Two identical blocks $A$ and $B$, each of mass $'m'$ resting on smooth floor are connected by a light spring of natural length $L$ and spring constant $K$, with the spring at its natural length. $A$ third identical block $'C'$ (mass $m$) moving with a speed $v$ along the line joining $A$ and $B$ collides with $A$. the maximum compression in the spring is
Two identical blocks $A$ and $B$, each of mass $'m'$ resting on smooth floor are connected by a light spring of natural length $L$ and spring constant $K$, with the spring at its natural length. $A$ third identical block $'C'$ (mass $m$) moving with a speed $v$ along the line joining $A$ and $B$ collides with $A$. the maximum compression in the spring is
$v\sqrt {\frac{m}{{2k}}} $
$m\sqrt {\frac{v}{{2k}}} $
$\sqrt {\frac{{mv}}{k}} $
$\frac{{mv}}{{2k}}$
Two identical blocks $A$ and $B$, each of mass $'m'$ resting on smooth floor are connected by a light spring of natural length $L$ and spring constant $K$, with the spring at its natural length. $A$ third identical block $'C'$ (mass $m$) moving with a speed $v$ along the line joining $A$ and $B$ collides with $A$. the maximum compression in the spring is
Initial momentum of the system (block $C$) $= mv$
After striking with $A$, the block $C$ comes to rest and now both block $A$ and $B$ moves with velocity $V$, when compression in spring is maximum.
By the law of conservation of linear momentum
$mv = (m + m) V$ ==> $V = \frac{v}{2}$
By the law of conservation of energy
$K.E.$ of block $C = K.E. $ of system $+ P.E.$ of system
$\frac{1}{2}m{v^2} = \frac{1}{2}(2m)\,{V^2} + \frac{1}{2}k{x^2}$
==> $\frac{1}{2}m{v^2} = \frac{1}{2}(2m)\;{\left( {\frac{v}{2}} \right)^2} + \frac{1}{2}k{x^2}$
==> $k{x^2} = \frac{1}{2}m{v^2}$
==> $x = v\sqrt {\frac{m}{{2k}}} $
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