(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.

Learning objectives

1.11Effects of forcesDescribe the effects of forces between bodies such as changes in speed, shape or direction 1.12Types of forceIdentify different types of force such as gravitational or electrostatic 1.13Vectors and scalarsUnderstand how vector quantities differ from scalar quantities 1.14Force as a vectorUnderstand that force is a vector quantity 1.15Resultant forceCalculate the resultant force of forces that act along a line 1.16FrictionKnow that friction is a force that opposes motion 1.17Force, mass and accelerationKnow and use unbalanced force = mass × acceleration, F = ma.1.18Weight equationKnow and use weight = mass × gravitational field strength, W = mg.1.19Stopping distanceKnow that the stopping distance of a vehicle is made up of the sum of the thinking distance and the braking distance 1.20Factors affecting stopping distanceDescribe the factors affecting vehicle stopping distance, including speed, mass, road condition and reaction time 1.21Falling objects and terminal velocityDescribe the forces acting on falling objects (and explain why falling objects reach a terminal velocity)1.22Force-extension practicalPractical: investigate how extension varies with applied force for helical springs, metal wires and rubber bands 1.23Hooke’s law regionKnow that the initial linear region of a force-extension graph is associated with Hooke’s law 1.24Elastic behaviourDescribe elastic behaviour as the ability of a material to recover its original shape after the forces causing deformation have been removed 125P Momentum equationKnow and use momentum = mass × velocity, p = mv.126P Momentum and safety featuresUse the idea of momentum to explain safety features 127P Conservation of momentumUse the conservation of momentum to calculate the mass, velocity or momentum of objects 128P Force and momentum changeUse force = change in momentum ÷ time taken, F = Δp/t = (mv − mu)/t.129P Newton’s third lawDemonstrate an understanding of Newton’s third law 130P Moments equationKnow and use moment = force × perpendicular distance from the pivot.131P Centre of gravityKnow that the weight of a body acts through its centre of gravity 132P Principle of momentsUse the principle of moments for a simple system of parallel forces acting in one plane 133P Forces on a supported beamUnderstand how the upward forces on a light beam, supported at its ends, vary with the position of a heavy object placed on the beam

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.

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—Pearson Edexcel International GCSE Physics (4PH1)Read

(c) Forces, movement, shape and momentum

Learning objectives

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

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