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A small block is connected to one end of a massless spring of un-stretched length $4.9 \mathrm{~m}$. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by $0.2 \mathrm{~m}$ and released from rest at $t=0$. It then executes simple harmonic motion with angular frequency $\omega=\pi / 3 \mathrm{rad} / \mathrm{s}$. Simultaneously at $t=0$, a small pebble is projected with speed $v$ form point $P$ at an angle of $45^{\circ}$ as shown in the figure. Point $P$ is at a horizontal distance of $10 \mathrm{~m}$ from $O$. If the pebble hits the block at $t=1 \mathrm{~s}$, the value of $v$ is (take $g$ $=10 \mathrm{~m} / \mathrm{s}^{2}$ )
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The correct answer is:
$\sqrt{50} \mathrm{~m} / \mathrm{s}$
Time of flight of projectile,
$T=\frac{2 V \sin \theta}{g}$
$\therefore \quad 1=\frac{2 V \sin 45^{\circ}}{g} \quad \therefore \quad V=\sqrt{50} \mathrm{~ms}^{-1}$
Hence pebble is projected with a speed $V=\sqrt{50} \mathrm{~ms}^{-1}$
$T=\frac{2 V \sin \theta}{g}$
$\therefore \quad 1=\frac{2 V \sin 45^{\circ}}{g} \quad \therefore \quad V=\sqrt{50} \mathrm{~ms}^{-1}$
Hence pebble is projected with a speed $V=\sqrt{50} \mathrm{~ms}^{-1}$
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