# Collisions in One and Two Dimensions

> Collisions in one and two dimensions for NEET Physics: momentum conservation, coefficient of restitution, elastic equal-mass swap and oblique impacts.

- Canonical URL: https://prepelephant.com/topics/neet-ug/physics/collisions-one-two-dimension-neet
- Exam / course: NEET-UG · Subject: Physics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Collisions in One and Two Dimensions", PrepElephant, https://prepelephant.com/topics/neet-ug/physics/collisions-one-two-dimension-neet

## Direct answer

In every collision — elastic, inelastic or perfectly inelastic — total linear momentum of the colliding system is conserved, because the collision forces are internal and equal-and-opposite. Kinetic energy survives only the elastic case; the coefficient of restitution e = (relative velocity of separation)/(relative velocity of approach) separates the regimes, with e = 1 elastic, 0 < e < 1 inelastic, and e = 0 perfectly inelastic (bodies stick and move together). In one dimension, an elastic collision between equal masses simply exchanges velocities, and for unequal masses the final velocities follow v_1 = [(m_1 − m_2)/(m_1 + m_2)]u_1 and v_2 = [2m_1/(m_1 + m_2)]u_1. In two dimensions, momentum conservation along two perpendicular axes gives two equations, which is all that is needed for the standard oblique-collision questions NEET asks.

## What you must remember

- **Always-conserved list:** momentum and total energy (including heat) in all collisions; kinetic energy only when e = 1; never assume KE conservation without the word "elastic" or "perfectly elastic" in the question.
- **Restitution in 1D:** v_2 − v_1 = e(u_1 − u_2); pair this with momentum conservation and any 1D collision is two linear equations.
- **Elastic equal-mass swap:** identical masses in a head-on elastic collision exchange velocities; a moving billiard ball stops dead after hitting a stationary identical one.
- **Special limits:** if m_2 >> m_1 (ball on wall), v_1 rebounds with −u_1 and momentum change is 2m_1u_1; if m_2 << m_1, the heavy body is unaffected (v_1 ≈ u_1) and the light one shoots off near 2u_1.
- **Perfectly inelastic:** common velocity V = (m_1u_1 + m_2u_2)/(m_1 + m_2), and the KE lost is ½ μ (u_1 − u_2)^2 where μ = m_1m_2/(m_1 + m_2) is the reduced mass — a compact formula worth knowing.
- **Oblique elastic collision with equal masses:** the two velocities after impact are perpendicular (90° rule), the single most reused 2D fact; it fails if the masses differ.
- **Ball bouncing off a floor:** e = √(h_2/h_1), heights of bounce after and before; energy ratio equals e^2.

## A worked one-dimensional collision

A 4 kg block at 5 m/s collides with a stationary 2 kg block; the collision is elastic, so both momentum and KE survive. The brute-force route is two equations: momentum gives 20 = 4v_1 + 2v_2, and kinetic energy gives 50 = 2v_1^2 + v_2^2. Substituting v_2 = 10 − 2v_1 from the first into the second yields 3v_1^2 − 20v_1 + 25 = 0, whose roots are v_1 = 5 (the pre-collision state, to be discarded) and v_1 = 5/3 ≈ 1.67 m/s, with v_2 = 10 − 2(1.67) = 6.67 m/s. The memorised formula reaches the same place without the quadratic: v_1 = [(m_1 − m_2)/(m_1 + m_2)]u_1 = (2/6)(5) = 1.67 m/s and v_2 = [2m_1/(m_1 + m_2)]u_1 = (8/6)(5) = 6.67 m/s. Two lessons travel with this number. First, elastic-collision quadratics always carry the trivial pre-collision root, and picking the wrong root under exam pressure is the standard error — the formula sidesteps it. Second, always verify with momentum: 4(1.67) + 2(6.67) = 20 exactly, a check costing five seconds.

## Where students slip

The deadliest slip is conserving kinetic energy in a question that never said "elastic" — NEET wording "the two blocks move together afterwards" signals perfectly inelastic, where the maximum possible KE is lost. The second is sign chaos in the restitution equation: relative velocities must be taken in a consistent order (separation = v_2 − v_1, approach = u_1 − u_2 for the same labelling), and mixing labels flips e's sign. In 2D, students forget that the y-momentum equation exists; writing only x-conservation and hoping is half a solution, since one equation cannot yield two unknown velocity components. Finally, remember the 90° perpendicularity rule after an oblique elastic collision holds only for equal masses — applying it to unequal masses is a planted distractor in options.

## Frequently asked questions

### Which quantities are conserved in all collisions?

Linear momentum and total energy are always conserved; kinetic energy is conserved only in perfectly elastic collisions (e = 1).

### What happens when two equal masses collide elastically in one dimension?

They exchange velocities: the moving one stops (if the other was at rest) and the struck one moves off with the original speed.

### How is the coefficient of restitution used in numericals?

Combine v_2 − v_1 = e(u_1 − u_2) with momentum conservation; e = 0 makes the bodies stick, e = 1 recovers the elastic case, and intermediate e gives inelastic finals.

### Why do the final velocities after an oblique elastic collision of identical spheres lie at 90 degrees?

Conservation of both momentum and kinetic energy for equal masses forces the velocity vectors' dot product to vanish, making them mutually perpendicular.

### How much kinetic energy is lost in a perfectly inelastic collision?

ΔK = ½ μ (u_1 − u_2)^2 with reduced mass μ = m_1m_2/(m_1 + m_2); the loss depends only on the relative speed and the masses, not on which body is heavier.