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(B) Let the mass of each block be $m = 250\,g = 0.25\,kg$ and the spring constant be $k = 2\,N/m$. At the moment of maximum elongation $x$,both blocks

Vedclass · Accepted Answer

$V \sqrt{\frac{m}{2k}}$

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(B) Let the mass of each block be $m = 250\,g = 0.25\,kg$ and the spring constant be $k = 2\,N/m$. At the moment of maximum elongation $x$,both blocks momentarily come to rest in the center-of-mass frame. By the law of conservation of energy,the initial kinetic energy of the system is converted into the elastic potential energy of the spring at maximum elongation. The total initial kinetic energy is $K_i = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$. The potential energy stored in the spring at maximum elongation $x$ is $U_f = \frac{1}{2}kx^2$. Equating the two: $mv^2 = \frac{1}{2}kx^2$. Solving for $x$: $x^2 = \frac{2mv^2}{k} \implies x = v \sqrt{\frac{2m}{k}}$. Substituting the given values $m = 0.25\,kg$ and $k = 2\,N/m$: $x = V \sqrt{\frac{2 	imes 0.25}{2}} = V \sqrt{0.25} = 0.5V = \frac{V}{2}$.

As per the given figure,two blocks each of mass $250\,g$ are connected to a spring of spring constant $2\,N/m$. If both are given velocity $V$ in opposite directions,then the maximum elongation of the spring is:

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