A. Space, time and motion

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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

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

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.

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.

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.

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.

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.

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.

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.

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.

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.

Mechanical shortcut

For a constant force parallel to velocity,

$$P=Fv$$

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.

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.

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.

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

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.

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.

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.

Link to linear motion

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

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.

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.

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.

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.