In a two-dimensional collision, momentum conservation must be applied as a vector equation. If the net external impulse during the collision is negligible,

$$m_1\mathbf u_1+m_2\mathbf u_2 =m_1\mathbf v_1+m_2\mathbf v_2.$$

Choosing $x$- and $y$-axes turns this into two scalar equations:

$$m_1u_{1x}+m_2u_{2x}=m_1v_{1x}+m_2v_{2x},$$

$$m_1u_{1y}+m_2u_{2y}=m_1v_{1y}+m_2v_{2y}.$$

These equations express momentum conservation independently in each direction. They do not, by themselves, determine every possible final speed and direction; the collision type and geometry provide additional constraints.

Perfectly inelastic case

If the objects stick together, they share one final velocity $\mathbf v_f$:

$$\mathbf v_f=\frac{m_1\mathbf u_1+m_2\mathbf u_2}{m_1+m_2}.$$

The same formula works in any number of dimensions because momentum is a vector.

Elastic case

For an elastic collision, total kinetic energy supplies an additional scalar equation:

$$\frac12m_1u_1^2+\frac12m_2u_2^2 =\frac12m_1v_1^2+\frac12m_2v_2^2.$$

The momentum components and energy equation can then be combined with any known scattering angle, symmetry, or contact geometry.

Equal masses with one initially at rest

A particularly useful result occurs when two equal masses collide elastically and the second is initially at rest. Let the initial velocity be $\mathbf u$. Momentum conservation gives

$$\mathbf u=\mathbf v_1+\mathbf v_2,$$

and energy conservation gives

$$u^2=v_1^2+v_2^2.$$

Taking the squared magnitude of the momentum equation,

$$u^2=v_1^2+v_2^2+2\mathbf v_1\cdot\mathbf v_2.$$

Comparison with the energy equation gives

$$\mathbf v_1\cdot\mathbf v_2=0.$$

Therefore, unless one object stops completely, the two outgoing velocity vectors are perpendicular.

Numerical example

Suppose one equal-mass ball approaches at $5,\mathrm{m/s}$ along the $x$-axis. After an elastic collision, the first ball moves at $3,\mathrm{m/s}$ with components

$$\mathbf v_1=(1.8,,2.4),\mathrm{m/s}.$$

Momentum conservation requires

$$\mathbf v_2=(5,0)-\mathbf v_1=(3.2,-2.4),\mathrm{m/s},$$

whose speed is $4,\mathrm{m/s}$. The kinetic-energy condition is satisfied because

$$5^2=3^2+4^2,$$

and

$$\mathbf v_1\cdot\mathbf v_2=(1.8)(3.2)+(2.4)(-2.4)=0.$$

The two final directions are therefore at right angles. The result is not a special rule to memorize; it follows from vector momentum conservation plus kinetic-energy conservation.