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$A$ & $B$ are blocks of same mass $m$ exactly equivalent to each other. Both are placed on frictionless surface connected by one spring. Natural length of spring is $L$ and force constant $K$. Initially spring is in natural length. Another equivalent block $C$ of mass $m$ travelling at speed $v$ along line joining $A$ & $B$ collide with $A$. In ideal condition maximum compression of spring is :-
$v \sqrt[]{\frac{m}{2K}}$
$ m \sqrt[]{\frac{v}{2K}}$
$\sqrt{\frac{mv}{K}}$
$\frac{mv}{2K}$
Solution
Just after collision $\mathrm{C}$ will stop $\&$. $A$ stant moving at speed $v$
Now apply conservation of momentum for $\mathrm{A} . \mathrm{B}$ system
${m v+0=(m+m) V'} $
${V'=\frac{v}{2}}$
Now apply energy balance method
$\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2}(\mathrm{m}+\mathrm{m}) \mathrm{V}^{2}+\frac{1}{2} \mathrm{Kx}_{\max }^{2}$
$\frac{1}{2} m v^{2}=\frac{1}{2} \times 2 m \cdot \frac{v^{2}}{4}+\frac{1}{2} K x^{2} \max$
$\frac{1}{4} \mathrm{mv}^{2}=\frac{1}{2} \mathrm{Kx}_{\max }^{2}$
$\mathrm{x}_{\max }=\mathrm{v} \sqrt{\frac{\mathrm{m}}{2 \mathrm{K}}}$
