Two-Dimensional Collisions

If the net external impulse on the pair during contact is negligible, the total momentum vector is the same before and after, so each component is conserved on its own. Put the x-axis along the incoming velocity. The initial y-momentum is then zero, and the whole y-equation says the outgoing sideways momenta cancel: m₁v₁′ sin θ = m₂v₂′ sin φ, with θ and φ measured on opposite sides of the axis. Take a 0.25 kg puck at 2.0 m s⁻¹ that strikes a 0.40 kg puck at rest and leaves at 1.2 m s⁻¹, 30° above the axis. Along y, the struck puck carries 0.25 × 1.2 × sin 30° = 0.150 kg m s⁻¹ downward, a y-velocity of 0.375 m s⁻¹. Along x it carries 0.50 − 0.25 × 1.2 × cos 30° = 0.240 kg m s⁻¹, an x-velocity of 0.600 m s⁻¹. So it leaves at 0.708 m s⁻¹, 32.0° below the axis. Isolation is a claim about the contact time: table friction acts, but over a few milliseconds its impulse is tiny beside 0.50 kg m s⁻¹.

With the target at rest, p = p₁′ + p₂′ is a closed triangle: p runs along the incoming line, and p₁′ and p₂′ chain from its tail to its head. It holds exactly what the two component equations hold, but trades components for trigonometry. The angle between p₁′ and p is the deflection θ, the angle between p₂′ and p is the recoil angle φ, and the angle opposite p is 180° − (θ + φ). The cosine rule gives p₂′² = p² + p₁′² − 2pp₁′ cos θ. For the puck above, p = 0.500 and p₁′ = 0.300 kg m s⁻¹ at θ = 30°, so p₂′ = √(0.250 + 0.090 − 0.260) = 0.283 kg m s⁻¹, and the law of sines gives sin φ = 0.300 × sin 30°/0.283, so φ = 32.0°: the same answer with no components written. When both bodies move beforehand the triangle becomes a quadrilateral, p₁ + p₂ = p₁′ + p₂′, and the rule is unchanged: the chain of momentum arrows before and the chain after must end at the same point.

Momentum closes the triangle whatever happens; kinetic energy decides which triangles can occur. Write K = p²/2m so energy uses the triangle's own sides: an elastic collision needs p²/m₁ = p₁′²/m₁ + p₂′²/m₂. With equal masses that is p² = p₁′² + p₂′², Pythagoras, so the outgoing paths are exactly 90° apart, and the opening angle becomes a free energy test for identical bodies: under 90° means kinetic energy was lost. For unequal masses, completing the square in the same condition puts the tip of p₁′ on a circle of radius m₂p/(m₁ + m₂), centred m₁/(m₁ + m₂) of the way along p, and outcomes that lose energy fall inside it. For the 0.25 kg and 0.40 kg pucks that circle has radius 0.615p, and the actual tip sits only 0.329p from its centre: well inside, as the energy confirms, 0.500 J before and 0.280 J after, a 44% loss.

Two bodies leaving a collision in a plane carry four unknown numbers: two speeds and two directions, or four velocity components. Momentum supplies two equations. If the bodies stick, the final state is a single velocity with two unknowns and momentum fixes it completely; run that count the other way and a wreck's measured velocity returns the speeds of two vehicles that met on known roads. If the collision is elastic, energy adds a third equation and one number is still missing. If it is neither, two are missing. That is physics, not a gap in the method: two pucks meeting at the same speed can leave in endlessly many ways, depending on where they touch. So before solving, ask which measured quantity closes the count, whether a deflection angle, one final speed, a recoil direction or the geometry of the contact, and notice when a problem gives more data than it needs, because the extra number is then a consistency check.

Smooth spheres push on each other only along the line joining their centres at impact. Two consequences follow. A target at rest receives its whole impulse along that line, so it leaves along it: its recoil angle φ is set by where it was struck. And the projectile's velocity component across the line meets no force, so it is unchanged. Along the line the event is a one-dimensional collision with approach speed u cos φ, and restitution gives v₂′ = (1 + e)m₁u cos φ/(m₁ + m₂). For identical billiard balls with e = 1 and φ = 30°, the object ball takes u cos 30° = 0.866u along 30°, and the cue ball keeps only its cross-line part, 0.500u, heading 60° to the other side: 90° apart, as energy demanded. With e = 0.90 the object ball gets 0.823u, the cue ball keeps 0.043u along the line, and the paths open to 85.1°. The rule fails when surfaces grip or spin: friction at the contact adds impulse across the line, and topspin lets table friction bend the cue ball's path after impact.

Measuring the projectile's deflection θ supplies the missing number, but not always uniquely. Eliminating p₂′ between the cosine rule and the elastic condition leaves a quadratic, (m₁ + m₂)v₁′² − 2m₁u cos θ v₁′ + (m₁ − m₂)u² = 0, with roots v₁′ = u[m₁cos θ ± √(m₂² − m₁²sin²θ)]/(m₁ + m₂). If m₁ is less than m₂ the minus root is negative and one outcome survives. If m₁ exceeds m₂, the root is real only while sin θ ≤ m₂/m₁, so a heavy projectile can never be turned through more than θₘₐₓ = arcsin(m₂/m₁): a 3.0 kg puck elastically striking a 1.0 kg puck at rest is deflected by at most 19.5°. Below that limit both roots are positive, a glancing and a fuller collision that send the projectile off at the same angle with different speeds. The honest statement is that the angle alone does not fix the outcome; the target's recoil direction, or either final speed, is needed to choose.