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A spring is compressed between two blocks of masses $m_{1}$ and $m_{2}$ placed on a horizontal frictionless surface as shown in Fig. When the blocks are released, they have initial velocity of $v_{1}$ and $v_{2}$ as shown. The blocks travel distances $x_{1}$ and $x_{2}$ respectively before coming to rest. The ratio $\frac{x_{1}}{x_{2}}$ is :

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$\frac{m_{2}}{m_{1}}$

$\sqrt{\frac{m_{1}}{m_{2}}}$

$\sqrt{\frac{m_{2}}{m_{1}}}$

$\frac{m_{1}}{m_{2}}$

answer is B.

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Detailed Solution

From the principle of conservation of momentum, we have

$m_{1} v_{1} = m_{2} v_{2} o r \frac{v_{1}}{v_{2}} = \frac{m_{2}}{m_{1}}$

$\Rightarrow \frac{v_{1}}{v_{2}} = \frac{\frac{d x_{1}}{d t}}{\frac{d x_{2}}{d t}} = \frac{m_{2}}{m_{1}} \Rightarrow \frac{d x_{1}}{d x_{2}} = \frac{m_{2}}{m_{1}} \Rightarrow \frac{x_{1}}{x_{2}} = \frac{m_{2}}{m_{1}}$

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