 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-2/16d/d/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

The

x

component of the velocity is

\overset{x}{˙} = \frac{c ^{2} p _{0}}{E ( t )} .

Thus

x (t) = \frac{c p _{0}}{qE} arsinh (\frac{c qEt}{E _{0}}) .

Solving this equation for

t

gives

\frac{c qEt}{E _{0}} = sinh (\frac{qE x}{c p _{0}}) .

Substitution into the expression for

z (t)

and the identity

1 + sinh^{2} u = cosh^{2} u

produce the trajectory

z (x) = \frac{E _{0}}{qE} [cosh (\frac{qE x}{c p _{0}}) - 1] .

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