A particle is projected with velocity $\mathrm{v}_{0}$ along $\mathrm{x}$-axis. A damping forceis acting on the particle which is proportional to the square of the distance from the origin i.e. $m a=-\alpha X^{2}$. The distance at which the particle stops :
Correct Option: , 3
(3)
$\mathrm{a}=\frac{\mathrm{vdv}}{\mathrm{dx}}$
$\int_{\mathrm{v}_{\mathrm{i}}}^{\mathrm{v}_{\mathrm{f}}} \mathrm{V} \mathrm{dv}=\int_{\mathrm{x}_{\mathrm{i}}}^{\mathrm{xf}} \mathrm{adx}$
Given : $-\mathrm{v}_{\mathrm{i}}=\mathrm{v}_{0}$
$\mathrm{V}_{\mathrm{f}}=0$
$X_{i}=0$
$\mathrm{X}_{\mathrm{f}}=\mathrm{x}$
From Damping Force : $\mathrm{a}=-\frac{\alpha \mathrm{X}^{2}}{\mathrm{~m}}$
$\int_{V_{0}}^{O} V d V=-\int_{0}^{x} \frac{\alpha x^{2}}{m} d x$
$-\frac{v_{0}^{2}}{2}=\frac{-\alpha}{m}\left[\frac{x^{3}}{3}\right]$
$x=\left\lceil\frac{3 m v_{0}^{2}}{2 \alpha}\right\rceil^{\frac{1}{3}}$