Past exam of the physics 1 paper of the University of Oxford / 2013 resit / 6 / b

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- Solution b

Solution

Applying the same work-energy theorem at the higher pair of speeds gives

W_{b} = 511 (\frac{1}{1 - 0.9 9 ^{2}} - \frac{1}{1 - 0.9 8 ^{2}}) keV ≃ 1055 keV ≃ 1.689 \times 1 0^{- 13} J .

The identical increment

0.01 c

now needs about two thousand times as much work. The growth follows from the rapidly increasing Lorentz factor near the speed of light.

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