A particle of mass m and initial velocity V is faced with a damping force
proportional to the square of its velocity and equal to a v2. In how much time will
the particle come to a stop?
2m/aV
2 ma/V
2 mV/a
Never stop
✅ Correct Answer
The particle never stops. The time required for the particle to come to a complete stop is infinite due to the nature of the damping force proportional to \(v^2\).
🎯 Problem Setup
A particle of mass \(m\) with initial velocity \(V\) experiences a damping force \(F = -a v^2\), where \(a > 0\) and \(v\) is the instantaneous velocity. By Newton’s second law, \(m \frac{dv}{dt} = -a v^2\), or \(\frac{dv}{dt} = -\frac{a}{m} v^2\).
📐 Detailed Solution
Separate variables: \(\frac{dv}{v^2} = -\frac{a}{m} dt\). Integrate from initial velocity \(V\) at \(t=0\) to velocity \(v\) at time \(t\):
\int_V^v \frac{dv’}{{v’}^2} = -\frac{a}{m} \int_0^t dt’ \implies \left[ -\frac{1}{v’} \right]_V^v = -\frac{a}{m} t \implies -\frac{1}{v} + \frac{1}{V} = -\frac{a}{m} t.
$$
Solving for \(v(t)\):
v(t) = \frac{V}{1 + \frac{a V}{m} t} = \frac{m V}{m + a V t}.
$$
Velocity approaches zero asymptotically as \(t \to \infty\), but never reaches exactly zero in finite time. Setting \(v(T) = 0\) yields division by infinity, confirming \(T = \infty\).
📊 Option Analysis
| Option | Expression | Analysis |
|---|---|---|
| A | \(\frac{2m}{aV}\) | Has dimensions of time \([M]/[ (F/v^2) \cdot (L/T) ] = T\), but assumes linear damping form \(\frac{dv}{dt} \propto -v\). Incorrect for \(v^2\) damping. |
| B | \(\frac{2 m a}{V}\) | Dimensions \([M \cdot (F/v^2)] / (L/T) = [M^2 L^0 T^{-1}]\), not time. Nonsensical. |
| C | \(\frac{2 m V}{a}\) | Dimensions \([M \cdot (L/T)] / (F/v^2) = [M L T^{-1}] / [M L^{-1} T^2] = T\), but derives from incorrect integration. Wrong for this case. |
| D | Never stops | Correct, as \(v(t) > 0\) for all finite \(t\), approaching zero only at \(t = \infty\). |
- Identify force-velocity dependence for separation of variables
- Check dimensions: time requires \([T]\); eliminate mismatches early
- Recognize asymptotic behavior in nonlinear ODEs
- Practice variants: for \(F = -b v^n\), stopping time scales as \(\Delta t \propto m^{1} b^{-1} V^{1-n}\)
