Two masses $m_{1} = 2 k g$ and $m_{2} = 5 k g$ are moving on a frictionless surface with velocities $10 \frac{m}{s}$ and $3 \frac{m}{s}$ respectively. $m_{2}$ is ahead of $m_{1}$ . An ideal spring constant $k = 1120 N / m$ is attached on the back side of $m_{1} .$ The maximum compression of the spring will be

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$0.72 m$

$0.25 m$

$0.062 m$

$0.51 m$

answer is C.

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Detailed Solution

Max. compression happen when both have same speed

$∴ 2 \times 10 + 5 \times 3 = 7 \times v ∴ v = \frac{35}{7} = 5 \frac{m}{s}$

Energy conservation

$\frac{1}{2} \times 2 \times 10^{2} + \frac{1}{2} \times 7 \times 3^{2} = \frac{1}{2} \times 7 \times 5^{2} + \frac{1}{2} 1120 x^{2} \Rightarrow x^{2} = 0.0625 ∴ x = 0.25 m$

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