Question 83 of 183

As per the given figure, two blocks each of mass $250g$ are connected to a spring of spring constant $2N{m}^{-1}$. If both are given velocity $v$ in opposite directions, then maximum elongation of the spring is

A$\dfrac{v}{2\sqrt{2}}$B$\dfrac{v}{2}$C$\dfrac{v}{4}$D$\dfrac{v}{\sqrt{2}}$

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Answer: (B) $\dfrac{v}{2}$

Let the maximum elongation of the spring be $x$.

Using energy conservation
Loss in kinetic energy of both blocks$=$Gain in spring energy
$\dfrac{1}{2}m{v}^{2}\times 2=\dfrac{1}{2}k{x}^{2}$
$\Rightarrow \left(0.25\right){v}^{2}=\dfrac{1}{2}\times 2\times {x}^{2}$
$\Rightarrow \dfrac{1}{4}{v}^{2}=\dfrac{1}{2}\times 2\times {x}^{2}$
$∴x=\dfrac{v}{2}$