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A mass of $0.5 \mathrm{~kg}$ moving with a speed of $1.5 \mathrm{~m} /$ $s$ on a horizontal smooth surface, collides with a nearly weightless spring of force constant $\mathrm{k}=$ $50 \mathrm{~N} / \mathrm{m}$. The maximum compression of the spring would be
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Verified Answer
The correct answer is:
$0.15 \mathrm{~m}$
Apply law of conservation of energy,
$$
\begin{array}{l}
\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \mathrm{kx}^{2} \\
\Rightarrow \mathrm{mv}^{2}=\mathrm{kx}^{2} \text { or } 0.5 \times(1.5)^{2}=50 \times \mathrm{x}^{2} \\
\therefore \mathrm{x}=0.15 \mathrm{~m}
\end{array}
$$
$$
\begin{array}{l}
\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \mathrm{kx}^{2} \\
\Rightarrow \mathrm{mv}^{2}=\mathrm{kx}^{2} \text { or } 0.5 \times(1.5)^{2}=50 \times \mathrm{x}^{2} \\
\therefore \mathrm{x}=0.15 \mathrm{~m}
\end{array}
$$
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