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A ball of mass 2 kg moving with velocity $3 \mathrm{~m} / \mathrm{s}$, collides with spring of natural length 2 m and force constant $144 \mathrm{~N} / \mathrm{m}$. What will be length of compressed spring?
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The correct answer is:
1.5 m
Let spring is compressed by a length $x$. Kinetic energy of ball
= Potential energy of spring
$i e, \quad \frac{1}{2} m v^2=\frac{1}{2} k x^2$
Given, $\quad m=2 \mathrm{~kg}, v=3 \mathrm{~m} / \mathrm{s}, k=144 \mathrm{~N} / \mathrm{m}$
$\therefore \quad \frac{1}{2} \times 2 \times(3)^2=\frac{1}{2} \times 144 \times x^2$
$\therefore \quad x=\sqrt{\frac{9}{72}}=\frac{1}{2 \sqrt{2}} \mathrm{~m}$
Hence, length of compressed spring
$=2-\frac{1}{2 \sqrt{2}}$
$=\frac{4 \sqrt{2}-1}{2 \sqrt{2}}$
$=1.5 \mathrm{~m}$
= Potential energy of spring
$i e, \quad \frac{1}{2} m v^2=\frac{1}{2} k x^2$
Given, $\quad m=2 \mathrm{~kg}, v=3 \mathrm{~m} / \mathrm{s}, k=144 \mathrm{~N} / \mathrm{m}$
$\therefore \quad \frac{1}{2} \times 2 \times(3)^2=\frac{1}{2} \times 144 \times x^2$
$\therefore \quad x=\sqrt{\frac{9}{72}}=\frac{1}{2 \sqrt{2}} \mathrm{~m}$
Hence, length of compressed spring
$=2-\frac{1}{2 \sqrt{2}}$
$=\frac{4 \sqrt{2}-1}{2 \sqrt{2}}$
$=1.5 \mathrm{~m}$
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