A.2 Forces and momentum

Three linked laws

If the resultant force is zero, velocity is constant.
A resultant force changes momentum; for constant mass, $\vec F_{net}=m\vec a$ .
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.

How it is assessed

The evidence tests an engine slowing a probe using Newton’s second/third law reasoning and asks for the direction of the net force on a projectile.

Command terms

Explain / Identify

What earns marks

Name the chosen object and the resultant force. For a rocket, explain the force pair or momentum transfer to expelled gas and connect the resulting force to deceleration or acceleration. For a projectile, use the net force direction, not the velocity direction.

Watch for

Treating the equal and opposite third-law forces as acting on the same object or confusing velocity direction with net-force direction.

Representative question

[Maximum number: 3]

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.

How it is assessed

The evidence asks why two current-carrying coils move together, rewarding an explanation based on the magnetic field produced by each coil and the resulting force on the other.

Command terms

Explain

What earns marks

Identify the two interacting bodies and use the relevant field or contact interaction to explain the force direction. For current-carrying coils, refer to the magnetic fields of the turns and state whether the resulting force is attractive or repulsive.

Watch for

Saying the coils attract because current exists, without identifying the mutual magnetic-field interaction or force direction.

Representative question

[Maximum number: 2]

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 $\sum F_x=ma_x$ and $\sum F_y=ma_y$ 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.

How it is assessed

The evidence asks for a labelled diagram of a ball supported by a tension, rewarding the correct force labels, directions and omission of non-forces.

Command terms

Draw

What earns marks

Choose the stated object, draw only external forces, label weight and tension/normal/contact forces, and orient them correctly. Resolve angled forces only after the free-body diagram is complete.

Watch for

Including velocity or acceleration as arrows, or drawing the reaction force on the supporting body instead of the force on the chosen ball.

Representative question

[Maximum number: 2]

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

Labelled vertical weight / $\mathrm{W} / \mathrm{mg} / \mathrm{F}_{\mathrm{g}}$ and tension / $\mathrm{T} / \mathrm{F}_{\mathrm{T}}$ in approximately correct direction

With vertical T component equal to weight

Ignore other forces drawn for MP1

[2]

Add force components

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

$\vec F_{net}=\sum\vec F$

Resolve angled forces into perpendicular components before adding.

Connect to acceleration

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

$\sum F_x=ma_x,\qquad \sum F_y=ma_y$

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.

How it is assessed

The evidence asks for the acceleration of a truck from a tension diagram, rewarding the correct component equation and trigonometric interpretation.

Command terms

Determine / Calculate

What earns marks

Resolve the angled tension or other force into components, identify the component that produces acceleration, and apply $F=ma$. Keep the component angle tied to the diagram; a complementary angle changes sine to cosine.

Watch for

Using the total tension rather than its horizontal component, or using sine/cosine for the wrong angle shown in the diagram.

Representative question

[Maximum number: 2]

Determine the acceleration of the truck.

$\begin{aligned} T \sin 30^{\circ}=m a \text { OR } T \cos 30^{\circ}=m g a=g \tan 30^{\circ}=5.66 \approx 5.7 \mathrm{~m} \mathrm{~s}^{-2} \end{aligned}$

Accept reversed trig functions if $60^{\circ}$ used.

Award [2] for CNA

[2]

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.

How it is assessed

The evidence tests a terminal-velocity free-body diagram and asks for a physical explanation of a discrepancy in an experiment.

Command terms

Identify / Explain

What earns marks

Identify every contact interaction present and draw its direction on the selected body. At terminal velocity, use zero resultant force but retain weight and drag; in an experiment, connect differences from accepted values to friction, air resistance or release conditions.

Watch for

Removing weight or drag because acceleration is zero at terminal velocity; zero resultant force does not mean zero individual forces.

Representative question

[Maximum number: 1]

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.

How it is assessed

The evidence asks for the normal force in a vertical loop and tests how the normal force changes as an incline angle increases.

Command terms

Determine / Identify

What earns marks

Draw the normal perpendicular to the local surface and apply Newton’s second law along that direction. In a loop, include the relevant component of weight and the centripetal term; on an incline, use the perpendicular component of weight.

Watch for

Setting N equal to weight without considering curvature, acceleration or the component of weight perpendicular to the surface.

Representative question

[Maximum number: 3]

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

$\begin{aligned} N+m g=\frac{m v^{2}}{r} N=0.12 \times\left(\frac{1.72^{2}}{0.15}-9.81\right) N=1.2 \text { «N» } \end{aligned}$

Allow 1.1 or 1.2 depending on g and rounding of v.

Award [0] for answers based on N=W.

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:

$F_f\leq\mu_sN$

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

Dynamic friction during sliding

Once surfaces slide, the model gives

$F_f=\mu_dN$

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.

How it is assessed

The evidence asks for the minimum force needed to start a box or move a stacked-block system, so the key decision is the static-friction threshold and the correct normal force.

Command terms

Show / Calculate / Determine

What earns marks

Decide whether the object is just about to move or is already sliding. At impending motion use the maximum static friction $\mu_sN$; during sliding use $\mu_dN$. Resolve the applied force and calculate the normal force for the actual contact.

Watch for

Using μd for a minimum-starting-force question or using the total weight as the normal force without checking which surfaces are in contact.

Representative question

[Maximum number: 2]

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

uses the static coefficient
$F=0.36 \times 1.2 \times 9.8=0.36 \times 11.76$

OR
$F=4.2 \text { «N» }$

Must see full substitution OR answer to 2 (or more) significant figures for

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 $\sum F=ma$ . 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.

How it is assessed

The evidence asks for the maximum tension in a string from a measured force or extension, so identify the string force and use the stated model before calculating.

Command terms

Calculate

What earns marks

Use the string geometry and the stated time or extension to determine tension. For a light inextensible string, connect the same tension to each body’s force balance; include units and check that the result is a pulling force.

Watch for

Using a force perpendicular to the string as tension or omitting the unit N.

Representative question

[Maximum number: 1]

Calculate the maximum tension in the string.

« $\frac{0.40}{0.012}=$ » 33.3 N

Restoring force

For an ideal elastic element within its proportional range,

$\vec F_H=-k\vec x$

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.

How it is assessed

The evidence asks for the spring constant from natural length, loaded length and mass, requiring the extension and the equilibrium force balance.

Command terms

Calculate / Identify

What earns marks

Use the spring’s extension $\Delta x=l-l_0$, not its total length, and apply the stated equilibrium or Hooke relationship. Rearrange symbolically before substituting and give $k$ in N m⁻¹.

Watch for

Using the loaded length instead of the extension when calculating $k$.

Representative question

[Maximum number: 1]

A spring of negligible mass and length $l_{0}$ 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 $k \Delta x$ , where k is a constant and $\Delta x$ is the extension of the spring. What is k ?

$\frac{m g}{l_{0}}$

$\frac{m g}{l}$

$\frac{m g}{l-l_{0}}$

$\frac{m g}{l_{0}-l}$

Stokes drag

For a small sphere moving slowly through a viscous fluid,

$F_d=6\pi\eta r v$

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

How it is assessed

The evidence asks why a droplet’s acceleration changes and asks for the shape of acceleration against velocity during a fall.

Command terms

Describe / Identify

What earns marks

As a droplet speeds up, use the given drag model to explain that drag increases, so the net force and acceleration change. At terminal speed, drag balances the driving force and acceleration is zero.

Watch for

Claiming that acceleration remains constant at g even after viscous drag becomes significant.

Representative question

[Maximum number: 2]

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,

$F_b=\rho_fV_{disp}g$

where $V_{disp}$ 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.

How it is assessed

The evidence asks for a numerical buoyant force and asks learners to derive a floating-object density or depth relationship by balancing buoyancy and weight.

Command terms

Show / Calculate / Derive

What earns marks

Use the displaced-fluid volume and fluid density in Fb=ρVg. For a floating object, set buoyancy equal to weight only after identifying equilibrium; for an immersed object, do not assume the object is fully submerged unless the diagram or wording says so.

Watch for

Using the object’s density in the buoyancy equation or equating buoyancy to weight when the object is accelerating.

Representative question

[Maximum number: 1]

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

$(0.001427-0.001208) \cdot 9.8$ OR 2.1 mN

Allow the calculation to give a result
in mN or N . Look for full substitution
or answer with at least two SF.

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.

How it is assessed

The evidence asks learners to identify fundamental forces and to list the forces acting on quarks, testing recognition of electric, weak, strong and gravitational interactions.

Command terms

List / Identify

What earns marks

Identify the relevant field interaction and state what the force acts on. Distinguish gravity, electric and magnetic forces by their source and by whether mass, charge or motion/current is required.

Watch for

Treating the three field forces as interchangeable or omitting the interaction condition that distinguishes magnetic force from electric force.

Representative question

[Maximum number: 1]

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

Strong nuclear, gravity, electromagnetic

Electromagnetic, strong nuclear, gravity

Strong nuclear, electromagnetic, gravity

Gravity, weak nuclear, electromagnetic

Weight is a force

Weight is the gravitational force on a mass:

$F_g=mg$

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.

How it is assessed

The evidence asks for the weight of a probe near an asteroid, so select the local value of g rather than automatically using Earth’s surface value.

Command terms

Calculate

What earns marks

Use Fg=mg with the local gravitational field strength and keep mass separate from weight. Check the requested location and report force in newtons.

Watch for

Using the object’s mass as its weight or using Earth’s g when the question gives a different local gravitational field.

Representative question

[Maximum number: 1]

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.

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.

Momentum

Linear momentum is

$\vec p=m\vec v$

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.

How it is assessed

The evidence uses a collision and a rod–particle system to test vector momentum conservation and the motion of the combined system after interaction.

Command terms

Predict / Calculate

What earns marks

Choose the system and a positive direction, then set vector total momentum before equal to vector total momentum after when external impulse is negligible. In collisions, keep each mass–velocity product and sign explicit.

Watch for

Conserving speed rather than signed momentum, or ignoring a non-negligible external force on the chosen system.

Representative question

[Maximum number: 1]

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

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

$1 \mathrm{~m} \mathrm{~s}^{-1}$ to the right

$1 \mathrm{~m} \mathrm{~s}^{-1}$ to the left

$2 \mathrm{~m} \mathrm{~s}^{-1}$ to the right

$2 \mathrm{~m} \mathrm{~s}^{-1}$ to the left

Impulse changes momentum

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

$\vec J=\vec F_{net}\Delta t=\Delta\vec p$

Use the momentum change

Calculate $\Delta\vec p=\vec p_f-\vec p_i$ , 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, $F\Delta t$ 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.

How it is assessed

The evidence asks for contact time from average force and momentum change, or tests the magnitude of impulse for a change in velocity.

Command terms

Determine / Calculate

What earns marks

Find the vector change in momentum and use J=Δp. For a rebound, choose a sign convention and subtract the initial momentum from the final momentum; then divide by contact time only if average force is requested.

Watch for

Adding the initial and final momentum magnitudes without accounting for their opposite directions during a rebound.

Representative question

[Maximum number: 2]

The ball rebounds from the ground with speed $7.8 \mathrm{~ms}^{-1}$ . 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.

ALTERNATE 1
$\begin{aligned} & \Delta p=\ll 2.7 \times 10^{-3} \times(7.8+9.5)=>0.0467<\mathrm{Ns}>d & \ll 1.1=\frac{\Delta p}{\Delta t} \text { so } T=\frac{0.0467}{1.1} \Rightarrow \gg 0.042 \ll s \gg \end{aligned}$

ALTERNATE 2
$\begin{aligned} & a=\ll \frac{F}{m} \Rightarrow \gg \frac{1.1}{2.7 \times 10^{-3}}=407 \ll \mathrm{~m} \mathrm{~s}^{-2} \gg & T=\ll \frac{9.5+7.8}{407}=\gg 0.042 \ll s \gg \end{aligned}$

Watch for ECF from incorrect value of v in cii).

Award [1] for $t=0.076 \ll s \gg$ using an impact speed of $23 \mathrm{~m} \mathrm{~s}^{-1}$ .

[2]

Impulse is external to the system

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

$\vec J_{ext}=\Delta\vec p_{system}$

Same momentum change, different force

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

$F_{avg}=\frac{\Delta p}{\Delta t}$

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.

How it is assessed

The evidence asks why a flexible safety net is less harmful than a rigid barrier, rewarding the link between increased stopping time, unchanged momentum change and reduced average force.

Command terms

Explain

What earns marks

State that the safety net increases the stopping time while the skier undergoes the same change in momentum. Then use Favg=Δp/Δt to conclude that the average force is smaller.

Watch for

Saying the net reduces the change in momentum instead of explaining that it increases the time over which the change occurs.

Representative question

[Maximum number: 2]

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

safety net extends stopping time

$F=\frac{\Delta p}{\Delta t}$ therefore F is smaller «with safety net»

OR

force is proportional to rate of change of momentum therefore F is smaller «with safety net»

Marking guidance:

Accept reverse argument.

Constant mass

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

$\vec F_{net}=m\vec a$

Use the resultant force, not one arbitrarily selected force.

General momentum form

The broader statement is

$\vec F_{net}=\frac{\Delta\vec p}{\Delta t}$

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, $\Delta p=m\Delta v$ , 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.

How it is assessed

The evidence tests acceleration from an electric force and tests a revised acceleration when an applied force changes while resistance remains fixed.

Command terms

Calculate / Identify

What earns marks

For constant mass, use Fnet=ma after finding the resultant force. For a charged particle, identify the force first, such as qE, then divide by mass. If mass changes, use the momentum-rate form and include the mass-flow contribution.

Watch for

Using the applied force instead of the resultant force when a resistive force remains.

Representative question

[Maximum number: 2]

Calculate the magnitude of the initial acceleration of the electron.

$\begin{aligned} F=q \times E \text { OR } F=1.6 \times 10^{-19} \times 3.4 \times 10^{8}=5.4 \times 10^{-11}<\mathrm{N} \gg a=<\frac{5.4 \times 10^{-11}}{9.1 \times 10^{-31}}=>5.9 \times 10^{19} \ll \mathrm{~ms}^{-2} \gg \end{aligned}$

Ignore any negative sign.

Award [1] for a calculation leading to $a=3.7 \times 10^{38}<\mathrm{ms}^{-2}$ »

Award [2] for bald correct answer

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.

How it is assessed

The evidence asks for the speed of a ship after an object joins it, requiring a shared final velocity and momentum conservation.

Command terms

Calculate / Show

What earns marks

Use momentum conservation for the final speed, especially when bodies stick together. To show that a collision is inelastic, compare initial and final total kinetic energy and identify the energy transferred to other forms.

Watch for

Using kinetic-energy conservation for a sticking collision or assigning separate final velocities after the bodies have joined.

Representative question

[Maximum number: 2]

Calculate the speed of the ship after the collision.

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

$\begin{aligned} \frac{1}{40} m 12=\left(\frac{1}{40} m+m\right) v v=0.29\left\langle\mathrm{~m} \mathrm{~s}^{-1}\right\rangle \end{aligned}$

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.

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.

How it is assessed

The evidence asks learners to show a collision is inelastic or explain why final kinetic energy is lower after a pellet penetrates a ball.

Command terms

Show / Suggest / Explain

What earns marks

To show a collision is inelastic, calculate or compare initial and final total kinetic energy and identify the energy transferred to deformation or other stores. Do not confuse conservation of total energy with conservation of kinetic energy.

Watch for

Saying energy is destroyed rather than identifying work done by contact forces or deformation as the transfer mechanism.

Representative question

[Maximum number: 3]

Show that the collision is inelastic.

initial energy 24 mJ and final energy 12 mJ energy is lost/unequal/change in energy is $12 \mathrm{~mJ}$ inelastic collisions occur when energy is lost

Radial acceleration

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

$a_c=\frac{v^2}{r}=\omega^2r=\frac{4\pi^2r}{T^2}$

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 $v^2/r$ when speed and radius are given, $\omega^2r$ when angular speed is given, or $4\pi^2r/T^2$ when period is given.

Common trap

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

How it is assessed

The evidence includes a fan-tip calculation and a comparison of points on wheels with different radii, testing the radius dependence and angular-speed conversion.

Command terms

Calculate / Identify

What earns marks

Select the version of the centripetal-acceleration equation matching the data, convert revolutions per minute to angular speed or period when needed, and give the radial direction if asked.

Watch for

Using tangential acceleration or forgetting to convert rotational frequency into angular speed before applying ω²r.

Representative question

[Maximum number: 2]

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

ALTERNATE 1

« $\omega=$ » $4 \pi \mathrm{rad} \mathrm{s}^{-1}$ « $a=r \omega^{2}=$ » 280 « $\mathrm{ms}^{-2}$ »

ALTERNATE 2
$<v=\frac{2 \pi r}{T}>=22.6 \mathrm{~m} \mathrm{~s}^{-1}$

« $a=\frac{v^{2}}{r}$ » =280 « $\mathrm{m} \mathrm{s}^{-2}$ »

Marking guidance:

Allow ECF from MP1 for wrong $\boldsymbol{\omega}(\mathbf{1 2 0}$ gives $\left.2.6 \times 10^{4}<\mathrm{m} \mathrm{s}^{-2} \gg\right)$

Allow ECF from MP1 for wrong T (2 s gives 18 « $\mathrm{m} \mathrm{s}^{-2}$ »)

Centripetal force is a resultant

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

$F_c=ma_c=\frac{mv^2}{r}$

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.

How it is assessed

The evidence asks why a planet needs centripetal force and asks for tension in a vertical-circle situation.

Command terms

Explain / Calculate

What earns marks

Explain that circular motion requires a resultant force toward the centre because velocity direction changes. In a vertical circle, combine the source force and the relevant component of weight to obtain the required inward resultant.

Watch for

Treating centripetal force as an additional force or saying that a constant speed means zero resultant force.

Representative question

[Maximum number: 2]

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.

How it is assessed

The evidence asks for the path after a string breaks and asks why a car remains in contact with a loop.

Command terms

State / Explain

What earns marks

If the inward force disappears, state that the object leaves along the tangent because its instantaneous velocity is tangent to the circle. For loop-contact questions, set the normal force condition and compare the actual speed with the minimum required speed.

Watch for

Choosing a radial path after release or claiming that the object stops when the centripetal force is removed.

Representative question

[Maximum number: 1]

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

line along a radius of the circle.

horizontal circle.

curve in a horizontal plane.

curve in a vertical plane.

Connect the descriptions

For uniform circular motion,

$v=\frac{2\pi r}{T}=\omega r$

Angular speed $\omega$ 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 $T$ , so $\omega=2\pi/T$ . 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.

How it is assessed

The evidence asks for ratios of linear speed and centripetal acceleration at two radii and asks for angular velocity from a 24-hour orbital period.

Command terms

Calculate / Identify

What earns marks

Use v=ωr and ω=2π/T. For a rigid disk, compare radii at the same angular speed; for an orbit, convert the period to seconds before calculating angular velocity.

Watch for

Using the same tangential speed at different radii on a rigid rotating disk or leaving a period in hours.

Representative question

[Maximum number: 1]

A disk of radius R rotates about its axis with angular speed $\omega$ . Point X is at a distance of $\frac{R}{2}$ 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 $v_{X}$ and its acceleration is $a_{X}$ ; the linear speed of Y is $v_{Y}$ and its acceleration is $a_{Y}$ .

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

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

$\frac{1}{2}$

$\frac{1}{4}$

$\frac{1}{2}$

$\frac{1}{4}$

$\frac{1}{2}$

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 $a_c=v^2/r=\omega^2r$ . It may come from tension, gravity, normal, friction or a field force. Angular and linear descriptions are linked by $v=\omega r=2\pi r/T$ .

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?

Objective notes

26 learning objectives

—A.2.1—Newton’s three laws of motion• Newton’s three laws of motion.View —A.2.2—Forces as interactions between bodies• Forces as interactions between bodies.View —A.2.3—Free-body diagrams• Forces acting on a body can be represented in a free-body diagram.View —A.2.4—Resultant force from diagrams• Free-body diagrams can be analysed to find the resultant force on a system.View —A.2.5—Contact forces• Contact forces include normal, friction, tension, elastic restoring force, viscous drag and buoyancy.View —A.2.6—Normal force• Normal force FN acts perpendicular to the contact surface.View —A.2.7—Frictional force• Friction acts parallel to contact.• Static: Ff <= μsFN; dynamic: Ff = μdFN.View —A.2.8—Tension• Tension.View —A.2.9—Hooke’s law restoring force• Elastic restoring force follows Hooke’s law: FH = -kx.View —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.View —A.2.11—Buoyancy• Buoyancy from displaced fluid: Fb = ρVg.View —A.2.12—Field forces• Know field forces: gravitational, electric and magnetic.View —A.2.13—Weight• Weight is gravitational force: Fg = mg.View —A.2.14—Electric force Fe• Electric force Fe.View —A.2.15—Magnetic force Fm• Magnetic force Fm.View —A.2.16—Linear momentum conservation• Linear momentum p=mv is conserved unless a resultant external force acts.View —A.2.17—Impulse• Impulse from resultant external force: J = FΔt = Δp.View —A.2.18—Impulse-momentum change• The applied external impulse equals the change in momentum of the system.View —A.2.19—Newton’s second law forms• Use F=ma for constant mass; use F=Δp/Δt when mass changes.View —A.2.20—Elastic and inelastic collisions• Elastic and inelastic collisions of two bodies.View —A.2.21—Explosions• Explosions.View —A.2.22—Collision energy• Compare energy in elastic collisions, inelastic collisions and explosions.View —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.View —A.2.24—Centripetal force• Circular motion is caused by a centripetal force acting perpendicular to the velocity.View —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.View —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.View

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