Question

A particle is projected with velocity $v_o$ 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. ${\left(\frac{2m{v}_o^2}{3\alpha }\right)}^{\frac{1}{3}}$
• B. ${\left(\frac{3m{v}_o^2}{2\alpha }\right)}^{\frac{1}{2}}$
• C. ${\left(\frac{3m{v}_o^2}{2\alpha }\right)}^{\frac{1}{3}}$
• D. ${\left(\frac{2m{v}_o^2}{3\alpha }\right)}^{\frac{1}{2}}$

Answer 1

The correct option is C ${\left(\frac{3m{v}_o^2}{2\alpha }\right)}^{\frac{1}{3}}$

For damping force, acceleration of the particle is,
$$a=\frac{-\alpha x^2}{m}$$
Here, acceleration of the particle is varying with distance.
So that,
$$a=v\frac{dv}{dx}=\frac{-\alpha x^2}{m}$$
$\Rightarrow vdv=\frac{-\alpha x^2}{m}dx$

$\Rightarrow {\int }_{v_i}^{v_f}vdv=\frac{-\alpha }{m}{\int }_{x_i}^{x_f}{x}^{2}dx$

$\Rightarrow \left[\frac{v_f^2-v_i^2}{2}\right]=\frac{-\alpha }{m}\left[\frac{x_f^3-x_i^3}{3}\right]$

Here,
$v_f=0,v_i=v_0$
$x_f=x,x_i=0$

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

$\Rightarrow \left[\frac{v_o^2}{2}\right]=\frac{\alpha }{m}\left[\frac{x^3}{3}\right]$

$\Rightarrow \left[\frac{3m{v}_o^2}{2\alpha }\right]=x^3$

$\Rightarrow x={\left(\frac{3m{v}_o^2}{2\alpha }\right)}^{\frac{1}{3}}$

Hence, $\left(C\right)$ is the correct answer.