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?

2.4.1 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?

2.4 Circular motion, inertia and momentum

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

Review 2.4 Circular motion, inertia and momentum by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

2.4 Circular motion, inertia and momentum Revision Guide

2.4 Circular motion, inertia and momentum

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

Review 2.4 Circular motion, inertia and momentum by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

2.4 Circular motion, inertia and momentum Revision Guide