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

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(A) $1$. Initial momentum of the system (block $C$) is $P_i = mv$.$2$. After the perfectly inelastic collision between $C$ and $A$,block $C$ comes to

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) $1$. Initial momentum of the system (block $C$) is $P_i = mv$.$2$. After the perfectly inelastic collision between $C$ and $A$,block $C$ comes to rest and block $A$ starts moving with velocity $v$. Now,the system consists of blocks $A$ and $B$ connected by a spring,with $A$ moving at velocity $v$ and $B$ at rest.$3$. At maximum compression $x$,both blocks $A$ and $B$ move with the same velocity $V$. By the law of conservation of linear momentum for the system $(A+B)$:$mv = (m + m)V \implies V = \frac{v}{2}$.$4$. By the law of conservation of energy for the system $(A+B)$:Initial kinetic energy of $A$ = Final kinetic energy of $(A+B)$ + Potential energy of the spring.$\frac{1}{2}mv^2 = \frac{1}{2}(2m)V^2 + \frac{1}{2}Kx^2$.$5$. Substituting $V = \frac{v}{2}$:$\frac{1}{2}mv^2 = m(\frac{v}{2})^2 + \frac{1}{2}Kx^2$$\frac{1}{2}mv^2 = \frac{1}{4}mv^2 + \frac{1}{2}Kx^2$$\frac{1}{4}mv^2 = \frac{1}{2}Kx^2$$Kx^2 = \frac{1}{2}mv^2$$x = v\sqrt{\frac{m}{2K}}$.

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