Solution

The work done along the trajectory is

W = \int_{t_{a}}^{t_{b}} F (x (t), t) \overset{x}{˙} (t) d t = \int_{a}^{b} F (x, t (x)) d x .

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By Newton's second law,

F = m \overset{x}{¨}

, so

W = m \int_{t_{a}}^{t_{b}} \overset{x}{¨} \overset{x}{˙} d t = [\frac{1}{2} m \overset{x}{˙}^{2}]_{t_{a}}^{t_{b}} .

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This is the work-energy theorem: the work equals the change in kinetic energy.

A force is conservative when its work between two points is independent of the path, equivalently when it is the negative gradient of a potential energy. Here a potential with

V (0) = 0

V (x) = F_{0} λ (1 - e^{- ∣ x ∣/ λ}) .

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Conservation of energy for

x > 0

gives

\frac{1}{2} m v^{2} + F_{0} λ (1 - e^{- x / λ}) = \frac{1}{2} m v_{0}^{2} .

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Thus

v (x) = v_{0}^{2} + v_{e}^{2} (e^{- ∣ x ∣/ λ} - 1), v_{e} = \frac{2 F _{0} λ}{m} .

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v_{0} > v_{e}

, the particle escapes with asymptotic speed

v_{0}^{2} - v_{e}^{2}

; the graph decreases from

v_{0}

to this horizontal asymptote. At the escape velocity

v_{0} = v_{e}

, it decreases as

v_{e} e^{- x / (2 λ)}

and approaches zero only at infinity. If

v_{0} < v_{e}

, it reaches the turning point

x_{m a x} = - λ lo g (1 - \frac{v _{0}^{2}}{v _{e}^{2}}),

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where

v = 0

, then returns on the negative branch of the same curve and oscillates symmetrically about the origin.

For oscillation put

w = 1 - v_{0}^{2} / v_{e}^{2}

. One quarter of the period is

\int_{0}^{x_{m a x}} \frac{d x}{v ( x )} = \frac{λ}{v _{e}} \int_{w}^{1} \frac{d u}{u u - w} = \frac{2 λ}{v _{e} w} cos^{- 1} w,

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where

u = e^{- x / λ}

. Therefore

T = \frac{8 λ}{v _{e} 1 - v _{0}^{2} / v _{e}^{2}} cos^{- 1} 1 - v_{0}^{2} / v_{e}^{2} .

(43)

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 4 / 9C / Solution

 Ancestors (10)