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How do elastic, inelastic and perfectly inelastic collisions differ, and how do we use momentum and energy conservation to analyze each?

Topic 4.4 Collisions: classify collisions as elastic, inelastic or perfectly inelastic, apply momentum conservation to all and kinetic-energy conservation to elastic collisions, in one and two dimensions.

A focused answer to AP Physics C: Mechanics Topic 4.4, covering the classification of collisions, momentum conservation in all collisions, kinetic-energy conservation only in elastic collisions, the combined-mass result for perfectly inelastic collisions, two-dimensional collisions by components, and the elastic one-dimensional relative-velocity result, with worked examples.

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  1. What this topic is asking
  2. Classifying collisions
  3. Perfectly inelastic collisions
  4. Elastic collisions
  5. Two-dimensional collisions
  6. Try this

What this topic is asking

The College Board (Topic 4.4) wants you to classify collisions as elastic, inelastic or perfectly inelastic, to apply momentum conservation to all of them and kinetic-energy conservation only to elastic ones, and to handle collisions in one and two dimensions. Collisions are the showcase application of Unit 4, combining the momentum tools with energy ideas, and they appear on essentially every AP Physics C exam.

Classifying collisions

The common thread is momentum conservation: the brief, large interaction forces are internal to the two-object system, so the total momentum is unchanged regardless of the collision type. What distinguishes the types is what happens to kinetic energy. Truly elastic collisions are idealisations (billiard balls and atoms come close); most real collisions are inelastic; cars that crumple together or balls of putty that stick are perfectly inelastic.

Perfectly inelastic collisions

When the objects stick and move together, there is a single final velocity. Momentum conservation alone determines it:

m1vβƒ—1+m2vβƒ—2=(m1+m2)vβƒ—f⟹vβƒ—f=m1vβƒ—1+m2vβƒ—2m1+m2.m_1\vec{v}_1 + m_2\vec{v}_2 = (m_1 + m_2)\vec{v}_f \quad\Longrightarrow\quad \vec{v}_f = \frac{m_1\vec{v}_1 + m_2\vec{v}_2}{m_1 + m_2}.

This is the simplest collision to analyze because the unknown is a single common velocity. The kinetic energy afterward is always less than before; the difference is the energy converted to heat and permanent deformation, which you find by computing the kinetic energy at the start and end and subtracting. The "lost" energy is real energy that has left the mechanical account, not energy that has vanished.

Elastic collisions

An elastic collision conserves both momentum and kinetic energy, giving two equations. For a one-dimensional elastic collision the algebra simplifies to a useful result: the relative velocity of approach equals the relative velocity of separation, reversed:

v1βˆ’v2=βˆ’(v1β€²βˆ’v2β€²).v_1 - v_2 = -(v_1' - v_2').

Pairing this with momentum conservation lets you solve for both final velocities without wrestling with the quadratic energy equation directly. Special cases are worth knowing: equal masses exchange velocities; a light object bouncing off a very heavy one reverses its velocity nearly unchanged in speed; a heavy object barely slows when it strikes a light one.

Two-dimensional collisions

When objects scatter in a plane (a glancing collision), momentum conservation applies componentwise: conserve pxp_x and pyp_y separately. Resolve each velocity into components, write a conservation equation for each axis, and solve them together with the energy condition if the collision is elastic. Two-dimensional collisions are common in particle physics and on the exam appear as pucks, balls or carts striking at an angle; the discipline of keeping the components separate is exactly as in the conservation topic.

Try this

Q1. A 0.200.20 kg ball at 4.04.0 m/s strikes an identical stationary ball elastically and head-on. State the velocities afterward. [2 points]

  • Cue. Equal masses exchange velocities: the first stops, the second moves off at 4.04.0 m/s.

Q2. A 3.03.0 kg lump of clay at 2.02.0 m/s hits and sticks to a 1.01.0 kg lump at rest. Calculate their common speed. [2 points]

  • Cue. vf=(3.0)(2.0)4.0=1.5v_f = \dfrac{(3.0)(2.0)}{4.0} = 1.5 m/s.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2024 (style)6 marksSection II (FRQ). A 2.02.0 kg cart moving at 3.03.0 m/s strikes a stationary 1.01.0 kg cart. (a) If they stick together, determine their common velocity and the kinetic energy lost. (b) If instead the collision is elastic, determine the velocity of each cart afterward. (c) State which type of collision conserves kinetic energy.
Show worked answer β†’

A 6-point collision FRQ comparing inelastic and elastic outcomes.

(a) Perfectly inelastic (3 points): momentum conservation m1v1=(m1+m2)vfm_1 v_1 = (m_1 + m_2)v_f, so vf=(2.0)(3.0)3.0=2.0v_f = \dfrac{(2.0)(3.0)}{3.0} = 2.0 m/s. Kinetic energy: before 12(2.0)(3.0)2=9.0\tfrac{1}{2}(2.0)(3.0)^2 = 9.0 J; after 12(3.0)(2.0)2=6.0\tfrac{1}{2}(3.0)(2.0)^2 = 6.0 J; lost =3.0= 3.0 J.
(b) Elastic (2 points): use momentum and the elastic relative-velocity result v1βˆ’v2=βˆ’(v1β€²βˆ’v2β€²)v_1 - v_2 = -(v_1' - v_2'). Solving, v1β€²=m1βˆ’m2m1+m2v1=1.03.0(3.0)=1.0v_1' = \dfrac{m_1 - m_2}{m_1 + m_2}v_1 = \dfrac{1.0}{3.0}(3.0) = 1.0 m/s; v2β€²=2m1m1+m2v1=4.03.0(3.0)β‹…11=4.0v_2' = \dfrac{2m_1}{m_1 + m_2}v_1 = \dfrac{4.0}{3.0}(3.0)\cdot\dfrac{1}{1} = 4.0 m/s. Check momentum: (2.0)(1.0)+(1.0)(4.0)=6.0(2.0)(1.0) + (1.0)(4.0) = 6.0 = before.
(c) Which conserves energy (1 point): the elastic collision conserves kinetic energy; the perfectly inelastic one does not.

Markers reward using the combined mass for the inelastic case and the relative-velocity reversal for the elastic case.

AP 2021 (style)1 marksSection I (multiple choice). In a perfectly inelastic collision between two objects, which quantity is always conserved? (A) kinetic energy only (B) momentum only (C) both momentum and kinetic energy (D) neither. Justify your reasoning.
Show worked answer β†’

A 1-point conceptual MCQ. The answer is (B).

In any collision with no net external force, momentum is conserved. A perfectly inelastic collision (the objects stick together) loses the maximum kinetic energy consistent with momentum conservation, converting it to heat and deformation, so kinetic energy is not conserved. Only the elastic collision conserves both. The trap (C) wrongly assumes energy is always conserved.

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