Conservation of Momentum said that with no net outside force, total momentum stays the same. Momentum is a vector, so in a plane that one statement is two equations, one for each direction.
Key idea
In a collision with negligible outside forces, $$\sum p_x \text{ before} = \sum p_x \text{ after}, \qquad \sum p_y \text{ before} = \sum p_y \text{ after}.$$ Solve them like any two equations; then rebuild speeds and angles from the components. Kinetic energy is conserved only if the collision is elastic.
Why it works
During the brief contact, the objects push on each other with equal and opposite forces (Newton's third law), so their impulses on each other cancel as vectors. Outside forces like friction act too, but over a few milliseconds their impulse is tiny. A vector that doesn't change has components that don't change, so each direction is conserved on its own.
Worked example: Cars at an intersection
A $1200\ \text{kg}$ car heading east at $15\ \text{m/s}$ and a $2000\ \text{kg}$ truck heading north at $10\ \text{m/s}$ collide and lock together. Find the speed and direction of the wreck.
- $p_x = 1200(15) = 18\,000\ \text{kg·m/s}$; $p_y = 2000(10) = 20\,000\ \text{kg·m/s}$.
- After: $3200\,\vec{V} = \langle 18\,000, 20\,000 \rangle$, so $$V = \dfrac{\sqrt{18\,000^2 + 20\,000^2}}{3200} = \dfrac{26\,900}{3200} = 8.41\ \text{m/s}$$
- Direction: $\tan\theta = 20\,000/18\,000$, so $\theta = 48.0^\circ$ north of east.
Answer: $8.41\ \text{m/s}$ at $48.0^\circ$ north of east.
Worked example: Equal masses, elastic
A billiard ball hits an identical ball at rest, off center and elastically. The first ball leaves at $30^\circ$ to its original line. In what direction does the second go?
- Momentum (the masses cancel): $\vec{v}_0 = \vec{v}_1 + \vec{v}_2$. Energy: $v_0^2 = v_1^2 + v_2^2$.
- Square the momentum equation: $v_0^2 = v_1^2 + v_2^2 + 2\vec{v}_1\cdot\vec{v}_2$. Compare: $\vec{v}_1\cdot\vec{v}_2 = 0$.
- The balls leave at right angles, so the second goes at $60^\circ$ on the other side of the original line.
Answer: $60^\circ$ on the other side.
Is energy conserved too? Compare: the car and truck had $\tfrac{1}{2}(1200)(15)^2 + \tfrac{1}{2}(2000)(10)^2 = 235\ \text{kJ}$ before and $\tfrac{1}{2}(3200)(8.41)^2 = 113\ \text{kJ}$ after. More than half went into crumpling metal, heat and sound, as expected when objects stick. Momentum is conserved in every collision; kinetic energy only in elastic ones.
Common mistake
Conserving the total speed, or adding magnitudes of momentum as if they were numbers. In the intersection crash, $18\,000 + 20\,000 = 38\,000\ \text{kg·m/s}$ would give $11.9\ \text{m/s}$, but the momenta are perpendicular and add like the sides of a right triangle. Check: write the $x$ and $y$ equations separately, with signs, and only combine them into a magnitude at the very end.