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

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Question

(A) Let the collision between $C$ and $A$ be perfectly inelastic. By conservation of linear momentum,the velocity of the combined system $(C+A)$ immed

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the collision between $C$ and $A$ be perfectly inelastic. By conservation of linear momentum,the velocity of the combined system $(C+A)$ immediately after collision is $v' = \frac{mv}{m+m} = \frac{v}{2}$.At the state of maximum compression $x$,the blocks $A$,$B$,and $C$ move with a common velocity $V$. By conservation of linear momentum,$mv = (m+m+m)V$,so $V = \frac{v}{3}$.By conservation of mechanical energy,the initial kinetic energy of the system $(C+A)$ equals the sum of the final kinetic energy and the potential energy stored in the spring:$\frac{1}{2}(2m)v'^2 = \frac{1}{2}(3m)V^2 + \frac{1}{2}kx^2$$m(\frac{v}{2})^2 = \frac{3}{2}m(\frac{v}{3})^2 + \frac{1}{2}kx^2$$\frac{mv^2}{4} = \frac{mv^2}{6} + \frac{1}{2}kx^2$$\frac{1}{2}kx^2 = \frac{mv^2}{4} - \frac{mv^2}{6} = \frac{mv^2}{12}$$x^2 = \frac{mv^2}{6k} \implies x = v \sqrt{\frac{m}{6k}}$.Since the question asks for proportionality,we look at the reduced mass system. The effective mass for the oscillation of $A$ and $B$ is $\mu = \frac{m \cdot m}{m+m} = \frac{m}{2}$. The maximum compression is proportional to $v \sqrt{\frac{\mu}{k}} = v \sqrt{\frac{m}{2k}}$.

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