 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2013-resit/6/a/solution

The external work equals the change in relativistic kinetic energy,

W = m_{e} c^{2} (γ_{f} - γ_{i})

. With

m_{e} c^{2} = 511 keV

W_{a} = 511 (\frac{1}{1 - 0.1 1 ^{2}} - \frac{1}{1 - 0.1 0 ^{2}}) keV ≃ 0.5456 keV ≃ 8.742 \times 1 0^{- 17} J .

The rest energy cancels between endpoints. At these speeds, the nonrelativistic estimate

\frac{1}{2} m_{e} c^{2} (0.1 1^{2} - 0.1 0^{2}) = 0.53655 keV

is close but slightly smaller.

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