A collision is an event in which two bodies exert mutual forces on each other over a short time interval. Stages: (1) Approach — bodies move toward each other; (2) Deformation — bodies deform at contact, storing elastic PE; (3) Recovery — deformation energy releases bodies apart.

• In a collision, the time of contact is very short — external forces (gravity, friction) are negligible during collision, so impulse from external forces ≈ 0.
• Because external impulse ≈ 0 during collision, the total momentum of the system is conserved.
• These laws are the fundamental laws of physics and applicable for any type of collision but this is not true for conservation of kinetic energy.

Momentum is always conserved in all collisions (m₁u₁+m₂u₂ = m₁v₁+m₂v₂). KE is conserved only in perfectly elastic collisions (e=1). In a collision 'total energy' is also always conserved — but total mechanical energy is not (some converts to heat/deformation).

• Momentum conservation: applies to all collisions as long as no external forces act during the collision interval.
• Total energy (including heat, sound, deformation) is conserved in all collisions — only the form changes.
• KE is conserved only in elastic collisions; in inelastic collisions, KE is partially lost (converted to heat/deformation energy).

Perfectly elastic: e=1, KE conserved; examples: atomic collisions, billiard balls. Inelastic: 0

• Coefficient of restitution e = (velocity of separation after collision)/(velocity of approach before collision) = (v₂−v₁)/(u₁−u₂). Range: 0 ≤ e ≤ 1.
• All real macroscopic collisions are partially inelastic (0 < e < 1 or e = 0) — perfectly elastic collisions occur at the atomic/sub-atomic scale.
• Common NEET trap: assuming that inelastic means perfectly inelastic (bodies stick together) — inelastic means only that KE is NOT conserved; bodies can still separate.

v₁ = [(m₁−m₂)/(m₁+m₂)]u₁ + [2m₂/(m₁+m₂)]u₂; v₂ = [2m₁/(m₁+m₂)]u₁ + [(m₂−m₁)/(m₁+m₂)]u₂. Derived from: momentum conservation + e = 1 (Newton's law of impact). Relative velocity of separation is equal to relative velocity of approach.

• Equal masses (m₁ = m₂): v₁ = u₂ and v₂ = u₁ — velocities exchange completely.
• Heavy ball (m₁ >> m₂) hits stationary light ball (u₂=0): v₁ ≈ u₁, v₂ ≈ 2u₁ (light ball moves forward at twice heavy ball's speed).
• Light ball (m₁ << m₂) hits heavy stationary ball (u₂=0): v₁ ≈ −u₁ (reverses), v₂ ≈ 0 (heavy ball barely moves).

Fraction of KE transferred from m₁ to m₂: ΔKE/KE₁ = 4m₁m₂/(m₁+m₂)². Maximum KE transfer occurs when m₁ = m₂ (100% transfer). For m₁ << m₂ or m₁ >> m₂, KE transfer fraction → 0.

• Use ΔKE/KE₁ = 4m₁m₂/(m₁+m₂)² to compute the fraction directly — NEET often asks 'what fraction of initial KE is transferred to target'.
• Maximum KE transfer (= 100%) when m₁ = m₂; for equal masses, the first ball completely stops and the second ball moves with the initial speed of the first.
• Common NEET application: a neutron (m₁ ≈ 1 amu) colliding with a hydrogen nucleus (m₂ ≈ 1 amu) — maximum KE transfer → maximum neutron slowdown (relevant in nuclear physics context).

For m₂ initially at rest (u₂=0), elastic collision: v₂ = 2m₁u₁/(m₁+m₂). KE of target after collision: KE₂ = 2m₁m₂u₁²/(m₁+m₂)² × 2 = 4m₁m₂KE₁/(m₁+m₂)². This is the fraction ΔKE/KE₁.

• Target acquires maximum velocity when m₁ >> m₂: v₂ → 2u₁ (target moves at double the projectile's initial speed).
• For equal masses, v₂ = u₁ (target acquires the full initial velocity of the projectile).
• These results are derived by substituting u₂ = 0 into the general elastic head-on formulae — always verify by checking momentum and KE conservation.

In oblique elastic collision, resolve velocities into components parallel and perpendicular to the line joining the centres of the two balls at contact. Only the component along the line of impact undergoes elastic collision (exchange/formula); the perpendicular components are unchanged.

• The line joining the centres of the two balls at the moment of contact is called the 'line of impact'.
• Perpendicular to line of impact: no force acts, so each ball's velocity component perpendicular to line of impact is unchanged after collision.
• Along line of impact: use elastic head-on collision formulae (treating these components as u₁ and u₂) to find the post-collision components along the line of impact.

For head-on inelastic collision with coefficient of restitution e (0 < e < 1): use momentum conservation (m₁u₁+m₂u₂ = m₁v₁+m₂v₂) and Newton's law of impact (v₂−v₁ = e(u₁−u₂)) simultaneously to find v₁ and v₂.

• Two equations (momentum + restitution) for two unknowns (v₁, v₂) — always fully determined; no need for energy equation.
• KE loss = ½m₁m₂(u₁−u₂)²(1−e²)/(m₁+m₂) — for elastic e=1: KE loss = 0; for perfectly inelastic e=0: maximum KE loss.
• Common NEET trap: applying elastic formulae (v₁, v₂ derivation using KE conservation) when e < 1 — always use the two-equation system (momentum + restitution).

A ball dropped from height H and rebounding to height h: e = sqrt(h/H). After n bounces, rebound height hₙ = e^(2n) × H. Velocity just before impact: v₁ = sqrt(2gH); just after: v₂ = e × v₁.

• Ground is treated as infinitely massive — the ground's velocity does not change. So Newton's law: e = v_after/v_before = v₂/v₁.
• After n bounces: height = H × e^(2n); velocity after nth bounce = e^n × sqrt(2gH).
• Common NEET trap: using e = h/H instead of e = sqrt(h/H) — the ratio h/H = e² (ratio of heights), not e directly.

Perfectly inelastic: e=0, bodies move together after collision. Common velocity: V = (m₁u₁+m₂u₂)/(m₁+m₂). KE loss = ½m₁m₂(u₁−u₂)²/(m₁+m₂). This is the maximum possible KE loss for a given pair of colliding bodies.

• KE loss = ½ × (reduced mass) × (relative velocity)² where reduced mass μ = m₁m₂/(m₁+m₂).
• Common velocity V = (m₁u₁+m₂u₂)/(m₁+m₂) — this is the velocity of the centre of mass, which is the same before and after collision (momentum conservation).
• Common NEET trap: computing KE using V in the formula ½(m₁+m₂)V² and comparing with initial KE to check — KE is NOT conserved but must be less than initial KE.

Bullet (mass m, velocity u) embeds in block (mass M at rest). Phase 1 (collision): momentum: mu = (m+M)V. Phase 2 (swing): energy: ½(m+M)V² = (m+M)gh → V = sqrt(2gh). Bullet speed: u = (m+M)/m × sqrt(2gh).

• Do NOT mix the two phases — energy is NOT conserved during the collision phase (KE is lost to bullet deformation and heat).
• The height h of swing is measurable experimentally, making this the 'ballistic pendulum' method for measuring bullet speed.
• The fraction of KE lost in the collision = M/(m+M) — a large block relative to bullet means most energy is lost.