A particle is projected with velocity $v_{0}$ along $x$-axis. A damping force is acting on the particle which is proportional to the square of the distance from the origin i.e., ma $=-\alpha x^{2} . .$ The distance at which the particle stops :
Correct Option: , 4
$\mathrm{F}=-\alpha \mathrm{x}^{2}$
$\mathrm{ma}=-\alpha \mathrm{x}^{2}$
$a=\frac{-\alpha x^{2}}{m}$
$\frac{\mathrm{vdv}}{\mathrm{dx}}=-\frac{\alpha}{m} \mathrm{x}^{2}$
$\int_{v_{0}}^{0} v d v=\int_{0}^{x}-\frac{\alpha}{m} x^{2} d x$
$\left(\frac{v^{2}}{2}\right)_{v_{0}}^{0}=-\frac{\alpha}{m}\left(\frac{x^{3}}{3}\right)_{0}^{x}$
$\frac{-v_{0}^{2}}{2}=-\frac{\alpha}{m} \frac{x^{3}}{3}$
$x=\left(\frac{3 m v_{0}^{2}}{2 \alpha}\right)^{\frac{1}{3}}$
Option(4) is most suitable option as (m) is not given in any option
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