Two blocks of masses $m_{1}$ and $m_{2}$ are connected by a spring of spring constant $k$ . The block of mass $m_{2}$ is given a sharp impulse so that it acquires a velocity $v_{0}$ towards right. Find the maximum elongation that the spring will suffer.

{[\frac{m_{1} m_{2}}{m_{1} + m_{2}}]}^{\frac{1}{2}} v_{0}

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(\frac{m_{1} + m_{2}}{m_{1} - m_{2}}) v_{0}

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{[\frac{m_{1} + m_{2}}{m_{1} - m_{2}}]}^{\frac{1}{2}} v_{0}

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{[\frac{2 m_{1} + m_{2}}{m_{1} m_{2}}]}^{\frac{1}{2}} v_{0}

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Solution

The correct option is B ${[\frac{m_{1} m_{2}}{m_{1} + m_{2}}]}^{\frac{1}{2}} v_{0}$
At maximum elongation the two blocks are moving with same speed, thus, the conservation of linear momentum and energy gives us:
$m_{2} v_{0} = (m_{1} + m_{2}) v \Rightarrow v = \frac{m_{2} v_{0}}{m_{1} + m_{2}} \frac{1}{2} m_{2} v_{0}^{2} = \frac{1}{2} (m_{1} + m_{2}) v^{2} + \frac{1}{2} k x^{2} \Rightarrow x = \sqrt{\frac{m_{1} m_{2}}{m_{1} + m_{2}}} v_{0}$

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Figure

Two blocks of masses $m_{1}$ and $m_{2}$ are connected by a spring of constant k (given figure).The block of mass $m_{2}$ is given a sharp impulse so that it acquires a velocity $v_{0}$ towards right.Find (a) the velocity of the centre of mass,(b) the maximum elongation that the spring will suffer.