Two blocks $A$ and $B$ each of mass m are connected by a massless spring of natural length L and spring constant $K$. The blocks are initially resting on a smooth horizontal floor with the spring at its natural length as shown in figure. A third identical block $C$ also of mass $m$ moves on the floor with a speed $v$ along the line joining $A$ and $B$ and collides with $A$. Then
IIT 1993, Diffcult
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(d) Let the velocity acquired by $A$ and $B$ be $V$, then
$mv = mV + mV $
$mv = mV + mV $
$\Rightarrow V = \frac{v}{2}$
Also $\frac{1}{2}m{v^2} = \frac{1}{2}m{V^2} + \frac{1}{2}m{V^2} + \frac{1}{2}k{x^2}$
Where $ x$ is the maximum compression of the spring. On solving the above equations, we get $x = v{\left( {\frac{m}{{2k}}} \right)^{1/2}}$
At maximum compression, kinetic energy of the
$A -B$ system $ = \frac{1}{2}m{V^2} + \frac{1}{2}m{V^2} = m{V^2} = \frac{{m{v^2}}}{4}$
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