# A particle is projected with velocity

Question:

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 :

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

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

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

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

Correct Option: , 3

Solution:

(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 : $\mathrm{a}=-\frac{\alpha \mathrm{X}^{2}}{\mathrm{~m}}$

$\int_{V_{0}}^{O} V d V=-\int_{0}^{x} \frac{\alpha x^{2}}{m} d x$

$-\frac{v_{0}^{2}}{2}=\frac{-\alpha}{m}\left[\frac{x^{3}}{3}\right]$

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