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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2023

Paper 2

The $x$ component of the <velocity> is
$$
\dot x=\frac{c^2p_0}{\mathcal E(t)}.
$$
Thus
$$
x(t)=\frac{cp_0}{qE}
\operatorname{arsinh}\left(\frac{cqEt}{\mathcal E_0}\right).
$$
Solving this equation for $t$ gives
$$
\frac{cqEt}{\mathcal E_0}
=\sinh\left(\frac{qEx}{cp_0}\right).
$$
Substitution into the expression for $z(t)$ and the identity $1+\sinh^2u=\cosh^2u$ produce the trajectory
$$
\boxed{
z(x)=\frac{\mathcal E_0}{qE}
\left[
\cosh\left(\frac{qEx}{cp_0}\right)-1
\right]
}.
$$
This is <relativistic motion in a constant electric field with transverse momentum>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 2 / 16D / d / Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 2 / 16D / d / Solution

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Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 2 / 16D / d / Solution

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