A.2 Forces and momentum

Syllabus
First assessment 2025
Topic
—
Level
HL

Learning objectives

—A.2.1—Newton’s three laws of motion• Newton’s three laws of motion.—A.2.2—Forces as interactions between bodies• Forces as interactions between bodies.—A.2.3—Free-body diagrams• Forces acting on a body can be represented in a free-body diagram.—A.2.4—Resultant force from diagrams• Free-body diagrams can be analysed to find the resultant force on a system.—A.2.5—Contact forces• Contact forces include normal, friction, tension, elastic restoring force, viscous drag and buoyancy.—A.2.6—Normal force• Normal force FN acts perpendicular to the contact surface.—A.2.7—Frictional force• Friction acts parallel to contact.• Static: Ff <= μsFN; dynamic: Ff = μdFN.—A.2.8—Tension• Tension.—A.2.9—Hooke’s law restoring force• Elastic restoring force follows Hooke’s law: FH = -kx.—A.2.10—Viscous drag• Viscous drag on a small sphere: Fd = 6πηrv, opposite motion.• η is fluid viscosity, r sphere radius, v speed through fluid.—A.2.11—Buoyancy• Buoyancy from displaced fluid: Fb = ρVg.—A.2.12—Field forces• Know field forces: gravitational, electric and magnetic.—A.2.13—Weight• Weight is gravitational force: Fg = mg.—A.2.14—Electric force Fe• Electric force Fe.—A.2.15—Magnetic force Fm• Magnetic force Fm.—A.2.16—Linear momentum conservation• Linear momentum p=mv is conserved unless a resultant external force acts.—A.2.17—Impulse• Impulse from resultant external force: J = FΔt = Δp.—A.2.18—Impulse-momentum change• The applied external impulse equals the change in momentum of the system.—A.2.19—Newton’s second law forms• Use F=ma for constant mass; use F=Δp/Δt when mass changes.—A.2.20—Elastic and inelastic collisions• Elastic and inelastic collisions of two bodies.—A.2.21—Explosions• Explosions.—A.2.22—Collision energy• Compare energy in elastic collisions, inelastic collisions and explosions.—A.2.23—Centripetal acceleration• Centripetal acceleration is radial: a=v^2/r=ω^2r=4π^2r/T^2.• Direction is radially toward the centre of the circle.—A.2.24—Centripetal force• Circular motion is caused by a centripetal force acting perpendicular to the velocity.—A.2.25—Direction change in circular motion• A centripetal force causes the body to change direction even if its magnitude of velocity may remain constant.—A.2.26—Angular and linear speed• Circular motion relation: v=2πr/T=ωr.• Use angular velocity ω and period T to link angular and linear descriptions.

Use Newton’s Three Laws

Three linked laws

  1. If the resultant force is zero, velocity is constant.
  2. A resultant force changes momentum; for constant mass, F⃗net=ma⃗\vec F_{net}=m\vec a.
  3. Forces between two bodies are equal in magnitude and opposite in direction, acting on different bodies.

Choose the system

Draw forces acting on the chosen object, then use the resultant force to predict its acceleration. For action–reaction pairs, identify the two different bodies before applying the third law.

Use interactions to explain motion

A rocket pushes gas backward; the gas exerts an equal and opposite force on the rocket. The rocket can therefore accelerate even in the absence of a supporting surface.

Common trap

The forces in a third-law pair do not cancel in one free-body diagram because they act on different objects.

A.2.1 Exam Analysis

3 marks

As the probe approaches the surface of the asteroid, a rocket engine is fired to slow its descent. Explain how the engine changes the speed of the probe.

Treat Force as an Interaction Between Bodies

A force needs an interaction

A force is an interaction between bodies. One body exerts the force and another body experiences it. Contact, gravitational, electric and magnetic interactions can all change momentum.

Name both bodies

When explaining a force, state the interacting pair and the direction of the force on the chosen body. The reaction force acts on the other body, not back on the same free-body diagram.

Fields can mediate interaction

Bodies do not need to touch for gravitational, electric or magnetic forces. For example, current-carrying coils interact through their magnetic fields, producing attraction or repulsion depending on the field arrangement.

Common trap

Do not describe a force as a property that an isolated object “has” without naming the other body or field involved.

A.2.2 Exam Analysis

2 marks

Explain why, when there is a current in the coil, the separation of X and Y decreases.

Draw a Labelled Free-Body Diagram

Isolate one body

A free-body diagram shows only the chosen body and the external forces acting on it. Replace the body with a point or simple shape and choose useful axes.

Draw actual forces

Use arrows from the body, label each interaction and draw the direction physically. Typical labels include weight mgmg, normal force NN, tension TT, friction and drag.

Resolve only when needed

If a force is angled, resolve it into the chosen axes. Then apply ∑Fx=max\sum F_x=ma_x and ∑Fy=may\sum F_y=ma_y to the same body.

Common trap

Do not draw velocity, acceleration or a force exerted by the chosen body on its surroundings as forces acting on the chosen body.

A.2.3 Exam Analysis

2 marks

Draw a labelled free-body diagram of the forces on the ball.

Find the Resultant Force from a Diagram

Add force components

The resultant force is the vector sum of all forces on the chosen body:

F⃗net=∑F⃗\vec F_{net}=\sum\vec F

Resolve angled forces into perpendicular components before adding.

Connect to acceleration

For constant mass, apply Newton’s second law along each axis:

∑Fx=max,∑Fy=may\sum F_x=ma_x,\qquad \sum F_y=ma_y

Use equilibrium correctly

If the resultant force is zero, acceleration is zero, but the object may still have constant non-zero velocity. A balanced vertical component does not imply every force is absent.

Common trap

Do not add force magnitudes without their directions. A component that balances another contributes zero only along the same axis.

A.2.4 Exam Analysis

2 marks

Determine the acceleration of the truck.

Classify Contact Forces

Contact-force family

Contact forces arise when bodies or a body and fluid interact: normal force, friction, tension, elastic restoring force, viscous drag and buoyancy.

Use the interaction geometry

Normal force is perpendicular to the surface; friction acts along the surface opposing relative motion or attempted motion; tension acts along a taut string; drag opposes motion through a fluid; buoyancy acts upward due to fluid pressure differences.

Check the condition

Friction can be static or kinetic, drag depends on speed and shape, and buoyancy depends on displaced fluid. The magnitudes are determined by the interaction and constraints, not by a memorized universal value.

Common trap

Do not include every possible contact force. Include only interactions actually present in the described situation.

A.2.5 Exam Analysis

1 mark

A ball is thrown from an aircraft in flight.

Which of the following shows the correct free-body diagram for the forces acting on the ball when terminal velocity is reached?

Model Normal Force Perpendicular to the Surface

Normal means perpendicular

The normal force NN is the contact force exerted by a surface perpendicular to that surface. Its direction follows the local surface normal, not necessarily the vertical direction.

Find it from the force balance

Use the component of Newton’s second law perpendicular to the surface. In a curved path, the normal force may combine with a component of weight to provide the required centripetal resultant.

Do not assume N=mgN=mg

N=mgN=mg applies only in situations where the perpendicular acceleration and other perpendicular force components make that balance valid. Inclines, lifts, loops and vertical acceleration change the normal force.

A.2.6 Exam Analysis

3 marks

Determine the normal force exerted by the loop on the car at P .

Model Static and Dynamic Friction

Friction follows the contact

Friction acts parallel to the contact surface and opposes relative motion or the tendency of surfaces to move relative to each other.

Static friction adapts

Before slipping, static friction has whatever value is needed up to a maximum:

Ff≤μsNF_f\leq\mu_sN

It is not automatically equal to μsN\mu_sN; that value occurs at impending motion.

Dynamic friction during sliding

Once surfaces slide, the model gives

Ff=μdNF_f=\mu_dN

Use the normal force for the actual contact and combine friction with the other forces along the surface.

Common trap

Do not use the dynamic coefficient before motion begins, or assume static friction is always at its maximum.

A.2.7 Exam Analysis

2 marks

Show that the minimum force needed to accelerate the box is about 4 N .

Trace Tension Along a String

Tension is a pull

Tension is the force exerted by a taut string, cable or rope on an attached body. It acts along the string and pulls away from the body.

Use the ideal-string model carefully

For a light, inextensible string over a frictionless pulley, tension has the same magnitude throughout. If the string, pulley or contact is non-ideal, tension can vary and must be found from each body’s force balance.

Connect tension to motion

Draw tension in the string direction, then use ∑F=ma\sum F=ma. A body can have non-zero tension while at rest if other forces balance it.

Common trap

A string can pull but not push. Do not draw tension toward the string’s far end through the body or assume its value equals the weight without a force balance.

A.2.8 Exam Analysis

1 mark

Calculate the maximum tension in the string.

Apply Hooke’s Law to Elastic Restoring Force

Restoring force

For an ideal elastic element within its proportional range,

F⃗H=−kx⃗\vec F_H=-k\vec x

The minus sign means the force acts opposite the displacement from equilibrium.

Use extension correctly

For a spring, xx is extension or compression relative to its natural length. In a vertical equilibrium, the spring tension can balance weight, but the extension is not the total spring length.

Respect the model boundary

Hooke’s law is a linear approximation. Beyond the limit of proportionality, the force–extension graph is no longer linear and the same kk cannot be used.

A.2.9 Exam Analysis

1 mark

A spring of negligible mass and length l0l_{0} hangs from a fixed point. When a mass m is attached to the free end of the spring, the length of the spring increases to l. The tension in the spring is equal to kΔxk \Delta x, where k is a constant and Δx\Delta x is the extension of the spring. What is k ?

Model Viscous Drag on a Small Sphere

Stokes drag

For a small sphere moving slowly through a viscous fluid,

Fd=6πηrvF_d=6\pi\eta r v

where η\eta is viscosity, rr is sphere radius and vv is speed relative to the fluid.

Drag opposes motion

The drag force points opposite the sphere’s velocity. As speed increases, drag increases linearly in this model, reducing the resultant force when the driving force is fixed.

Approach to terminal speed

For a falling sphere, weight drives the motion and viscous drag grows with speed. When drag balances the effective weight, acceleration becomes zero and terminal speed is reached.

Common trap

Do not treat viscosity η\eta as the same quantity as drag force, and do not forget that the formula applies to the stated small-sphere, viscous-flow model.

A.2.10 Exam Analysis

2 marks

Describe why the acceleration of the oil droplet changes.

Calculate Buoyant Force from Displaced Fluid

Buoyancy from pressure difference

A fluid exerts a net upward buoyant force on an immersed object because pressure is greater at greater depth. In the IB model,

Fb=ρfVdispgF_b=\rho_fV_{disp}g

where VdispV_{disp} is the displaced fluid volume.

Separate buoyancy from net force

The buoyant force is one force in the free-body diagram. The net force is found after combining it with weight, tension, drag or other forces.

Floating condition

For an object at rest on the fluid, buoyancy balances its weight. This gives a useful density or submerged-volume relationship, but only after the equilibrium assumption is stated.

Common trap

Use the density of the displaced fluid and the displaced volume, not automatically the object’s total volume or density.

A.2.11 Exam Analysis

1 mark

Show that FbF_{\mathrm{b}} is about 2 mN .

Separate Gravitational, Electric and Magnetic Forces

Three field interactions

Gravitational, electric and magnetic forces are field forces: bodies can interact without contact. Identify the source of the field, the object acted on and the force direction.

Keep the mechanisms distinct

Gravity acts on mass, electric force acts on charge, and magnetic force acts on moving charges or currents in a magnetic field. Their equations and direction rules are not interchangeable.

Use the force relevant to the system

A free-body diagram may contain more than one field force. Add them as vectors and apply Newton’s second law to the selected body.

Common trap

Do not call every non-contact force “electromagnetic”; gravitational attraction is a separate interaction.

A.2.12 Exam Analysis

1 mark

What are three fundamental forces listed in decreasing order of strength?

Calculate Weight from Mass

Weight is a force

Weight is the gravitational force on a mass:

Fg=mgF_g=mg

The direction is toward the local gravitational field source.

Use local gg

The value of gg depends on location. Use the value stated or the local field strength appropriate to the body’s position; mass does not change when the object is moved.

Common trap

Mass is measured in kilograms and is not a force. Weight is measured in newtons and can change when gg changes.

A.2.13 Exam Analysis

1 mark

The probe is carried to the asteroid on board a spacecraft.

Calculate the weight of the probe when close to the surface of the asteroid.

Identify Electric Force

Electric interaction

Electric force acts between charged bodies. Its direction depends on the signs of the charges: like charges repel and unlike charges attract.

Use the electric field

A positive test charge is pushed in the electric-field direction; a negative charge experiences force opposite to the field. Keep field direction and force direction separate when the charge sign matters.

Common trap

Do not reverse the force direction for a positive charge, and do not treat electric force as a contact force.

A.2.14 Exam Analysis

This exam question is unavailable.

Identify Magnetic Force

Magnetic interaction

A magnetic force acts on a moving charge or current in a magnetic field. Its direction is perpendicular to the relevant velocity/current and magnetic-field directions.

Apply the direction rule

Use the stated right-hand rule or vector relationship, then reverse the result for a negative charge. Parallel motion and field give zero magnetic force in the ideal model.

Common trap

A magnetic field can change the direction of velocity without doing work on an ideal moving charge; do not automatically infer a speed change from a magnetic force.

A.2.15 Exam Analysis

This exam question is unavailable.

Conserve Linear Momentum

Momentum

Linear momentum is

p⃗=mv⃗\vec p=m\vec v

It is a vector. For an isolated system, total momentum is conserved before and after an interaction.

Check the system

Momentum is conserved when the resultant external impulse on the chosen system is negligible. Internal forces can change individual momenta while leaving the vector total unchanged.

Use signs or components

Choose a positive direction and conserve momentum component-by-component. A negative final velocity means motion opposite to the chosen positive direction.

Common trap

Do not conserve kinetic energy automatically. Momentum conservation and kinetic-energy conservation are separate claims.

A.2.16 Exam Analysis

1 mark

Cart X , of mass 2 kg , is moving at a speed of 3 m s−13 \mathrm{~m} \mathrm{~s}^{-1} to the right and collides on a horizontal track with cart Y of mass 1 kg.Y1 \mathrm{~kg} . Y is initially stationary.

The velocity of Y immediately after the collision is 4 m s−14 \mathrm{~m} \mathrm{~s}^{-1} to the right. What is the velocity of X immediately after the collision?

Calculate Impulse from Force and Time

Impulse changes momentum

Impulse is the integral of resultant force over time. For a constant average force,

J⃗=F⃗netΔt=Δp⃗\vec J=\vec F_{net}\Delta t=\Delta\vec p

Use the momentum change

Calculate Δp⃗=p⃗f−p⃗i\Delta\vec p=\vec p_f-\vec p_i, including direction. A rebound reverses the velocity component and can make the momentum change larger than either momentum magnitude alone.

Average force

If the force varies, FΔtF\Delta t represents average resultant force over the contact interval. Use consistent units for impulse in N s or kg m s⁻¹.

Common trap

Do not use the initial momentum alone when the object rebounds or ends with a non-zero final velocity.

A.2.17 Exam Analysis

2 marks

The ball rebounds from the ground with speed 7.8 ms−17.8 \mathrm{~ms}^{-1}. The ball is in contact with the ground for a time T. The average resultant force on the ball during this time is 1.1 N .
Determine T.

Link External Impulse to Momentum Change

Impulse is external to the system

For a chosen system, the net external impulse equals the system’s change in total momentum:

J⃗ext=Δp⃗system\vec J_{ext}=\Delta\vec p_{system}

Same momentum change, different force

If an object must undergo the same Δp\Delta p, increasing the stopping time reduces the average resultant force:

Favg=ΔpΔtF_{avg}=\frac{\Delta p}{\Delta t}

Apply to safety systems

A flexible safety net, airbag or crumple zone extends the interaction time while producing the required momentum change, reducing the average force on the person or vehicle.

Common trap

Extending the stopping time does not make the momentum change disappear; it changes the rate at which that change occurs.

A.2.18 Exam Analysis

2 marks

Explain, with reference to change in momentum, why a flexible safety net is less likely to harm the skier than a rigid barrier.

Choose the Momentum Form of Newton’s Second Law

Constant mass

For a body of constant mass, Newton’s second law becomes

F⃗net=ma⃗\vec F_{net}=m\vec a

Use the resultant force, not one arbitrarily selected force.

General momentum form

The broader statement is

F⃗net=Δp⃗Δt\vec F_{net}=\frac{\Delta\vec p}{\Delta t}

or its instantaneous form. This is the safer form when mass changes or when momentum is the quantity given.

Check what changes

If mass is constant, Δp=mΔv\Delta p=m\Delta v, so the two forms agree. If mass enters or leaves the system, include the momentum carried by that mass and define the system carefully.

Common trap

Do not double the acceleration simply because an applied force doubles when a fixed resistive force remains; calculate the new resultant force first.

A.2.19 Exam Analysis

2 marks

Calculate the magnitude of the initial acceleration of the electron.

Distinguish Elastic and Inelastic Collisions

Momentum first

In an isolated collision, total linear momentum is conserved for both elastic and inelastic collisions.

Kinetic energy distinguishes them

In an elastic collision, total kinetic energy is also conserved. In an inelastic collision, some kinetic energy is transferred to internal energy, sound or deformation; in a perfectly inelastic collision the bodies move together afterward.

Use the right conservation law

Apply momentum conservation to find final velocities, then compare initial and final kinetic energy if the collision type is required.

Common trap

“Inelastic” does not mean momentum is lost. It means kinetic energy is not conserved.

A.2.20 Exam Analysis

2 marks

Calculate the speed of the ship after the collision.

Ice in a still lake will usually form in a single layer on the surface.

Model an Explosion with Momentum Conservation

Explosion model

An explosion is an interaction in which an initially combined system separates into parts. If the external impulse is negligible, total momentum before and after is equal.

Use a sign convention

For an object initially at rest, the vector momenta after the explosion sum to zero. In one dimension, equal and opposite momenta can give different speeds when the masses differ.

Energy is separate

The chemical, elastic or other internal energy released can increase total kinetic energy while momentum remains conserved.

Common trap

Do not assume the fragments have equal speeds. Momentum magnitudes are equal and opposite only when the initial total momentum is zero.

A.2.21 Exam Analysis

This exam question is unavailable.

Track Energy in Collisions and Explosions

Track the energy store

Total energy is conserved, but kinetic energy may be transferred to internal energy, sound, deformation or chemical energy during an interaction.

Collision comparison

Elastic collisions conserve total kinetic energy as well as momentum. Inelastic collisions conserve momentum but have a lower final total kinetic energy.

Explosion comparison

An explosion can convert internal energy into kinetic energy, so final kinetic energy can exceed the initial kinetic energy while total momentum remains conserved.

Common trap

“Kinetic energy is lost” is shorthand for transferred to other stores; it is not destroyed.

A.2.22 Exam Analysis

3 marks

Show that the collision is inelastic.

Calculate Centripetal Acceleration

Radial acceleration

For uniform circular motion, the centripetal acceleration is directed toward the centre:

ac=v2r=ω2r=4π2rT2a_c=\frac{v^2}{r}=\omega^2r=\frac{4\pi^2r}{T^2}

Velocity can be constant in magnitude

Even when speed is constant, the velocity direction changes continuously. That directional change produces inward acceleration.

Choose the matching data

Use v2/rv^2/r when speed and radius are given, ω2r\omega^2r when angular speed is given, or 4π2r/T24\pi^2r/T^2 when period is given.

Common trap

Centripetal acceleration is not tangential and does not point along the instantaneous velocity.

A.2.23 Exam Analysis

2 marks

The fan is rotating at 120 revolutions every minute. Calculate the centripetal acceleration of the tip of a fan blade.

Find the Centripetal Force

Centripetal force is a resultant

Centripetal force is the name for the net inward force required for circular motion:

Fc=mac=mv2rF_c=ma_c=\frac{mv^2}{r}

Identify its physical source

Centripetal force is not an extra force. It may be supplied by tension, gravity, friction, normal force, electric force or a combination of forces.

Keep the direction clear

The required resultant points toward the centre and is perpendicular to instantaneous velocity in uniform circular motion.

Common trap

Do not add a separate “centripetal force” arrow to a free-body diagram unless the question explicitly uses it as a shorthand for the inward resultant.

A.2.24 Exam Analysis

2 marks

Explain why a centripetal force is needed for the planet to be in a circular orbit.

Explain How Centripetal Force Changes Direction

Velocity direction changes

In circular motion, the inward centripetal acceleration changes the direction of the velocity. If speed is constant, the magnitude of velocity stays constant while its direction changes.

What happens if the inward force disappears

If the centripetal interaction is removed, the object continues along the tangent at the release point, consistent with Newton’s first law.

Maintain contact

In a vertical loop, the inward resultant must be sufficient to maintain the required radial acceleration. At the limiting contact condition, the normal force can fall to zero.

Common trap

The released object does not move along the radius; its instantaneous path is tangent to the circle.

A.2.25 Exam Analysis

1 mark

A mass at the end of a string is swung in a horizontal circle at increasing speed until the string breaks.

The subsequent path taken by the mass is a

Link Angular and Linear Speed

Connect the descriptions

For uniform circular motion,

v=2πrT=ωrv=\frac{2\pi r}{T}=\omega r

Angular speed ω\omega is the same for all points on a rigid rotating body, while linear speed increases with distance from the axis.

Use the period

One revolution takes period TT, so ω=2π/T\omega=2\pi/T. Keep radians and seconds consistent.

Compare points on one disk

If one point is twice as far from the centre, its linear speed is twice as large at the same angular speed; its centripetal acceleration is also twice as large.

Common trap

Do not assume equal linear speeds for all points on a rotating disk. Equal angular speed does not mean equal tangential speed.

A.2.26 Exam Analysis

1 mark

A disk of radius R rotates about its axis with angular speed ω\omega. Point X is at a distance of R2\frac{R}{2} from the centre and point Y is on the circumference.

What are the ratios of the linear speeds and the centripetal acceleration of X to Y.
The linear speed of X is vXv_{X} and its acceleration is aXa_{X}; the linear speed of Y is vYv_{Y} and its acceleration is aYa_{Y}.

Linear speeds vXvY\frac{\boldsymbol{v}_{\mathbf{X}}}{\boldsymbol{v}_{\mathbf{Y}}}

Acceleration aXaY\frac{\mathbf{a}_{\mathbf{X}}}{\mathbf{a}_{\mathbf{Y}}}

12\frac{1}{2}

14\frac{1}{4}

12\frac{1}{2}

12\frac{1}{2}

1

14\frac{1}{4}

1

12\frac{1}{2}

Retrieve the A.2 Forces and Momentum Model

Build the force model

Choose the system, draw a labelled free-body diagram, classify the interactions and resolve components. Apply Newton’s laws with the correct boundary: contact forces, field forces, friction, tension, buoyancy and restoring forces each have their own direction and conditions.

Track momentum

Use ec p=m ec v, ec J=\Delta ec p and momentum conservation only after checking external impulse. Distinguish elastic and inelastic collisions, explosions and energy transfer.

Track circular motion

The inward resultant provides ac=v2/r=ω2ra_c=v^2/r=\omega^2r. It may come from tension, gravity, normal, friction or a field force. Angular and linear descriptions are linked by v=ωr=2πr/Tv=\omega r=2\pi r/T.

Final checks

Ask: Which body is the system? Which forces are external? Is mass constant? Is acceleration uniform or radial? Is kinetic energy conserved, transferred or increased?