# A particle is projected

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 :

1. $\left(\frac{3 v_{0}^{2}}{2 \alpha}\right)^{\frac{1}{2}}$

2. $\left(\frac{2 v_{0}}{3 \alpha}\right)^{\frac{1}{3}}$

3. $\left(\frac{2 v_{0}^{2}}{3 \alpha}\right)^{\frac{1}{2}}$

4. $\left(\frac{3 v_{0}^{2}}{2 \alpha}\right)^{\frac{1}{3}}$

Correct Option: , 4

Solution:

$\mathrm{F}=-\alpha \mathrm{x}^{2}$

$\mathrm{ma}=-\alpha \mathrm{x}^{2}$

$a=\frac{-\alpha x^{2}}{m}$

$\frac{\mathrm{vdv}}{\mathrm{dx}}=-\frac{\alpha}{m} \mathrm{x}^{2}$

$\int_{v_{0}}^{0} v d v=\int_{0}^{x}-\frac{\alpha}{m} x^{2} d x$

$\left(\frac{v^{2}}{2}\right)_{v_{0}}^{0}=-\frac{\alpha}{m}\left(\frac{x^{3}}{3}\right)_{0}^{x}$

$\frac{-v_{0}^{2}}{2}=-\frac{\alpha}{m} \frac{x^{3}}{3}$

$x=\left(\frac{3 m v_{0}^{2}}{2 \alpha}\right)^{\frac{1}{3}}$

Option(4) is most suitable option as (m) is not given in any option