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Step-by-Step Solution
Step 1: Identify the Forces and Equations of Motion
A particle of mass $m$ is projected along the x-axis with initial speed $v_0$. A damping force proportional to the square of the distance from the origin acts on it, given by
$F = -\alpha x^2.$
According to Newton’s Second Law, $F = ma$, so
$ma = -\alpha x^2 \quad \Longrightarrow \quad a = \frac{dv}{dt} = -\frac{\alpha}{m} x^2.$
Since the motion is along one dimension (the x-axis), we can use the relation
$a = v\,\frac{dv}{dx}$.
Step 2: Express Acceleration in Terms of $v$ and $x$
From
$a = v \frac{dv}{dx}$,
substitute
$a = -\frac{\alpha}{m} x^2$
to get
$v \frac{dv}{dx} = -\frac{\alpha}{m} x^2.$
Step 3: Separate Variables and Integrate
Rearranging, we get
$v\,dv = -\frac{\alpha}{m} \,x^2\,dx.
We integrate both sides from the initial conditions to the final:
• Velocity: from $v = v_0$ to $v = 0$ (since the particle stops).
• Position: from $x = 0$ to $x = X$, where $X$ is the distance at which the particle stops.
Thus,
$\displaystyle \int_{v_0}^{0} v\,dv = \int_{0}^{X} \left(-\frac{\alpha}{m} x^2\right)\,dx.$
Step 4: Perform the Integrals
The left-hand side:
$\displaystyle \int_{v_0}^{0} v\,dv = \left[ \frac{v^2}{2} \right]_{v_0}^{0} = \left( 0 - \frac{v_0^2}{2} \right) = -\frac{v_0^2}{2}.$
The right-hand side:
$\displaystyle \int_{0}^{X} \left(-\frac{\alpha}{m} x^2\right)\,dx = -\frac{\alpha}{m} \left[ \frac{x^3}{3} \right]_{0}^{X} = -\frac{\alpha}{m}\left(\frac{X^3}{3}\right).$
Equating these results:
$\displaystyle -\frac{v_0^2}{2} = -\frac{\alpha}{m} \cdot \frac{X^3}{3}.$
Step 5: Solve for $X$
Cancel the negative signs on both sides and rearrange to get:
$\displaystyle \frac{v_0^2}{2} = \frac{\alpha}{m} \frac{X^3}{3} \quad \Longrightarrow \quad X^3 = \frac{3 m v_0^2}{2 \alpha}.$
Therefore,
$\displaystyle X = \left( \frac{3 m v_0^2}{2 \alpha} \right)^{\frac{1}{3}}.$
This is the distance at which the particle comes to rest.
