1. Motion, forces and energy

For perpendicular components A and B, use Pythagoras to find the magnitude. Use trigonometry to find the angle, then state the angle from a named direction so the vector is complete.

Use a distance and time that refer to the same part of the journey, convert them to compatible units, substitute, and give the speed with its unit.

On a distance–time graph, gradient represents speed. On a speed–time graph, gradient represents acceleration and area under the graph represents distance travelled.

Multiplying m/s by s gives m, confirming that the area represents distance.

Its velocity changes by about 9.8 m/s every second in the downward direction. The same $g$ applies to different masses at the same location when resistance is negligible.

Choose a positive direction, keep velocity signs consistent, subtract initial velocity from final velocity, then divide by the elapsed time.

For a curve, imagine tangents at successive points. A tangent that becomes steeper means the magnitude of acceleration increases; one that becomes less steep means it decreases.

Positive gradient gives positive acceleration, zero gradient gives zero acceleration, and negative gradient gives negative acceleration.

Keep the sign when the question asks for acceleration. If it asks for the magnitude of deceleration, report the positive size of that negative acceleration.

Without air or liquid resistance, the object continues to accelerate downward at approximately constant $g$ near Earth's surface and does not reach terminal velocity.

If matter is added, mass increases; if matter is removed, mass decreases. Compressing an object without losing matter changes its volume but not its mass.

A force meter or newton meter measures weight directly in newtons. A balance compares masses or weights rather than giving a direct force reading.

Gravitational field strength is numerically equivalent to the acceleration of free fall at that location: 1 N/kg is equivalent to 1 m/s$^2$.

Select the equation form for the unknown, convert mass to kilograms, substitute the local value of $g$, and keep weight in newtons.

A beam or pan balance compares masses. A spring balance or newton meter responds to force and is used to measure weight in newtons.

Weight acts in the direction of the gravitational field. Near a planet, that direction is towards the planet's centre.

An astronaut on the Moon has the same mass as on Earth but less weight because the Moon's gravitational field is weaker.

Choose the form that makes the unknown the subject: $m=\rho V$ for mass and $V=m/\rho$ for volume. Substitute only after the units are consistent.

A block of mass 100 g and volume 40 cm$^3$ has density $100/40=2.5$ g/cm$^3$. The units come from the mass and volume used in the calculation.

For a liquid, keep the measuring cylinder upright and read the scale at eye level. Use the appropriate part of the meniscus, then divide the liquid mass—not the mass of liquid plus container—by its volume.

For a regular solid, measure every required dimension with a ruler or calipers. For a sinking irregular solid, lower it gently until it is fully submerged, remove trapped air and record the rise in liquid volume.

Zero or tare the balance, use suitable scale ranges, repeat measurements when practical and average consistent results. Record all readings with units before calculating.

If density is not given, calculate it using $\rho=m/V$. Put both densities in the same units, compare their numerical values, then state the outcome and support it with the comparison.

A sealed object of mass 80 g and volume 100 cm$^3$ has average density 0.80 g/cm$^3$. In a liquid of density 0.88 g/cm$^3$, it floats because 0.80 is less than 0.88.

Use the average density of the whole object, including enclosed air or combined parts. Joining a dense object to a low-density object can make their combined average density lower than the liquid's density.

Suppose three immiscible liquids have densities 0.60, 0.83 and 1.19 g/cm$^3$. Their final order is 0.60 at the top, 0.83 in the middle and 1.19 at the bottom.

State both the position and the comparison: for example, liquid P is above liquid Q because $\rho_P<\rho_Q$. Equal volumes or total masses are not required for this comparison.

Deformation commonly results from forces acting at different points or in different directions, such as pulling both ends of a spring or squeezing opposite sides of a sponge.

Use sensible linear scales, label axes with units, plot points accurately and draw a best-fit line or smooth curve. A straight line through the origin shows extension proportional to load over that region.

Read extension, not loaded length, unless the graph explicitly asks for length. If original length is known, convert with loaded length = original length + extension.

Secure the clamp stand, keep the load close to the bench, and do not add loads beyond the safe range of the solid.

Choose one direction as positive, give every collinear force a sign, add the signed values, then report the magnitude and direction indicated by the sign.

For 9 N right, 3 N left and 2 N left, the resultant is $9-3-2=4$ N to the right.

Forces can act while the resultant is zero: equal opposing forces are balanced. For example, driving force can equal total resistance while a car moves at constant speed.

During a sudden stop, an unrestrained passenger tends to continue moving forwards; a seat belt provides the resultant force needed to change that motion.

Velocity includes both speed and direction, so an object moving at constant speed around a curve still has changing velocity.

Identify the resultant direction before predicting the change; individual forces do not determine the motion independently.

Friction acts parallel to the contact surfaces and opposes the relative motion or tendency to move between them.

Surface condition matters: water, oil or ice can reduce useful friction, while rougher contact may increase it.

Drag acts against the object's motion relative to the liquid. A ship driven forwards therefore experiences a backward water-resistance force.

To accelerate forwards, the driving force must exceed the liquid drag; at constant velocity the horizontal forces are balanced.

Air resistance acts against motion relative to the air. For a vehicle moving forwards through still air, it acts backwards.

If driving force stays constant while speed rises, increasing air resistance reduces the resultant force and therefore reduces the acceleration.

Find extension by subtracting original length from loaded length, convert it to the unit required for $k$, then calculate $F/x$. Rearrangements are $F=kx$ and $x=F/k$.

For a force-against-extension graph in the proportional region, $k$ is the gradient $\Delta F/\Delta x$.

On a load–extension graph, identify the end of the initial straight-line proportional region—the point where the graph first begins to curve away from that line.

Below this limit, doubling the load doubles the extension and $F/x$ is constant. Beyond it, equal increases in load no longer produce equal increases in extension.

When asked for a range that obeys proportionality, give values from zero up to and including the identified limit, using the graph's units.

First combine all forces to find the resultant. Then use $a=F/m$, $F=ma$, or $m=F/a$, keeping direction or a consistent sign convention.

A 800 kg car accelerates at 1.0 m/s$^2$, so its resultant force is 800 N in the acceleration direction. If the engine force is 5000 N forwards, resistive forces total 4200 N backwards.

The force continuously changes the direction of velocity. An object can therefore accelerate while its speed remains constant.

At any point, draw the motion tangent to the circle and the resultant force radially inwards; these directions are perpendicular.

A larger force or a force acting farther from the pivot usually produces a greater turning effect.

Extend the force's line of action if necessary, draw the shortest perpendicular from the pivot to that line, convert the distance to metres, then multiply and state clockwise or anticlockwise.

A 20 N force with a perpendicular distance of 0.30 m produces a moment of $20\times0.30=6.0$ N m.

Choose the pivot, label each force's turning direction, calculate each force × perpendicular distance, equate the two moments and solve for the unknown.

Include the beam's own weight if it is not negligible; for a uniform beam it acts at the beam's centre.

A smaller force can balance a larger force if its perpendicular distance from the pivot is proportionally larger.

Resolve or compare all forces so upward equals downward and left equals right; then check clockwise moments equal anticlockwise moments about any point.

Equilibrium can be static or dynamic: an object may remain at rest, or continue with constant velocity and constant rotational motion state.

For a bridge or beam with two supports, moments about one support can find the other reaction; vertical force balance then finds the first.

Add clockwise moments and anticlockwise moments separately. Within measurement uncertainty, the two totals should agree, so the resultant moment is zero.

Use a sharp low-friction pivot, read positions at eye level, keep strings vertical, repeat with other masses and distances, and include the rule's weight if it was not initially balanced at its centre of gravity.

Gravity acts throughout the object, but for forces and moments we represent the combined gravitational effect by one downward force—the object's weight—acting through its centre of gravity.

For a uniform, symmetrical object the centre of gravity is at its geometrical centre. For an irregular shape or uneven mass distribution, it need not be at the geometrical centre and can even lie outside the material of the object.

Each drawn line must pass through the centre of gravity because the centre of gravity lies directly below that suspension point. Two such lines therefore locate the same point by intersection.

Use a third suspension point to check that all three lines meet close to one point. Let the lamina hang without touching the stand, use a fine plumb line, and mark each line without parallax.

Imagine a vertical line downward through the centre of gravity. While that line falls inside the base, the object's weight produces a restoring effect. At the edge is the tipping point; beyond the edge, the weight produces an overturning moment and the object topples.

Place heavy loads low in a vehicle or ship and keep supports far apart when greater stability is needed. Compare both centre-of-gravity height and base width before deciding which object is most stable.

Momentum is a vector and points in the same direction as velocity. In one dimension, choose one direction as positive and give motion in the opposite direction a negative velocity and momentum.

For constant mass, calculate change in momentum as final minus initial: $\Delta p=mv-mu=m(v-u)$. Keep the velocity signs when an object reverses direction.

Impulse is measured in newton seconds (N s), which is equivalent to kg m/s. Convert milliseconds to seconds before using $F\Delta t$.

For the same change in momentum, increasing the collision or stopping time reduces the average force. This is why crumple zones, padding and moving the hands backwards while catching reduce injury.

If two objects stick together, their final masses combine: $m_1u_1+m_2u_2=(m_1+m_2)v$. For recoil or separation from rest, the two final momenta are equal in magnitude and opposite in direction.

Apply the principle only to the stated one-dimensional system during the event. External forces such as friction must be negligible over the collision or explosion time.

A larger momentum change in the same time gives a larger average resultant force. For the same momentum change, a longer time gives a smaller average resultant force.

Rearranging gives $F\Delta t=\Delta p$, so the area represented by force × time is the impulse. For constant mass, this result is consistent with $F=ma$.

Name the store and the object that has it: for example, the gravitational potential store of raised water or the elastic store of a compressed spring.

Use the structure: energy is transferred from the [initial] store of [object], by [pathway], to the [final] store of [object]. Include more than one destination when energy is dissipated to the surroundings.

As a falling object slows in air, its gravitational potential store decreases; energy is transferred mechanically to its kinetic store and by heating to the internal stores of the object and air.

In a simple flow diagram, every output branch is part of the energy account. The sum of the output energies must equal the input energy.

Dissipated energy has not disappeared: it has spread, usually into internal stores of the surroundings, and is less available for a useful transfer.

Convert mass to kilograms, square the speed, multiply by the mass, then divide by two. To find speed, rearrange to $v=\sqrt{2E_k/m}$ and take the positive speed.

At constant mass, doubling speed makes kinetic energy four times larger; tripling speed makes it nine times larger. At constant speed, kinetic energy is directly proportional to mass.

Choose a reference level, calculate final height minus initial height, then substitute. A rise gives positive $\Delta h$ and an energy gain; a fall gives negative $\Delta h$ and an energy loss.

Use only the vertical height change, not the length of a ramp or the distance travelled along a curved path.

In a Sankey diagram, arrow width represents energy. The main forward arrow usually shows useful output and side branches show other outputs; their widths and energy values must add to the input width and value.

When a scale is given, count the width units perpendicular to the arrow, convert each width to energy, and add all relevant wasted branches. Do not measure arrow length.

Work done on an object transfers energy to it; work done by an object transfers energy away from one of its stores. Work and energy are both measured in joules (J).

A force can act without doing mechanical work if there is no displacement in the force's direction. A heavy load hanging motionless has a force on it but no mechanical work is being done on the load.

Select the force doing the work, use the displacement in that force's direction, convert to SI units, multiply, and name the associated energy transfer. Rearrangements are $F=W/d$ and $d=W/F$.

When several forces act, calculate their work separately or use the resultant force for the net change in kinetic energy. Work against friction transfers energy to internal stores; lifting at constant speed transfers energy to the gravitational potential store.

A turbine is turned by moving fluid: steam, water or air. A generator is driven by the turbine and produces electrical power. A boiler supplies steam by transferring thermal energy to water; it is not needed by solar cells, wind turbines or hydroelectric schemes.

For a justified decision, name the criterion, connect it to the named resource and the stated location or demand, then explain the consequence. A feature such as 'renewable' is not by itself a complete advantage unless its effect is stated.

Energy is conserved: total input energy equals useful output energy plus wasted output energy. Wasted energy is usually dissipated to the surroundings and becomes less available for useful transfer.

The syllabus exceptions are geothermal, nuclear and tidal energy. Geothermal comes from Earth's internal thermal energy; nuclear comes from changes in atomic nuclei; tides arise from gravitational interactions, principally with the Moon.

Fusion combines light nuclei. Fission splits a heavy nucleus. Both are nuclear processes, but the Sun's energy in this syllabus is attributed to fusion.

The key syllabus claim is the research status: fusion is being investigated as a possible route to large-scale electrical energy. This is different from saying it is already a routine commercial source.

For a decimal efficiency $\eta$, useful output $=\eta imes$ total input and total input $=$ useful output $/\eta$. Convert a percentage to a decimal before these rearrangements: for example, 20% is 0.20.

One watt is one joule per second: $1\,\text{W}=1\,\text{J}/\text{s}$. A kilowatt is $1000\,\text{W}$, a megawatt is $10^6\,\text{W}$ and one minute is $60\,\text{s}$.

Identify the work done or the relevant energy change, convert energy to joules and time to seconds, then divide by the time. Use $W=Pt$ or $\Delta E=Pt$ to find energy, and $t=W/P$ or $t=\Delta E/P$ to find time.

Example: a machine transfers $18\,\text{kJ}$ in $60\,\text{s}$. Convert $18\,\text{kJ}=18{,}000\,\text{J}$, so $P=18{,}000/60=300\,\text{W}$.

One pascal is one newton per square metre: $1\,\text{Pa}=1\,\text{N}/\text{m}^2$. Convert areas before dividing: $1\,\text{cm}^2=10^{-4}\,\text{m}^2$.

Identify the force acting on the named surface, identify its actual contact area, convert to SI units, then divide. Rearrangements are $F=pA$ and $A=F/p$.

For an object resting on a horizontal surface with no other vertical forces, the contact force equals its weight, not its mass. If mass is given, first find weight using $F=mg$.

A complete explanation names what stays constant, states how the contact area or force changes, and concludes how pressure changes. For example: the person's weight is unchanged, lying down increases contact area, so pressure on the ice decreases.

A deeper point has a taller column of liquid above it, so a greater liquid weight acts per unit area. A denser liquid has more mass—and therefore more weight—in the same volume, so its pressure rises more rapidly with depth.

For one uniform liquid at constant gravitational field strength, pressure due to the liquid rises as a straight line with depth. In layered liquids, the graph remains continuous but becomes steeper in the denser layer.

At the same horizontal level in a connected liquid at rest, pressure is the same. Container volume, base area and sloping sides do not by themselves change pressure at a specified depth.

Choose the two levels, measure their vertical separation, convert density and height to SI units, then multiply. Rearrangements are $\rho=\Delta p/(g\Delta h)$ and $\Delta h=\Delta p/(\rho g)$.

If the reference is the liquid surface, pressure due to the liquid at depth $h$ is $p=\rho gh$. If total pressure is requested and the surface pressure is known, add it: $p_{\text{total}}=p_{\text{surface}}+\rho gh$.

For several unmixed layers, calculate $\rho g\Delta h$ for each layer crossed and add the pressure changes. The container's cross-sectional area is not part of the equation.