2.3.1 Newton's first law

Resultant force

Definition

Resultant force

Resultant force is the single force that has the same effect as all the forces acting on an object, after their sizes and directions have been taken into account.

A force is a vector quantity, so its direction matters as well as its size. Forces acting in the same direction are added. Forces acting in opposite directions are subtracted.
For forces along one line, the direction of the larger total determines the direction of the resultant force. For example, $N700\ \text{N}$ forwards and $N250\ \text{N}$ backwards give $N450\ \text{N}$ forwards.
If forces are equal in size and opposite in direction, the resultant force is $N0\ \text{N}$ . The forces are balanced, but forces are still acting on the object.

Example

Balancing forces

A car travels at constant velocity. Its engine provides a driving force of $N1800\ \text{N}$ forwards.
Constant velocity means the resultant force is $N0\ \text{N}$ .
Therefore $force=0\text{driving force}-\text{resistive force}=0$ .
Substitution gives $N1800\ \text{N}-\text{resistive force}=0\ \text{N}$ .
The resistive force is $N1800\ \text{N}$ backwards.

Newton's first law

Definition

Newton's first law

Newton's first law states that an object remains at rest or continues to move at constant velocity unless a non-zero resultant force acts on it.

A constant velocity means constant speed in a constant direction. Motion at constant speed around a bend is not constant velocity because the direction changes.
Newton's first law does not say that a moving object needs a forward resultant force. A resultant force is needed to change velocity, not to maintain a constant velocity.
A bicycle slows when the rider stops pedalling because friction and air resistance provide a resultant force opposite to its motion. If no resultant force acted, it would continue at constant velocity.

Zero resultant force

Definition

Balanced forces

Balanced forces are forces whose vector sum is zero, so they produce no change in velocity.

When the resultant force is zero, acceleration is zero. An object can therefore remain at rest or move at constant velocity.
A book at rest on a table has weight acting downwards and an equal normal contact force acting upwards. The two forces balance, so the book remains at rest.
A vehicle moving at constant velocity has a driving force forwards equal to the total friction and drag backwards. The forces balance, so its speed and direction remain constant.
A falling object at terminal velocity has weight downwards equal to air resistance upwards. Its resultant force and acceleration are zero, so it continues downwards at constant speed.

Exam technique

Explaining balanced motion

Link the force condition to the motion: the forces are balanced, so the resultant force is zero, therefore the object remains at rest or moves at constant velocity.
Do not write that there are no forces. State that the forces are equal and opposite or that their resultant is zero.
When a value and direction are required, include both, such as $N450\ \text{N}$ forwards.

Non-zero resultant force

Definition

Acceleration

Acceleration is the rate of change of velocity.

When the resultant force is not zero, the object's velocity changes. The acceleration is in the same direction as the resultant force.
A resultant force in the direction of motion makes the object speed up. A resultant force opposite to the direction of motion makes it slow down.
A resultant force at an angle to the motion changes the direction of the velocity. The speed can stay constant while the velocity changes.

Example

Using an unbalanced force

A car has a driving force of $N4200\ \text{N}$ forwards and resistive forces totalling $N3000\ \text{N}$ backwards.
The resultant force is $N4200\ \text{N}-3000\ \text{N}=1200\ \text{N}$ forwards.
The resultant force is non-zero, so the car accelerates.
The resultant force acts in the direction of motion, so the car speeds up.

Common Mistake

Common force errors

Zero resultant force does not mean zero individual forces.
Constant speed only proves zero acceleration if the direction is also constant.
Do not say that a force is used up. Forces act on objects and can change their velocity.

Self review

State Newton's first law.
Explain why an object moving at constant velocity has a resultant force of zero.
Calculate the resultant force when $N650\ \text{N}$ acts forwards and $N280\ \text{N}$ acts backwards.
Explain why zero resultant force does not mean that no forces are acting.
Describe how a non-zero resultant force changes an object's velocity.

2.3.2 Newton's second law

Newton's second law

Definition

Newton's second law

Newton's second law states that the resultant force on an object equals its mass multiplied by its acceleration.

The relationship is $F = ma$ , where $F$ is resultant force in newtons, $m$ is mass in kilograms and $a$ is acceleration in metres per second squared.
The acceleration is in the same direction as the resultant force. A larger resultant force causes a larger acceleration in that direction.
For constant mass, acceleration is directly proportional to resultant force: $a∝Fa\propto F$ . Doubling $F$ doubles $a$ .
For constant resultant force, acceleration is inversely proportional to mass: $a∝1ma\propto\dfrac{1}{m}$ . Doubling $m$ halves $a$ .
One newton is the resultant force that gives a mass of $kg1\ \text{kg}$ an acceleration of $s−21\ \text{m s}^{-2}$ , so $s−21\ \text{N}=1\ \text{kg m s}^{-2}$ .

Using the force equation

Use the equation $F = ma$ . Its rearranged forms are $a=Fma=\dfrac{F}{m}$ and $m=Fam=\dfrac{F}{a}$ .
The force in this equation is the resultant force. Subtract opposing forces before substitution.
Convert mass to kilograms before using the equation. For example, $kg750\ \text{g}=0.750\ \text{kg}$ .
If velocity data are given, first calculate acceleration using $a=v−uta=\dfrac{v-u}{t}$ , then use $F = ma$ .

Example

Calculating acceleration

A van of mass $kg1500\ \text{kg}$ has a driving force of $N4200\ \text{N}$ and resistive forces of $N1200\ \text{N}$ .
The resultant force is $NF=4200-1200=3000\ \text{N}$ .
Rearranging $F = ma$ gives $a=Fma=\dfrac{F}{m}$ .
Substitution gives $s−2a=\dfrac{3000}{1500}=2.0\ \text{m s}^{-2}$ .
The acceleration is $s−22.0\ \text{m s}^{-2}$ forwards.

Example

Finding resultant force

A $kg900\ \text{kg}$ car increases its velocity from $s−18.0\ \text{m s}^{-1}$ to $s−120\ \text{m s}^{-1}$ in $s6.0\ \text{s}$ .
First calculate acceleration: $s−2a=\dfrac{v-u}{t}=\dfrac{20-8.0}{6.0}=2.0\ \text{m s}^{-2}$ .
Then use $F = ma$ : $NF=900\times2.0=1800\ \text{N}$ .
The resultant force is $N1800\ \text{N}$ in the direction of the acceleration.

Practical

Force, mass and acceleration

Aim: to investigate how acceleration depends on resultant force and total moving mass using a trolley system.
Apparatus: dynamics trolley, level or gently inclined runway, pulley and clamp, light string, mass hanger, slotted masses, additional trolley masses, two light gates with a data logger or a motion sensor, interrupt card, balance, metre rule and stop block.
Variables, mass investigation: total moving mass is the independent variable, acceleration is the dependent variable, and the hanging mass, pulling force, release position, runway angle, light-gate positions and pulley arrangement are controlled.
Method, set-up:
- Measure the mass of the trolley, hanger, string and every slotted mass that will move. Record the total moving mass in kilograms.
- Attach the trolley to the hanger using a light string over the pulley. Keep the string parallel to the runway and secure the pulley firmly.
- Fit the interrupt card, align and zero the light gates or motion sensor, mark a fixed release position and place a stop block before the pulley.
- Check whether the runway is level. If friction compensation is used, raise one end slightly until the trolley moves at nearly constant velocity after a gentle push, then leave the angle unchanged.
Method, changing mass at constant pulling force:
- Place a fixed mass on the hanger so its weight provides an approximately constant pulling force, $F=mhgF=m_{\mathrm{h}}g$ .
- Release the trolley from the marked point without pushing it. Record the velocity-time data and determine the acceleration.
- Repeat at least three times and calculate the mean acceleration. Investigate anomalous readings rather than removing them automatically.
- Add a known mass to the trolley while leaving the hanging mass unchanged. Recalculate the total moving mass and repeat for at least five masses.
Method, changing force at constant total mass:
- Place several slotted masses on the trolley and keep the total moving mass fixed throughout.
- Transfer one mass at a time from the trolley to the hanger. Each transfer increases the pulling force without changing the total mass being accelerated.
- For each pulling force, release the trolley from the same point, record at least three accelerations and calculate a mean.
Results: increasing total mass decreases acceleration when the pulling force is constant. Increasing resultant force increases acceleration when total mass is constant.
Maths: calculate acceleration using $a=v−uta=\dfrac{v-u}{t}$ where required. Plot $a$ against $m$ for the mass investigation and $a$ against $F$ for the force investigation. A plot of $a$ against $1m\dfrac{1}{m}$ should be linear for constant force, while $a$ against $F$ should be a straight line through the origin when mass is constant.
Watch out: use the mass of the whole moving system, including the hanger. Friction and pulley resistance mean the hanging weight is not exactly the resultant force. Slack string, a sloping runway, inconsistent release and misaligned light gates can also distort the results.
Safety: secure the runway and pulley, use a stop block, keep feet clear of the falling hanger and use modest masses so the trolley remains controlled.

Example

Using a graph gradient

An acceleration against force graph passes through $s−2)(0.20\ \text{N},0.25\ \text{m s}^{-2})$ and $s−2)(1.00\ \text{N},1.25\ \text{m s}^{-2})$ .
The gradient is $N−1\dfrac{1.25-0.25}{1.00-0.20}=1.25\ \text{m s}^{-2}\text{ N}^{-1}$ .
Since $a=Fma=\dfrac{F}{m}$ , the gradient equals $1m\dfrac{1}{m}$ .
Therefore $kgm=\dfrac{1}{1.25}=0.80\ \text{kg}$ .

Exam technique

Force calculation method

Write $F = ma$ before substituting, then show the resultant-force calculation separately.
Use the unit $N\text{N}$ for force, $kg\text{kg}$ for mass and $s−2\text{m s}^{-2}$ for acceleration.
For a practical improvement, name the change and link it to its effect, such as using an automatic release so the trolley is not given an extra push.
Use directly proportional only when the graph is a straight line through the origin.

Common Mistake

Common force errors

Do not use the driving or pulling force as $F$ without considering opposing forces.
Do not put weight in newtons into the mass position. Calculate mass in kilograms first.
Do not change force and mass at the same time unless total moving mass is deliberately kept constant by transferring masses.

Self review

State Newton's second law and give the unit of each quantity in $F = ma$ .
Calculate the acceleration of a $kg1200\ \text{kg}$ car acted on by a resultant force of $N3600\ \text{N}$ .
Explain how acceleration changes when mass doubles at constant resultant force.
Describe how the practical changes force while keeping total moving mass constant.
Explain why repeats, a mean and friction control improve the investigation.

2.3.3 Weight and gravitational field strength

Weight and gravity

Definition

Weight

Weight is the force acting on an object because of a gravitational field.

Weight is a vector force. Near the surface of a planet or moon it acts towards the centre of that body, described locally as vertically downwards.
The symbol for weight is $W$ and its unit is the newton, $N\text{N}$ . Weight must not be stated in kilograms.
An object has weight whenever it is in a gravitational field. An astronaut in orbit still has weight because gravity provides the force that keeps the astronaut in orbit.

Gravitational field strength

Definition

Gravitational field strength

Gravitational field strength is the force per unit mass acting on an object placed in a gravitational field, measured in newtons per kilogram (N/kg).

The symbol is $g$ and the unit is newtons per kilogram, $kg−1\text{N kg}^{-1}$ . A field strength of $kg−110\ \text{N kg}^{-1}$ means that every kilogram experiences a gravitational force of $N10\ \text{N}$ .
Near Earth's surface, use $kg−1g=10\ \text{N kg}^{-1}$ unless a different value is supplied. Values such as $kg−19.8\ \text{N kg}^{-1}$ may be used when given.
Gravitational field strength depends on location. It is different on different planets and moons, and becomes weaker with increasing distance from the body producing the field.

Mass and weight

Definition

Mass

Mass is a measure of the amount of matter in an object and is measured in kilograms.

Mass is a scalar and does not change when an object moves from Earth to the Moon. Weight changes because gravitational field strength changes.
A $kg50\ \text{kg}$ person has the same mass everywhere. On Earth, using $kg−1g=9.8\ \text{N kg}^{-1}$ , the weight is $N490\ \text{N}$ . On the Moon, using $kg−1g=1.6\ \text{N kg}^{-1}$ , the weight is $N80\ \text{N}$ .

A diagram comparing the mass and weight of a person on Earth, Mars, and the Moon. The mass remains constant at 50 kg, while the weight changes based on the gravitational field strength (g) of each body: 490 N on Earth (g=9.8 m/s²), 185 N on Mars (g=3.7 m/s²), and 80 N on the Moon (g=1.6 m/s²).

The weight equation

The relationship is $W = m g$ , where $W$ is weight in newtons, $m$ is mass in kilograms and $g$ is gravitational field strength in newtons per kilogram.
The rearranged forms are $m=Wgm=\dfrac{W}{g}$ and $g=Wmg=\dfrac{W}{m}$ .
For a fixed mass, $W∝gW\propto g$ . Doubling the gravitational field strength doubles the weight.
For a fixed gravitational field strength, $W∝mW\propto m$ . Doubling the mass doubles the weight.

Example

Calculating weight

A student has a mass of $kg58\ \text{kg}$ where $kg−1g=10\ \text{N kg}^{-1}$ .
Use $W = m g$ .
Substitution gives $NW=58\times10=580\ \text{N}$ .
The student's weight is $N580\ \text{N}$ .

Example

Finding mass and weight

A rock weighs $N96\ \text{N}$ on the Moon, where $kg−1g=1.6\ \text{N kg}^{-1}$ .
Rearrange $W = m g$ to $m=Wgm=\dfrac{W}{g}$ .
Substitution gives $kgm=\dfrac{96}{1.6}=60\ \text{kg}$ .
The mass remains $kg60\ \text{kg}$ on Earth. Using $kg−1g=10\ \text{N kg}^{-1}$ , $NW=60\times10=600\ \text{N}$ .

Measuring weight

Definition

Newtonmeter

A newtonmeter is a calibrated force meter that measures force in newtons by the extension of a spring.

Check that the newtonmeter reads zero before attaching the object. Correct the zero setting where possible.
Select a newtonmeter with a range large enough for the expected weight and a scale fine enough to give a useful resolution.
Hold or clamp the newtonmeter vertically, attach the object securely and allow it to hang freely without touching another surface.
Wait for oscillations to stop, then read the scale at eye level to reduce parallax error.
Record the value in newtons. A balance measures mass in kilograms, whereas a newtonmeter measures weight in newtons.
The spring stretches because the attached weight pulls on it. The scale is calibrated so the extension gives the force directly.

Weight in different fields

The same object weighs less where $g$ is smaller because $W = m g$ and its mass remains constant.
A value of $kg−1g=1.6\ \text{N kg}^{-1}$ on the Moon gives a weight about one sixth of the value on Earth when $kg−1g\approx9.8\ \text{N kg}^{-1}$ .
The strength of a gravitational field decreases with distance from the body producing it. This reduces weight, although mass remains unchanged.

Exam technique

Explaining changes in weight

Give the complete causal chain: the gravitational field strength is smaller, mass is unchanged, and $W = m g$ , so weight is smaller.
For calculations, write the equation, substitute values with consistent units, evaluate and state the final unit.
Use mass for kilograms and weight for newtons. Calling kilograms a unit of weight loses precision and can lose the mark.
When describing measurement, name the newtonmeter, state that it is read in newtons and include a method detail such as zeroing or reading at eye level.

Common Mistake

Common gravity errors

Do not say that mass changes with location. The gravitational field strength and weight change.
Do not use $s−2\text{m s}^{-2}$ when the required quantity is gravitational field strength. Use $kg−1\text{N kg}^{-1}$ .
Do not claim that orbit means no gravity. Orbiting objects are continuously falling under the gravitational force.

Self review

Define weight and gravitational field strength.
State the instrument used to measure weight and describe how it is read accurately.
Calculate the weight of a $kg6.0\ \text{kg}$ object where $kg−1g=10\ \text{N kg}^{-1}$ .
Explain why an object's mass stays constant but its weight changes on the Moon.
Rearrange $W = m g$ to make $g$ the subject.

2.4.1 Circular motion and centripetal force

Speed and velocity

Definition

Velocity

Velocity is speed in a stated direction, measured in metres per second.

Speed is a scalar quantity. Velocity is a vector quantity. An object can therefore have constant speed while its velocity changes.
An object travelling in a circle constantly changes its direction of motion. Its velocity changes even if its speed remains constant.
At any point on the circular path, the instantaneous velocity is directed along a tangent to the circle. The tangent direction changes from point to point.

A diagram of circular motion showing velocity vectors tangent to the circular path at two different points, illustrating how the direction of motion changes even if speed is constant.

Acceleration in a circle

Definition

Acceleration

Acceleration is the rate of change of velocity.

The relationship $a=v−uta=\dfrac{v-u}{t}$ describes change in velocity. In circular motion, the change is in direction rather than necessarily in speed.
An object moving in a circle at constant speed is accelerating because its velocity direction changes continuously.
A changing velocity requires a non-zero resultant force. Without that force, Newton's first law predicts motion at constant velocity in a straight line.

Centripetal force

Definition

Centripetal force

Centripetal force is the resultant force acting on an object moving in a circle, directed towards the centre of the circle.

The centripetal force acts towards the centre at every point on the path. It is perpendicular to the instantaneous velocity in uniform circular motion.
The inward force continually changes the direction of the velocity, producing circular motion. It does not point along the tangent.
If the centripetal force suddenly disappears, the object initially moves along the tangent to the circle at the point where the force stopped acting.

Forces that provide centripetal force

Centripetal force is not an additional type of force. It is the name given to whichever real force, or resultant of forces, acts towards the centre.
For a satellite or moon in orbit, gravitational force provides the centripetal force towards the centre of the planet.
For a ball moving in a circle on a string, tension in the string provides the centripetal force towards the hand or pivot.
For a car turning on a level road, friction between the tyres and road provides the centripetal force towards the centre of the bend.
For clothes moving in a circle inside a washing-machine drum, the normal contact force from the drum provides the inward force.

Example

Explaining an orbit

A satellite moves around Earth in a circular orbit at constant speed.
Its direction changes continuously, so its velocity changes.
A change in velocity means the satellite is accelerating even though its speed is constant.
The gravitational force acts towards the centre of Earth and provides the centripetal resultant force.

Example

A string breaks

A ball moves in a horizontal circle on a string.
The tension acts towards the centre and provides the centripetal force.
When the string breaks, the tension disappears and there is no inward force.
The ball initially continues in a straight line along the tangent at the breaking point, in accordance with Newton's first law.

Exam technique

Explaining circular motion

Use the linked chain: direction changes, so velocity changes, therefore the object accelerates.
State that the resultant force acts towards the centre of the circle. Words such as sideways or inwards are less precise.
For a named situation, identify the real force providing the centripetal force, such as gravity, tension, friction or normal contact force.
Keep the explanation qualitative. No centripetal-force calculation is required for this content.

Common Mistake

Common circular-motion errors

Constant speed does not mean constant velocity when direction changes.
Do not draw an outward force on the moving object. The resultant force required for circular motion is towards the centre.
Do not say that an orbiting object has no gravity. Gravity supplies the centripetal force.

Self review

Explain why an object moving in a circle at constant speed is accelerating.
State the direction of the centripetal resultant force.
Name the force providing centripetal force for a satellite, a ball on a string and a car on a bend.
Describe the motion of a ball immediately after its string breaks.
Explain why centripetal force is not an additional type of force.

2.4.2 Inertial mass

Inertial mass measures resistance to acceleration

Definition

Inertia

Inertia is the tendency of an object to remain at rest or continue moving at constant velocity unless a resultant force acts on it.

Definition

Inertial mass

Inertial mass is the ratio of the resultant force acting on an object to the acceleration it produces, measured in kilograms.

Inertial mass measures how difficult it is to change an object's velocity, including how difficult it is to start an object moving from rest.
The defining equation is $m=Fam=\dfrac{F}{a}$ , where $m$ is inertial mass in kilograms, $F$ is resultant force in newtons and $a$ is acceleration in metres per second squared.
A large inertial mass means that the same resultant force produces a smaller acceleration because $a=Fma=\dfrac{F}{m}$ .
A small inertial mass means that the same resultant force produces a larger acceleration.

Resultant force must be used

Use the resultant force, not one individual force, because opposing forces may partly or completely cancel.
If the resultant force is zero, the acceleration is zero, so the object remains at rest or continues at constant velocity.
Inertial mass is a property of the object and does not depend on the strength of the gravitational field.

Common Mistake

Do not describe inertia as a force because inertia is a tendency, not an interaction acting on the object.
Do not use weight in $m=Fam=\dfrac{F}{a}$ unless weight is the resultant force in the situation.

Example

Given: a resultant force of $N18\ \text{N}$ produces an acceleration of $m/s23.0\ \text{m/s}^2$ .
Rearrange Newton's second law: $m=Fam=\dfrac{F}{a}$ .
Substitute: $kgm=\dfrac{18\ \text{N}}{3.0\ \text{m/s}^2}=6.0\ \text{kg}$ .
The object's inertial mass is $kg6.0\ \text{kg}$ .

Comparing objects from data

For the same resultant force, compare accelerations using $m=Fam=\dfrac{F}{a}$ ; the object with the smaller acceleration has the greater inertial mass.
For the same acceleration, the object that requires the larger resultant force has the greater inertial mass.
On a graph of resultant force against acceleration, $F = ma$ , so the gradient equals the inertial mass.

Exam technique

When explaining a comparison, link the same force to the different accelerations and then state which object has the greater inertial mass.
For calculations, identify the resultant force before substituting and include the unit $kg\text{kg}$ .

Force and acceleration data reveal mass

A straight-line graph through the origin shows that resultant force is directly proportional to acceleration when mass is constant.
A steeper $F$ against $a$ graph represents a larger inertial mass because its gradient $ΔFΔa\dfrac{\Delta F}{\Delta a}$ is larger.

Self review

What does inertial mass measure?
State the equation that defines inertial mass.
Why must resultant force be used in the equation?
What does the gradient of a force against acceleration graph represent?

2.4.3 Newton's third law and momentum in collisions

Newton's third law links interaction forces

Definition

Newton's third law

Newton's third law states that when two objects interact, they exert forces on each other that are equal in magnitude and opposite in direction.

Definition

Momentum

Momentum is the product of an object's mass and velocity, measured in kilogram metres per second (kg m/s).

A Newton's third-law pair acts on two different objects during the same interaction.
The forces are the same type, equal in magnitude, opposite in direction and present for the same time.
The pair does not cancel because the two forces act on different objects.
For a book resting on a table, the book pushes down on the table and the table pushes up on the book with an equal force.

Common Mistake

Do not pair the weight of a book with the table's normal contact force because both forces act on the book.
The third-law partner of the book's weight is the gravitational force that the book exerts on Earth.

Momentum includes direction

Momentum is calculated using $p = m v$ , where $p$ is momentum in $m/s\text{kg m/s}$ , $m$ is mass in $kg\text{kg}$ and $v$ is velocity in $m/s\text{m/s}$ .
Momentum is a vector, so choose one direction as positive and give motion in the opposite direction a negative velocity and momentum.
An object at rest has zero momentum because $v = 0$ .

Example

Moving car: a $kg1200\ \text{kg}$ car travels east at $m/s15\ \text{m/s}$ .
Taking east as positive, $m/sp=mv=1200\times15=1.8\times10^4\ \text{kg m/s}$ .
Its momentum is $m/s1.8\times10^4\ \text{kg m/s}$ east.

Momentum is conserved in a closed system

Definition

Conservation of momentum

Conservation of momentum states that in a closed system the total momentum before an interaction equals the total momentum after the interaction.

During a collision, each object experiences an equal and opposite force for the same time.
Each object therefore has an equal and opposite change in momentum, so the total momentum of the system does not change.
Write conservation as $∑pbefore=∑pafter\sum p_{\text{before}}=\sum p_{\text{after}}$ and include signs for direction.
Kinetic energy does not have to be conserved in a collision because some energy may be transferred to thermal stores, sound or deformation.

Example

Before: a $kg2.0\ \text{kg}$ trolley moving right at $m/s4.0\ \text{m/s}$ hits a stationary $kg3.0\ \text{kg}$ trolley and they stick together.
Initial momentum: $m/sp=(2.0\times4.0)+(3.0\times0)=8.0\ \text{kg m/s}$ .
Combined mass: $kgm=2.0+3.0=5.0\ \text{kg}$ .
Conservation gives $8.0 = 5.0 v$ , so $m/sv=1.6\ \text{m/s}$ to the right.

Explosions also conserve momentum

If two objects start at rest, their total initial momentum is zero.
After they push apart, their momenta are equal in magnitude and opposite in direction, so their vector sum remains zero.

Exam technique

State a positive direction before calculating and keep negative signs until the final answer.
For a third-law explanation, name both objects, both forces, and state that the forces are equal in magnitude and opposite in direction.
For conservation, calculate the total momentum of the whole system before and after the interaction.

Interaction forces explain momentum changes

A larger force or a longer interaction time produces a larger change in momentum.
Equal and opposite changes in momentum preserve the total momentum of a closed system.

Self review

State Newton's third law.
Define momentum and give its unit.
Why do third-law forces not cancel each other?
What condition is needed for total momentum to be conserved?
How should direction be represented in a momentum calculation?

2.4.4 Force and rate of change of momentum

Force changes momentum over time

Definition

Rate of change of momentum

Rate of change of momentum is the change in an object's momentum divided by the time taken for that change.

Definition

Impulse

Impulse is the product of a force and the time for which it acts, and it equals the change in momentum it produces.

Newton's second law can be written as $F=Δpt=mv−mutF=\dfrac{\Delta p}{t}=\dfrac{mv-mu}{t}$ for constant mass.
$F$ is the resultant force in newtons, $m$ is mass in kilograms, $u$ is initial velocity, $v$ is final velocity and $t$ is time in seconds.
The change in momentum is $Δp=pfinal−pinitial=mv−mu\Delta p=p_{\text{final}}-p_{\text{initial}}=mv-mu$ .
The direction of the resultant force is the direction of the change in momentum, not necessarily the direction in which the object is moving.

Signs show changes in direction

Choose a positive direction before substituting velocities.
A negative velocity represents motion in the opposite direction, so a rebound usually gives a larger momentum change than stopping.

Example

Stopping: a $kg0.20\ \text{kg}$ ball moving at $m/s12\ \text{m/s}$ is brought to rest in $s0.040\ \text{s}$ .
$m/s\Delta p=m(v-u)=0.20(0-12)=-2.4\ \text{kg m/s}$ .
$NF=\dfrac{-2.4}{0.040}=-60\ \text{N}$ .
The mean force has magnitude $N60\ \text{N}$ and acts opposite to the ball's initial motion.

A longer stopping time reduces force

Rearranging gives $FΔt=ΔpF\Delta t=\Delta p$ , so the same momentum change spread over a longer time produces a smaller mean force.
Crumple zones, airbags, seat belts, crash mats and bending the knees on landing increase the time over which momentum changes.
The change in momentum is fixed by the initial and final velocities, so increasing time reduces the rate of change of momentum and therefore the force.

Example

Safety comparison: a passenger's momentum changes by $m/s420\ \text{kg m/s}$ .
Without an airbag, $NF=\dfrac{420}{0.030}=1.4\times10^4\ \text{N}$ .
With an airbag, $NF=\dfrac{420}{0.12}=3.5\times10^3\ \text{N}$ .
Increasing the stopping time by a factor of four reduces the mean force by a factor of four.

Force-time graphs give impulse

The area under a force-time graph equals impulse, $J=FΔtJ=F\Delta t$ , when the force is constant.
For a changing force, estimate or calculate the total area under the graph to find $Δp\Delta p$ .
The unit of impulse is $s\text{N s}$ , which is equivalent to $m/s\text{kg m/s}$ .

Common Mistake

Do not calculate $m v + m u$ unless the chosen velocity signs make that the correct subtraction.
Do not use speed without direction when an object rebounds.
Use the resultant or mean force specified by the data, not an unrelated contact force.

Multi-stage calculations need a clear route

If force and time are given, find momentum change using $Δp=FΔt\Delta p=F\Delta t$ .
If momentum change and one velocity are known, use $m(v−u)=Δpm(v-u)=\Delta p$ to find the missing velocity.

Exam technique

Write $Δp=m(v−u)\Delta p=m(v-u)$ before dividing by time.
Keep signs throughout the calculation, then describe the direction of a negative answer in words.
When explaining a safety feature, state that it increases the stopping time for the same momentum change, which reduces the mean force.

Self review

Write Newton's second law in momentum form.
Why can a rebound produce a large change in momentum?
How does an airbag reduce the force on a passenger?
What does the area under a force-time graph represent?

2.3.1 Newton's first law

Resultant force

Definition

Resultant force

Resultant force is the single force that has the same effect as all the forces acting on an object, after their sizes and directions have been taken into account.

A force is a vector quantity, so its direction matters as well as its size. Forces acting in the same direction are added. Forces acting in opposite directions are subtracted.
For forces along one line, the direction of the larger total determines the direction of the resultant force. For example, $N700\ \text{N}$ forwards and $N250\ \text{N}$ backwards give $N450\ \text{N}$ forwards.
If forces are equal in size and opposite in direction, the resultant force is $N0\ \text{N}$ . The forces are balanced, but forces are still acting on the object.

Example

Balancing forces

A car travels at constant velocity. Its engine provides a driving force of $N1800\ \text{N}$ forwards.
Constant velocity means the resultant force is $N0\ \text{N}$ .
Therefore $force=0\text{driving force}-\text{resistive force}=0$ .
Substitution gives $N1800\ \text{N}-\text{resistive force}=0\ \text{N}$ .
The resistive force is $N1800\ \text{N}$ backwards.

Newton's first law

Definition

Newton's first law

Newton's first law states that an object remains at rest or continues to move at constant velocity unless a non-zero resultant force acts on it.

A constant velocity means constant speed in a constant direction. Motion at constant speed around a bend is not constant velocity because the direction changes.
Newton's first law does not say that a moving object needs a forward resultant force. A resultant force is needed to change velocity, not to maintain a constant velocity.
A bicycle slows when the rider stops pedalling because friction and air resistance provide a resultant force opposite to its motion. If no resultant force acted, it would continue at constant velocity.

Zero resultant force

Definition

Balanced forces

Balanced forces are forces whose vector sum is zero, so they produce no change in velocity.

When the resultant force is zero, acceleration is zero. An object can therefore remain at rest or move at constant velocity.
A book at rest on a table has weight acting downwards and an equal normal contact force acting upwards. The two forces balance, so the book remains at rest.
A vehicle moving at constant velocity has a driving force forwards equal to the total friction and drag backwards. The forces balance, so its speed and direction remain constant.
A falling object at terminal velocity has weight downwards equal to air resistance upwards. Its resultant force and acceleration are zero, so it continues downwards at constant speed.

Exam technique

Explaining balanced motion

Link the force condition to the motion: the forces are balanced, so the resultant force is zero, therefore the object remains at rest or moves at constant velocity.
Do not write that there are no forces. State that the forces are equal and opposite or that their resultant is zero.
When a value and direction are required, include both, such as $N450\ \text{N}$ forwards.

Non-zero resultant force

Definition

Acceleration

Acceleration is the rate of change of velocity.

When the resultant force is not zero, the object's velocity changes. The acceleration is in the same direction as the resultant force.
A resultant force in the direction of motion makes the object speed up. A resultant force opposite to the direction of motion makes it slow down.
A resultant force at an angle to the motion changes the direction of the velocity. The speed can stay constant while the velocity changes.

Example

Using an unbalanced force

A car has a driving force of $N4200\ \text{N}$ forwards and resistive forces totalling $N3000\ \text{N}$ backwards.
The resultant force is $N4200\ \text{N}-3000\ \text{N}=1200\ \text{N}$ forwards.
The resultant force is non-zero, so the car accelerates.
The resultant force acts in the direction of motion, so the car speeds up.

Common Mistake

Common force errors

Zero resultant force does not mean zero individual forces.
Constant speed only proves zero acceleration if the direction is also constant.
Do not say that a force is used up. Forces act on objects and can change their velocity.

Self review

State Newton's first law.
Explain why an object moving at constant velocity has a resultant force of zero.
Calculate the resultant force when $N650\ \text{N}$ acts forwards and $N280\ \text{N}$ acts backwards.
Explain why zero resultant force does not mean that no forces are acting.
Describe how a non-zero resultant force changes an object's velocity.

2.3.2 Newton's second law

Newton's second law

Definition

Newton's second law

Newton's second law states that the resultant force on an object equals its mass multiplied by its acceleration.

The relationship is $F = ma$ , where $F$ is resultant force in newtons, $m$ is mass in kilograms and $a$ is acceleration in metres per second squared.
The acceleration is in the same direction as the resultant force. A larger resultant force causes a larger acceleration in that direction.
For constant mass, acceleration is directly proportional to resultant force: $a∝Fa\propto F$ . Doubling $F$ doubles $a$ .
For constant resultant force, acceleration is inversely proportional to mass: $a∝1ma\propto\dfrac{1}{m}$ . Doubling $m$ halves $a$ .
One newton is the resultant force that gives a mass of $kg1\ \text{kg}$ an acceleration of $s−21\ \text{m s}^{-2}$ , so $s−21\ \text{N}=1\ \text{kg m s}^{-2}$ .

Using the force equation

Use the equation $F = ma$ . Its rearranged forms are $a=Fma=\dfrac{F}{m}$ and $m=Fam=\dfrac{F}{a}$ .
The force in this equation is the resultant force. Subtract opposing forces before substitution.
Convert mass to kilograms before using the equation. For example, $kg750\ \text{g}=0.750\ \text{kg}$ .
If velocity data are given, first calculate acceleration using $a=v−uta=\dfrac{v-u}{t}$ , then use $F = ma$ .

Example

Calculating acceleration

A van of mass $kg1500\ \text{kg}$ has a driving force of $N4200\ \text{N}$ and resistive forces of $N1200\ \text{N}$ .
The resultant force is $NF=4200-1200=3000\ \text{N}$ .
Rearranging $F = ma$ gives $a=Fma=\dfrac{F}{m}$ .
Substitution gives $s−2a=\dfrac{3000}{1500}=2.0\ \text{m s}^{-2}$ .
The acceleration is $s−22.0\ \text{m s}^{-2}$ forwards.

Example

Finding resultant force

A $kg900\ \text{kg}$ car increases its velocity from $s−18.0\ \text{m s}^{-1}$ to $s−120\ \text{m s}^{-1}$ in $s6.0\ \text{s}$ .
First calculate acceleration: $s−2a=\dfrac{v-u}{t}=\dfrac{20-8.0}{6.0}=2.0\ \text{m s}^{-2}$ .
Then use $F = ma$ : $NF=900\times2.0=1800\ \text{N}$ .
The resultant force is $N1800\ \text{N}$ in the direction of the acceleration.

Practical

Force, mass and acceleration

Aim: to investigate how acceleration depends on resultant force and total moving mass using a trolley system.
Apparatus: dynamics trolley, level or gently inclined runway, pulley and clamp, light string, mass hanger, slotted masses, additional trolley masses, two light gates with a data logger or a motion sensor, interrupt card, balance, metre rule and stop block.
Variables, mass investigation: total moving mass is the independent variable, acceleration is the dependent variable, and the hanging mass, pulling force, release position, runway angle, light-gate positions and pulley arrangement are controlled.
Method, set-up:
- Measure the mass of the trolley, hanger, string and every slotted mass that will move. Record the total moving mass in kilograms.
- Attach the trolley to the hanger using a light string over the pulley. Keep the string parallel to the runway and secure the pulley firmly.
- Fit the interrupt card, align and zero the light gates or motion sensor, mark a fixed release position and place a stop block before the pulley.
- Check whether the runway is level. If friction compensation is used, raise one end slightly until the trolley moves at nearly constant velocity after a gentle push, then leave the angle unchanged.
Method, changing mass at constant pulling force:
- Place a fixed mass on the hanger so its weight provides an approximately constant pulling force, $F=mhgF=m_{\mathrm{h}}g$ .
- Release the trolley from the marked point without pushing it. Record the velocity-time data and determine the acceleration.
- Repeat at least three times and calculate the mean acceleration. Investigate anomalous readings rather than removing them automatically.
- Add a known mass to the trolley while leaving the hanging mass unchanged. Recalculate the total moving mass and repeat for at least five masses.
Method, changing force at constant total mass:
- Place several slotted masses on the trolley and keep the total moving mass fixed throughout.
- Transfer one mass at a time from the trolley to the hanger. Each transfer increases the pulling force without changing the total mass being accelerated.
- For each pulling force, release the trolley from the same point, record at least three accelerations and calculate a mean.
Results: increasing total mass decreases acceleration when the pulling force is constant. Increasing resultant force increases acceleration when total mass is constant.
Maths: calculate acceleration using $a=v−uta=\dfrac{v-u}{t}$ where required. Plot $a$ against $m$ for the mass investigation and $a$ against $F$ for the force investigation. A plot of $a$ against $1m\dfrac{1}{m}$ should be linear for constant force, while $a$ against $F$ should be a straight line through the origin when mass is constant.
Watch out: use the mass of the whole moving system, including the hanger. Friction and pulley resistance mean the hanging weight is not exactly the resultant force. Slack string, a sloping runway, inconsistent release and misaligned light gates can also distort the results.
Safety: secure the runway and pulley, use a stop block, keep feet clear of the falling hanger and use modest masses so the trolley remains controlled.

Example

Using a graph gradient

An acceleration against force graph passes through $s−2)(0.20\ \text{N},0.25\ \text{m s}^{-2})$ and $s−2)(1.00\ \text{N},1.25\ \text{m s}^{-2})$ .
The gradient is $N−1\dfrac{1.25-0.25}{1.00-0.20}=1.25\ \text{m s}^{-2}\text{ N}^{-1}$ .
Since $a=Fma=\dfrac{F}{m}$ , the gradient equals $1m\dfrac{1}{m}$ .
Therefore $kgm=\dfrac{1}{1.25}=0.80\ \text{kg}$ .

Exam technique

Force calculation method

Write $F = ma$ before substituting, then show the resultant-force calculation separately.
Use the unit $N\text{N}$ for force, $kg\text{kg}$ for mass and $s−2\text{m s}^{-2}$ for acceleration.
For a practical improvement, name the change and link it to its effect, such as using an automatic release so the trolley is not given an extra push.
Use directly proportional only when the graph is a straight line through the origin.

Common Mistake

Common force errors

Do not use the driving or pulling force as $F$ without considering opposing forces.
Do not put weight in newtons into the mass position. Calculate mass in kilograms first.
Do not change force and mass at the same time unless total moving mass is deliberately kept constant by transferring masses.

Self review

State Newton's second law and give the unit of each quantity in $F = ma$ .
Calculate the acceleration of a $kg1200\ \text{kg}$ car acted on by a resultant force of $N3600\ \text{N}$ .
Explain how acceleration changes when mass doubles at constant resultant force.
Describe how the practical changes force while keeping total moving mass constant.
Explain why repeats, a mean and friction control improve the investigation.

2.3.3 Weight and gravitational field strength

Weight and gravity

Definition

Weight

Weight is the force acting on an object because of a gravitational field.

Weight is a vector force. Near the surface of a planet or moon it acts towards the centre of that body, described locally as vertically downwards.
The symbol for weight is $W$ and its unit is the newton, $N\text{N}$ . Weight must not be stated in kilograms.
An object has weight whenever it is in a gravitational field. An astronaut in orbit still has weight because gravity provides the force that keeps the astronaut in orbit.

Gravitational field strength

Definition

Gravitational field strength

Gravitational field strength is the force per unit mass acting on an object placed in a gravitational field, measured in newtons per kilogram (N/kg).

The symbol is $g$ and the unit is newtons per kilogram, $kg−1\text{N kg}^{-1}$ . A field strength of $kg−110\ \text{N kg}^{-1}$ means that every kilogram experiences a gravitational force of $N10\ \text{N}$ .
Near Earth's surface, use $kg−1g=10\ \text{N kg}^{-1}$ unless a different value is supplied. Values such as $kg−19.8\ \text{N kg}^{-1}$ may be used when given.
Gravitational field strength depends on location. It is different on different planets and moons, and becomes weaker with increasing distance from the body producing the field.

Mass and weight

Definition

Mass

Mass is a measure of the amount of matter in an object and is measured in kilograms.

Mass is a scalar and does not change when an object moves from Earth to the Moon. Weight changes because gravitational field strength changes.
A $kg50\ \text{kg}$ person has the same mass everywhere. On Earth, using $kg−1g=9.8\ \text{N kg}^{-1}$ , the weight is $N490\ \text{N}$ . On the Moon, using $kg−1g=1.6\ \text{N kg}^{-1}$ , the weight is $N80\ \text{N}$ .

The weight equation

The relationship is $W = m g$ , where $W$ is weight in newtons, $m$ is mass in kilograms and $g$ is gravitational field strength in newtons per kilogram.
The rearranged forms are $m=Wgm=\dfrac{W}{g}$ and $g=Wmg=\dfrac{W}{m}$ .
For a fixed mass, $W∝gW\propto g$ . Doubling the gravitational field strength doubles the weight.
For a fixed gravitational field strength, $W∝mW\propto m$ . Doubling the mass doubles the weight.

Example

Calculating weight

A student has a mass of $kg58\ \text{kg}$ where $kg−1g=10\ \text{N kg}^{-1}$ .
Use $W = m g$ .
Substitution gives $NW=58\times10=580\ \text{N}$ .
The student's weight is $N580\ \text{N}$ .

Example

Finding mass and weight

A rock weighs $N96\ \text{N}$ on the Moon, where $kg−1g=1.6\ \text{N kg}^{-1}$ .
Rearrange $W = m g$ to $m=Wgm=\dfrac{W}{g}$ .
Substitution gives $kgm=\dfrac{96}{1.6}=60\ \text{kg}$ .
The mass remains $kg60\ \text{kg}$ on Earth. Using $kg−1g=10\ \text{N kg}^{-1}$ , $NW=60\times10=600\ \text{N}$ .

Measuring weight

Definition

Newtonmeter

A newtonmeter is a calibrated force meter that measures force in newtons by the extension of a spring.

Check that the newtonmeter reads zero before attaching the object. Correct the zero setting where possible.
Select a newtonmeter with a range large enough for the expected weight and a scale fine enough to give a useful resolution.
Hold or clamp the newtonmeter vertically, attach the object securely and allow it to hang freely without touching another surface.
Wait for oscillations to stop, then read the scale at eye level to reduce parallax error.
Record the value in newtons. A balance measures mass in kilograms, whereas a newtonmeter measures weight in newtons.
The spring stretches because the attached weight pulls on it. The scale is calibrated so the extension gives the force directly.

Weight in different fields

The same object weighs less where $g$ is smaller because $W = m g$ and its mass remains constant.
A value of $kg−1g=1.6\ \text{N kg}^{-1}$ on the Moon gives a weight about one sixth of the value on Earth when $kg−1g\approx9.8\ \text{N kg}^{-1}$ .
The strength of a gravitational field decreases with distance from the body producing it. This reduces weight, although mass remains unchanged.

Exam technique

Explaining changes in weight

Give the complete causal chain: the gravitational field strength is smaller, mass is unchanged, and $W = m g$ , so weight is smaller.
For calculations, write the equation, substitute values with consistent units, evaluate and state the final unit.
Use mass for kilograms and weight for newtons. Calling kilograms a unit of weight loses precision and can lose the mark.
When describing measurement, name the newtonmeter, state that it is read in newtons and include a method detail such as zeroing or reading at eye level.

Common Mistake

Common gravity errors

Do not say that mass changes with location. The gravitational field strength and weight change.
Do not use $s−2\text{m s}^{-2}$ when the required quantity is gravitational field strength. Use $kg−1\text{N kg}^{-1}$ .
Do not claim that orbit means no gravity. Orbiting objects are continuously falling under the gravitational force.

Self review

Define weight and gravitational field strength.
State the instrument used to measure weight and describe how it is read accurately.
Calculate the weight of a $kg6.0\ \text{kg}$ object where $kg−1g=10\ \text{N kg}^{-1}$ .
Explain why an object's mass stays constant but its weight changes on the Moon.
Rearrange $W = m g$ to make $g$ the subject.

2.4.1 Circular motion and centripetal force

Speed and velocity

Definition

Velocity

Velocity is speed in a stated direction, measured in metres per second.

Speed is a scalar quantity. Velocity is a vector quantity. An object can therefore have constant speed while its velocity changes.
An object travelling in a circle constantly changes its direction of motion. Its velocity changes even if its speed remains constant.
At any point on the circular path, the instantaneous velocity is directed along a tangent to the circle. The tangent direction changes from point to point.

A diagram of circular motion showing velocity vectors tangent to the circular path at two different points, illustrating how the direction of motion changes even if speed is constant.

Acceleration in a circle

Definition

Acceleration

Acceleration is the rate of change of velocity.

The relationship $a=v−uta=\dfrac{v-u}{t}$ describes change in velocity. In circular motion, the change is in direction rather than necessarily in speed.
An object moving in a circle at constant speed is accelerating because its velocity direction changes continuously.
A changing velocity requires a non-zero resultant force. Without that force, Newton's first law predicts motion at constant velocity in a straight line.

Centripetal force

Definition

Centripetal force

Centripetal force is the resultant force acting on an object moving in a circle, directed towards the centre of the circle.

The centripetal force acts towards the centre at every point on the path. It is perpendicular to the instantaneous velocity in uniform circular motion.
The inward force continually changes the direction of the velocity, producing circular motion. It does not point along the tangent.
If the centripetal force suddenly disappears, the object initially moves along the tangent to the circle at the point where the force stopped acting.

Forces that provide centripetal force

Centripetal force is not an additional type of force. It is the name given to whichever real force, or resultant of forces, acts towards the centre.
For a satellite or moon in orbit, gravitational force provides the centripetal force towards the centre of the planet.
For a ball moving in a circle on a string, tension in the string provides the centripetal force towards the hand or pivot.
For a car turning on a level road, friction between the tyres and road provides the centripetal force towards the centre of the bend.
For clothes moving in a circle inside a washing-machine drum, the normal contact force from the drum provides the inward force.

Example

Explaining an orbit

A satellite moves around Earth in a circular orbit at constant speed.
Its direction changes continuously, so its velocity changes.
A change in velocity means the satellite is accelerating even though its speed is constant.
The gravitational force acts towards the centre of Earth and provides the centripetal resultant force.

Example

A string breaks

A ball moves in a horizontal circle on a string.
The tension acts towards the centre and provides the centripetal force.
When the string breaks, the tension disappears and there is no inward force.
The ball initially continues in a straight line along the tangent at the breaking point, in accordance with Newton's first law.

Exam technique

Explaining circular motion

Use the linked chain: direction changes, so velocity changes, therefore the object accelerates.
State that the resultant force acts towards the centre of the circle. Words such as sideways or inwards are less precise.
For a named situation, identify the real force providing the centripetal force, such as gravity, tension, friction or normal contact force.
Keep the explanation qualitative. No centripetal-force calculation is required for this content.

Common Mistake

Common circular-motion errors

Constant speed does not mean constant velocity when direction changes.
Do not draw an outward force on the moving object. The resultant force required for circular motion is towards the centre.
Do not say that an orbiting object has no gravity. Gravity supplies the centripetal force.

Self review

Explain why an object moving in a circle at constant speed is accelerating.
State the direction of the centripetal resultant force.
Name the force providing centripetal force for a satellite, a ball on a string and a car on a bend.
Describe the motion of a ball immediately after its string breaks.
Explain why centripetal force is not an additional type of force.

2.4.2 Inertial mass

Inertial mass measures resistance to acceleration

Definition

Inertia

Inertia is the tendency of an object to remain at rest or continue moving at constant velocity unless a resultant force acts on it.

Definition

Inertial mass

Inertial mass is the ratio of the resultant force acting on an object to the acceleration it produces, measured in kilograms.

Inertial mass measures how difficult it is to change an object's velocity, including how difficult it is to start an object moving from rest.
The defining equation is $m=Fam=\dfrac{F}{a}$ , where $m$ is inertial mass in kilograms, $F$ is resultant force in newtons and $a$ is acceleration in metres per second squared.
A large inertial mass means that the same resultant force produces a smaller acceleration because $a=Fma=\dfrac{F}{m}$ .
A small inertial mass means that the same resultant force produces a larger acceleration.

Resultant force must be used

Use the resultant force, not one individual force, because opposing forces may partly or completely cancel.
If the resultant force is zero, the acceleration is zero, so the object remains at rest or continues at constant velocity.
Inertial mass is a property of the object and does not depend on the strength of the gravitational field.

Common Mistake

Do not describe inertia as a force because inertia is a tendency, not an interaction acting on the object.
Do not use weight in $m=Fam=\dfrac{F}{a}$ unless weight is the resultant force in the situation.

Example

Given: a resultant force of $N18\ \text{N}$ produces an acceleration of $m/s23.0\ \text{m/s}^2$ .
Rearrange Newton's second law: $m=Fam=\dfrac{F}{a}$ .
Substitute: $kgm=\dfrac{18\ \text{N}}{3.0\ \text{m/s}^2}=6.0\ \text{kg}$ .
The object's inertial mass is $kg6.0\ \text{kg}$ .

Comparing objects from data

For the same resultant force, compare accelerations using $m=Fam=\dfrac{F}{a}$ ; the object with the smaller acceleration has the greater inertial mass.
For the same acceleration, the object that requires the larger resultant force has the greater inertial mass.
On a graph of resultant force against acceleration, $F = ma$ , so the gradient equals the inertial mass.

Exam technique

When explaining a comparison, link the same force to the different accelerations and then state which object has the greater inertial mass.
For calculations, identify the resultant force before substituting and include the unit $kg\text{kg}$ .

Force and acceleration data reveal mass

A straight-line graph through the origin shows that resultant force is directly proportional to acceleration when mass is constant.
A steeper $F$ against $a$ graph represents a larger inertial mass because its gradient $ΔFΔa\dfrac{\Delta F}{\Delta a}$ is larger.

Self review

What does inertial mass measure?
State the equation that defines inertial mass.
Why must resultant force be used in the equation?
What does the gradient of a force against acceleration graph represent?

2.4.3 Newton's third law and momentum in collisions

Newton's third law links interaction forces

Definition

Newton's third law

Newton's third law states that when two objects interact, they exert forces on each other that are equal in magnitude and opposite in direction.

Definition

Momentum

Momentum is the product of an object's mass and velocity, measured in kilogram metres per second (kg m/s).

A Newton's third-law pair acts on two different objects during the same interaction.
The forces are the same type, equal in magnitude, opposite in direction and present for the same time.
The pair does not cancel because the two forces act on different objects.
For a book resting on a table, the book pushes down on the table and the table pushes up on the book with an equal force.

Common Mistake

Do not pair the weight of a book with the table's normal contact force because both forces act on the book.
The third-law partner of the book's weight is the gravitational force that the book exerts on Earth.

Momentum includes direction

Momentum is calculated using $p = m v$ , where $p$ is momentum in $m/s\text{kg m/s}$ , $m$ is mass in $kg\text{kg}$ and $v$ is velocity in $m/s\text{m/s}$ .
Momentum is a vector, so choose one direction as positive and give motion in the opposite direction a negative velocity and momentum.
An object at rest has zero momentum because $v = 0$ .

Example

Moving car: a $kg1200\ \text{kg}$ car travels east at $m/s15\ \text{m/s}$ .
Taking east as positive, $m/sp=mv=1200\times15=1.8\times10^4\ \text{kg m/s}$ .
Its momentum is $m/s1.8\times10^4\ \text{kg m/s}$ east.

Momentum is conserved in a closed system

Definition

Conservation of momentum

Conservation of momentum states that in a closed system the total momentum before an interaction equals the total momentum after the interaction.

During a collision, each object experiences an equal and opposite force for the same time.
Each object therefore has an equal and opposite change in momentum, so the total momentum of the system does not change.
Write conservation as $∑pbefore=∑pafter\sum p_{\text{before}}=\sum p_{\text{after}}$ and include signs for direction.
Kinetic energy does not have to be conserved in a collision because some energy may be transferred to thermal stores, sound or deformation.

Example

Before: a $kg2.0\ \text{kg}$ trolley moving right at $m/s4.0\ \text{m/s}$ hits a stationary $kg3.0\ \text{kg}$ trolley and they stick together.
Initial momentum: $m/sp=(2.0\times4.0)+(3.0\times0)=8.0\ \text{kg m/s}$ .
Combined mass: $kgm=2.0+3.0=5.0\ \text{kg}$ .
Conservation gives $8.0 = 5.0 v$ , so $m/sv=1.6\ \text{m/s}$ to the right.

Explosions also conserve momentum

If two objects start at rest, their total initial momentum is zero.
After they push apart, their momenta are equal in magnitude and opposite in direction, so their vector sum remains zero.

Exam technique

State a positive direction before calculating and keep negative signs until the final answer.
For a third-law explanation, name both objects, both forces, and state that the forces are equal in magnitude and opposite in direction.
For conservation, calculate the total momentum of the whole system before and after the interaction.

Interaction forces explain momentum changes

A larger force or a longer interaction time produces a larger change in momentum.
Equal and opposite changes in momentum preserve the total momentum of a closed system.

Self review

State Newton's third law.
Define momentum and give its unit.
Why do third-law forces not cancel each other?
What condition is needed for total momentum to be conserved?
How should direction be represented in a momentum calculation?

2.4.4 Force and rate of change of momentum

Force changes momentum over time

Definition

Rate of change of momentum

Rate of change of momentum is the change in an object's momentum divided by the time taken for that change.

Definition

Impulse

Impulse is the product of a force and the time for which it acts, and it equals the change in momentum it produces.

Newton's second law can be written as $F=Δpt=mv−mutF=\dfrac{\Delta p}{t}=\dfrac{mv-mu}{t}$ for constant mass.
$F$ is the resultant force in newtons, $m$ is mass in kilograms, $u$ is initial velocity, $v$ is final velocity and $t$ is time in seconds.
The change in momentum is $Δp=pfinal−pinitial=mv−mu\Delta p=p_{\text{final}}-p_{\text{initial}}=mv-mu$ .
The direction of the resultant force is the direction of the change in momentum, not necessarily the direction in which the object is moving.

Signs show changes in direction

Choose a positive direction before substituting velocities.
A negative velocity represents motion in the opposite direction, so a rebound usually gives a larger momentum change than stopping.

Example

Stopping: a $kg0.20\ \text{kg}$ ball moving at $m/s12\ \text{m/s}$ is brought to rest in $s0.040\ \text{s}$ .
$m/s\Delta p=m(v-u)=0.20(0-12)=-2.4\ \text{kg m/s}$ .
$NF=\dfrac{-2.4}{0.040}=-60\ \text{N}$ .
The mean force has magnitude $N60\ \text{N}$ and acts opposite to the ball's initial motion.

A longer stopping time reduces force

Rearranging gives $FΔt=ΔpF\Delta t=\Delta p$ , so the same momentum change spread over a longer time produces a smaller mean force.
Crumple zones, airbags, seat belts, crash mats and bending the knees on landing increase the time over which momentum changes.
The change in momentum is fixed by the initial and final velocities, so increasing time reduces the rate of change of momentum and therefore the force.

Example

Safety comparison: a passenger's momentum changes by $m/s420\ \text{kg m/s}$ .
Without an airbag, $NF=\dfrac{420}{0.030}=1.4\times10^4\ \text{N}$ .
With an airbag, $NF=\dfrac{420}{0.12}=3.5\times10^3\ \text{N}$ .
Increasing the stopping time by a factor of four reduces the mean force by a factor of four.

Force-time graphs give impulse

The area under a force-time graph equals impulse, $J=FΔtJ=F\Delta t$ , when the force is constant.
For a changing force, estimate or calculate the total area under the graph to find $Δp\Delta p$ .
The unit of impulse is $s\text{N s}$ , which is equivalent to $m/s\text{kg m/s}$ .

Common Mistake

Do not calculate $m v + m u$ unless the chosen velocity signs make that the correct subtraction.
Do not use speed without direction when an object rebounds.
Use the resultant or mean force specified by the data, not an unrelated contact force.

Multi-stage calculations need a clear route

If force and time are given, find momentum change using $Δp=FΔt\Delta p=F\Delta t$ .
If momentum change and one velocity are known, use $m(v−u)=Δpm(v-u)=\Delta p$ to find the missing velocity.

Exam technique

Write $Δp=m(v−u)\Delta p=m(v-u)$ before dividing by time.
Keep signs throughout the calculation, then describe the direction of a negative answer in words.
When explaining a safety feature, state that it increases the stopping time for the same momentum change, which reduces the mean force.

Self review

Write Newton's second law in momentum form.
Why can a rebound produce a large change in momentum?
How does an airbag reduce the force on a passenger?
What does the area under a force-time graph represent?

Newton's laws and momentum

2.3.1 Newton's first law

Resultant force

Balancing forces

Newton's first law

Zero resultant force

Explaining balanced motion

Non-zero resultant force

Using an unbalanced force

Common force errors

2.3.2 Newton's second law

Newton's second law

Using the force equation

Calculating acceleration

Finding resultant force

Force, mass and acceleration

Using a graph gradient

Force calculation method

Common force errors

2.3.3 Weight and gravitational field strength

Weight and gravity

Gravitational field strength

Mass and weight

The weight equation

Calculating weight

Finding mass and weight

Measuring weight

Weight in different fields

Explaining changes in weight

Common gravity errors

2.4.1 Circular motion and centripetal force

Speed and velocity

Acceleration in a circle

Centripetal force

Forces that provide centripetal force

Explaining an orbit

A string breaks

Explaining circular motion

Common circular-motion errors

2.4.2 Inertial mass

Inertial mass measures resistance to acceleration

Resultant force must be used

Comparing objects from data

Force and acceleration data reveal mass

2.4.3 Newton's third law and momentum in collisions

Newton's third law links interaction forces

Momentum includes direction

Momentum is conserved in a closed system

Explosions also conserve momentum

Interaction forces explain momentum changes

2.4.4 Force and rate of change of momentum

Force changes momentum over time

Signs show changes in direction

A longer stopping time reduces force

Force-time graphs give impulse

Multi-stage calculations need a clear route

Recap questions

Review 2.3 Newton's laws and momentum by teaching Genie

Recap your knowledge with an interactive lesson

Put it into practice with exam-style questions

Remember key concepts with flashcards

2.3 Newton's laws and momentum Revision Guide

Newton's laws and momentum

2.3.1 Newton's first law

Resultant force

Balancing forces

Newton's first law

Zero resultant force

Explaining balanced motion

Non-zero resultant force

Using an unbalanced force

Common force errors

2.3.2 Newton's second law

Newton's second law

Using the force equation

Calculating acceleration

Finding resultant force

Force, mass and acceleration

Using a graph gradient

Force calculation method