Oblique Elastic Collision

The situation

In a 2D (oblique) elastic collision, the bodies do not move along the same line. After collision, they move in different directions in the plane.

The conservation laws still hold — now applied as vector equations.

Conservation laws in 2D

Momentum — x direction:

$m_1 u_{1x} + m_2 u_{2x} = m_1 v_{1x} + m_2 v_{2x}$

Momentum — y direction:

$m_1 u_{1y} + m_2 u_{2y} = m_1 v_{1y} + m_2 v_{2y}$

Kinetic energy:

$\frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$

Three equations. Four unknowns (two velocity components each for $v_1$ and $v_2$). The system is underdetermined without additional information (e.g., the collision geometry — impact parameter, or the angle one body makes after collision).

Special case: equal masses, target at rest

$m_1 = m_2$, $m_2$ initially at rest.

After elastic collision, momentum conservation and KE conservation give:

$\vec{v}_1 + \vec{v}_2 = \vec{u}_1 \quad \text{(momentum, dividing by } m\text{)}$

$v_1^2 + v_2^2 = u_1^2 \quad \text{(KE, dividing by } \frac{1}{2}m\text{)}$

From the first: $|\vec{v}_1 + \vec{v}_2|^2 = u_1^2$

$v_1^2 + v_2^2 + 2\vec{v}_1 \cdot \vec{v}_2 = u_1^2$

Substituting $v_1^2 + v_2^2 = u_1^2$:

$2\vec{v}_1 \cdot \vec{v}_2 = 0 \implies \vec{v}_1 \perp \vec{v}_2$

After an elastic collision between equal masses where one is at rest, the two bodies always move at 90° to each other.

This is a beautiful result — the angle between the two outgoing bodies is always exactly $90°$, regardless of the impact geometry.

Verification

This is why billiard balls always separate at right angles (approximately, since real billiard balls are not perfectly elastic and have friction/spin). Snooker players use this principle instinctively.

General case: unequal masses

No such clean angle result exists. The problem requires the specific collision geometry (or one final angle) as additional information. Set up the three conservation equations and solve for the remaining unknowns.