Question

A force F= -F₀ e^(-x/λ ) (where F₀ and λ are positive constants) acts on a particle of mass m that is initially at x = x₀ and moving with velocity v₀ (> 0). Show that the velocity of the particle is given by,

v(x) = ± ( v₀² + (2F₀λ/m)(e^-x/λ - 1) ) ^1/2 ,

where the upper (lower) sign corresponds to the motion in the positive (negative) x direction. Consider first the upper sign. For simplicity, define v_e = ?(2F₀ λ/m)^1/2. Then, show that the asymptotic velocity (limiting velocity as x→∞) is given by v_infinity = ( v₀² - v_e² )^1/2 . Note that v_infinity exists if v₀ ≥ v_e. Sketch the graph of v(x) in this case. Analyse the problem when v₀ < v_e by taking into account of the lower sign in the above solution. Sketch the graph of v(x) in this case. Show that the particle comes to rest (v(x) = 0) at a finite value of x given by x_m= −λ ln(1- (v₀² / v_e²)).

Answer 1