= Solution
Choose the initial incident <velocity> as $u_1\hat{\mathbf x}$. The <centre-of-mass velocity> in the <laboratory frame> is
$$
\mathbf V=\frac{m_1u_1}{m_1+m_2}\hat{\mathbf x}.
$$
In the <centre-of-mass frame>, the initial <velocities> are $\mathbf u_1^*=m_2u_1\hat{\mathbf x}/(m_1+m_2)$ and $\mathbf u_2^*=-m_1u_1\hat{\mathbf x}/(m_1+m_2)$. Conservation of <momentum> and <kinetic energy> in an <elastic collision> preserves these two <speed> magnitudes. Let $\theta^*$ be the outgoing angle of particle 1 in that frame, measured from the initial positive $x$ direction. Then
$$
\mathbf v_1^*=\frac{m_2u_1}{m_1+m_2}(\cos\theta^*,\sin\theta^*),\qquad
\mathbf v_2^*=-\frac{m_1u_1}{m_1+m_2}(\cos\theta^*,\sin\theta^*).
$$
Adding $\mathbf V$ to transform back to the <laboratory frame> gives
$$
\mathbf v_1=\frac{u_1}{m_1+m_2}(m_1+m_2\cos\theta^*,m_2\sin\theta^*),\qquad
\mathbf v_2=\frac{m_1u_1}{m_1+m_2}(1-\cos\theta^*,-\sin\theta^*).
$$
The nonnegative laboratory recoil angle $\theta_2$ is measured below the incident axis in this choice of scattering plane. Hence the <recoil angle in a two-body elastic collision> obeys
$$
\boxed{\tan\theta_2=\frac{|v_{2y}|}{v_{2x}}
=\frac{\sin\theta^*}{1-\cos\theta^*}
=\cot\frac{\theta^*}{2},\qquad
\theta_2=\frac{\pi-\theta^*}{2}\quad(0<\theta^*\leq\pi).}
$$
The source identity therefore uses $\theta^*$ for the projectile's outgoing direction in the <centre-of-mass frame>. If instead it denotes particle 2's outgoing direction, the corresponding expression has $1+\cos\theta^*$ in the denominator. A signed laboratory angle for the drawn target trajectory would carry a minus sign. At zero deflection the target remains at rest, so its direction is undefined; $\theta_2\to\pi/2$ is a limit.
\Image[/past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2018-collision-velocity-diagrams.png]
{title=Outgoing velocity vectors in the laboratory and centre-of-mass frames}
{description=An illustrative mass ratio $m_1:m_2=2:1$ and projectile deflection $\theta^*=70^\circ$ display the angle conventions. The two outgoing <momenta> cancel in the <centre-of-mass frame>, although the velocity magnitudes differ. Adding the common boost $\mathbf V$ gives the laboratory vectors.}
Back to article page
