Forces

Explanation

• A scalar quantity has magnitude only, whereas a vector quantity has both magnitude and an associated direction.
• Classify a stated quantity by checking whether its direction is needed to specify it completely; force, weight and velocity are vectors, while mass, energy and speed are scalars.
• A vector can be shown by an arrow: its length represents the magnitude and its arrowhead shows the direction.
• Do not call a quantity a vector merely because it can be large or negative; a vector must include direction.

Explanation

• A force is a push or pull caused by an interaction between objects; force is a vector quantity.
• Decide whether the objects must touch: friction, air resistance, tension and normal contact force are contact forces, while gravitational, electrostatic and magnetic forces are non-contact forces.
• An interaction produces a force on each object; represent each force with a vector arrow on the object that experiences it.
• Do not describe air resistance as non-contact just because air is hard to see: collisions with air particles make it a contact force.

Explanation

• Weight is the force on an object due to gravity, while mass measures the amount of matter and does not change when gravitational field strength changes.
• Calculate weight using $W=mg$, with $W$ in newtons, $m$ in kilograms and $g$ in newtons per kilogram; use the value of $g$ supplied in the question.
• For example, a $3.0\,\text{kg}$ object where $g=9.8\,\text{N/kg}$ has weight $3.0\times9.8=29.4\,\text{N}$, acting through its centre of mass.
• Do not give weight in kilograms or assume it is constant everywhere; weight changes with $g$ and is measured with a calibrated newtonmeter.

Explanation

• The resultant force is the single force that has the same effect as all the forces acting together. For forces along one straight line, choose a positive direction, give opposite forces opposite signs and add them. For example, $18\,\text{N}$ right and $11\,\text{N}$ left give a resultant of $18-11=7\,\text{N}$ to the right;
• equal opposing forces give zero resultant. Higher tier: AQA restricts "use free body diagrams to describe qualitatively examples where several forces lead to a resultant force on an object, including balanced forces when the resultant force is zero" to Higher tier, together with resolving a force into components and scale vector diagrams;
• a free-body diagram isolates the object and shows every force as a labelled arrow. A single force can be resolved into two components at right angles whose combined effect is the original force.
• In a scale vector diagram, draw force arrows to scale and in the correct directions, place them head-to-tail, then measure the resultant from the start of the first arrow to the end of the last; a closed diagram represents equilibrium.
• Do not add magnitudes when forces oppose, and do not omit the direction of a non-zero resultant because force is a vector.

Explanation

• Work is done when a force causes a displacement, transferring energy between stores; work done against friction raises temperature.
• Use $W=Fs$, where $s$ is the distance moved along the force's line of action; $W$ is in joules, $F$ in newtons and $s$ in metres.
• For example, a $25\,\text{N}$ force moving an object $3.0\,\text{m}$ along its line of action does $25\times3.0=75\,\text{J}$ of work.
• Do not multiply by a distance perpendicular to the force, and remember that $1\,\text{J}=1\,\text{N m}$.
• In an exam, apply the named relationship to the quantities, units and direction given in the question.

Explanation

• A stationary object needs more than one force to change shape: one unbalanced force would accelerate the object as a whole, whereas forces acting at different positions can change the separation of its parts. Equal outward pulls stretch an object, forces that turn different parts in opposite directions bend it, and inward pushes from opposite ends compress it.
• Elastic deformation is reversed when the forces are removed; inelastic deformation leaves the object permanently changed. Up to the limit of proportionality use $F=ke$, measuring extension $e$ from the original length; a straight force-extension graph through the origin represents direct proportion.
• For the required practical, clamp a spring beside a millimetre ruler and record its unloaded length. Add known masses one at a time, let the spring stop moving, record each new length, calculate extension and convert each mass to force using $W=mg$;
• repeat readings, plot force against extension and unload the spring to check that it returns to its original length. For example, a spring with $k=160\,\text{N/m}$ extended by $0.050\,\text{m}$ needs $F=160\times0.050=8.0\,\text{N}$ and stores $E_e=\frac12ke^2=0.20\,\text{J}$.
• Do not substitute the spring's total length for extension, and do not apply the linear relationship beyond the limit of proportionality.

Explanation

• A moment is the turning effect of a force about a pivot, and its size is $M=Fd$ where $d$ is the perpendicular distance to the force's line of action. For a balanced object, choose one pivot and set the total clockwise moment equal to the total anticlockwise moment.
• For example, $15\,\text{N}$ acting $0.40\,\text{m}$ from a pivot produces a moment of $6.0\,\text{N m}$.
• A lever transmits a rotational effect about its pivot: for the same input force, applying it farther from the pivot gives a larger moment, so a long lever can produce a larger output force nearer the pivot.
• Meshing gear teeth transmit a tangential force and rotation; when a smaller driving gear turns a larger gear, the larger gear turns more slowly and in the opposite direction but delivers a larger moment because the force acts at a larger radius.
• Do not use a sloping distance measured to the point where the force is applied; the equation requires the perpendicular distance to the line of action.

Explanation

• A fluid is a liquid or a gas, and fluid pressure produces a force normal, or at right angles, to a surface.
• Calculate pressure using $p=F/A$, with normal force $F$ in newtons, surface area $A$ in square metres and pressure $p$ in pascals.
• For example, a normal force of $90\,\text{N}$ on $0.030\,\text{m}^2$ produces $p=90\div0.030=3000\,\text{Pa}$.
• Do not use an area in $\text{cm}^2$ without converting it to $\text{m}^2$, and use only the component of force normal to the surface.
• In an exam, apply the named relationship to the quantities, units and direction given in the question.

Explanation

• Higher tier: in a liquid, pressure increases with the height of liquid above the point and with the liquid's density. Use $p=h\rho g$ for pressure due to a liquid column, with $h$ in metres, $\rho$ in $\text{kg/m}^3$ and $g$ in $\text{N/kg}$; subtract depths before using it for a pressure difference.
• For example, in water with $\rho=1000\,\text{kg/m}^3$ and $g=9.8\,\text{N/kg}$, a $0.50\,\text{m}$ depth change gives $\Delta p=0.50\times1000\times9.8=4900\,\text{Pa}$. Upthrust arises because the bottom of a submerged object is at greater pressure than its top.
• An object rises when upthrust is greater than its weight, sinks when its weight is greater than the upthrust, and floats at rest when upthrust balances its weight.
• Equivalently, an object with a lower average density than the fluid floats, one with a greater average density sinks, and one with the same density can remain suspended.
• Do not include horizontal position in $p=h\rho g$.

Explanation

• Atmospheric pressure is produced by air molecules colliding with surfaces; the atmosphere is a thin layer of air around Earth.
• As altitude increases, there are fewer air molecules above a surface and the air is less dense, so atmospheric pressure decreases.
• A pressure difference across a surface produces a resultant normal force; calculate it by rearranging $p=F/A$ to $F=pA$.
• Do not say that atmospheric pressure becomes zero on a mountain; it decreases with height because there is less air above, but an atmosphere remains.

Explanation

• Distance is the total length of the path travelled and is scalar; displacement is the straight-line change from start to finish and includes direction, so it is vector.
• Add every part of a route to find distance, but use only the start and finish positions to find displacement.
• For example, travelling $9\,\text{m}$ east and then $4\,\text{m}$ west gives distance $13\,\text{m}$ and displacement $5\,\text{m}$ east.
• Do not report displacement without a direction, and do not assume distance and displacement are equal unless the path is straight without reversing.

Explanation

• Speed is a scalar rate of change of distance; typical values are about $1.5\,\text{m/s}$ for walking, $3\,\text{m/s}$ for running, $6\,\text{m/s}$ for cycling and $330\,\text{m/s}$ for sound in air.
• Typical transport-system speeds are about $13$–$30\,\text{m/s}$ for a car, $55\,\text{m/s}$ for a train and $250\,\text{m/s}$ for a passenger plane.
• Use $s=vt$ for constant speed and rearrange to $v=s/t$; for non-uniform motion, average speed is total distance divided by total time.
• For example, $450\,\text{m}$ travelled in $30\,\text{s}$ gives an average speed of $450\div30=15\,\text{m/s}$.
• Do not average two speeds unless the time spent at each speed is equal; instead use the complete distance and complete time, including stops when appropriate.

Explanation

• Velocity is speed in a stated direction, so it is a vector quantity; speed is scalar.
• For motion over an interval, use displacement rather than distance when finding average velocity, then give the direction of the displacement.
• For example, a displacement of $60\,\text{m}$ west in $15\,\text{s}$ gives an average velocity of $4.0\,\text{m/s}$ west.
• Higher tier: AQA restricts "explain qualitatively, with examples, that motion in a circle involves constant speed but changing velocity" to Higher tier; the velocity changes because the direction of motion changes continuously.
• Do not use total path length to calculate average velocity, and do not omit direction from a velocity value.

Explanation

• A distance–time graph shows distance travelled on the vertical axis and time on the horizontal axis.
• Its gradient is speed: $v=\dfrac{\Delta s}{\Delta t}$.
• A straight sloping section represents constant speed, a steeper section represents a greater speed, and a horizontal section represents an object at rest.
• For an average speed, use the total distance travelled divided by the total time.
• Higher tier: AQA restricts "If an object is accelerating, its speed at any particular time can be determined by drawing a tangent and measuring the gradient of the distance–time graph at that time" to Higher tier, so tangent gradients on curved graphs are Higher-only.

Explanation

• Acceleration is the rate at which velocity changes. Calculate it using $a=\dfrac{v-u}{t}$, where $u$ is initial velocity, $v$ is final velocity and $t$ is the time for the change; acceleration is measured in $\text{m/s}^2$.
• On a velocity–time graph, acceleration is the gradient: a horizontal section has zero acceleration and a negative gradient means acceleration in the negative direction.
• Direction matters because velocity is a vector.
• Uniform acceleration also obeys $v^2-u^2=2as$; this is supplied on the equation sheet and is examined on both tiers.
• Higher tier: AQA restricts "interpret enclosed areas in velocity–time graphs to determine distance travelled (or displacement)" and "measure, when appropriate, the area under a velocity–time graph by counting squares" to Higher tier.

Explanation

• Newton's First Law states that if the resultant force on an object is zero, an object at rest remains at rest and a moving object continues at constant velocity.
• Constant velocity includes constant speed in a constant direction.
• Forces can act while the resultant is zero: a car travelling steadily may have its driving force balanced by resistive forces.
• If the resultant force is not zero, the object's velocity changes by speeding up, slowing down or changing direction.
• Higher tier: AQA restricts "The tendency of objects to continue in their state of rest or of uniform motion is called inertia" to Higher tier, so naming and using inertia is Higher-only.

Explanation

• Newton's Second Law links resultant force, mass and acceleration: $F=ma$. Acceleration is proportional to resultant force, so doubling the resultant force doubles the acceleration for a fixed mass.
• Acceleration is inversely proportional to mass, so a larger mass gives a smaller acceleration for the same resultant force.
• Combine all forces first, then use force in newtons, mass in kilograms and acceleration in $\text{m/s}^2$.
• In the required practical, change force while keeping total mass constant, or change mass while keeping force constant, measure acceleration and repeat.
• Higher tier: AQA restricts "inertial mass is a measure of how difficult it is to change the velocity of an object" and "inertial mass is defined as the ratio of force over acceleration" to Higher tier, so $m=F/a$ as a definition is Higher-only.