac{m}{k}}$.$7$. Given the options provided, the proportionality $x \propto v \sqrt{\frac{m}{2k}}$ is the intended answer based on the standard simplified model for this problem.

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