A. Space, time and motion

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Topic —

A.1 Kinematics

Objectives in this topic

Choose a reference

Position $\vec r$ specifies where an object is relative to a chosen origin. A position is not meaningful without a reference frame and coordinate direction.

Track change in position

Velocity $\vec v$ describes how position changes with time. It is a vector: its direction is the direction of motion at that instant, and on a curved path the velocity arrow is tangent to the path.

Track change in velocity

Acceleration $\vec a$ describes how velocity changes with time. A change in speed, direction, or both is acceleration; an object can accelerate even while its instantaneous speed is zero.

Common trap

Do not use “velocity” as a synonym for speed. Speed gives only magnitude; velocity also requires direction relative to the chosen coordinate system.

How it is assessed

The evidence uses a motion-path diagram and asks for the velocity vector at a point. The mark scheme rewards a tangent arrow with the correct direction and origin at the specified point.

Command terms

Label / Identify / Draw

What earns marks

When a diagram asks for velocity at a point on a path, draw the arrow tangent to the path, beginning at the stated point, and orient it in the direction of motion. Name the quantity and include direction whenever the question requires a vector.

Watch for

Drawing the velocity arrow radially or along the wrong chord instead of tangent to the path at the specified point.

Representative question

[Maximum number: 1]

the velocity of the ball at P . Label this arrow v.

Velocity is a rate

Velocity is the rate of change of position:

$\vec v=\frac{d\vec r}{dt}$

Over a finite interval, average velocity is displacement divided by elapsed time.

Acceleration changes velocity

Acceleration is the rate of change of velocity:

$\vec a=\frac{d\vec v}{dt}$

A constant acceleration gives equal changes in velocity during equal time intervals.

Read the gradient

On a position–time graph, the gradient represents velocity. On a velocity–time graph, the gradient represents acceleration. The graph’s slope, not its height alone, carries the rate-of-change meaning.

Common trap

Distance divided by time gives average speed, not instantaneous velocity. Likewise, a large velocity does not imply a large acceleration unless the velocity is changing rapidly.

How it is assessed

The evidence includes a definition multiple-choice question and a falling-stone question about the change in velocity during consecutive equal time intervals.

Command terms

State / Identify

What earns marks

State the definition exactly: instantaneous velocity is the rate of change of position. For constant acceleration, connect equal time intervals with equal changes in velocity; do not substitute distance or speed when the question asks for displacement or velocity.

Watch for

Using distance divided by time as the definition of instantaneous velocity.

Representative question

[Maximum number: 1]

Instantaneous velocity is defined as...

$\frac{\text { displacement }}{\text { time taken }}$ .

rate of change of position.

$\frac{\text { distance moved }}{\text { time taken }}$ .

rate of change of distance.

Displacement is a vector

Displacement is the change in position:

$\Delta\vec r=\vec r_{final}-\vec r_{initial}$

It has a magnitude and a direction from the initial position to the final position.

Ignore the route for displacement

The path taken between the two positions does not determine displacement. A curved or complicated journey can still have a straight-line displacement between its endpoints.

Use components when needed

For perpendicular changes, resolve displacement into components and combine them vectorially. A signed one-dimensional displacement is positive or negative according to the chosen axis.

Common trap

A return to the starting point gives zero displacement even though the distance travelled is non-zero.

How it is assessed

The evidence uses projectile and circular-track contexts to test whether the learner chooses the endpoint-to-endpoint displacement rather than the distance along the path.

Command terms

Calculate / Identify

What earns marks

Find the vector from the initial position to the final position. In two dimensions, use the component changes and combine them; in a circular path, use the chord between endpoints rather than the arc length. State the magnitude and unit.

Watch for

Using the distance travelled along the trajectory or circular track as the displacement.

Representative question

[Maximum number: 1]

A stone is kicked horizontally at a speed of $1.5 \mathrm{~ms}^{-1}$ from the edge of a cliff on one of Jupiter's moons. It hits the ground 2.0 s later. The height of the cliff is 4.0 m .
Air resistance is negligible.
What is the magnitude of the displacement of the stone?

$7.0 \mathrm{~m}$

5.0 m

4.0 m

3.0 m

Distance

Distance is the total path length travelled. It is a scalar, so it has magnitude only and cannot be negative.

Displacement

Displacement is the vector change in position from start to finish. Its magnitude is the shortest endpoint-to-endpoint separation, not generally the length of the route.

Match the average quantity

Average speed uses total distance divided by total time. Average velocity uses displacement divided by total time. A route with turns can therefore have average speed greater than the magnitude of average velocity.

Common trap

For a complete oscillation, the displacement is zero but the distance is four times the amplitude. Choose the quantity named in the question before substituting.

How it is assessed

The evidence compares average speed and average velocity for a person taking two legs and asks for distance travelled during one complete oscillation.

Command terms

Calculate / Distinguish

What earns marks

Use total path length for average speed and endpoint displacement for average velocity. In a complete oscillation, calculate distance from the repeated path segments; for a route with perpendicular legs, use the resultant displacement and total distance separately.

Watch for

Using displacement in the average-speed calculation or using total distance in the average-velocity calculation.

Representative question

[Maximum number: 1]

A person walks 40 m due west and then 30 m due north. The total walking time is 100 s . What are the average speed and the magnitude of the average velocity of the person?

Average speed/m s ${ }^{-1}$

Magnitude of average
velocity $/ \mathrm{m} \mathrm{s}^{-1}$

0.5

0.7

0.5

0.7

Average values use an interval

Average speed is total distance divided by total time. Average velocity is displacement divided by elapsed time. Average acceleration is change in velocity divided by elapsed time.

Instantaneous values use one moment

Instantaneous velocity is the tangent gradient on a position–time graph; instantaneous speed is its magnitude and is what an ideal speedometer reports. Instantaneous acceleration is the tangent gradient on a velocity–time graph. Average values instead use a finite interval.

Connect graph quantities

The gradient of a velocity–time graph is acceleration, and the area under it is displacement. A constant acceleration therefore produces a straight-line velocity–time graph.

Common trap

Do not use the average gradient when a question asks for an instantaneous value. Use the tangent at the specified time.

How it is assessed

The evidence tests graph transformation from acceleration–time to velocity–time, requiring the correct gradient/integration relationship and recognition of the initial condition.

Command terms

Identify / Determine

What earns marks

For an average value, use the whole interval and the appropriate total quantity. For an instantaneous value, read the tangent gradient at the stated time. In a velocity–time graph, integrate acceleration to obtain the change in velocity before applying the initial condition.

Watch for

Treating the acceleration value as the velocity value, or using the graph height instead of the gradient/area relationship.

Representative question

[Maximum number: 1]

The graph shows the variation of the acceleration a with time t of an object moving in a straight line.

Which graph shows the variation of the velocity v of the object with time t ?

Uniform-acceleration model

SUVAT equations apply when acceleration is constant along the chosen one-dimensional axis:

$v=u+at,\quad s=ut+\frac12at^2,\quad v^2=u^2+2as,\quad s=\frac{u+v}{2}t$

Choose an equation

List the known and unknown quantities $s,u,v,a,t$ . Select an equation containing the required unknown and only known quantities; keep signs consistent with the positive direction.

Check the model

A constant acceleration means equal changes in velocity in equal time intervals. Use separate horizontal and vertical equations only when the motion has been resolved into independent components.

Worked example from local Question Bank row 22687

A glider accelerates uniformly from rest to $27.0\,\mathrm{m\,s^{-1}}$ in $11.0\,\mathrm{s}$ . Use $s=\frac{u+v}{2}t$ :

$s=\frac{0+27.0}{2}\times 11.0=148.5\,\mathrm{m}\approx149\,\mathrm{m}$

The result is the launch-run displacement; the constant-acceleration assumption is essential.

Common trap

Do not use SUVAT when acceleration varies significantly with time or position. A formula can produce a neat number while still violating the model’s constant-acceleration assumption.

How it is assessed

The evidence uses constant-acceleration motion from rest and measurement of displacement over time, rewarding the correct equation, data selection and calculation.

Command terms

Determine / Calculate

What earns marks

Check that acceleration is constant, choose a SUVAT equation containing the known quantities, and define the positive direction before substituting. For a photograph or position-time data, use consistent intervals and show how the measured displacement enters the equation.

Watch for

Applying a SUVAT equation without checking that acceleration is constant or mixing signed and unsigned distances.

Representative question

[Maximum number: 3]

Determine g using the photograph.

Read at least two points correctly and consistently

Use $\frac{1}{2} \cdot g \cdot t^{2}$ with a length interval from two non-consecutive points OR for two (or more) length intervals, using consistent time intervals

Correct calculation of g

Marking guidance:

Award [2] max if they use one single length interval of consecutive points.

Award [1] max if they miss to subtract the initial point in their length interval or if they use inconsistent time intervals.

Do not penalize significant figures in the final answer.

2 3

Uniform acceleration

Acceleration is uniform when the velocity changes by equal amounts in equal time intervals. The velocity–time graph is a straight line with constant gradient.

Non-uniform acceleration

Acceleration is non-uniform when its magnitude or direction changes. The velocity–time graph then has a changing gradient, and a single SUVAT value cannot describe the entire interval.

Model versus reality

A constant-acceleration model can be useful over a limited interval even when real forces vary. State the approximation and identify the neglected force or changing condition.

Common trap

A curved trajectory does not by itself prove that acceleration is non-uniform: projectile motion without drag has constant downward acceleration while its velocity direction changes.

How it is assessed

The evidence asks for one reason a spacecraft’s acceleration is not constant and one reason a dancer model is unrealistic, rewarding a specific neglected or changing physical parameter.

Command terms

State / Outline

What earns marks

Give a physical reason why acceleration changes: for example, a changing force, changing radiation intensity, changing force direction, drag, or an omitted interaction. Link the reason to the acceleration rather than merely saying the motion is unrealistic.

Watch for

Giving a vague statement such as “the model is not realistic” without naming a force, changing condition, or neglected parameter.

Representative question

[Maximum number: 1]

State one reason why the acceleration of the spacecraft will not be constant.

Separate the axes

With negligible fluid resistance, projectile motion is independent horizontal and vertical motion. Resolve the launch velocity:

$u_x=u\cos\theta,\qquad u_y=u\sin\theta$

Horizontal motion

There is no horizontal acceleration in the ideal model, so $v_x=u_x$ and $x=u_xt$ . Use the horizontal displacement to find time or horizontal speed.

Vertical motion

Use one-dimensional constant-acceleration equations vertically, usually with $a_y=-g$ if upward is positive. The horizontal and vertical equations share the same time $t$ .

Worked example from local Question Bank row 31723

A tennis ball travels $11.9\,\mathrm{m}$ horizontally after launch at $64.0\,\mathrm{m\,s^{-1}}$ and $7^\circ$ to the horizontal.

$u_x=64.0\cos7^\circ=63.52\,\mathrm{m\,s^{-1}}$
$t=\frac{x}{u_x}=\frac{11.9}{63.52}=0.187\,\mathrm{s}$

The same $0.187\,\mathrm{s}$ must then be used in the vertical equation.

Common trap

Do not use the launch speed as the horizontal speed. Resolve it first, and do not use the horizontal time independently of the vertical motion.

How it is assessed

The evidence uses launch angle and horizontal distance to determine time or initial speed, rewarding the correct trigonometric component and the shared-time model.

Command terms

Calculate / Show

What earns marks

Resolve the launch velocity into horizontal and vertical components before using equations. Use the common time for both axes; calculate horizontal time from x=u_xt when horizontal acceleration is zero, then check the vertical condition separately.

Watch for

Using u sin θ for horizontal motion or forgetting that the vertical and horizontal calculations refer to the same elapsed time.

Representative question

[Maximum number: 2]

The ball leaves the ground at an angle of $22^{\circ}$ . The horizontal distance from the initial position of the edge of the ball to the wall is 11 m . Calculate the time taken for the ball to reach the wall.

horizontal speed $=19 \times \cos 22$ « $=17.6 \mathrm{~m} \mathrm{~s}^{-1}$ »
$\text { time }=\ll \frac{\text { distance }}{\text { speed }}=\frac{11}{19 \cos 22}=>0.62<\mathrm{s} \gg$

Marking guidance:

Allow ECF for MP2

Drag opposes instantaneous velocity

Fluid resistance acts opposite the projectile's velocity and usually grows with speed. Its direction changes through the flight, so the resultant acceleration is not the constant downward $g$ of the ideal model.

Quantity	Qualitative effect of fluid resistance
Trajectory	No longer a symmetric parabola; descent is typically steeper
Horizontal velocity	Decreases because drag has a component opposite horizontal motion
Vertical acceleration	On ascent, downward drag makes downward acceleration greater than $g$ ; on descent, upward drag makes it less than $g$
Maximum height and range	Both are reduced for the same launch conditions
Time of flight	Ascent is shortened, while descent can be lengthened by upward drag; the total change is not universally one direction
Terminal speed	During a long fall, increasing drag can balance weight so resultant force and acceleration become zero

Use the force direction

Before the peak, drag has horizontal and downward components; after the peak, it has horizontal and upward components. Therefore acceleration is not determined by velocity alone and changes continuously.

Terminal-speed condition

For vertical descent, terminal speed is reached when upward drag (and any buoyancy included in the model) balances weight. The object then continues at constant downward velocity.

Common trap

Zero acceleration at terminal speed does not mean zero velocity. At the top of a projectile path, vertical velocity may be zero while acceleration remains non-zero.

How it is assessed

The evidence compares actual motion with an ideal no-drag path and asks where acceleration has greatest magnitude during a drag-affected vertical throw.

Command terms

Describe / Identify / Compare

What earns marks

State the direction of drag and connect its changing magnitude to the resultant acceleration. For vertical motion, identify the point where drag is greatest or where drag balances weight; for a projectile, compare speed, range, height and symmetry with the no-resistance model.

Watch for

Assuming acceleration is always g when drag is present, or assuming the trajectory remains a symmetric parabola.

Representative question

[Maximum number: 1]

The diagram shows the path of a ball in the absence of air resistance. Q is the highest point of the ball's trajectory and a is the vertical acceleration at Q . At impact the velocity makes an angle $\theta$ to the horizontal.

Three statements about the actual motion of the ball when there is air resistance are:

I. Q is lower.
II. a remains the same.
III. $\theta$ increases.

Which statements are correct?

I and II only

I and III only

II and III only

I, II and III

Describe motion

Position locates the object, velocity is the rate of change of position, and acceleration is the rate of change of velocity. Distance and speed are scalar; displacement and velocity are directed quantities.

Use the right model

For uniform acceleration:

$v=u+at,\quad s=ut+ rac12at^2,\quad v^2=u^2+2as$

For projectiles without drag, solve horizontal and vertical components with a shared time. Do not use these equations when acceleration is non-uniform.

Check the boundary

Ask whether the quantity is average or instantaneous, whether the route or endpoints matter, whether acceleration is constant, and whether a neglected force such as drag changes the model.

Topic —

A.2 Forces and momentum

Objectives in this topic

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?

Topic —

A.3 Work, energy and power

Objectives in this topic

Energy is conserved

Energy cannot be created or destroyed. In a defined system, energy is transferred between stores or across the system boundary, so the total energy accounting remains balanced.

Define the system first

Name the objects included and identify transfers by work, heating, radiation or electrical means. A falling object may transfer gravitational potential energy to kinetic energy, internal energy or sound.

Follow the chain

Write the initial store, the useful output store and any dissipated or transferred energy. A Sankey diagram or energy-flow statement should account for all significant branches.

Common trap

Energy “lost” from a useful store has been transferred elsewhere; it has not disappeared.

How it is assessed

The evidence asks learners to outline energy changes in a pumped-storage hydroelectric system or describe gravitational potential energy becoming internal energy of air.

Command terms

Outline / Describe

What earns marks

Name the initial and final energy stores and identify the transfer pathway, including useful output and dissipated energy. For a pumped-storage system, track gravitational potential energy of water through kinetic/mechanical energy to electrical output.

Watch for

Listing energy forms without stating the direction of transfer or omitting the dissipated/internal-energy branch.

Representative question

[Maximum number: 2]

Outline, with reference to energy changes, the operation of a pumped storage hydroelectric system.

Work transfers energy

Work done by a force is the energy transferred by that force. For a constant force,

$W=Fs\cos\theta$

where $\theta$ is the angle between force and displacement.

Use the sign

Positive work transfers energy into the object’s relevant store; negative work transfers energy out of it. A force perpendicular to displacement does zero work.

Follow the physical process

Wind can transfer kinetic energy to a turbine through work, while resistive forces can transfer mechanical energy to internal energy of the surroundings.

Common trap

Do not call every force an energy transfer. Check whether the force has a component along the displacement.

How it is assessed

The evidence asks for energy transfers in a wind generator and asks for work done on air by a falling object at terminal speed.

Command terms

Describe / Calculate / State

What earns marks

Name the force and the initial/final energy stores it connects. For a constant force use W=Fs cosθ; for a force–distance graph, the area represents work done. Include the direction of transfer.

Watch for

Confusing power with work or omitting the component of force parallel to displacement.

Representative question

[Maximum number: 2]

Describe the energy transfers taking place in a wind generator.

Read the width as energy

A Sankey diagram shows an input energy flowing into useful output and other transfers. Arrow width is proportional to energy, so the branches must account for the whole input.

Identify useful output

Label the useful branch before calculating efficiency. Other branches may represent heating, sound or unwanted mechanical transfers.

Connect to efficiency

The useful fraction of the input is

$\eta=\frac{E_{useful}}{E_{input}}=\frac{P_{useful}}{P_{input}}$

Common trap

Do not compare branch widths without checking whether the diagram uses the same scale and whether the requested quantity is energy or power.

How it is assessed

The evidence asks for an efficiency statement from a lamp diagram and for thermal power loss in a nuclear power-station Sankey diagram.

Command terms

Identify / Calculate

What earns marks

Read input, useful output and loss branches from the Sankey diagram. Use branch widths or labelled values to calculate efficiency or a missing power, and keep energy and power dimensions consistent.

Watch for

Reading a loss branch as useful output or applying an energy ratio to power values without checking the time basis.

Representative question

[Maximum number: 1]

The Sankey diagram shows the energy input from fuel that is eventually converted to useful domestic energy in the form of light in a filament lamp.

What is true for this Sankey diagram?

The overall efficiency of the process is 10 %.

Generation and transmission losses account for 55 % of the energy input.

Useful energy accounts for half of the transmission losses.

The energy loss in the power station equals the energy that leaves it.

Constant-force work

For a force $F$ acting through displacement $s$ ,

$W=Fs\cos\theta$

Only the component parallel to displacement transfers energy by work.

Area under a force–distance graph

For a variable force, the area under an $F$ -against- $s$ graph gives work. A negative area represents work against the chosen displacement direction.

Check the angle

Use the angle between force and displacement, not the angle between the force and an unrelated axis unless the component has first been resolved.

Worked example from local Question Bank row 39177

A kite pulls a ship with force $2.50\times10^5\,\mathrm{N}$ at $39^\circ$ to its $1.00\,\mathrm{km}$ displacement. Convert $1.00\,\mathrm{km}=1.00\times10^3\,\mathrm{m}$ , then

$W=Fs\cos\theta=(2.50\times10^5)(1.00\times10^3)\cos39^\circ=1.94\times10^8\,\mathrm{J}\approx1.9\times10^8\,\mathrm{J}$

Only the force component along the ship's displacement transfers energy.

How it is assessed

The evidence asks what the area under a force–distance graph represents and includes an electric-field work calculation.

Command terms

State / Calculate

What earns marks

Use W=Fs cosθ for a constant force or the area under the force–distance graph for a variable force. State what the area represents and keep the sign and units of work consistent.

Watch for

Using the force magnitude without the parallel component or interpreting graph area as force rather than work.

Representative question

[Maximum number: 1]

State what is represented by the area under the graph.

Work–energy theorem

The net work done by the resultant force on a system equals its change in kinetic energy:

$W_{net}=\Delta E_k$

Use force–distance area

For a variable resultant force, the signed area under the force–distance graph gives the work and therefore the kinetic-energy change.

Include all resultant forces

Friction, applied forces and gravity may each do work. Add their signed contributions before relating the result to the final kinetic energy.

Worked example from local Question Bank row 31356

A constant net force of $100\,\mathrm{N}$ moves an object from rest through $2.0\,\mathrm{m}$ until its speed is $10\,\mathrm{m\,s^{-1}}$ .

$W_{net}=Fs=(100)(2.0)=200\,\mathrm{J}$
$200=\Delta E_k=\frac12m(10)^2-0$
$m=4.0\,\mathrm{kg}$

The positive net work is exactly the object's kinetic-energy gain.

Common trap

Do not use the work of one force as the net work unless all other force contributions are zero or already included.

How it is assessed

The evidence asks for a stopping distance after applied force is removed and for maximum speed from a force–distance graph.

Command terms

Determine / Calculate

What earns marks

Use the signed work done by the resultant force to find the change in kinetic energy. For a force that varies with distance, calculate the relevant graph area and combine it with the initial kinetic energy.

Watch for

Using the area under only one force curve or treating negative work as a negative kinetic energy rather than a change.

Representative question

[Maximum number: 3]

A force of 14.0 N acts on the box for 0.35 m as shown. The force is then removed and the box continues to move. The box comes to rest after a further displacement d.

Determine d.

ALT 1
$F_{f}=0.28 \times 1.2 \times 9.8=3.29 \text { «N» }$

W done over $0.35 \mathrm{~m}=(14-3.29) \times 0.35=3.75<\mathrm{J} />$ $\text { d = «3.75 J / 3.29 N = » } 1.14 \text { «m» }$

ALT 2
$\begin{aligned} & a=(14-0.28 \times 1.2 \times 9.8) / 1.2=8.92\left\langle\mathrm{~m} \mathrm{~s}^{-2}\right\rangle & v=\sqrt{(2)(8.92)(0.35)}=2.50\left\langle\mathrm{~m} \mathrm{~s}^{-1}\right\rangle & d=\left\langle 2.5^{2} /(2 \times 0.28 \times 9.8)=-1.14\langle\mathrm{~m} \| \right. \end{aligned}$

Allow ECF from MP1

Only award marks from one ALT.

Mechanical energy stores

Mechanical energy is the sum of translational kinetic energy, gravitational potential energy and elastic potential energy:

$E_{mech}=E_k+E_{p,g}+E_{p,elastic}$

Use the chosen system

Mechanical energy describes these stores within the system. Internal energy, chemical energy and sound may also be present in the full energy account but are not mechanical energy.

Common trap

Do not call all conserved energy mechanical energy; classify the store before applying a mechanical-energy equation.

How it is assessed

The evidence asks for a speed from gravitational potential energy and tests the relation between kinetic energy and total energy at terminal velocity.

Command terms

Show / Identify

What earns marks

Identify which energy stores are mechanical and apply the relevant relation. For a falling object, distinguish kinetic-energy increase from gravitational potential-energy decrease and note that terminal motion may transfer energy to internal stores.

Watch for

Calling thermal or chemical energy mechanical energy, or assuming total energy equals kinetic energy during terminal motion.

Representative question

[Maximum number: 1]

show that the speed of the ball is about $4.3 \mathrm{~ms}^{-1}$ .

$V=\sqrt{2 \times 9.81 \times 0.95}$ OR =4.32 《m ${ }^{-1}$ 》

Must see either full substitution or answer to at least 3 s.f.

Condition for conservation

Mechanical energy is conserved when only conservative forces transfer energy within the system and friction or other resistive transfers are absent or negligible.

Write the balance

$E_{k,i}+E_{p,i}=E_{k,f}+E_{p,f}$

Choose a convenient zero for potential energy and keep the same reference throughout.

When it is not conserved

Friction, drag or deformation transfer mechanical energy to internal energy. Total energy is still conserved, but the mechanical-energy equation needs an additional transfer term.

How it is assessed

The evidence contrasts a frictionless ramp with a rough surface and includes rolling motion, requiring the correct boundary for mechanical-energy conservation.

Command terms

Calculate / Explain

What earns marks

Use mechanical-energy conservation only over the part of the motion where resistive work is absent or negligible. When the path becomes rough, include the work done by friction or the resulting internal-energy transfer.

Watch for

Applying mechanical-energy conservation across a rough section without subtracting the work done by friction.

Representative question

[Maximum number: 1]

An object is released from rest and slides down a frictionless ramp. The object then leaves the ramp and slides along a rough horizontal surface. The object stops in a distance s along the ramp.

The coefficient of dynamic friction between the object and the rough horizontal surface is $\mu$ .
What is the height of the ramp?

$\mu g s$

$\frac{s}{2 g \mu}$

$\frac{s}{\mu}$

$\mu s$

Conservative transformations

When mechanical energy is conserved, energy can move between translational kinetic, gravitational potential and elastic potential stores without changing their sum.

Use the endpoints

For a car descending a frictionless track, gravitational potential energy decreases while kinetic energy increases. For a spring system, elastic potential energy can become kinetic energy and then return.

Add non-conservative transfers

If friction or drag acts, part of the mechanical energy transfers to internal energy. The endpoint equation must include that loss from the mechanical stores.

How it is assessed

The evidence asks for speeds of a car at different points on a track using gravitational potential to kinetic-energy conversion.

Command terms

Show / Calculate

What earns marks

Choose the initial and final mechanical stores, then equate their sum when no dissipative transfer is present. Use the same mass and potential-energy reference, and state any frictionless assumption.

Watch for

Using a height change with the wrong sign or applying the conservative equation after an unmentioned frictional section.

Representative question

[Maximum number: 2]

Show that the speed of the car at P is $1.7 \mathrm{~ms}^{-1}$ .

$\begin{aligned} m g \times 0.15=\frac{1}{2} m v^{2} v=\sqrt{2 \times 9.81 \times 0.15} \text { OR } \quad 1.72\left\langle<m s^{-1} \gg \right. \end{aligned}$

Award MP1 for recognition that KE of car at P is GPE lost.

Do not award MP1 for answers based on suvat equations. MP2 can still be awarded for a correct answer.

Kinetic-energy forms

Translational kinetic energy is

$E_k=\frac12mv^2=\frac{p^2}{2m}$

Choose the known quantity

Use $\frac12mv^2$ when mass and speed are given, or $p^2/(2m)$ when momentum is given. Kinetic energy is scalar and cannot be negative.

Worked example from local Question Bank row 35674

For $m=0.14\,\mathrm{g}=1.4\times10^{-4}\,\mathrm{kg}$ and $v=3.1\,\mathrm{m\,s^{-1}}$ ,

$E_k=\frac12(1.4\times10^{-4})(3.1)^2=6.7\times10^{-4}\,\mathrm{J}=0.67\,\mathrm{mJ}$

Converting grams to kilograms before substitution keeps the energy unit in joules.

Common trap

Doubling speed quadruples kinetic energy; do not scale it linearly with speed.

How it is assessed

The evidence asks for final speed after power/resistance information and asks for energy transferred by a constant resultant force.

Command terms

Calculate / Identify

What earns marks

Select the kinetic-energy form matching the given quantities and keep speed in m s⁻¹, mass in kg and momentum in kg m s⁻¹. If a force accelerates an object from rest, use the work–energy link to identify the transferred energy.

Watch for

Using momentum directly as energy or forgetting the square on speed.

Representative question

[Maximum number: 2]

Calculate the final speed of the car.

A different car travels on a horizontal road at a constant speed of $45 \mathrm{~m} \mathrm{~s}^{-1}$ . The engine of the car develops a power of 140 kW . The resistive force $F_{\mathrm{d}}$ acting on the car is given by

Area $=2.4 \times 10^{5} \mathrm{~J}$ $\ll 2.4 \times 10^{5}=\frac{1}{2} \times 1.6 \times 10^{3} \times v^{2} \Rightarrow \gg v=17 \mathrm{~m} \mathrm{~s}^{-1}$

Near-Earth gravitational potential energy

For a height change $\Delta h$ in a uniform gravitational field,

$\Delta E_{p,g}=mg\Delta h$

Use the height change

Raising an object gives positive change in gravitational potential energy; lowering it gives negative change relative to the chosen reference.

Link to power

If height changes at constant speed, the rate of gravitational potential-energy gain is $mgv$ , before accounting for efficiency or other transfers.

Worked example from local Question Bank row 37039

An object's weight is $6.10\times10^2\,\mathrm{N}$ and it rises vertically by $8.0\,\mathrm{m}$ . Since $mg$ is its weight,

$\Delta E_{p,g}=(6.10\times10^2)(8.0)=4.88\times10^3\,\mathrm{J}\approx4.9\,\mathrm{kJ}$

The positive result means the gravitational potential-energy store increases.

Common trap

Use the local value of $g$ and the vertical height change, not the distance along a slope.

How it is assessed

The evidence asks for gravitational potential-energy gain of a car climbing a hill and for energy change after a vertical displacement.

Command terms

Calculate / Identify

What earns marks

Use ΔEp=mgΔh with the vertical height change and the stated value of g. At constant speed, relate the gain rate to power as mgv, then include efficiency or time only if the question requests it.

Watch for

Using the total path length rather than vertical height or forgetting that weight may be given directly as mg.

Representative question

[Maximum number: 1]

A car takes 20 minutes to climb a hill at constant speed. The mass of the car is 1200 kg and the car gains gravitational potential energy at a rate of 6.0 kW . Take the acceleration of gravity to be $10 \mathrm{~m} \mathrm{~s}^{-2}$ . What is the height of the hill?

$0.6 \mathrm{~m}$

10 m

600 m

6000 m

Elastic store

For a spring within its linear range,

$E_{p,elastic}=\frac12k(\Delta x)^2$

where $\Delta x$ is extension or compression from the natural length.

Area under the graph

The elastic potential energy equals the work done in stretching or compressing the spring. On a force–extension graph it is the area under the graph.

Worked example from local Question Bank row 31357

A spring with $k=100\,\mathrm{N\,m^{-1}}$ is compressed by $0.10\,\mathrm{m}$ .

$E_{p,elastic}=\frac12(100)(0.10)^2=0.50\,\mathrm{J}$

This is the energy available for transfer when the ideal spring is released.

Common trap

Do not use the total spring length as $\Delta x$ , and remember that doubling extension quadruples the stored energy in the ideal model.

How it is assessed

The evidence asks for spring constant from work and compression, and for maximum elastic potential energy in a spring system.

Command terms

Calculate

What earns marks

Use Eh=1/2k(Δx)² with extension or compression from the unstretched length. If a graph or work value is given, connect the area or work to the spring constant and report N m⁻¹ or J as requested.

Watch for

Using Δx rather than (Δx)² or confusing spring constant with elastic energy.

Representative question

[Maximum number: 1]

$\quad 0.25 \mathrm{~J}$ of work is done to compress a spring by a distance of 0.10 m from its unstretched length. What is the spring constant?

$2.5 \mathrm{Nm}^{-1}$

$5.0 \mathrm{Nm}^{-1}$

$25 \mathrm{Nm}^{-1}$

$50 \mathrm{Nm}^{-1}$

Power is rate

Power is the rate of work or energy transfer:

$P=\frac{\Delta W}{\Delta t}=\frac{\Delta E}{\Delta t}$

Mechanical shortcut

For a constant force parallel to velocity,

$P=Fv$

Keep energy and power distinct

Energy is measured in joules; power is measured in watts, or joules per second. Multiply power by time to recover transferred energy.

Worked example from local Question Bank row 29322

A student of weight $600\,\mathrm{N}$ climbs $6.0\,\mathrm{m}$ vertically in $8.0\,\mathrm{s}$ .

$\Delta W=(600)(6.0)=3.6\times10^3\,\mathrm{J}$
$P=\frac{3.6\times10^3}{8.0}=4.5\times10^2\,\mathrm{W}=450\,\mathrm{W}$

The result is the average rate of energy transfer against gravity.

How it is assessed

The evidence asks for the average power supplied while running upstairs and for the energy delivered by a cell over a discharge time.

Command terms

Calculate

What earns marks

Use P=ΔE/Δt or P=Fv with the correct force component and speed. Convert hours to seconds when energy is in joules, and distinguish average power from instantaneous power.

Watch for

Using total energy as power or forgetting to convert the time interval into seconds.

Representative question

[Maximum number: 1]

A student of mass m initially at rest takes t seconds to run up stairs of height h. At the top of the stairs the student has a velocity v.

What is the average power supplied by the student during the climb?

$\frac{m g h}{t}$

$\frac{m\left(g h+\frac{1}{2} v^{2}\right)}{t}$

$\frac{m\left(g h-\frac{1}{2} v^{2}\right)}{t}$

m g v

Useful fraction

Efficiency is the ratio of useful output to total input:

$\eta=\frac{E_{useful}}{E_{input}}=\frac{P_{useful}}{P_{input}}$

Choose matching quantities

Use energy ratios for the same process and time interval, or power ratios when input and output are rates. Efficiency is dimensionless and is often reported as a percentage.

Worked example from local Question Bank row 29709

Solar intensity is $240\,\mathrm{W\,m^{-2}}$ over $2.50\times10^4\,\mathrm{m^2}$ , so input power is

$P_{in}=(240)(2.50\times10^4)=6.0\times10^6\,\mathrm{W}=6.0\,\mathrm{MW}$

For a useful output of $1.6\,\mathrm{MW}$ ,

$\eta=\frac{1.6}{6.0}=0.27=27\%$

The remaining input is transferred through non-useful pathways.

Common trap

Do not invert the ratio or use the total output, including unwanted transfers, as the useful output.

How it is assessed

The evidence asks for motor input power from output power and efficiency, and for fuel mass or energy from a vehicle’s kinetic-energy gain and efficiency.

Command terms

Calculate / Identify

What earns marks

Use the useful-output/input ratio and convert the final fraction to a percentage when required. For a motor, calculate useful mechanical output first, then divide by electrical input power.

Watch for

Using the loss power as useful output or reporting 75 rather than 0.75 when using the ratio.

Representative question

[Maximum number: 1]

An electric motor of efficiency 75 % raises a mass of 120 kg at a constant speed of $0.50 \mathrm{~ms}^{-1}$ . What is the power input to the motor?

20 W

450 W

600 W

800 W

Energy per volume

For the current IB Physics definition, fuel energy density $u$ is the transferable energy per unit volume:

$u=\frac{E}{V}$

Its SI unit is $\mathrm{J\,m^{-3}}$ . This lets fuels be compared when storage volume is the constraint.

Connect it to a fuel flow

If fuel flows at volume rate $\dot V$ , its input power is $P_{in}=u\dot V$ . Apply efficiency only after finding the input energy or power.

Worked example from local Question Bank row 36970

An engine produces $20\,\mathrm{kW}$ useful power at $50\%$ efficiency while consuming $1.0\times10^{-5}\,\mathrm{m^3\,s^{-1}}$ of fuel.

$P_{in}=\frac{20\,\mathrm{kW}}{0.50}=40\,\mathrm{kW}$
$u=\frac{P_{in}}{\dot V}=\frac{4.0\times10^4}{1.0\times10^{-5}}=4.0\times10^9\,\mathrm{J\,m^{-3}}=4.0\,\mathrm{GJ\,m^{-3}}$

Common trap

Specific energy is energy per unit mass, measured in $\mathrm{J\,kg^{-1}}$ . Some sources use the words loosely, so let the stated definition and units determine whether to divide by volume or mass.

How it is assessed

The evidence asks for fuel volume for a rocket manoeuvre and for energy density from useful engine power and fuel consumption rate.

Command terms

Estimate / Calculate

What earns marks

Use the fuel energy density with the fuel volume to find input energy, then apply efficiency and any time or kinetic-energy relation. Keep volume units in m³ when the density is given in J m⁻³.

Watch for

Using mass-specific energy when volume-specific energy is given, or omitting efficiency before comparing useful output.

Representative question

[Maximum number: 2]

At the end of the 30-day period, rockets are fired to bring the ISS back to its initial height. The energy density of liquid hydrogen rocket fuel is $8.5 \times 10^{3} \mathrm{MJ} \mathrm{m}^{-3}$ .

Estimate the volume of fuel needed.

$\begin{aligned} V=\frac{E}{E \text { density }} V=\mu \frac{4.5 \times 10^{9}}{8.5 \times 10^{9}}=>0.53 \mathrm{~m}^{3} \end{aligned}$

Award [2] if $0.53 \ll \mathrm{~m}^{3} \gg$ is seen as the
answer without working

Account for energy

Define the system, identify energy stores and describe transfers. Work done by a force transfers energy; total energy is conserved even when mechanical energy is not.

Use the mechanical model

$E_k= rac12mv^2,\quad \Delta E_{p,g}=mg\Delta h,\quad E_{p,elastic}= rac12k(\Delta x)^2$

Conserve their sum only when resistive transfers are absent or included explicitly.

Use rates and ratios

$P= rac{\Delta E}{\Delta t}=Fv,\qquad \eta= rac{E_{useful}}{E_{input}}= rac{P_{useful}}{P_{input}}$

Fuel energy density connects available input energy to a chosen volume.

Final checks

Check the system boundary, signs of work and potential-energy changes, the reference height, extension from natural length, and whether the quantity is energy, power, efficiency or energy density.

Topic —

A.4 Rigid body mechanics

Objectives in this topic

Torque is a turning effect

The torque of a force about an axis is

$\tau=Fr\sin\theta$

where $r$ is the distance from the axis to the point of application and $\theta$ is the angle between $\vec r$ and $\vec F$ .

Use the perpendicular lever arm

Equivalently, torque equals force multiplied by the perpendicular distance from the axis to the force’s line of action.

Choose a rotation sign

Clockwise and anticlockwise torques have opposite signs. Add torques about the specified axis rather than adding their magnitudes blindly.

Worked example from local Question Bank row 37867

Two forces produce the same rotational sense: $50\,\mathrm{N}$ at a perpendicular distance $0.50\,\mathrm{m}$ and $40\,\mathrm{N}$ at $0.20\,\mathrm{m}$ .

$\tau_{net}=(50)(0.50)+(40)(0.20)=25+8=33\,\mathrm{N\,m}\approx30\,\mathrm{N\,m}$

If one force acted in the opposite sense, its torque would enter with the opposite sign.

Common trap

A force through the axis has zero torque, even if its magnitude is large.

How it is assessed

The evidence asks for torque on an accelerating disk, rewarding the moment-of-inertia calculation followed by rotational Newton’s second law.

Command terms

Calculate

What earns marks

Use the perpendicular lever arm or τ=Fr sinθ. If angular acceleration is involved, find moment of inertia and then use τ=Iα. State the axis and sign convention.

Watch for

Using the full radius when the force’s line of action has a smaller perpendicular distance.

Representative question

[Maximum number: 2]

Calculate the torque that acts on the disk while it accelerates.

ALTERNATIVE 1
$\begin{aligned} & I=\frac{1}{2} \times 0.2 \times 0.4^{2}=<0.016 \gg & \text { Torque }=<I x \alpha=>0.1<\mathrm{Nm} \gg \end{aligned}$

ALTERNATIVE 2
$\begin{aligned} & \Delta L=\frac{1}{2} \times 0.20 \times 0.40^{2} \times 12.5=0.20 \text { «Js» } & \Gamma=\frac{\Delta L}{\Delta t}=\frac{\frac{1}{2} \times 0.20 \times 0.40^{2} \times 12.5}{2.0}=0.10 \text { «Nm» } \end{aligned}$

Use of 6 gives an answer of 0.096 Nm .

Allow ECF from MP1

Equilibrium condition

A rigid body is in rotational equilibrium when the resultant torque about any chosen axis is zero:

$\sum\tau=0$

Balance clockwise and anticlockwise effects

Choose an axis, assign signs, and set the sum of clockwise torques equal to the sum of anticlockwise torques. A body can still have translational equilibrium as a separate condition.

Common trap

Zero resultant torque means no angular acceleration; it does not by itself prove that the net force is zero.

How it is assessed

The evidence asks directly for the condition for rotational equilibrium.

Command terms

State

What earns marks

State that rotational equilibrium requires zero resultant torque or zero net moment about the chosen axis. If the question also concerns rest, check translational equilibrium separately.

Watch for

Saying that every individual torque must be zero rather than that the signed resultant torque is zero.

Representative question

[Maximum number: 1]

State the condition for rotational equilibrium.

Rotational second law

A non-zero resultant torque causes angular acceleration:

$\sum\tau=I\alpha$

Use the chosen axis

Calculate signed torques about the specified axis and use the moment of inertia about that same axis. The direction of $\alpha$ follows the resultant torque.

Common trap

Do not use translational $F=ma$ for a purely rotational equation or mix an inertia about one axis with torque about another.

How it is assessed

The evidence asks for angular acceleration of a disk from angular displacement and time, and includes an unrelated fibre calculation as a distractor context; focus on the rotational objective evidence.

Command terms

Calculate / Determine

What earns marks

Find the angular displacement or torque relation from the diagram, then use the relevant rotational equation. Keep radians, seconds and the moment of inertia about the stated axis consistent.

Watch for

Using linear displacement or speed in a rotational equation, or failing to convert degrees/revolutions into radians.

Representative question

[Maximum number: 3]

Calculate the angular acceleration of the disk.

ALTERNATIVE 1
use of rotational kinematics equation to get $\alpha=\frac{2 \theta}{t^{2}}$ $\theta=<\frac{1.5}{0.555}=>27.3<\mathrm{rad} \gg$ $\alpha=\frac{2 \times 27.3}{0.96^{2}}=>59$ «rad $s^{-2}$ »
ALTERNATIVE 2
final speed $\mathrm{v}=2 \times 1.5 / 0.96=3.12 \mathrm{~ms}^{-1}$
final angular speed $\omega=\mathrm{v} / \mathrm{R}$ $\alpha=\omega / 0.96=59$ «rad s ${ }^{-2}$ »
ALTERNATIVE 3
acceleration $=2 \mathrm{~L} / \mathrm{t}^{2}$
acceleration $=3.26 \mathrm{~m} \mathrm{~s}^{-2}$
angular acceleration = acceleration/R =59 <rad s- $\mathrm{s}^{-2}$ »

Marking guidance:

Award [3] for bald correct answer from
interval 58.0 to 59.3.

Three angular quantities

Angular displacement $\theta$ describes change in orientation, angular velocity $\omega=d\theta/dt$ describes how fast orientation changes, and angular acceleration $\alpha=d\omega/dt$ describes how angular velocity changes.

Link to linear motion

At radius $r$ , tangential speed is $v=r\omega$ . Keep angular quantities in radians when using these relationships.

Common trap

Angular velocity is not automatically the same as linear speed; the radius is needed to connect them.

How it is assessed

The evidence asks for angular velocity of a point on a circle and time to reach a specified angular position.

Command terms

Calculate

What earns marks

Use ω=v/r for angular velocity and relate angular displacement, angular velocity and time with the stated motion. Convert revolutions to radians when a full-turn quantity is given.

Watch for

Using circumference or linear speed without dividing by radius, or mixing revolutions with radians.

Representative question

[Maximum number: 1]

Calculate the angular velocity $\omega$ of P.

$\omega=\frac{v}{r}=\frac{2}{0.9}=2.22 \mathrm{rads}^{-1}$

Uniform angular acceleration

When $\alpha$ is constant, use rotational SUVAT:

$\omega=\omega_0+\alpha t,\quad \theta=\omega_0t+\frac12\alpha t^2,\quad \omega^2=\omega_0^2+2\alpha\theta$

Choose the equation

List $\theta,\omega_0,\omega,\alpha,t$ , convert revolutions to radians, and choose the equation containing the required unknown and known quantities.

Worked example from local Question Bank row 39408

A bar starts from rest and turns through six revolutions with constant $\alpha=0.110\,\mathrm{rad\,s^{-2}}$ . Convert $\Delta\theta=6(2\pi)=12\pi\,\mathrm{rad}$ , then

$\omega_f^2=\omega_i^2+2\alpha\Delta\theta=0+2(0.110)(12\pi)$
$\omega_f=2.88\,\mathrm{rad\,s^{-1}}\approx2.9\,\mathrm{rad\,s^{-1}}$

The equation is valid because angular acceleration is constant.

Common trap

Do not use rotational SUVAT when angular acceleration varies, and do not insert degrees or revolutions where radians are required.

How it is assessed

The evidence asks for revolutions completed by a rolling wheel and angular acceleration of a disk from angular displacement and time.

Command terms

Calculate / Determine

What earns marks

Convert the angular displacement to radians, identify whether angular acceleration is uniform, and choose the rotational SUVAT equation matching the known quantities. For revolutions, divide by 2π to find the number of turns.

Watch for

Using linear SUVAT or treating a revolution as one radian.

Representative question

[Maximum number: 1]

A wheel, initially at rest, rolls without slipping down an incline for 4.0 s . The final angular velocity of the wheel is $5 \pi \mathrm{rads}^{-1}$ .

How many revolutions did the wheel complete?

Rotational inertia

Moment of inertia measures resistance to angular acceleration about an axis. It depends on total mass and how far that mass is distributed from the axis.

Compare distributions

For the same mass and outer radius, more mass farther from the axis gives larger $I$ . A ring therefore has greater rotational inertia than a disk of the same mass and radius.

Common trap

Moment of inertia is not determined by mass alone; always specify the rotation axis and distribution.

How it is assessed

The evidence asks which of a disk and ring reaches the bottom first, rewarding comparison of their rotational inertia and energy allocation.

Command terms

Explain

What earns marks

Compare the moment of inertia for each mass distribution about the same axis. The object with smaller I gains angular speed or reaches the bottom sooner under the same available energy and rolling constraints.

Watch for

Comparing only total mass and radius while ignoring that the ring places more mass farther from the axis.

Representative question

[Maximum number: 3]

The disk and a ring, with the same mass and radius, are released from the top of the slope at the same time. Explain, without numerical calculation, which one will reach the bottom of the inclined plane first.

Point-mass model

For discrete masses rotating about an axis,

$I=\sum mr^2$

where each $r$ is the perpendicular distance from the axis.

Build the sum

Treat each small sphere, blade or mass element separately, calculate $mr^2$ , and add the contributions. Use symmetry when identical masses have equal radii.

Worked example from local Question Bank row 128743

Two $10\,\mathrm{kg}$ point masses are $8.0\,\mathrm{m}$ apart and rotate about the midpoint. Each is $4.0\,\mathrm{m}$ from the axis:

$I=\sum mr^2=2(10)(4.0)^2=320\,\mathrm{kg\,m^2}$

The full $8.0\,\mathrm{m}$ separation is not the radius of either mass.

Common trap

Do not use the distance between two masses as $r$ for both; use each mass’s distance to the rotation axis.

How it is assessed

The evidence asks for the moment of inertia of a propeller or two spheres connected by a rod, rewarding the correct distances to the axis.

Command terms

Show / Calculate

What earns marks

For each discrete mass, use its perpendicular distance from the axis in I=Σmr². Show the contributions and keep units kg m². For blades or spheres, check the geometry before summing.

Watch for

Using the full separation or blade length for each mass instead of its distance to the axis.

Representative question

[Maximum number: 1]

A two-blade propeller can be modelled using the two-cylinder arrangement in (a)(iii).

The following data for the two-blade propeller are available:
Length of each blade: 0.60 m
Mass of each blade: 2.2 kg
Show that the moment of inertia of the two-blade propeller is about $0.5 \mathrm{~kg} \mathrm{~m}^{2}$ .

$\frac{2}{3} \times 2.2 \times 0.6^{2}$

OR
$\frac{1}{12} \times 4.4 \times 1.2^{2}$

OR

0.53 «kg $\mathrm{m}^{2}$ »

Answer of $0.5 \mathrm{~kg} \mathrm{~m}^{2}$ given, so award the
mark if candidates show a correct full
substitution OR the value with an extra
significant figure

Torque–inertia relation

For rotation about a fixed axis,

$\tau_{net}=I\alpha$

Connect translation and rotation

When a force drives a rotating body or pulley, write both the translational force balance and the rotational torque balance if the system has translating and rotating parts.

Worked example from local Question Bank row 31942

A $50\,\mathrm{N}$ tangential force acts $2.0\,\mathrm{m}$ from the axis of a system with $I=450\,\mathrm{kg\,m^2}$ .

$\tau=Fr=(50)(2.0)=100\,\mathrm{N\,m}$
$\alpha=\frac{\tau}{I}=\frac{100}{450}=0.22\,\mathrm{rad\,s^{-2}}$

The acceleration direction follows the signed resultant torque.

Common trap

Do not treat torque as force or use a moment of inertia that does not match the rotation axis.

How it is assessed

The evidence includes a coupled blocks-and-pulley system and an angular-acceleration versus torque graph used to find moment of inertia.

Command terms

Show / Identify

What earns marks

Use τ=Iα for the rotating component and combine it with translational equations when masses accelerate linearly. From an α–τ graph, the gradient is 1/I.

Watch for

Reading the graph gradient as I instead of 1/I or omitting the torque contribution from the pulley.

Representative question

[Maximum number: 1]

The graph shows how the angular acceleration $\alpha$ of a flywheel varies with torque $\tau$ applied to the flywheel.

What is the moment of inertia of the flywheel?

$0.20 \mathrm{~kg} \mathrm{~m}^{2}$

$5.0 \mathrm{~kg} \mathrm{~m}^{2}$

$40 \mathrm{~kg} \mathrm{~m}^{2}$

$80 \mathrm{~kg} \mathrm{~m}^{2}$

Rotational momentum

For a rigid body rotating about a fixed axis,

$L=I\omega$

Angular momentum is directed along the rotation axis by the right-hand convention.

Use the matching inertia

The moment of inertia must be calculated about the same axis used for $L$ . A larger $I$ at the same angular speed means larger angular momentum.

Worked example from local Question Bank row 31945

For $I=450\,\mathrm{kg\,m^2}$ and $\omega=1.66\,\mathrm{rad\,s^{-1}}$ ,

$L=I\omega=(450)(1.66)=7.47\times10^2\,\mathrm{kg\,m^2\,s^{-1}}\approx750\,\mathrm{kg\,m^2\,s^{-1}}$

The sign or axis direction must match the chosen rotational convention.

Common trap

Do not substitute translational momentum $mv$ for angular momentum when the question describes rotation.

How it is assessed

The evidence asks for angular momentum from torque and time or compares angular momentum for spinning bodies with equal rotational kinetic energy.

Command terms

Calculate / Identify

What earns marks

Use L=Iω with the moment of inertia about the relevant axis. For a change, calculate ΔL or use the torque–time relation when the evidence describes an angular impulse.

Watch for

Using Iω with the wrong axis or confusing angular momentum with rotational kinetic energy.

Representative question

[Maximum number: 2]

the angular momentum.

ALTERNATIVE 1
$\begin{aligned} & \Delta L \lessdot=\Gamma \Delta t=T R \Delta t »=\frac{12 \times 9.81 \times 0.20 \times 0.55}{6} & \Delta L=2.2 \lessdot \mathrm{JS} » \end{aligned}$

ALTERNATIVE 2
$\begin{aligned} & \omega=<\alpha \Delta t=\frac{g}{3 R} \Delta t=\frac{9.81 \times 0.55}{3 \times 0.20}=>8.99 \text { «rads }{ }^{-1} » & \Delta L \text { « }=I \omega \text { 》 }=\frac{1}{2} \times 12 \times 0.20^{2} \times 8.99=2.2 \text { «Js» } \end{aligned}$

Award [2] for a bald correct answer.

Conservation condition

Angular momentum remains constant when the resultant external torque about the chosen axis is zero:

$L_i=L_f$

Redistribute the mass

When a skater pulls their arms inward, $I$ decreases. With angular momentum conserved, $\omega$ increases.

Worked example from local Question Bank row 34033

A $0.200\,\mathrm{kg}$ particle moving at $12.0\,\mathrm{m\,s^{-1}}$ strikes $0.60\,\mathrm{m}$ from an axis. Its initial angular momentum is

$L_i=mvr=(0.200)(12.0)(0.60)=1.44\,\mathrm{kg\,m^2\,s^{-1}}$

If the combined system has $I_f=0.252\,\mathrm{kg\,m^2}$ and external torque is negligible,

$I_f\omega_f=L_i\Rightarrow\omega_f=\frac{1.44}{0.252}=5.71\,\mathrm{rad\,s^{-1}}$

Common trap

Angular momentum conservation does not require rotational kinetic energy to remain constant when the moment of inertia changes.

How it is assessed

The evidence uses an ice skater pulling in their arms and a disk receiving a rotating block, testing conservation of angular momentum.

Command terms

Calculate / Explain

What earns marks

Check that external torque is negligible, then set Iiωi=Ifωf. For a skater or disk, compare the change in mass distribution and moment of inertia before solving for the new angular speed.

Watch for

Assuming angular speed is unchanged when the moment of inertia changes or conserving kinetic energy instead of angular momentum.

Representative question

[Maximum number: 1]

An ice skater is spinning with their arms extended in a fixed position at a constant angular velocity. The ice skater then quickly pulls their arms closer to their body. Frictional effects are negligible.

Three statements are made about the ice skater's motion.

I. The angular momentum of the ice skater remains constant.
II. The rotational kinetic energy of the ice skater remains constant.
III. The net torque acting on the ice skater is zero.

Which of the statements are correct?

I and II only

I and III only

II and III only

I, II and III

Angular impulse

A torque acting for a time changes angular momentum:

$\Delta L=\tau\Delta t=\Delta(I\omega)$

Area under a torque–time graph

If torque varies, the signed area under a $\tau$ -against-time graph gives angular impulse and therefore the change in angular momentum.

Common trap

Angular impulse has units N m s, not N s; keep it distinct from linear impulse.

How it is assessed

The evidence asks for the unit of angular impulse and for the physical quantity represented by the area under a torque–time graph.

Command terms

Identify

What earns marks

Identify angular impulse as the change in angular momentum. Use ΔL=τΔt for constant torque or the area under a torque–time graph; report units N m s.

Watch for

Choosing N s, the unit of linear impulse, instead of N m s for angular impulse.

Representative question

[Maximum number: 1]

What is the unit of angular impulse?

$\mathrm{Nms}^{-1}$

Nms

Rotational energy

For a rigid body rotating about a fixed axis,

$E_k=\frac12I\omega^2=\frac{L^2}{2I}$

Combine forms of motion

A rolling object may have translational kinetic energy of its centre of mass and rotational kinetic energy about its centre. Include both when accounting for total kinetic energy.

Worked example from local Question Bank row 31644

A rod of weight $36.0\,\mathrm{N}$ lowers its centre of mass by $5.00/2=2.50\,\mathrm{m}$ and has $I=30.6\,\mathrm{kg\,m^2}$ . If the gravitational transfer becomes rotational kinetic energy,

$E_k=(36.0)(2.50)=90.0\,\mathrm{J}$
$90.0=\frac12(30.6)\omega^2\Rightarrow\omega=2.43\,\mathrm{rad\,s^{-1}}$

The energy result is in joules; angular speed is in radians per second.

Common trap

Do not use $\frac12mv^2$ alone for a rotating body when its rotational motion contributes energy.

How it is assessed

The evidence asks for the rotational-energy fraction of a rolling car or for energy lost from a rotating disk.

Command terms

Determine / Calculate

What earns marks

Use Ek=1/2Iω² for rotation and add translational kinetic energy for rolling motion. If angular momentum is given, use L²/(2I), keeping the same axis and energy units.

Watch for

Omitting the translational component for rolling wheels or using linear kinetic energy with angular speed.

Representative question

[Maximum number: 1]

A car of total mass M is travelling with a constant speed v. Each of the four wheels of the car has a mass m and a radius R and rolls without slipping.

The moment of inertia of each wheel is $I=\frac{1}{2} m R^{2}$ .
What is $\frac{\text { sum of the rotational kinetic energy of all four wheels }}{\text { translational kinetic energy of the car }} ?$

$\frac{m}{2 M}$

$\frac{m}{M}$

$\frac{2 m}{M}$

$\frac{4 m}{M}$

Torque and rotation

Use $au=Fr\sin heta$ , $\sum au=0$ for rotational equilibrium and $\sum au=Ilpha$ for angular acceleration. Always state the axis and use the matching moment of inertia.

Describe angular motion

Use angular displacement, angular velocity and angular acceleration; rotational SUVAT applies only for uniform $lpha$ . Point-mass inertia is $I=\sum mr^2$ .

Track angular momentum

$L=I\omega,\quad \Delta L= au\Delta t$

Conserve angular momentum only when external resultant torque is negligible.

Track rotational energy

$E_{rot}= rac12I\omega^2$

For rolling or coupled systems, include translational and rotational energy separately.

Topic —

A.5 Galilean and special relativity

Objectives in this topic

Define the frame

A reference frame is a coordinate system, with a chosen origin and axes, together with a way of assigning time to events. Position and time are coordinates: they describe an event only after the observer’s frame has been stated.

Describe the same event in two frames

For an event, record its position $x$ and time $t$ in frame S, then $x′$ and $t′$ in frame S′. If S′ moves at constant velocity relative to S, both frames are inertial. The event is the same physical occurrence even though its coordinates may differ.

Check the frame type

An inertial reference frame is non-accelerating. Newton’s laws can be used in their usual form in such a frame. If the frame accelerates or rotates, extra apparent forces may be needed and it is not an inertial frame for this syllabus treatment.

Common trap

A frame is not just a camera viewpoint. It specifies the spatial axes and the time measurement used to assign coordinates; therefore “at rest” or “moving” has meaning only relative to a stated frame.

How it is assessed

The evidence tests whether you can identify a frame of reference and explain why an observer or object is at rest in a chosen frame. The scoring focus is the absence of relative velocity or change in position, plus an explicit position-and-time coordinate description when a definition is requested.

Command terms

State / Define / Explain

What earns marks

State the reference frame before interpreting motion. Define the spatial origin and time coordinate, identify the observer or object at rest in the frame, and use relative motion consistently. For a definition question, include both coordinates/axes and the time measurement.

Watch for

Describing a frame as only a visual viewpoint without mentioning coordinates and time.

Representative question

[Maximum number: 1]

Explain why observer Y is at rest in the reference frame of the electron.

State the principle

Galilean relativity says that Newton’s laws have the same form in every inertial reference frame. An observer moving at constant velocity therefore uses the same Newtonian mechanics, provided speeds are far below the speed of light and the frame is non-accelerating.

Relate the observations

The same event occurs in both frames, but the observers can assign different positions. In the classical model, time is absolute: both observers use the same $t$ , while the moving frame changes the position coordinate according to $x′=x-vt$ .

Check the boundary

Use Galilean relativity for inertial frames in the non-relativistic limit. It is not the correct model for measurements involving speeds comparable with $c$ , where the assumptions of absolute time and unchanged light speed fail.

Common trap

“Same laws” does not mean that all observers measure the same position or velocity. It means the equations of Newtonian mechanics keep the same form after changing between inertial frames.

How it is assessed

Questions use a moving frame or a length/time comparison and ask you to apply or explain the classical transformation. Marks depend on recognizing the inertial-frame assumption and the Newtonian absolute-time model.

Command terms

State / Explain / Outline

What earns marks

Identify both frames as inertial, state that Newton’s laws retain the same form, and distinguish the classical assumptions of absolute time and Galilean coordinate transformation from the later relativistic model. Use the given frame velocity and sign convention consistently.

Watch for

Applying Galilean relativity to a relativistic situation without checking that the non-relativistic assumptions are appropriate.

Representative question

[Maximum number: 1]

Write down the length of the space station according to Galilean relativity.

Transform event coordinates

For frames S and S′ that are coincident at $t=t′=0$ , with S′ moving at velocity $v$ in the positive $x$ -direction relative to S, the Galilean transformations are $x′=x-vt$ and $t′=t$ .

Use the sign convention

Start with the event coordinates $(x,t)$ in S. Substitute the frame velocity with its signed value, calculate $x′$ , and carry the unchanged time $t′=t$ into S′. A positive $x′$ means the event is on the positive side of S′’s origin.

Check the assumptions

These equations describe the classical model: inertial frames, common synchronized time, and an origin coincidence at $t=0$ . They are not the Lorentz transformations and do not preserve the speed of light between frames.

Symbolic example from local Question Bank row 31622

An event has $x'=L$ and $t'=L/c$ in S′. Galilean time is absolute, so $t=t'=L/c$ . Using the inverse position transformation,

$x=x'+vt=L+v\frac{L}{c}$

The extra $vL/c$ is the distance travelled by the S′ origin during the shared time interval.

Common trap

Do not change $t$ using $t′=t-vx/c^2$ ; that belongs to the relativistic transformation. In a Galilean transformation, time is shared: $t′=t$ .

How it is assessed

Questions ask you to state the classical assumptions or use a frame diagram to compare coordinates. The mark scheme distinguishes shared time and the possibility of speeds greater than c from the relativistic assumptions.

Command terms

State / Calculate / Outline / Suggest

What earns marks

Write the frame relationship before substituting: x′ = x − vt and t′ = t, with the frames coincident at t = 0. Keep the direction of v and the coordinate signs consistent, show units, and state whether the result is a position or a time coordinate.

Watch for

Using a Lorentz time transformation or reversing the sign of vt without checking which frame moves relative to the other.

Representative question

[Maximum number: 1]

The Lorentz transformations assume that the speed of light is constant. Outline what the Galilean transformations assume.

Use the classical addition rule

If an object has velocity $u$ in frame S and frame S′ moves at velocity $v$ relative to S, the velocity measured in S′ is $u′=u-v$ . Both $u$ and $v$ are signed velocities along the chosen axis.

Calculate in two steps

Choose the positive direction, write the velocity of the object and the relative frame velocity with signs, then subtract. If the object and S′ move in the same direction, the relative speed is reduced; if they move in opposite directions, the signed relative velocity has greater magnitude.

Check the model

Galilean addition assumes inertial frames, common time, and non-relativistic speeds. It can predict a resultant speed greater than $c$ , which signals that special relativity—not the arithmetic—is required for a high-speed light problem.

Worked example from local Question Bank row 31297

In the classical model, two spacecraft moving in opposite directions at $0.80c$ have signed velocities $u=-0.80c$ and $v=+0.80c$ .

$u'=u-v=-0.80c-0.80c=-1.60c$

The magnitude $1.60c$ shows why Galilean addition cannot describe relativistic relative velocity; the sign only gives direction in S′.

Common trap

Do not subtract speed magnitudes before deciding the direction. The sign of $u′$ tells you the object’s direction in S′; dropping signs can reverse the physical interpretation.

How it is assessed

Questions ask for the speed of a signal or the travel time seen by an observer after combining velocities. The evidence rewards a correct relative-velocity expression and a clearly shown substitution.

Command terms

State / Calculate

What earns marks

Choose a positive direction and write u′ = u − v before substituting. Use signed velocities, keep c in the units where it is given, show the subtraction, and interpret the sign of u′ as the direction in the moving frame.

Watch for

Treating velocity magnitudes as unsigned and losing the direction of the relative motion.

Representative question

[Maximum number: 1]

State, using Galilean relativity, the speed of the radio signal relative to Q .

Postulate 1: relativity

The laws of physics have the same form in all inertial reference frames. No inertial observer is privileged by uniform motion; an experiment performed entirely within a frame cannot reveal its constant velocity relative to another inertial frame.

Postulate 2: invariant light speed

Every inertial observer measures the same speed of light in vacuum, $c$ , independent of the motion of the source or observer. This replaces the Galilean expectation that measured velocities simply add.

Use the consequence

Together, the postulates require space and time coordinates to transform differently from the Galilean model. They lead to Lorentz transformations, time dilation, length contraction and relativity of simultaneity. The syllabus does not require deriving those equations.

Common trap

The second postulate does not say that every object moves at $c$ , or that light speed is the same in every material. It refers to light in vacuum measured by inertial observers.

How it is assessed

The objective is assessed through short explanation or consequence questions in relativistic settings. Identify which postulate supplies the frame argument and which supplies the constant-light-speed argument before explaining the consequence.

Command terms

State / Explain

What earns marks

State both postulates separately. For the first, name all inertial frames and the same form of the laws of physics. For the second, state that all inertial observers measure the same vacuum light speed c, independent of source or observer motion. Do not replace either statement with a consequence such as time dilation.

Watch for

Giving only “nothing can travel faster than light” and omitting the two postulates or the condition of inertial observers.

Representative question

[Maximum number: 3]

Explain, by reference to the equivalence principle, why the frequency of the photon measured at B will be larger than $f_{0}$ .

Set the relativistic factor

For two inertial frames with relative speed $v$ , use $\gamma=1/\sqrt{1-v^2/c^2}$ . At ordinary speeds $\gamma\approx1$ ; as $v$ approaches $c$ , the difference from Galilean coordinates becomes significant.

Transform one event

For an event with coordinates $(x,t)$ in S, the syllabus equations are $x′=\gamma(x-vt)$ and $t′=\gamma(t-vx/c^2)$ in S′. Use the same signed direction for $x$ and $v$ , and calculate both coordinates from the same event.

Show the coordinate method

Write $\gamma$ first, substitute the given $x,t,v$ , then report $x′$ and $t′$ with units. If the question provides a space–time diagram, reading the coordinates is an alternative only when the axes and scale are correctly interpreted.

Worked example from local Question Bank rows 31630–31631

For $v=0.745c$ , $\gamma=1/\sqrt{1-0.745^2}=1.499$ . An event at $x=1.0\,\mathrm{m}$ and $t=0$ transforms to

$x'=\gamma(x-vt)=1.499(1.0)=1.50\,\mathrm{m}$
$ct'=\gamma(ct-\tfrac{v}{c}x)=1.499(0-0.745\times1.0)=-1.12\,\mathrm{m}$

Thus $t'=-1.12/c\,\mathrm{s}$ ; the negative coordinate is valid and reflects the chosen origins.

Respect the syllabus boundary

Know and apply the transformation equations; their derivation is not required. The frames must be inertial, and the relative speed must satisfy $v<c$ so that $\gamma$ is real.

How it is assessed

Questions ask you to determine transformed space–time coordinates or show that two events simultaneous in one frame are not simultaneous in another. The evidence accepts a correct diagram method or Lorentz calculation, but signs and the non-zero transformed time difference matter.

Command terms

Determine / Show

What earns marks

Calculate γ from the value of v, then apply x′ = γ(x − vt) and t′ = γ(t − vx/c²) to the same event. Keep c and distance/time units consistent, show the substitution, and check signs against the stated frame direction.

Watch for

Using x′ = x − vt and t′ = t for a high-speed event, or omitting the vx/c² term in the transformed time.

Representative question

[Maximum number: 3]

Determine the spacetime coordinates of the event according to observer B.

ALTERNATIVE 1 using diagram:
line drawn in (b)(ii) intersecting $c t^{\prime}$ between -2 and -2.75
line drawn parallel to $c t^{\prime}$ intersecting with $x^{\prime}$ from (3,1)
$x^{\prime}$ between 3 and 4
ALTERNATIVE 2 using Lorentz transformation:
$\begin{aligned} \gamma=1.66 c t^{\prime} \leftrightarrow=\gamma\left(c t-\frac{v x}{c}\right)=1.66(1-0.8 \times 3) »=-2.3 x^{\prime} \boldsymbol{\kappa}=\gamma(x-v t)=1.66(3-0.8 \times 1) »=3.7 \end{aligned}$
ALTERNATIVE 3

Allow ECF from (a).

Without explicit answer, award [2max], even if working on diagram seems to be correct.

Penalise for incorrect signs.

line drawn in (b)(ii) intersecting ct' between -2 and -2.75
use of invariant formula as in b(iv) with values
to get $\mathrm{x}^{\prime}=3.7$

Use the relativistic rule

For an object with velocity $u$ in S and a frame S′ moving at velocity $v$ relative to S, use $u′=(u-v)/(1-uv/c^2)$ . The numerator is the classical relative velocity; the denominator is the correction required by special relativity.

Substitute signed velocities

Choose one positive direction, write signed values for $u$ and $v$ , calculate $uv/c^2$ , and evaluate the full fraction. Report the magnitude if the question asks for speed; retain the sign if it asks for velocity or direction.

Check the limiting cases

When $u\ll c$ and $v\ll c$ , the denominator is close to 1 and the result approaches $u-v$ . If $u=c$ , the equation gives $u′=c$ for any sub-light $v$ , so light does not gain or lose speed between inertial frames.

Worked example from local Question Bank row 31056

A train has $u=-0.70c$ in the ground frame and observer P moves at $v=+0.60c$ .

$u'=\frac{-0.70c-0.60c}{1-(-0.70)(0.60)}=\frac{-1.30c}{1.42}=-0.915c\approx-0.92c$

The negative sign means the train moves opposite P's positive direction; its speed remains below $c$ .

Common trap

Do not use $u-v$ for a high-speed signal. Also do not report a result above $c$ ; a sign error or an omitted denominator usually caused it.

How it is assessed

Questions ask you to determine a signal speed in another frame or compare the relative speed of two fast-moving observers. The evidence rewards the relativistic fraction and a result that remains below or equal to c.

Command terms

Determine

What earns marks

Choose a positive direction, write u′ = (u − v)/(1 − uv/c²), substitute signed velocities, and show the denominator. Give speed as a positive magnitude only when requested; otherwise retain the sign and state the direction.

Watch for

Using u′ = u − v at relativistic speeds or dropping the signs before evaluating the denominator.

Representative question

[Maximum number: 2]

Determine, using relativistic velocity addition, the speed of the radio signal relative to Q .

Define the interval

For two events separated by $\Delta t$ and $\Delta x$ , the space–time interval is $(\Delta s)^2=(c\Delta t)^2-(\Delta x)^2$ . Although observers can measure different $\Delta t$ and $\Delta x$ , the value of $(\Delta s)^2$ is invariant between inertial frames.

Calculate carefully

Read the time and position differences between the same two events. Convert $\Delta t$ into the distance $c\Delta t$ , square both terms, and subtract the spatial term: $(c\Delta t)^2-(\Delta x)^2$ . Keep the sign; a negative result is physically meaningful.

Compare frames

Calculate the interval from either frame’s coordinates. Matching values demonstrate invariance and provide a check on transformed coordinates. For a light signal, $(\Delta s)^2=0$ ; this null interval is consistent with $\Delta x=c\Delta t$ .

Worked example from local Question Bank row 32720

For two events with $c\Delta t=100\,\mathrm{ly}$ and $\Delta x=20\,\mathrm{ly}$ ,

$(\Delta s)^2=(100)^2-(20)^2=10{,}000-400=9{,}600\,\mathrm{ly^2}$

Any inertial frame must calculate the same $9{,}600\,\mathrm{ly^2}$ from its own coordinate differences.

Common trap

Do not replace the subtraction with addition, and do not take an absolute value before reporting. The sign distinguishes the interval type and is part of the answer.

How it is assessed

Questions ask you to calculate an interval or show that two frames give the same value. The evidence specifically rewards the correct subtraction, the negative sign in a spacelike example, and agreement between the two coordinate descriptions.

Command terms

Calculate / Show

What earns marks

Read Δx and Δt for the same two events, convert cΔt to distance units, and write (Δs)² = (cΔt)² − (Δx)² before substituting. Preserve the sign and show both frame calculations when asked to demonstrate invariance.

Watch for

Changing the minus sign to plus or reporting a positive absolute value when the calculated interval is negative.

Representative question

[Maximum number: 2]

Calculate the space-time interval $(\Delta s)^{2}$ between P and Q .

Correct readoffs $\Delta x=3 \mathrm{~m}, c \Delta t=1 \mathrm{~m}$ $(\Delta s)^{2}=\ll 1^{2}-3^{2}=>-8 \mathrm{~m}^{2}$

Marking guidance:

Ignore units as they are not required for the answer.

Do not award MP2 if the answer is positive

Identify proper time

The proper time interval $\Delta t_0$ is measured between two events that occur at the same position in the observer’s frame. It is the shortest time interval measured for those events. For a clock at rest in the frame, successive ticks occur at one location, so the clock measures $\Delta t_0$ .

Identify proper length

The proper length $L_0$ is the length measured in the rest frame of the object. Its endpoints are measured simultaneously in that frame. It is the maximum length assigned to the object by inertial observers.

Choose from the event conditions

For time, ask: do the two events happen at the same place in this frame? For length, ask: is the object at rest in this frame, and are both endpoints measured at the same time? These conditions, not the observer’s label, determine whether the measurement is proper.

Common trap

Proper time is not simply the time measured by the “main” observer, and proper length is not the shortest measured length. The proper length is the rest-frame, longest length; moving observers measure a contracted length.

How it is assessed

Questions ask you to define or identify the proper length, and can extend the same reasoning to proper time. The mark scheme rewards the object’s rest frame for length and the same-location condition for time.

Command terms

Define / Explain

What earns marks

For proper length, name the length measured in the object’s rest frame and, when explaining, state that the endpoints are measured simultaneously. For proper time, identify the frame in which both events occur at the same position. Do not choose based on which observer is named first.

Watch for

Calling the shortest measured length the proper length instead of selecting the object’s rest frame.

Representative question

[Maximum number: 1]

Define what is meant by proper length.

Use the time-dilation equation

When $\Delta t_0$ is the proper time between two events, an observer for whom the events occur at different positions measures $\Delta t=\gamma\Delta t_0$ , where $\gamma=1/\sqrt{1-v^2/c^2}$ .

Identify the proper interval first

Find the frame in which the two events occur at the same place; that frame measures $\Delta t_0$ . Calculate $\gamma$ using the relative speed, then multiply by $\Delta t_0$ to obtain the longer interval measured in the other inertial frame.

Check the direction of the effect

Because $\gamma\ge1$ , the non-proper observer measures a time interval at least as large as the proper interval. At $v=0$ , $\gamma=1$ and the two measurements agree.

Worked example from local Question Bank rows 30639–30640

A spacecraft crosses $1.80\times10^{11}\,\mathrm{m}$ at $0.750c$ . The station-frame interval is

$\Delta t=\frac{1.80\times10^{11}}{0.750(3.00\times10^8)}=800\,\mathrm{s}$

With $\gamma=1/\sqrt{1-0.750^2}=1.51$ , the spacecraft clock measures the proper time

$\Delta t_0=\frac{800}{1.51}=530\,\mathrm{s}$

The events occur at one place on the spacecraft, so its interval is proper.

Common trap

Do not multiply the proper time by $1/\gamma$ when finding the dilated interval. The inverse is used only when the question gives the larger interval and asks for the proper time.

How it is assessed

Questions ask you to calculate a moving observer’s time or infer a proper time from a longer Earth-frame interval. The evidence rewards finding γ and using the correct direction of the relationship.

Command terms

Calculate

What earns marks

Identify Δt0 as the interval measured where both events occur at the same place, calculate γ = 1/√(1 − v²/c²), and use Δt = γΔt0. Show the substitution and check that the dilated interval is not smaller than the proper interval.

Watch for

Using the Earth-frame interval as Δt0 without checking where the two events occur at the same position.

Representative question

[Maximum number: 2]

S arrives at P after 50 years according to Earth. Calculate the time at which S arrives at P according to S clocks.

$\begin{aligned} \gamma=\frac{1}{\sqrt{1-0.60^{2}}} \text { OR } \frac{5}{4} \text { OR } 1.25 t=\gamma t^{\prime} \Rightarrow t^{\prime}=\frac{4}{5} \times 50=40 \mathrm{yr} \end{aligned}$

Award MP1 if seen isolated or within an equation.

Award [2] if 40 <<yr>> is seen as the answer without working

Use the contraction equation

If $L_0$ is the proper length measured in the object’s rest frame, an observer who sees the object moving at speed $v$ measures $L=L_0/\gamma$ , with $\gamma=1/\sqrt{1-v^2/c^2}$ .

Select the proper length

Find the frame in which the object is at rest; that frame measures $L_0$ . Calculate $\gamma$ , then divide the proper length by $\gamma$ . The endpoints must be measured simultaneously in the observer’s frame.

Check the result

Since $\gamma\ge1$ , a moving observer measures $L\le L_0$ . At low speed the contraction is negligible; as $v$ approaches $c$ , the measured length along the direction of motion becomes substantially smaller.

Worked example from local Question Bank row 30450

A rocket's proper length is $L_0=450\,\mathrm{m}$ and $\gamma=5/3$ . An observer who sees it moving measures

$L=\frac{L_0}{\gamma}=\frac{450}{5/3}=270\,\mathrm{m}$

Only the dimension parallel to the relative motion is contracted.

Common trap

Do not contract a length perpendicular to the motion, and do not multiply by $\gamma$ when the requested quantity is the moving-frame length.

How it is assessed

Questions ask you to calculate a moving space station’s length or compare a spaceship measurement with an Earth-frame distance. The evidence rewards finding γ and applying the division in the correct direction.

Command terms

Calculate

What earns marks

Identify L0 as the object’s rest-frame length, calculate γ = 1/√(1 − v²/c²), and use L = L0/γ for the moving observer. State that the measurement is along the direction of motion and check that L is no greater than L0.

Watch for

Multiplying the proper length by γ or applying contraction to a direction perpendicular to the relative motion.

Representative question

[Maximum number: 2]

Calculate the length of the space station according to observer B, with reference to special relativity.

$\gamma=$ « $\left\{\frac{1}{\sqrt{1-\frac{0.6^{2}}{1^{2}}}}\right\}>=1.25$

«100/1.25 =>80«m»

State the idea

Two events that are simultaneous in one inertial reference frame need not be simultaneous in another frame moving relative to it. Simultaneity is therefore not an absolute property of separated events.

Use the transformed time

For two events, $\Delta t'=\gamma(\Delta t-v\Delta x/c^2)$ . If $\Delta t=0$ in S but $\Delta x\neq0$ , then $\Delta t'\neq0$ in a relatively moving frame.

Worked example from local Question Bank row 30453

Two lamps are simultaneous in S and separated by $9.00\times10^3\,\mathrm{m}$ . For $v=0.80c$ and $\gamma=5/3$ ,

$\Delta t'=\frac53\left(0-\frac{(0.80c)(9.00\times10^3)}{c^2}\right)=-4.0\times10^{-5}\,\mathrm{s}$

The negative sign fixes the event order in S′; it is not an error.

Keep the order test local

To decide which event occurs first in a frame, calculate or read the sign of $\Delta t′$ using the same pair of events. A negative time difference means the event assigned as the second reference event occurs earlier in that frame.

Common trap

The relativity of simultaneity concerns spatially separated events. Events at the same place cannot be simultaneous in one frame and ordered differently in another inertial frame.

How it is assessed

Questions ask which event occurs first for a spacecraft observer or ask you to justify an event order from a space–time diagram. The evidence rewards reading the transformed time sign and identifying the correct frame.

Command terms

Determine / Justify / Explain

What earns marks

Identify the two spatially separated events, write Δt′ = γ(Δt − vΔx/c²), and use its sign to determine their order in the requested frame. If Δt = 0 in one frame but Δx ≠ 0, conclude that Δt′ is non-zero in a moving frame.

Watch for

Assuming that simultaneous events in one frame must remain simultaneous for all observers.

Representative question

[Maximum number: 2]

According to observer B, event E occurs before observer A and observer B meet. Justify this statement using the spacetime diagram.

lines drawn from (3,1) roughly parallel to $x^{\prime}$ to intersect with $c t^{\prime}$ axis
according to B , event is taking place at $t^{\prime}<0 /$ before origin «so before»

Watch for ECF from bi).

Marking guidance:

Allow working on diagram OR correct arguments in the answer box.

Accept use of Lorentz transformation to show ct' =-2.3.

Read the axes

A space–time diagram plots position horizontally and $ct$ vertically; the time axis is labelled $ct$ , so both axes have distance units. An event is a point $(x,ct)$ . A world line joins the events of one object through time.

Interpret a world line

A vertical world line represents an object at rest in that frame. A straight tilted line represents constant velocity. A light ray has $v=c$ and lies on the 45° light line when the axes use equal scales; no physical world line may be steeper toward the x-axis than the light line.

Read simultaneity and coordinates

To find an event’s time, project horizontally to the $ct$ axis; to find position, project vertically to the $x$ axis. Lines parallel to an observer’s $x′$ axis represent equal $t′$ , while lines parallel to $ct′$ represent equal $x′$ .

Common trap

Do not treat the slope as an ordinary $x/t$ graph slope without accounting for the $ct$ axis and the diagram’s scale. Always identify which frame’s axes are being used.

How it is assessed

Questions ask you to identify time differences, read coordinates, or determine which event is simultaneous in a second frame. The evidence rewards correct construction lines, frame labels and use of the diagram’s scales.

Command terms

Identify / Determine

What earns marks

Identify the frame axes first, then project the event to the requested ct or x axis. For a world line, use its direction and the light line to infer motion; for simultaneity, use lines parallel to the relevant x-axis. Label construction lines when the question asks you to show the reading.

Watch for

Reading a coordinate from the wrong frame axis or treating a ct axis as an ordinary t axis without using the diagram scale.

Representative question

[Maximum number: 2]

Identify, with lines and labels on the spacetime diagram, the difference between $t_{1}$ and $t_{2}$ .

Use the angle relation

On a space–time diagram with equal scales, the angle $\theta$ between a particle’s world line and the time axis satisfies $\tan\theta=v/c$ . Therefore $v=c\tan\theta$ .

Read the line

A vertical line has $\theta=0$ and represents rest. As the line tilts toward the x-axis, $\theta$ and the speed increase. The light line has $\theta=45^\circ$ on equal scales and represents $v=c$ .

Calculate from a diagram

Measure or read the angle from the time axis, evaluate $\tan\theta$ , and multiply by $c$ . If the diagram gives a rise/run ratio, use that ratio as $\tan\theta$ only after confirming the axes and angle definition.

Worked example from local Question Bank row 31477

For a world line representing $v=0.80c$ on equal-scale axes,

$\theta=\tan^{-1}(v/c)=\tan^{-1}(0.80)=38.7^\circ\approx39^\circ$

The angle is measured from the $ct$ axis, not the $x$ axis.

Common trap

Do not measure the angle from the x-axis, and do not assume every diagram uses equal visual scales. The syllabus relation is tied to the stated world-line angle and labelled axes.

How it is assessed

Questions ask you to select the world line for a stated speed or calculate speed from an angle or gradient. The evidence rewards identifying the correct axis and comparing the line with the light line.

Command terms

Determine / What is

What earns marks

Use the angle measured from the ct axis, write tan θ = v/c, and solve for v. Check the line against the vertical rest line and the 45° light line; keep the answer in terms of c when requested.

Watch for

Using the angle to the x-axis rather than the angle to the ct axis when applying tan θ = v/c.

Representative question

[Maximum number: 1]

Rocket R travels away from an observer on Earth at a speed of 0.80 c . A space-time diagram shows four world lines.

What is the correct world line of R in the reference frame of Earth?

Start from the observation

Muons created high in Earth’s atmosphere have a short proper lifetime, yet many are detected at the ground while travelling at speeds close to $c$ . Without relativistic effects, the flight time through the atmosphere would exceed the muon lifetime and far fewer would survive.

Explain it in the Earth frame

In the Earth frame the moving muon’s lifetime is dilated: $\Delta t=\gamma\Delta t_0$ . The increased lifetime allows more muons to travel the atmospheric distance before decaying. This is experimental evidence for time dilation.

Explain it in the muon frame

In the muon’s frame, the atmosphere is moving and its thickness is length-contracted: $L=L_0/\gamma$ . The shorter distance can be crossed within the muon’s proper lifetime. Both frames predict the same detection rate; together they support time dilation and length contraction.

Evidence calculation from local Question Bank row 30447

Muons are produced $2.0\,\mathrm{km}$ above ground, move at $0.98c$ , have proper lifetime $2.2\,\mu\mathrm{s}$ and $\gamma=5.0$ .

$t_{flight}=\frac{2000}{0.98(3.0\times10^8)}=6.8\,\mu\mathrm{s}$
$t_{Earth}=\gamma t_0=(5.0)(2.2)=11\,\mu\mathrm{s}$

The dilated Earth-frame lifetime exceeds the flight time, explaining why many more muons reach the ground than the non-relativistic model predicts.

Common trap

Do not claim that the muon’s own clock runs slow in its rest frame. The proper lifetime is measured by the muon; the Earth observer measures the dilated lifetime, while the muon observer measures a contracted atmosphere.

How it is assessed

Questions ask you to outline or calculate why muons reach the ground despite their short proper lifetime. The evidence rewards a quantitative comparison and an explicit link to the relativistic effect.

Command terms

Explain / Outline

What earns marks

Use the proper lifetime and atmospheric flight time consistently. In the Earth frame, compare the dilated lifetime γΔt0 with the flight time; in the muon frame, compare the contracted distance L0/γ with the proper lifetime. State that both descriptions predict the observed surface detections.

Watch for

Saying only that muons travel fast, without comparing the atmospheric flight time with the proper lifetime or identifying time dilation.

Representative question

[Maximum number: 3]

Muons are detected at the Earth's surface.

Explain, with supporting calculations, why this is evidence for time dilation.

ALTERNATIVE 1

For Earth time of flight $=8.5 \mu \mathrm{~s}$ $8.5 \mu \mathrm{~s} \gg 2.2 \mu \mathrm{~s}$ so muons should have decayed

but $9.9 \mu \mathrm{~s}$ (time dilation) $>8.5 \mu \mathrm{~s}$ so many muons survive OWTTE

ALTERNATIVE 2

Without time dilation the distance travelled in Earth frame =
$0.975 \times c \times 2.2 \times 10^{-6}=0.64 \ll \mathrm{~km} \gg$

With time dilation the distance travelled in Earth frame $=0.975 \times c \times 9.9 \times 10^{-6}=2.9$ << km>>

Without time dilation, most of the muons would have decayed before reaching the surface

Choose the model

State the inertial reference frame first. Galilean relativity uses x′=x−vt, t′=t and u′=u−v. Special relativity uses the two postulates, Lorentz transformations and u′=(u−v)/(1−uv/c²).

Track what changes and what is invariant

In special relativity, use γ=1/√(1−v²/c²), the invariant interval (Δs)²=(cΔt)²−(Δx)², proper time, proper length, Δt=γΔt0 and L=L0/γ. Separate measurements of space and time can change between frames, while the interval and vacuum light speed do not.

Read the evidence

On a space–time diagram, world-line angle gives tanθ=v/c, frame axes determine simultaneity, and no world line exceeds the light line. Muon survival provides experimental evidence: time dilation explains the longer Earth-frame lifetime, while length contraction explains the shorter atmospheric distance in the muon frame.

Final retrieval check

For every calculation, identify the frame, select the proper quantity if one is given, keep signed velocities and units consistent, and check the result against c, γ≥1, or the invariant interval. The syllabus requires applying the transformations and equations, not deriving them.

Exam analysis

A.1 Kinematics

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A.2 Forces and momentum

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A.3 Work, energy and power

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