A.2 Forces and momentum
- Syllabus
- First assessment 2025
- Topic
- —
- Level
- HL
Three linked laws
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.
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.
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.
2 marks
Explain why, when there is a current in the coil, the separation of X and Y decreases.
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 mg, normal force N, tension T, friction and drag.
Resolve only when needed
If a force is angled, resolve it into the chosen axes. Then apply ∑Fx=max and ∑Fy=may 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.
2 marks
Draw a labelled free-body diagram of the forces on the ball.
Add force components
The resultant force is the vector sum of all forces on the chosen body:
Fnet=∑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
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.
2 marks
Determine the acceleration of the truck.
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.
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?
Normal means perpendicular
The normal force N 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=mg
N=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.
3 marks
Determine the normal force exerted by the loop on the car at P .
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≤μsN
It is not automatically equal to μsN; that value occurs at impending motion.
Dynamic friction during sliding
Once surfaces slide, the model gives
Ff=μdN
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.
2 marks
Show that the minimum force needed to accelerate the box is about 4 N .
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. 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.
1 mark
Calculate the maximum tension in the string.
Restoring force
For an ideal elastic element within its proportional range,
FH=−kx
The minus sign means the force acts opposite the displacement from equilibrium.
Use extension correctly
For a spring, x 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 k cannot be used.
1 mark
A spring of negligible mass and length l0 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 kΔx, where k is a constant and Δx is the extension of the spring. What is k ?
Stokes drag
For a small sphere moving slowly through a viscous fluid,
Fd=6πηrv
where η is viscosity, r is sphere radius and v 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 as the same quantity as drag force, and do not forget that the formula applies to the stated small-sphere, viscous-flow model.
2 marks
Describe why the acceleration of the oil droplet changes.
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=ρfVdispg
where Vdisp 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.
1 mark
Show that Fb is about 2 mN .
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.
1 mark
What are three fundamental forces listed in decreasing order of strength?
Weight is a force
Weight is the gravitational force on a mass:
Fg=mg
The direction is toward the local gravitational field source.
Use local g
The value of g 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 g changes.
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.
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.
This exam question is unavailable.
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.
This exam question is unavailable.
Momentum
Linear momentum is
p=mv
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.
1 mark
Cart X , of mass 2 kg , is moving at a speed of 3 m s−1 to the right and collides on a horizontal track with cart Y of mass 1 kg.Y is initially stationary.
The velocity of Y immediately after the collision is 4 m s−1 to the right. What is the velocity of X immediately after the collision?
Impulse changes momentum
Impulse is the integral of resultant force over time. For a constant average force,
J=FnetΔt=Δp
Use the momentum change
Calculate Δp=pf−pi, 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, FΔt 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.
2 marks
The ball rebounds from the ground with speed 7.8 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.
Impulse is external to the system
For a chosen system, the net external impulse equals the system’s change in total momentum:
Jext=Δpsystem
Same momentum change, different force
If an object must undergo the same Δp, increasing the stopping time reduces the average resultant force:
Favg=ΔtΔp
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.
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.
Constant mass
For a body of constant mass, Newton’s second law becomes
Fnet=ma
Use the resultant force, not one arbitrarily selected force.
General momentum form
The broader statement is
Fnet=ΔtΔp
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, 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.
2 marks
Calculate the magnitude of the initial acceleration of the electron.
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.
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.
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.
This exam question is unavailable.
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.
3 marks
Show that the collision is inelastic.
Radial acceleration
For uniform circular motion, the centripetal acceleration is directed toward the centre:
ac=rv2=ω2r=T24π2r
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/r when speed and radius are given, ω2r when angular speed is given, or 4π2r/T2 when period is given.
Common trap
Centripetal acceleration is not tangential and does not point along the instantaneous velocity.
2 marks
The fan is rotating at 120 revolutions every minute. Calculate the centripetal acceleration of the tip of a fan blade.
Centripetal force is a resultant
Centripetal force is the name for the net inward force required for circular motion:
Fc=mac=rmv2
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.
2 marks
Explain why a centripetal force is needed for the planet to be in a circular orbit.
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.
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
Connect the descriptions
For uniform circular motion,
v=T2πr=ωr
Angular speed ω 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 T, so ω=2π/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.
1 mark
A disk of radius R rotates about its axis with angular speed ω. Point X is at a distance of 2R 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 vX and its acceleration is aX; the linear speed of Y is vY and its acceleration is aY.
Linear speeds vYvX
Acceleration aYaX
21
41
21
21
1
41
1
21
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=mec v, ec J=\Deltaec 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=ω2r. It may come from tension, gravity, normal, friction or a field force. Angular and linear descriptions are linked by v=ωr=2πr/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?