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A mass of 0.5 kg moving with a speed of 1.5 m/s on a horizontal smooth surface collides with a nearly weightless spring of force constant k = 50 N/m. The maximum compression of the spring would be:

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0.15 m

1.5 m

0.12 m

0.5 m

answer is A.

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Detailed Solution

By the law of conservation of energy,

$\begin{array}{ll} i.e., ΔK + ΔU = 0 \\ (0 - \frac{1}{2} {mv}^{2}) + (\frac{1}{2} {kx}^{2} - 0) = 0 \end{array}$

$\begin{aligned} \frac{1}{2} {mv}^{2} = \frac{1}{2} {kx}^{2} \\ \Rightarrow x^{2} = \frac{{mv}^{2}}{k} \end{aligned}$

$\Rightarrow x = \sqrt{(\frac{{mv}^{2}}{k})} \Rightarrow x = \sqrt{(\frac{0 .5 \times 1 .5 \times 1 .5}{50})}$

$∴ x = 0.15 m$

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