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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:
$\left(\frac{3 v_{0}^{2}}{2 \alpha}\right)^{\frac{1}{2}}$
$\left(\frac{2 v_{0}}{3 \alpha}\right)^{\frac{1}{3}}$
$\left(\frac{2 v_{0}^{2}}{3 \alpha}\right)^{\frac{1}{2}}$
$\left(\frac{3 v_{0}^{2}}{2 \alpha}\right)^{\frac{1}{3}}$
Solution
$F=-\alpha x^{2}$
$m a=-\alpha x^{2}$
$a=\frac{-\alpha x^{2}}{m}$
$\frac{v d v}{d x}=-\frac{\alpha}{m} x^{2}$
$\int_{v_{0}}^{0} v d v=\int_{0}^{x}-\frac{\alpha}{m} x^{2} d x$
$\left(\frac{v^{2}}{2}\right)^{0}=-\frac{\alpha}{m}\left(\frac{x^{3}}{3}\right)_{0}^{x}$
$\frac{-v_{0}^{2}}{2}=-\frac{\alpha}{m} \frac{x^{3}}{3}$
$x=\left(\frac{3 m v_{0}^{2}}{2 \alpha}\right)^{\frac{1}{3}}$
