 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2014/12/d/solution

Parametrize any actual particle trajectory by time, with

v = \dot{r}

F = \dot{v}

. The chain rule gives

\frac{d}{d t} (\frac{1}{2} ∣ v ∣^{2}) = v \cdot \dot{v} = F \cdot \dot{r} .

Integrating from the origin to the endpoint gives

I = \int F \cdot d r = K_{A} - K_{O}, Δ K = 1.

Here

K

is specific kinetic energy, because the source defines it without a mass factor. Equivalently,

∣ v_{A} ∣^{2} - ∣ v_{O} ∣^{2} = 2

in the chosen units. The individual initial or final speeds are not fixed by this result. For a particle of mass

m

, the change of ordinary kinetic energy would be

m I

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