Conservation of Momentum

In three steps: first, fix a positive direction and write all velocities with signs. Second, form the total momentum before the collision: p_before = m₁v₁ + m₂v₂. Third, set p_before = p_after and solve for the unknown. In a perfectly inelastic collision (the bodies stick together) there is only one final velocity: v' = (m₁v₁ + m₂v₂)/(m₁ + m₂). Example: a cart (2 kg, 3 m/s) couples to a cart at rest (1 kg): v' = 6/3 = 2 m/s. In an elastic collision kinetic energy is conserved as well; then you need both equations to determine the two unknown final velocities.

Momentum is conserved in both cases; that is the common core. The difference lies in kinetic energy: in an (ideally) elastic collision it is fully conserved and the bodies bounce off each other, like billiard balls or gas molecules approximately do. In an inelastic collision part of it converts into deformation, heat and sound; in a perfectly inelastic collision the bodies stick together afterwards and move as one. Car crashes are deliberately highly inelastic: the crumple zone is meant to put as much kinetic energy into deformation as possible so it does not reach the occupants. In problems the phrase "stick together" or "couple" always signals the perfectly inelastic case.

Momentum conservation follows directly from Newton third law: during the collision the bodies act on each other with exactly opposite, equal forces (action equals reaction). Their momentum changes are therefore opposite and equal and cancel in the sum, no matter how complicated the collision is in detail. Kinetic energy has no such protection: it can convert into other forms such as deformation work, heat or sound, and in every real collision it partly does. The energy is not lost, it merely leaves the mechanical motion. Hence: momentum conservation always (in a closed system), conservation of kinetic energy only in the elastic ideal case.

Before firing, the system is at rest and the total momentum is zero, and zero it must remain. What flies forward pushes the rest backward: 0 = m₁v₁' + m₂v₂', so v₂' = −(m₁/m₂)·v₁'. Example: a cannon (200 kg) fires a ball (2 kg) at 100 m/s. The recoil is v₂' = −(2/200)·100 = −1 m/s; the cannon rolls back at 1 m/s. The mass ratio decides everything: the heavier the cannon, the smaller its recoil. A rocket works on the same principle continuously: it constantly expels gas backward at high speed and thereby gains forward momentum, entirely without air to "push against".

The law only holds when no external forces act on the system, or when the collision is so brief that external forces transfer negligible momentum during it. A car crashing into a firmly anchored wall is the classic counterexample: the wall is connected to the Earth, the "missing" momentum goes to the Earth, and the car system alone does not keep it. Long processes with heavy friction (a rolling cart coasting to a stop) also distort the balance through the external friction force. The way out: choose the system large enough (car + Earth) or set up the balance only over the extremely short collision time. In exams: collision problems yes, prolonged friction processes no.