Q 11-02-105JEE MainJEE Main 2021 (24 Feb, Shift 2)Medium
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:
Answer: (B) $\left(\frac{3mv_0^2}{2\alpha}\right)^{\frac13}$
Write $a = v\dfrac{dv}{dx}$:
$$mv\,dv = -\alpha x^2\,dx \;\Rightarrow\; \int_{v_0}^{0} mv\,dv = -\int_0^x \alpha x^2\,dx \;\Rightarrow\; \frac{mv_0^2}{2} = \frac{\alpha x^3}{3}$$
$$x = \left(\frac{3mv_0^2}{2\alpha}\right)^{1/3}$$
Solution by Sreeraj P, M.Sc Physics