= Solution
\b[Energy–momentum transformations and invariant.] Let the primed <inertial frame> move at signed <velocity> $u$ along the unprimed $x$ axis. With $\Gamma=(1-u^2/c^2)^{-1/2}$, a <Lorentz transformation> of <four-momentum> gives
$$
\boxed{E'=\Gamma(E-up_x),\qquad p_x'=\Gamma\left(p_x-\frac{uE}{c^2}\right),\qquad p_y'=p_y,\quad p_z'=p_z.}
$$
Use $p^2=p_x^2+p_y^2+p_z^2$. The longitudinal terms satisfy
$$
E'^2-c^2p_x'^2
=\Gamma^2\left[(E-up_x)^2-c^2\left(p_x-\frac{uE}{c^2}\right)^2\right]
=\Gamma^2(1-u^2/c^2)(E^2-c^2p_x^2)
=E^2-c^2p_x^2.
$$
The transverse <momenta> are unchanged, so
$$
\boxed{E'^2-c^2p'^2=E^2-c^2p^2.}
$$
For one <electron>, the <relativistic energy-momentum relation> identifies this <Lorentz invariant> as $E_0^2=m_e^2c^4$.
\b[Symmetric electron scattering.] Assume an isolated <elastic collision> of the two <electrons>, with no emitted radiation. The incident electron has total <energy> $E$ and <momentum> magnitude $p$, so $p^2c^2=E^2-E_0^2$; the target electron initially contributes <rest energy> $E_0$ and zero <momentum>. In the nondegenerate configuration with final track <angles> $+\theta$ and $-\theta$, transverse <momentum conservation> gives
$$
p_1\sin\theta=p_2\sin\theta\quad\Longrightarrow\quad p_1=p_2=p_f.
$$
Because both final particles have the same <rest mass>, they also have equal total <energies>. <Conservation of energy> gives
$$
E_f=\frac{E+E_0}{2},\qquad
p_f^2c^2=E_f^2-E_0^2=\frac{(E-E_0)(E+3E_0)}{4}.
$$
Longitudinal <momentum conservation> is $p=2p_f\cos\theta$. Combining these expressions, with $E>E_0$, proves the <symmetric relativistic equal-mass scattering angle> relation:
$$
\cos^2\theta=\frac{E^2-E_0^2}{(E-E_0)(E+3E_0)}=\frac{E+E_0}{E+3E_0},
\qquad
\boxed{\theta=\arccos\sqrt{\frac{E+E_0}{E+3E_0}}
=\arctan\sqrt{\frac{2E_0}{E+E_0}}.}
$$
The low-<kinetic energy> limit has $\theta\to45^\circ$, recovering the familiar right angle between the outgoing tracks in nonrelativistic equal-<mass> scattering. Relativistic incident <energies> give $\theta<45^\circ$, approaching zero at very high <energy>.
\b[Accelerating potential.] For $\theta=30^\circ$, $\tan^2\theta=1/3=2E_0/(E+E_0)$, so $E=5E_0$. The acceleration from rest supplies <kinetic energy>, not total <energy>. If $e>0$ is the <elementary charge>, the required accelerating <electric potential difference> has magnitude
$$
\boxed{|V|=\frac{E-E_0}{e}=\frac{4E_0}{e}\simeq2.044\times10^6\,\mathrm V=2.044\,\mathrm{MV},}
$$
using electron <rest energy> $E_0\simeq0.511\,\mathrm{MeV}$. The <electric potential difference> is oriented to accelerate the negative <electron>; the result above specifies its magnitude.
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