A particle is projected with velocity v0 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 = -αx². The distance at which the particle stops is:

Question

Accepted Answer

(3v0²/2α)^(1/3). With m v dv/dx = -alpha x^2, integrate: (1/2)v0^2 = (alpha/3m) x^3 (taking m as given), so the stopping distance x = (3 v0^2 / (2 alpha))^(1/3).

A particle is projected with velocity v0 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 = -αx². The distance at which the particle stops is:

Solution

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