1 Forces and motion

Exam analysis

Most tested topics

Forces and motion units

Mechanics units connect each quantity to a measurable physical meaning, from base units for motion to compound units for force, field strength and momentum.

A number in mechanics is complete only when its unit identifies the physical quantity. The unit symbol is case-sensitive: $m$ is metre, while $N$ is newton.

Quantity	Unit	Symbol	What the form tells you
mass	kilogram	kg	amount of mass
length or distance	metre	m	position or distance travelled
time	second	s	duration
speed or velocity	metre per second	m/s	metres travelled each second
acceleration	metre per second squared	m/s²	change in velocity each second
force	newton	N	size of a force
gravitational field strength	newton per kilogram	N/kg	force on each kilogram of mass

$m/s$ and $m/s²$ describe different quantities: acceleration has an extra “per second” because velocity itself changes with time. $N/kg$ is used for gravitational field strength, not as another way to label every acceleration.

A compound unit follows the quantities in the relationship. Multiplication places unit symbols side by side; division keeps the divisor below the line or after a slash.

Quantity	How its unit is built	Required unit
moment of a force	force × perpendicular distance	newton metre, $N\,m$
momentum	mass × velocity	kilogram metre per second, $kg\,m/s$

For momentum, multiplying $kg$ by $m/s$ gives $kg\,m/s$ . For a moment, multiplying $N$ by $m$ gives $N\,m$ . These unit structures provide a quick check that the chosen relationship uses the correct quantities.

$N\,m$ means newton multiplied by metre, not $N/m$ . Momentum uses $kg\,m/s$ ; $kg\,m/s²$ instead has the same base-unit structure as a newton and is not the unit of momentum.

(b) Movement and position

Movement and position links graphs, equations and practical methods so motion can be measured, calculated and explained from evidence reliably.

A distance-time graph shows elapsed time on the horizontal axis and distance travelled on the vertical axis. Its gradient is the rate at which distance changes, so it represents speed.

$\text{speed}=\dfrac{\text{change in distance}}{\text{change in time}}$

Segment feature	Motion it represents
straight rising line	constant speed
steeper straight line	greater constant speed
horizontal line	stationary: distance does not change
curve getting steeper	speed is increasing
curve becoming less steep	speed is decreasing

To plot a journey, place each measured pair $(t,d)$ at its coordinate and join points in time order. Explain a segment by comparing its gradient, not simply its vertical position.

A higher point means more distance has been travelled; it does not by itself mean the object is moving faster. A horizontal segment has zero gradient, so the object is stationary, not moving at constant non-zero speed.

Average speed describes the whole journey: divide the total distance moved by the total elapsed time. Time spent stationary is part of the elapsed time.

$\text{average speed}=\dfrac{\text{total distance moved}}{\text{total time taken}}$

A toy car travels $18\,m$ in $6.0\,s$ , pauses for $2.0\,s$ , then travels $12\,m$ in $4.0\,s$ . Total distance $=30\,m$ and total time $=12.0\,s$ , so average speed $=30/12.0=2.5\,m/s$ .

Use compatible units before dividing. For an answer in $m/s$ , convert every distance to metres and every time to seconds; then include $m/s$ in the result.

Do not take the arithmetic mean of two speeds unless the time spent at each speed makes that method valid. The reliable method is always total distance divided by total elapsed time.

A motion investigation needs a measured distance and time, one deliberately changed independent variable, and other relevant conditions kept constant.

Role	Example for a toy car on a ramp
independent variable	release height or ramp angle
dependent variable	average speed over a fixed section
controls	same car, surface, start method and measured section

Mark a fixed start and finish and measure their separation with a metre rule or tape. 2. Release the same car from rest without pushing. 3. Measure travel time with light gates and a timer/data logger, or use video timestamps. 4. Calculate average speed as distance divided by time. 5. Repeat at least three times at each chosen independent-variable value, identify anomalies and calculate a mean from valid repeats. 6. Test a suitable range of at least five values and record values with units.

Light gates or frame-by-frame video reduce reaction-time uncertainty. A longer measured section increases the travel time and can reduce the percentage timing uncertainty, provided the intended motion is still being measured.

Repeating one uncontrolled run does not make the investigation valid. Reliability comes from consistent repeats; validity also requires changing only the intended independent variable and controlling other factors.

Acceleration is the change in velocity per unit time. Let $u$ be initial velocity, $v$ final velocity, $t$ elapsed time and $a$ acceleration.

$a=\dfrac{v-u}{t}$

A car changes velocity from $8.0\,m/s$ to $20.0\,m/s$ in $4.0\,s$ . First find $v-u=20.0-8.0=12.0\,m/s$ ; then $a=12.0/4.0=3.0\,m/s²$ .

Choose a positive direction and keep velocity signs consistent. If positive velocity falls from $20\,m/s$ to $8\,m/s$ in $4\,s$ , $a=(8-20)/4=-3\,m/s²$ : the negative sign shows acceleration opposite to the chosen positive direction.

Use the change $v-u$ , not just the final velocity. The unit is $(m/s)/s=m/s²$ , not $m/s$ .

A velocity-time graph shows elapsed time on the horizontal axis and velocity on the vertical axis. The vertical coordinate gives velocity at that instant; the gradient gives acceleration.

Segment feature	Velocity	Acceleration
horizontal line above the time axis	constant positive velocity	zero
straight line rising	velocity increases uniformly	constant positive
straight line falling	velocity decreases uniformly	constant negative
line on the time axis	zero: stationary	zero if horizontal
line below the time axis	motion in the negative direction	set by its gradient

Plot each measured $(t,v)$ pair in time order. A straight sloping segment represents constant acceleration because equal time intervals produce equal velocity changes; a curve represents changing acceleration.

A horizontal line above the time axis does not mean stationary: its height shows a constant non-zero velocity. Do not confuse graph height (velocity) with graph gradient (acceleration).

$\text{acceleration}=\text{gradient}=\dfrac{\Delta v}{\Delta t}=\dfrac{v_2-v_1}{t_2-t_1}$

Choose two well-separated points on the same straight segment. If velocity changes from $6.0\,m/s$ at $2.0\,s$ to $18.0\,m/s$ at $8.0\,s$ , then $a=(18.0-6.0)/(8.0-2.0)=12.0/6.0=2.0\,m/s²$ .

For acceleration at one instant on a curved graph, draw a tangent at that point and calculate the tangent's gradient using a large triangle. For mean acceleration over an interval, use the two interval endpoints.

Do not calculate $v/t$ from one arbitrary point unless the straight segment passes through the origin. A falling segment has a negative gradient under the chosen sign convention.

Distance travelled is found from the area between a velocity-time graph and the time axis. The units confirm the relationship: $(m/s)\times s=m$ .

$\text{distance travelled}=\text{area between the velocity-time graph and the time axis}$

A car accelerates uniformly from $4\,m/s$ to $10\,m/s$ over $3.0\,s$ . The area is a rectangle plus a triangle: $(4\times3.0)+[\tfrac12\times(10-4)\times3.0]=12+9=21\,m$ . Equivalently, use the trapezium area $\tfrac12(4+10)\times3.0$ .

Split a piecewise graph at every change of segment, calculate rectangle, triangle or trapezium areas, then add them. Use the axis scale rather than counting visual grid spaces without values.

Gradient gives acceleration, not distance. If velocity is below the time axis, the signed area is negative displacement; distance travelled uses the magnitude of each area between the graph and the axis.

For motion with constant acceleration, use the time-free equation when initial speed $u$ , final speed $v$ , acceleration $a$ and distance moved $s$ are involved. Choose one direction as positive before assigning signs.

$$v²=u²+2as$$

A car moving at $20\,m/s$ brakes with constant acceleration $-5.0\,m/s²$ over $30\,m$ . Taking forward as positive: $v²=20²+2(-5.0)(30)=100$ , so $v=\sqrt{100}=10\,m/s$ . Use the positive root because the question asks for speed.

Write the equation, identify the unknown, convert units, substitute signed values, rearrange, and only then take a square root if finding $u$ or $v$ . Check that the result matches the motion: braking should reduce speed.

The equation assumes constant acceleration. During braking, $a$ is negative if forward is positive; using a positive value would incorrectly make $v²$ increase. Do not report $v²$ as the final speed.

(c) Forces, movement, shape and momentum

Forces can change motion, shape and turning effect, with momentum and moments explaining collisions, safety features and balanced systems in mechanics.

A force is an interaction between bodies that can change an object's motion or deform it. A change in motion means a change in speed, direction, or both.

Effect	What changes	Example
speed up or slow down	magnitude of velocity	a push accelerates a trolley
turn	direction of velocity	a sideways force bends a moving ball's path
deform	shape or dimensions	a force stretches a spring

Whether motion changes depends on the resultant force. Balanced forces give zero resultant force, so they do not cause acceleration; an unbalanced resultant force does.

A force is not required to keep an object moving at constant velocity. It is required to change velocity or shape; friction or drag may make a continuing driving force necessary in everyday motion.

Identify a force by the interacting bodies and the direction in which it acts. Contact forces require bodies or fluids to touch; non-contact forces act across a separation.

Force	Contact?	Typical direction or role
weight (gravitational force)	no	toward the attracting body's centre
electrostatic force	no	attraction or repulsion between charges
magnetic force	no	attraction or repulsion involving magnets or magnetic fields
normal contact force	yes	perpendicular to a surface
friction or drag	yes	opposes relative motion
tension	yes	along a stretched string or cable
upthrust	yes	upward force from a fluid
thrust	yes	driving force from an engine or expelled fluid

Name the force, not merely the situation: use “weight” or “gravitational force” rather than “gravity”. An arrow's direction must match what the named force does.

A scalar quantity has magnitude only. A vector quantity has both magnitude and direction, so its direction must be included for a complete description.

Scalars	Vectors
distance, speed, time, mass, temperature, energy, power	displacement, velocity, acceleration, force, weight, momentum

“ $12\,m/s$ ” is a speed and is scalar. “ $12\,m/s$ east” is a velocity and is vector because east supplies the direction.

A negative sign can encode direction along a chosen axis, but it does not make every signed number a vector. The physical quantity must possess direction.

Force is a vector: a complete force states its magnitude in newtons and its direction. In a force arrow, the arrowhead gives direction and the labelled length can represent magnitude.

Choose and state a positive direction for one-dimensional problems. A $6\,N$ force to the right can be written $+6\,N$ ; a $4\,N$ force to the left is $-4\,N$ under that convention.

The arrow should lie along the force's line of action. For weight, draw vertically downward through the body's centre of gravity; for a contact force, use the physically correct direction at the contact.

A label without a direction is incomplete. Equal arrow lengths in opposite directions represent equal-magnitude forces, but they only balance if they act on the same body.

The resultant force is the single force with the same overall effect as all the forces acting on one body. Along one line, choose a positive direction and add forces with signs.

$F_{\text{resultant}}=\sum F$

Take right as positive. If $12\,N$ acts right and $7\,N$ acts left, $F_R=+12+(-7)=+5\,N$ , so the resultant is $5\,N$ to the right. If the forces were $7\,N$ in each direction, the resultant would be $0\,N$ .

Do not add magnitudes when forces oppose one another. A zero resultant means zero acceleration, not necessarily zero velocity.

Friction is a contact force that opposes relative motion, or the tendency for relative motion, between surfaces. Its direction is opposite to the sliding or attempted sliding.

For a puck sliding right across a surface, friction on the puck acts left and reduces its speed. For a tyre pushing the road backward, friction from the road on the tyre can act forward; friction does not always point opposite to the object's overall motion.

Drag is a frictional force from a fluid such as air or water. It acts opposite to the body's motion relative to the fluid and often increases as speed increases.

Friction opposes relative motion at a contact, not every applied force. Its magnitude is not automatically equal to the driving force.

An unbalanced resultant force causes acceleration. For a given mass, a larger resultant force produces a larger acceleration; for a given resultant force, a larger mass produces a smaller acceleration.

$$F=ma$$

$F$ is resultant force in newtons, $m$ is mass in kilograms, and $a$ is acceleration in $m/s²$ . The direction of $a$ is the direction of the resultant force.

A $0.160\,kg$ firework has an upward resultant force of $26.4\,N$ . $a=F/m=26.4/0.160=165\,m/s²$ upward. Converting grams to kilograms before substitution is essential.

Use the resultant force, not one force chosen from the diagram. Balanced forces give $F=0$ , so $a=0$ even if the object is already moving.

Weight is the gravitational force on a mass. It acts toward the attracting body's centre and changes if gravitational field strength changes; mass is the amount of matter and does not change with location.

$$W=mg$$

$W$ is weight in newtons, $m$ is mass in kilograms, and $g$ is gravitational field strength in $N/kg$ .

For a $24\,kg$ mass where $g=10\,N/kg$ , $W=24\times10=240\,N$ . Conversely, an object weighing $520\,N$ has mass $m=520/10=52\,kg$ .

Do not give weight in kilograms. Kilogram is the unit of mass; newton is the unit of force, including weight.

Stopping begins when the driver sees a hazard and ends when the vehicle is stationary. It contains a thinking stage before the brakes act and a braking stage after they act.

$\text{stopping distance}=\text{thinking distance}+\text{braking distance}$

Stage	Begins	Ends
thinking distance	driver sees the hazard	driver applies the brakes
braking distance	brakes are applied	vehicle stops

If thinking distance is $7\,m$ and braking distance is $20\,m$ , stopping distance is $7+20=27\,m$ .

Thinking distance is not the whole distance from seeing a hazard to stopping; that complete distance is the stopping distance.

A factor changes stopping distance by changing thinking distance, braking distance, or both. Keep those mechanisms separate.

Factor increases	Main effect	Why
speed	both distances increase	more distance is travelled during reaction time and more braking is required
reaction time from tiredness, alcohol, drugs or distraction	thinking distance increases	brakes are applied later
vehicle mass	braking distance increases	more momentum must be reduced for the same speed
wet or icy road, worn tyres, poor brakes	braking distance increases	available braking force or grip is reduced
downhill slope	braking distance increases	a component of weight acts along the motion

Driver tiredness changes thinking distance but does not directly change the vehicle's braking performance. Road condition changes braking distance, not the driver's reaction time.

A falling object has weight downward and drag upward. Terminal velocity is reached when these forces become equal, making resultant force and acceleration zero.

Just after release, speed and drag are small, so weight is greater than drag and the object accelerates downward. 2. As speed increases, drag increases, so the downward resultant and acceleration decrease. 3. When drag equals weight, resultant force is zero. The object then continues at constant terminal velocity.

Opening a parachute greatly increases drag. Drag may initially exceed weight, producing an upward resultant force that reduces downward velocity. As the object slows, drag falls until it again equals weight at a lower terminal velocity.

At terminal velocity the object is not stationary. It moves at constant velocity because the forces balance and acceleration is zero.

Investigate how extension changes with applied force by measuring the original length and the loaded length of the same sample.

$\text{extension}=\text{loaded length}-\text{original length}$

Clamp the spring, wire or rubber band beside a vertical ruler and mark fixed measurement points. 2. Measure original length with no added load. 3. Add known masses in equal steps; convert each mass to weight using $W=mg$ . 4. After oscillations stop, measure loaded length at eye level and calculate extension. 5. Repeat readings and calculate means; then remove masses in steps to check whether the sample returns to its original length. 6. Plot extension against force.

Use a pointer and set square, read the ruler perpendicular to its scale, and keep the ruler fixed. Do not exceed a safe load: secure the stand and keep feet clear of falling masses.

Measure extension, not loaded length alone. Reusing different starting points or changing the sample would make the comparison invalid.

In the initial Hooke's law region, extension is directly proportional to applied force. Doubling the force doubles the extension while the relationship remains valid.

$F\propto x$

On a force-extension graph, direct proportionality appears as an initial straight line through the origin. A constant gradient shows a constant ratio between force and extension; curvature shows that proportionality no longer holds.

A graph that merely rises does not prove Hooke's law. The relevant region must be linear and pass through the origin when extension, rather than total length, is plotted.

A material behaves elastically when it returns to its original shape and dimensions after the deforming forces are removed.

After the force is removed	Behaviour
returns to original length or shape	elastic
retains a permanent extension or deformation	not fully elastic

On an extension-force graph, complete unloading back to zero extension shows recovery of the original length. If unloading ends at a positive extension, permanent deformation remains.

Elastic behaviour is about recovery after unloading. It is not the same claim as Hooke's law: a material can recover its shape even if force and extension were not directly proportional throughout.

Momentum measures motion using both mass and velocity. Because velocity has direction, momentum is also a vector and must follow the chosen sign convention.

$$p=mv$$

$p$ is momentum in $kg\,m/s$ , $m$ is mass in $kg$ , and $v$ is velocity in $m/s$ .

A $170\,g$ ball moving right at $5.2\,m/s$ has mass $0.170\,kg$ and momentum $p=0.170\times5.2=0.884\,kg\,m/s$ to the right.

Use velocity, including direction, rather than unsigned speed in multi-object problems. Convert grams to kilograms before calculating momentum in SI units.

In a collision, a person's or object's momentum must change. For the same momentum change, increasing the stopping time reduces the mean force.

$F=\dfrac{\Delta p}{\Delta t}$

Safety feature	How it acts	Effect
airbag or crumple zone	deforms while stopping the occupant or vehicle	increases stopping time, reducing mean force
padding or thick carpet	compresses during impact	increases impact time, reducing mean force
seat belt	restrains the occupant over a controlled time and area	prevents a much more abrupt stop against the interior

The feature does not need to reduce the required momentum change. It reduces force mainly by spreading that change over a longer time; spreading force over a larger area can additionally reduce pressure.

For an isolated system with no external resultant force, total momentum before an interaction equals total momentum after it. Choose one direction as positive and keep velocity signs.

$\sum p_{\text{before}}=\sum p_{\text{after}}$

A $0.20\,kg$ cart moving right at $3.0\,m/s$ sticks to a stationary $0.10\,kg$ cart. Before: $p=(0.20)(3.0)=0.60\,kg\,m/s$ . After: $(0.20+0.10)v=0.60$ , so $v=2.0\,m/s$ right.

Define the system, write every initial momentum with sign, equate the total to the signed final total, and then solve for the unknown mass or velocity.

Momentum is conserved for the total isolated system, not necessarily for each object. Do not conserve speed or kinetic energy unless another principle justifies it.

Mean force measures how quickly momentum changes. A larger momentum change in the same time, or the same change in less time, requires a larger mean force.

$F=\dfrac{\Delta p}{t}=\dfrac{mv-mu}{t}$

Take right as positive. A ball's momentum changes from $-4.2$ to $+6.7\,kg\,m/s$ in $0.012\,s$ . $\Delta p=6.7-(-4.2)=10.9\,kg\,m/s$ , so $F=10.9/0.012=9.1\times10²\,N$ to the right.

Convert milliseconds to seconds before dividing. Since $(kg\,m/s)/s=kg\,m/s²=N$ , the unit check leads to newtons.

For a rebound, initial and final momenta have opposite signs, so the change is the difference of signed values and can exceed either magnitude alone.

When body A exerts a force on body B, body B simultaneously exerts an equal-magnitude force in the opposite direction on body A.

Test for a third-law pair	Requirement
bodies	the two forces act on different bodies
interaction	both forces come from the same interaction
magnitude	equal
direction	opposite
timing	simultaneous

If a bat exerts $80\,N$ to the right on a ball, the ball exerts $80\,N$ to the left on the bat. The two forces do not cancel because they act on different objects.

Weight and the normal force on one resting object may balance, but they are not a third-law pair because both act on the same object and arise from different interactions.

The moment of a force measures its turning effect about a pivot. It depends on force and the perpendicular distance from the pivot to the force's line of action.

$M=F d_{\perp}$

$M$ is moment in $N\,m$ , $F$ is force in $N$ , and $d_{\perp}$ is perpendicular distance in $m$ .

A $28\,N$ force acts perpendicular to a wrench $0.15\,m$ from the nut. $M=28\times0.15=4.2\,N\,m$ . A larger force or a larger perpendicular distance produces a larger moment.

Do not automatically use the sloping length from pivot to application point. Use the shortest perpendicular distance to the force's line of action.

The whole weight of a body can be represented as one downward force acting through its centre of gravity.

Draw the weight arrow vertically downward with its line of action through the centre of gravity. For a uniform symmetric object the centre of gravity is at its geometric centre; for an irregular or non-uniform body it may not be.

The centre of gravity matters in turning problems because its perpendicular distance from a pivot determines the moment of the body's weight.

The centre of gravity is a point through which weight acts; it is not necessarily a support point or a point where material is physically concentrated.

For a body in rotational equilibrium, the total clockwise moment about any pivot equals the total anticlockwise moment about that pivot. The resultant moment is therefore zero.

$\sum M_{\text{clockwise}}=\sum M_{\text{anticlockwise}}$

A $230\,N$ weight acts $0.37\,m$ from a pivot and is balanced by force $F$ acting $0.98\,m$ away on the other side. $230(0.37)=F(0.98)$ , so $F=86.8\,N$ .

Choose a pivot that removes unknown forces passing through it, label clockwise and anticlockwise moments, use perpendicular distances in consistent units, equate totals, and solve.

Equal forces are not required for balance. A smaller force can balance a larger force when it acts at a proportionally greater perpendicular distance.

For a light horizontal beam supported at both ends, the upward support forces together equal the downward weight of the heavy object when the beam is in equilibrium.

$$R_A+R_B=W$$

Moving the object toward support B increases $R_B$ and decreases $R_A$ . Taking moments about A: $R_B L=W x$ , where $L$ is the distance between supports and $x$ is the object's distance from A. Thus $R_B=W x/L$ and $R_A=W-R_B$ .

For a $300\,N$ object on a $2.0\,m$ beam, $0.50\,m$ from A: $R_B=(300)(0.50)/2.0=75\,N$ and $R_A=300-75=225\,N$ . The nearer support carries the larger share.

The total upward force stays equal to the object's weight for a light beam; moving the object redistributes that total between supports. If beam weight were not negligible, its own weight would also need to be included.

1 Forces and motion

Exam analysis

Most tested topics

Forces and motion units

Choose the correct unit in mechanics

Build units for moments and momentum

(b) Movement and position

Read distance-time graphs

Calculate average speed

Investigate motion reliably

Calculate acceleration from velocity change

Read velocity-time graphs

Find acceleration from graph gradient

Find distance from area on a velocity-time graph

Use the time-free motion equation

(c) Forces, movement, shape and momentum

Recognise what a force can change

Identify common forces

Distinguish vectors from scalars

Represent force as a vector

Calculate a resultant force along a line

Explain friction as an opposing force

Use resultant force, mass and acceleration

Calculate weight from mass and field strength

Build stopping distance from two stages

Explain factors that change stopping distance

Explain terminal velocity from changing forces

Investigate force and extension

Recognise the Hooke's law region

Describe elastic behaviour

Calculate momentum

Explain how safety features reduce force

Apply conservation of momentum

Relate force to momentum change

Identify a Newton's third-law pair

Calculate the moment of a force

Locate the line of action of weight

Use the principle of moments

Explain support forces on a light beam