Question

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 :

• A. $\left(\frac{3 v_{0}^{2}}{2 \alpha}\right)^{\frac{1}{2}}$
• B. $\left(\frac{2 v_{0}}{3 \alpha}\right)^{\frac{1}{3}}$
• C. $\left(\frac{2 v_{0}^{2}}{3 \alpha}\right)^{\frac{1}{2}}$
• D. $\left(\frac{3 v_{0}^{2}}{2 \alpha}\right)^{\frac{1}{3}}$

Answer 1

Answer: (D)

$F =-\alpha x ^{2}$
$ma =-\alpha x ^{2}$
$a =\frac{-\alpha x ^{2}}{ m }$
$\frac{ vd v }{ dx }=-\frac{\alpha}{ m } x ^{2}$
$\int_\limits{ v _{0}}^{0} v d v =\int_\limits{0}^{ x }-\frac{\alpha}{ m } x ^{2} d x$
$\left(\frac{ v ^{2}}{2}\right)_{ v _{0}}^{0}=-\frac{\alpha}{ m }\left(\frac{ x ^{3}}{3}\right)_{0}^{ x }$
$\frac{- v _{0}^{2}}{2}=-\frac{\alpha}{ m } \frac{ x ^{3}}{3}$
$x =\left(\frac{\left.3 mv _{0}^{2}\right)^{\frac{1}{3}}}{2 \alpha}\right)$