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:

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Answer & Solution

Correct Answer

(D) \(\left(\frac{3 v_{0}^{2}}{2 \alpha}\right)^{\frac{1}{3}}\)

Detailed explanation

\(F=-\alpha x^{2}\)

\(m a=-\alpha x^{2}\)

\(a=\frac{-\alpha x^{2}}{m}\)

\(\frac{v d v}{d x}=-\frac{\alpha}{m} x^{2}\)

\(\int_{v_{0}}^{0} v d v=\int_{0}^{x}-\frac{\alpha}{m} x^{2} d x\)

\(\left(\frac{v^{2}}{2}\right)^{0}=-\frac{\alpha}{m}\left(\frac{x^{3}}{3}\right)_{0}^{x}\)

\(\frac{-v_{0}^{2}}{2}=-\frac{\alpha}{m} \frac{x^{3}}{3}\)

\(x=\left(\frac{3 m v_{0}^{2}}{2 \alpha}\right)^{\frac{1}{3}}\)

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