Classical Mechanics Domain Guide

Concise answer

Mechanics is three tools applied to one system. Newton's laws answer 'what is the acceleration right now'; the work-energy theorem and energy conservation answer 'what is the speed after this displacement'; momentum and angular momentum conservation answer 'what happened across this short, violent interaction'. Nearly every mechanics item on this exam is decided by choosing the right tool in the first ten seconds, because each tool converts the question into arithmetic you already know, and the wrong tool converts it into an integral you have no time for.

Definitions

Inertial frame: A frame in which a body with no net force on it moves at constant velocity. Newton's second law in the form $\vec{F}_{\text{net}} = m\vec{a}$ holds only here; in an accelerating frame you must add fictitious forces such as $-m\vec{a}_{\text{frame}}$ or the centrifugal and Coriolis terms.
Conservative force: A force whose work between two points is independent of the path, so it can be written as $\vec{F} = -\nabla U$ . Gravity and ideal springs qualify; kinetic friction and air drag do not, which is exactly why energy 'goes missing' when they act.
Impulse: $\vec{J} = \int \vec{F}\,dt = \Delta \vec{p}$ . It is the bridge between a force acting for a short time and a finite change in momentum, and it is why a collision can be analysed without knowing the force law inside the collision.
Moment of inertia: $I = \sum m_i r_i^{2}$ , the rotational analogue of mass, measured about a stated axis. For rigid bodies of mass $M$ and radius $R$ it is written $I = \beta M R^{2}$ with $\beta = 2/5$ (solid sphere), $1/2$ (solid cylinder or disc), $2/3$ (thin spherical shell), and $1$ (hoop or thin cylindrical shell).
Angular momentum: $\vec{L} = \vec{r} \times \vec{p}$ for a particle and $\vec{L} = I\vec{\omega}$ for a rigid body rotating about a symmetry axis. It is conserved when the net external torque about the chosen origin is zero — a condition that can hold even when linear momentum is not conserved.
Effective potential: $U_{\text{eff}}(r) = U(r) + \dfrac{L^{2}}{2\mu r^{2}}$ for central-force motion. Writing the orbit problem this way turns a two-dimensional problem into one-dimensional motion in a potential well, so circular orbits are its minima and bound orbits are the region where $E < 0$ .

Intuition

The exam almost never asks you to integrate Newton's second law. It asks for a quantity at one instant (use forces), a speed after a displacement (use energy), or a velocity after an interaction (use momentum). Reading the question for which of those three it is, before writing anything, is the single largest source of time saved in this domain — and time is the binding constraint, not knowledge.

Conservation laws are permissions, not facts about the universe as stated on the page. Momentum is conserved when no net external force acts on the system you drew; energy is conserved when no non-conservative force does work on it; angular momentum is conserved when no net torque acts about the origin you chose. Every one of these can be made true or false by redrawing the system boundary, which is why naming the system first is not a formality.

Rotation is not a new subject; it is the same subject with $m \to I$ , $\vec{F} \to \vec{\tau}$ , $\vec{p} \to \vec{L}$ , and $v \to \omega$ . A body that rolls has both stores of kinetic energy at once, and the fixed ratio between them — set entirely by $\beta$ — is why a hollow object always loses a race to a solid one of the same shape, regardless of mass or radius.

Concept walkthrough

Start where the exam starts. ETS lists classical mechanics as the largest content area, and our learning model groups Newtonian dynamics, energy and momentum, rotation, oscillations, gravitation, and fluids underneath it — that grouping is our own reading of the outline, not an official sub-division. Because it is the largest block, mechanics is also where the exam most reliably rewards procedure over cleverness: a fixed opening move that works on twenty percent of the paper is worth more than a brilliant technique that works on one item.

The opening move is always the same. Name the system. Draw only the forces external to it. Choose axes along the motion, not along the page — on an incline that means one axis down the slope, which makes the normal force $N = mg\cos\theta$ and the driving component $mg\sin\theta$ fall out with no trigonometry left over. Then apply $\vec{F}_{\text{net}} = m\vec{a}$ component by component. Two constraints do most of the remaining work: a taut inextensible string gives every connected body the same magnitude of acceleration, and a surface in contact gives $a_{\perp} = 0$ , which is what determines $N$ rather than any separate rule. Note that the normal force is not a constant $mg$ waiting to be recalled: in a lift accelerating upward it is $m(g+a)$ , and at the top of a loop it can be zero, which is exactly the condition that sets the minimum speed there, $v_{\min} = \sqrt{gR}$ .

Energy takes over the moment the question mentions a displacement, a height, a spring compression, or a speed 'at the bottom'. The work-energy theorem says the net work equals the change in kinetic energy, and when only conservative forces do work this becomes $K_i + U_i = K_f + U_f$ . The practical skill is bookkeeping non-conservative work honestly: friction over a distance $d$ removes $\mu_k N d$ , and that term belongs on the correct side of the equation. Springs contribute $U = \tfrac{1}{2}kx^{2}$ measured from the natural length, and gravity contributes $mgh$ only in the uniform-field approximation — near a planet, use $U = -GMm/r$ instead, where the zero is at infinity and every bound state therefore has negative total energy.

Momentum answers questions about interactions too fast to model. During a collision the internal forces are enormous and the external ones (gravity, friction) are comparatively negligible, so the total momentum of the colliding bodies is conserved even though nothing else is. Kinetic energy is a separate question: it is conserved only in an elastic collision. Two facts make one-dimensional collisions almost instant. First, for a perfectly inelastic collision the bodies leave together, so $v_f = (m_1 u_1 + m_2 u_2)/(m_1 + m_2)$ and the lost kinetic energy is $\tfrac{1}{2}\mu(u_1-u_2)^{2}$ with reduced mass $\mu = m_1 m_2/(m_1+m_2)$ . Second, for a perfectly elastic collision the relative velocity of approach equals the relative velocity of separation, $u_1 - u_2 = -(v_1 - v_2)$ — a linear relation that replaces the quadratic energy equation and removes the algebra entirely.

Rotation reuses all of it. Torque $\tau = rF\sin\theta$ drives angular acceleration through $\tau = I\alpha$ ; kinetic energy gains a term $\tfrac{1}{2}I\omega^{2}$ ; angular momentum $L = I\omega$ is conserved under zero net torque, which is why a skater speeds up when she pulls her arms in and why her kinetic energy rises rather than staying fixed — she did work. Rolling without slipping is the single constraint $v = \omega R$ , and it has two consequences students routinely miss: the contact point is instantaneously at rest, and the static friction that enables rolling does no work, so energy conservation is still available. The parallel-axis theorem $I = I_{\text{cm}} + Md^{2}$ handles any axis that is not through the centre of mass, and forgetting the $Md^{2}$ term is the most common rotational error on timed tests.

Oscillations and gravitation close the domain, and both reward recognition over derivation. Any system whose restoring force is linear in the displacement oscillates simple-harmonically with $\omega = \sqrt{\text{stiffness}/\text{inertia}}$ : $\sqrt{k/m}$ for a spring, $\sqrt{g/L}$ for a simple pendulum in the small-angle limit, $\sqrt{mgd/I}$ for a physical pendulum. The period is independent of amplitude, which is precisely the property that fails once the amplitude is large enough that $\sin\theta \neq \theta$ . In gravitation, circular orbits satisfy $GMm/r^{2} = mv^{2}/r$ , giving $v = \sqrt{GM/r}$ and total energy $E = -GMm/2r$ — half the potential energy, and negative, as any bound state must be. Kepler's third law, $T^{2} \propto a^{3}$ , then answers ratio questions in one line. Fluids follow the same static-then-dynamic split: $P = P_0 + \rho g h$ and Archimedes' $F_B = \rho_{\text{fluid}} V_{\text{displaced}} g$ for statics, continuity $A_1 v_1 = A_2 v_2$ and Bernoulli's $P + \tfrac{1}{2}\rho v^{2} + \rho g h = \text{constant}$ for steady, incompressible, non-viscous flow along a streamline.

After this page, you should be able to

Choose between a force analysis, an energy analysis, and a momentum analysis from the wording of the question rather than by trial.
Draw a free-body diagram that shows only external forces on a clearly named system, and read the constraint (incline, pulley, contact) off the geometry.
Apply momentum conservation across collisions and energy conservation between configurations, and state out loud which one fails and why in an inelastic collision.
Compute rotational quantities with $I = \beta M R^{2}$ and the parallel-axis theorem, and handle rolling without slipping through the single constraint $v = \omega R$ .
Recognise any oscillator by linearising its restoring force, then read the angular frequency straight off $\omega = \sqrt{\text{stiffness}/\text{inertia}}$ .
Use $U = -GMm/r$ , the effective potential, and Kepler's third law to answer orbital questions without integrating an equation of motion.
Apply hydrostatic pressure, buoyancy, continuity, and Bernoulli's equation, and state the conditions under which Bernoulli is legal.

Formulas and assumptions

Newton's second law and the constraint forces it determines

\sum \vec{F}_{\text{ext}} = m\vec{a}, \qquad N = mg\cos\theta, \qquad F_{\parallel} = mg\sin\theta, \qquad N_{\text{lift}} = m(g + a)

Variables

F_ext: forces external to the chosen system
N: normal force, a response force with no formula of its own
theta: angle of the incline from the horizontal

Assumptions

The frame is inertial; in an accelerating frame add the fictitious force -m a_frame.
The normal force is whatever the perpendicular equation requires, so it equals mg only on a level surface with no vertical acceleration.

Work-energy theorem and mechanical energy

W_{\text{net}} = \Delta K, \qquad K_i + U_i = K_f + U_f, \qquad W_{\text{friction}} = -\mu_k N d

Variables

K: kinetic energy
U: potential energy, mgh near the surface or -GMm/r in general
mu_k: coefficient of kinetic friction

Assumptions

Mechanical energy conservation requires that no non-conservative force does work on the system.
Static friction in rolling without slipping does no work, so energy conservation remains available there.

Collisions: momentum always, kinetic energy only if elastic

m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2, \qquad v_f = \frac{m_1u_1 + m_2u_2}{m_1+m_2}, \qquad \Delta K = -\tfrac{1}{2}\mu(u_1-u_2)^{2}, \qquad u_1 - u_2 = -(v_1 - v_2)

Variables

u: velocity before the collision
v: velocity after the collision
mu: reduced mass m1 m2 / (m1 + m2)

Assumptions

Momentum conservation needs the net external impulse on the pair to be negligible during the contact time.
The relative-velocity relation holds only for a perfectly elastic collision and replaces the quadratic energy equation.

Rotation, rolling, and the parallel-axis theorem

\tau = I\alpha, \quad L = I\omega, \quad K_{\text{rot}} = \tfrac{1}{2}I\omega^{2}, \quad I = \beta MR^{2}, \quad v = \omega R, \quad I = I_{\text{cm}} + Md^{2}

Variables

I: moment of inertia about the stated axis
beta: shape factor, dimensionless
d: distance from the centre-of-mass axis to the parallel axis

Assumptions

L = I omega applies to rotation about a symmetry axis; in general L and omega need not be parallel.
The parallel-axis theorem requires the reference axis to pass through the centre of mass.

Simple harmonic motion in every disguise

\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{g}{L}} = \sqrt{\frac{Mgd}{I}}, \qquad T = \frac{2\pi}{\omega}, \qquad x = A\cos(\omega t + \phi), \qquad E = \tfrac{1}{2}kA^{2}

Variables

A: amplitude
d: distance from the pivot to the centre of mass (physical pendulum)
phi: phase set by the initial conditions

Assumptions

The restoring force must be linear in the displacement; a pendulum qualifies only for small angles.
The period is independent of amplitude, which is what makes SHM identifiable from a single sentence.

Damped and driven oscillation

m\ddot{x} + b\dot{x} + kx = 0, \qquad A(t) = A_0 e^{-bt/2m}, \qquad \omega' = \sqrt{\omega_0^{2} - \left(\frac{b}{2m}\right)^{2}}

Variables

b: damping coefficient
omega0: undamped angular frequency sqrt(k/m)
Q: quality factor, roughly the number of radians before the energy falls by a factor e

Assumptions

Underdamped motion requires b < 2 sqrt(mk); at equality the system is critically damped and does not oscillate.
Damping always lowers the oscillation frequency below the undamped value, never raises it.

Gravitation and orbits

F = \frac{GMm}{r^{2}}, \quad U = -\frac{GMm}{r}, \quad v_{\text{orb}} = \sqrt{\frac{GM}{r}}, \quad E = -\frac{GMm}{2r}, \quad v_{\text{esc}} = \sqrt{\frac{2GM}{R}}, \quad T^{2} \propto a^{3}

Variables

G: gravitational constant, 6.674 x 10^-11 N m^2 / kg^2
a: semi-major axis of the orbit
E: total mechanical energy, negative for any bound orbit

Assumptions

U = -GMm/r takes the zero of potential energy at infinite separation; mgh is only its local linearisation.
Escape speed is independent of the escaping body's mass and of the launch direction, ignoring atmosphere and rotation.

Fluid statics and steady flow

P = P_0 + \rho g h, \quad F_B = \rho_{\text{fluid}} V_{\text{disp}} g, \quad A_1v_1 = A_2v_2, \quad P + \tfrac{1}{2}\rho v^{2} + \rho gh = \text{constant}

Variables

rho: density of the fluid
V_disp: volume of fluid displaced, not the volume of the object unless it is fully submerged
P0: pressure at the reference depth

Assumptions

Bernoulli's equation requires steady, incompressible, non-viscous flow evaluated along one streamline.
Continuity forces a narrower pipe to carry faster flow, which by Bernoulli means lower pressure — the two equations must be used together.

Worked example

The rolling race, and why mass and radius cancel

A solid sphere, a solid disc, and a hoop — each of mass $M$ and radius $R$ — are released from rest and roll without slipping down an incline of height $h$ and angle $\theta$ . Rank their speeds at the bottom, give each speed, find the acceleration of the sphere, and find the smallest coefficient of static friction that keeps the sphere rolling.

1Choose the tool. The question asks for a speed after a displacement with no non-conservative work — static friction in rolling without slipping acts at a point that is instantaneously at rest, so it does no work. That is an energy question, not a force question.
2Write the energy statement with both kinetic stores. $Mgh = \tfrac{1}{2}Mv^{2} + \tfrac{1}{2}I\omega^{2}$ . Apply the rolling constraint $\omega = v/R$ and write $I = \beta MR^{2}$ , so $\tfrac{1}{2}I\omega^{2} = \tfrac{1}{2}\beta Mv^{2}$ .
3Solve. $Mgh = \tfrac{1}{2}(1+\beta)Mv^{2}$ , so $v = \sqrt{\dfrac{2gh}{1+\beta}}$ . Both $M$ and $R$ have cancelled — the result depends on the shape factor alone, which is why the race has the same winner for a marble and a cannonball.
4Evaluate for each shape. Sphere, $\beta = 2/5$ : $v = \sqrt{10gh/7} \approx 1.195\sqrt{gh}$ . Disc, $\beta = 1/2$ : $v = \sqrt{4gh/3} \approx 1.155\sqrt{gh}$ . Hoop, $\beta = 1$ : $v = \sqrt{gh}$ . The ranking is sphere, then disc, then hoop — smallest $\beta$ wins, because it stores the least energy in rotation and leaves the most for translation.
5Get the acceleration by differentiating the same relation along the slope, or directly: with $x$ measured down the slope, $v^{2} = 2ax$ and $h = x\sin\theta$ give $a = \dfrac{g\sin\theta}{1+\beta}$ . For the sphere this is $\tfrac{5}{7}g\sin\theta$ ; for the disc $\tfrac{2}{3}g\sin\theta$ ; for the hoop $\tfrac{1}{2}g\sin\theta$ . A frictionless block sliding down the same incline would have $g\sin\theta$ — every rolling body is slower, and none of them is slower because of energy 'lost'.
6Find the friction requirement. Torque about the centre of mass comes only from friction: $fR = I\alpha = \beta MR^{2}(a/R)$ , so $f = \beta Ma$ . Substituting $a$ gives $f = \dfrac{\beta}{1+\beta}Mg\sin\theta$ . Rolling persists while $f \le \mu_s N = \mu_s Mg\cos\theta$ , so $\mu_s \ge \dfrac{\beta}{1+\beta}\tan\theta$ . For the sphere, $\beta/(1+\beta) = (2/5)/(7/5) = 2/7$ , giving $\mu_s \ge \tfrac{2}{7}\tan\theta$ .

Sphere $v = \sqrt{10gh/7} \approx 1.195\sqrt{gh}$ , disc $v = \sqrt{4gh/3} \approx 1.155\sqrt{gh}$ , hoop $v = \sqrt{gh}$ — independent of $M$ and $R$ . The sphere accelerates at $\tfrac{5}{7}g\sin\theta$ and needs $\mu_s \ge \tfrac{2}{7}\tan\theta$ to keep rolling. Answering $v = \sqrt{2gh} \approx 1.414\sqrt{gh}$ for any of them means the rotational kinetic energy was dropped.

Common traps

Treating the normal force as $mg$ by default. It is a response force set by the perpendicular equation: $mg\cos\theta$ on an incline, $m(g+a)$ in an accelerating lift, and zero at the top of a loop travelled at the minimum speed.
Using $v = \sqrt{2gh}$ for a body that rolls. That is the point-particle answer; a rolling body must divide the same energy between translation and rotation.
Applying momentum conservation when a significant external impulse acts — a ball rebounding from a wall, or a cart struck while a brake is engaged. Conservation is a permission granted by the system boundary you drew.
Assuming kinetic energy is conserved in every collision. It is conserved only in elastic collisions; 'the bodies stick together' is the signature of maximum kinetic-energy loss.
Forgetting the $Md^{2}$ term when the rotation axis does not pass through the centre of mass, or quoting a moment of inertia without saying which axis it belongs to.
Using $T = 2\pi\sqrt{L/g}$ for a pendulum released at a large angle. The amplitude-independence of the period is a property of the small-angle linearisation, not of pendulums.
Writing gravitational potential energy as positive $GMm/r$ , or mixing $mgh$ with $-GMm/r$ in one equation. With the zero at infinity, every bound orbit has $E < 0$ , and $E = -GMm/2r$ for a circular orbit.
Applying Bernoulli's equation across a pump, a viscous pipe, or between two different streamlines, or forgetting that continuity already fixes the speed ratio before Bernoulli gives the pressure difference.
Assuming buoyancy depends on the object's volume. It depends on the volume of fluid displaced, which equals the object's volume only when the object is fully submerged.

Energy Conservation on a Frictionless Track Mass-Spring Oscillation Period Net Force and Acceleration Perfectly Inelastic Collision Energy and Momentum Conservation Newton's Second Law and Free-Body Diagrams GRE Physics overview Simple Harmonic Motion GRE Physics reference sheet Sign in for current practice

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

Sources

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.
University Physics Volume 1, Section 5.3: Newton's Second Law — OpenStax. Accessed 2026-07-06. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
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.
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.
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.
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.
University Physics Volume 1, Section 15.1: Simple Harmonic Motion — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
University Physics Volume 1, Section 15.4: Pendulums — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
University Physics Volume 1, Section 15.5: Damped Oscillations — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.

Sources and corrections

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Classical Mechanics Domain Guide

Concise answer

Definitions

Intuition

Concept walkthrough

After this page, you should be able to

Formulas and assumptions

Newton's second law and the constraint forces it determines

Work-energy theorem and mechanical energy

Collisions: momentum always, kinetic energy only if elastic

Rotation, rolling, and the parallel-axis theorem

Simple harmonic motion in every disguise

Damped and driven oscillation

Gravitation and orbits

Fluid statics and steady flow

Worked example

The rolling race, and why mass and radius cancel

Common traps

Related pages and practice

Sources

Sources and corrections