 Solution

ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/section-a/2/solution

K = (γ - 1) M c^{2}, γ = \frac{1}{1 - v ^{2} / c ^{2}} .

Here

γ

c

b = v / c

, gives

(1 - b^{2})^{- 1/2} = 1 + \frac{1}{2} b^{2} + \frac{3}{8} b^{4} + O (b^{6}), K = \frac{1}{2} M v^{2} + \frac{3}{8} M \frac{v ^{4}}{c ^{2}} + O (M \frac{v ^{6}}{c ^{4}}) .

For

v ≪ c

, the leading term is the classical kinetic energy

M v^{2} /2

. The ratio of the fourth-order term to that leading term is

3 v^{2} / (4 c^{2})

. Consequently

\frac{3}{4} \frac{v ^{2}}{c ^{2}} = 0.01 ⟹ \frac{v}{c} = \frac{0.04}{3} = 0.11547 ≃ 11.5%.

The condition concerns one term of the binomial series; it is not the condition that the entire relativistic correction be exactly

1%

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