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(B) $1$. Conservation of Mechanical Energy: The potential energy stored in the compressed spring is converted into the kinetic energy of the block as

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$\sqrt{2}\,m$

Vedclass · Answer

(B) $1$. Conservation of Mechanical Energy: The potential energy stored in the compressed spring is converted into the kinetic energy of the block as it leaves the spring.$\frac{1}{2} kx^2 = \frac{1}{2} mv^2$$v = \sqrt{\frac{k}{m}} x$Given: $k = 500\,N/m$,$x = 5.0\,cm = 0.05\,m$,$m = 500\,g = 0.5\,kg$.$v = \sqrt{\frac{500}{0.5}} 	imes 0.05 = \sqrt{1000} 	imes 0.05 = 10\sqrt{10} 	imes 0.05 = 0.5\sqrt{10}\,m/s$.$2$. Projectile Motion: The block leaves the table horizontally at height $H = 4\,m$.The time of flight $t$ is given by $H = \frac{1}{2} gt^2 \implies t = \sqrt{\frac{2H}{g}}$.$t = \sqrt{\frac{2 	imes 4}{10}} = \sqrt{0.8} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}\,s$.$3$. Horizontal Distance $(R)$: $R = v 	imes t$$R = (0.5\sqrt{10}) 	imes (\frac{2}{\sqrt{5}}) = 0.5 	imes 2 	imes \sqrt{\frac{10}{5}} = 1 	imes \sqrt{2} = \sqrt{2}\,m$.Therefore,the correct option is $B$.

$Column-I$	$Column-II$
$(A)$ Angle of projection	$(p)$ $20 \, m$
$(B)$ Angle of velocity with horizontal after $4 \, s$	$(q)$ $80 \, m$
$(C)$ Maximum height	$(r)$ $45^{\circ}$
$(D)$ Horizontal range	$(s)$ $\tan^{-1}(1/2)$

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