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GRE Physics overview

Public topic · GRE Physics

Energy and Momentum Conservation

A free GRE Physics mechanics note on the work-energy theorem, mechanical energy conservation, momentum conservation, and elastic versus inelastic collisions.

Concise answer

Track two separate bookkeeping systems: momentum, conserved whenever the net external force is zero, and mechanical energy, conserved only when no nonconservative force does work. Collisions almost always conserve the first and often not the second.

Definitions

Kinetic energy
The energy of motion, $\tfrac{1}{2}mv^2$ for a particle of mass $m$ and speed $v$.
Work-energy theorem
The statement that the net work done on a particle equals the change in its kinetic energy.
Mechanical energy
The sum of a system's kinetic energy and potential energy.
Linear momentum
The product of an object's mass and velocity, $\vec{p} = m\vec{v}$; a vector quantity.
Elastic collision
A collision in which both total momentum and total kinetic energy are conserved.
Perfectly inelastic collision
A collision in which the objects stick together afterward; momentum is conserved but the maximum possible kinetic energy is lost.

Intuition

Momentum conservation comes from Newton's third law: internal forces between colliding objects cancel in pairs, so they cannot change the total momentum of the pair. Only external forces can.

Energy is a scalar and momentum is a vector, so they carry independent information. A system can lose kinetic energy in a collision while its total momentum stays exactly the same.

Concept walkthrough

The work-energy theorem says the net work done on a particle equals its change in kinetic energy. When only conservative forces such as gravity or an ideal spring do work, that work can be folded into a potential energy, and the total mechanical energy K+UK + U stays constant. Friction and other nonconservative forces break this bookkeeping and must be added as separate work terms.

Momentum conservation is a different and more robust statement: if the net external force on a system is zero, the system's total momentum cannot change, no matter how complicated the internal forces are. During a brief collision, external forces are usually negligible compared with the large internal impact forces, so momentum is conserved across the collision even when kinetic energy is not.

GRE Physics collision questions usually turn on knowing which quantity to conserve. Elastic collisions conserve both momentum and kinetic energy; inelastic collisions conserve only momentum; perfectly inelastic collisions additionally share one final velocity because the objects stick together.

After this page, you should be able to

  • Apply the work-energy theorem to relate net work to a change in speed.
  • Decide when mechanical energy is conserved and when nonconservative work must be accounted for.
  • Write momentum conservation as a vector equation for an isolated system.
  • Classify a collision as elastic, inelastic, or perfectly inelastic and choose the right conserved quantities.

Formulas and assumptions

Work-energy theorem

Wnet=12mvf2−12mvi2W_{\text{net}} = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2

Variables

  • W_net: net work in joules
  • m: mass in kilograms
  • v_i, v_f: initial and final speeds in meters per second

Assumptions

  • The object is treated as a particle.
  • W_net is the work done by the net force, summed over all forces.

Conservation of mechanical energy

K1+U1=K2+U2K_1 + U_1 = K_2 + U_2

Variables

  • K: kinetic energy in joules
  • U: total potential energy in joules
  • subscripts 1, 2: any two moments

Assumptions

  • Only conservative forces do work.
  • If friction or another nonconservative force acts, its work appears as an extra term instead.

Linear momentum

p⃗=mv⃗\vec{p} = m\vec{v}

Variables

  • p: momentum in kilogram meters per second
  • m: mass in kilograms
  • v: velocity in meters per second

Assumptions

  • Speeds are nonrelativistic.
  • Momentum is a vector; components are tracked separately.

Momentum conservation for an isolated system

∑p⃗before=∑p⃗after\sum \vec{p}_{\text{before}} = \sum \vec{p}_{\text{after}}

Variables

  • sum p: vector sum of the momenta of every object in the system

Assumptions

  • The net external force on the system is zero, or the interaction is brief enough that external impulses are negligible.

Perfectly inelastic collision

m1v⃗1+m2v⃗2=(m1+m2)v⃗fm_1\vec{v}_1 + m_2\vec{v}_2 = (m_1 + m_2)\vec{v}_f

Variables

  • m1, m2: colliding masses
  • v1, v2: velocities before the collision
  • v_f: shared velocity after the collision

Assumptions

  • The objects stick together and move with one final velocity.
  • External impulses during the collision are negligible.

Worked example

Carts that stick together

A 3.0 kg3.0\,\text{kg} cart moving at 4.0 m/s4.0\,\text{m/s} to the right collides with a stationary 1.0 kg1.0\,\text{kg} cart and the two couple together. Find their common final speed and the kinetic energy lost.

  1. 1Take the two carts as the system; during the brief collision, external impulses are negligible, so momentum is conserved.
  2. 2Initial momentum: pi=(3.0 kg)(4.0 m/s)+0=12 kg⋅m/sp_i = (3.0\,\text{kg})(4.0\,\text{m/s}) + 0 = 12\,\text{kg}\cdot\text{m/s}.
  3. 3Set pi=(m1+m2)vfp_i = (m_1 + m_2)v_f: 12=(4.0 kg) vf12 = (4.0\,\text{kg})\,v_f, so vf=3.0 m/sv_f = 3.0\,\text{m/s}.
  4. 4Kinetic energy before: 12(3.0)(4.0)2=24 J\tfrac{1}{2}(3.0)(4.0)^2 = 24\,\text{J}. After: 12(4.0)(3.0)2=18 J\tfrac{1}{2}(4.0)(3.0)^2 = 18\,\text{J}.
  5. 5The collision converts 24−18=6.0 J24 - 18 = 6.0\,\text{J} of kinetic energy into other forms.

The coupled carts move at 3.0 m/s3.0\,\text{m/s} to the right, and 6.0 J6.0\,\text{J} of kinetic energy is lost.

Common traps

  • Assuming kinetic energy is conserved in every collision; only elastic collisions conserve it.
  • Conserving speed or kinetic energy componentwise instead of momentum; momentum is the vector quantity that adds by components.
  • Applying mechanical energy conservation across a stretch where friction acts without adding the friction work term.
  • Dropping signs on momenta in one dimension; opposite directions must enter with opposite signs.
  • Using the elastic-collision shortcut formulas on a problem that never says the collision is elastic.

Related pages and practice

GRE Physics overviewNewton's Second Law and Free-Body DiagramsSimple Harmonic MotionSign in for current practice

Question depth and domain coverage vary by exam. Practice answers are checked after submission.

Sources

  1. GRE Subject Test Content and Structure — ETS. Accessed 2026-07-06. Use as a cited source for exam facts; do not imply affiliation or reproduce protected test material.
  2. University Physics Volume 1, Section 7.3: Work-Energy Theorem — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
  3. University Physics Volume 1, Section 8.3: Conservation of Energy — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
  4. University Physics Volume 1, Section 9.3: Conservation of Linear Momentum — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
  5. University Physics Volume 1, Section 9.4: Types of Collisions — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.

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Recheck ETS content areas and the OpenStax energy and momentum references before each major GRE Physics preparation cycle.

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