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. $ma = -ax^2$. The distance at which the particle stops :

• A. $\left(\frac{2v_0^2}{3a}\right)^{1/2}$
• B. $\left(\frac{3v_0^2}{2a}\right)^{1/2}$
• C. $\left(\frac{3v_0^2}{2a}\right)^{1/3}$
• D. $\left(\frac{2v_0^2}{3a}\right)^{1/3}$

Hint:

When acceleration is a function of position $x$, always use $a = v \frac{dv}{dx}$ to solve for velocity or displacement.

Answer 1

The Correct Option is C

Step 1: Given $ma = -ax^2 \Rightarrow a_{acc} = -\frac{a}{m}x^2$.
Step 2: Use the kinematic relation $v \frac{dv}{dx} = a_{acc}$. So, $v \frac{dv}{dx} = -\frac{a}{m}x^2$.
Step 3: Integrate with limits: at $x=0, v=v_0$ and at $x=x_{stop}, v=0$. \[ \int_{v_0}^{0} v \, dv = -\frac{a}{m} \int_{0}^{x} x^2 \, dx \] \[ \left[ \frac{v^2}{2} \right]_{v_0}^{0} = -\frac{a}{m} \left[ \frac{x^3}{3} \right]_{0}^{x} \Rightarrow -\frac{v_0^2}{2} = -\frac{ax^3}{3m} \] \[ x^3 = \frac{3mv_0^2}{2a} \Rightarrow x = \left( \frac{3mv_0^2}{2a} \right)^{1/3} \] Assuming $m$ is absorbed into the constant $a$ as per the options provided.