IB DP Physics · Theme A · Space, time and motion

Forces and momentum.

A.2 · Newton’s laws, the forces, momentum, circular motion
Standard level and higher level

SL + HL · 10 hours Collision Lab ↗

The journey

Five parts.

Part 01

Newton’s laws

Free-body diagrams and the resultant force

Part 02

The forces

Contact forces and field forces

Part 03

Momentum

Impulse and the full second law

Part 04

Collisions

Conservation, energy and explosions

Part 05

Circular motion

Centripetal acceleration and force

A.2 has no additional higher level content, but calculations on collisions in two dimensions are for HL: those two slides carry an HL only flag.

A.2 FORCES AND MOMENTUM · SL + HL

Part 01 · SL + HL

Newton’s laws.

Draw the forces, find the resultant, predict the motion

A.2 FORCES AND MOMENTUM · SL + HL

What is a force?

Forces are interactions.

A force is a push or a pull that one body exerts on another. A vector, measured in newtons
Every force needs two bodies: one that exerts it and one that it acts on

Contact forces

The bodies touch

Normal force, friction, tension, the force of a spring, viscous drag, buoyancy

Field forces

They act at a distance

Gravitational force (weight), electric force, magnetic force

A.2 FORCES AND MOMENTUM · SL + HL

A test for any force

Finish the sentence “the force of … on …”. The weight of a book is the force of the Earth on the book. The tension is the force of the string on the load. If you cannot name the body that exerts a force, the force is not real: there is no “force of motion” that keeps a moving body going.

Why the two families are not so different

Contact forces only look like touching. When a book rests on a table, the electrons in the two surfaces repel each other without ever meeting. Every contact force in this topic is the electric force acting between a very large number of atoms.

One newton is the force that gives a mass of 1 kg an acceleration of 1 m s⁻²: about the weight of a small apple.

Representing forces

Free-body diagrams.

One body only, drawn on its own
Every force that acts on it, as an arrow from where it acts
Length shows the size; label each with its name or symbol
Leave out the forces the body exerts on other things

A BLOCK PULLED ALONG A TABLE

A.2 FORCES AND MOMENTUM · SL + HL

A way to find every force

Start with the weight: anything with mass near the Earth has one. Then go round the outline of the body and stop at each thing that touches it. Each contact can supply a force: a surface gives a normal force and perhaps friction, a string gives a tension, a fluid gives buoyancy and drag. When you have been all the way round, you have them all.

Where each arrow starts

Weight acts at the center of mass, so its arrow starts at the center of the body. A contact force starts where the contact is: the normal force and friction at the surface the body rests on, a tension where the string is attached, a push or pull where it is applied.

Reading this diagram

The two vertical arrows are the same length, so there is no resultant force up or down and the block stays on the table. The applied force is longer than the friction, so there is a resultant force forwards and the block speeds up.

Adding an arrow for “the force of the motion” in the direction the body is moving. Velocity is not a force and never appears on a free-body diagram.

Adding forces

The resultant force.

The one force that has the same effect as all the forces together
Along a line: add with signs
In two dimensions: add as vectors

\(F = \sqrt{F_{x}^{2} + F_{y}^{2}} = \sqrt{4.0^{2} + 3.0^{2}} = \class{res}{5.0 \,\mathrm{N}}\) \(\tan \theta = \dfrac{F_{y}}{F_{x}} = \dfrac{3.0}{4.0}\) so \(\theta = \class{res}{37^{\circ}}\)

TWO FORCES AT RIGHT ANGLES

A.2 FORCES AND MOMENTUM · SL + HL

No resultant force

Newton’s first law.

With no resultant force, a body stays at rest or keeps moving at constant velocity
Such a body is in translational equilibrium

Vertical forces balance \(2T \sin \theta = mg\) \(T = \dfrac{mg}{2 \sin \theta} = \dfrac{49}{2 \sin 40^{\circ}} = \class{res}{38 \,\mathrm{N}}\)

A 5.0 kg SIGN HANGING AT REST

A.2 FORCES AND MOMENTUM · SL + HL

Equilibrium does not mean at rest

A parachutist falling at a steady speed and a car cruising along a straight motorway are both in equilibrium: the forces on each add to zero. A force is needed to change a velocity, not to keep one.

Why each string pulls with more than half the weight

Only the vertical part of each tension, \(T \sin 40^{\circ}\), holds the sign up. The horizontal parts pull against each other and cancel. The flatter the strings, the smaller the vertical part of each tension, so the larger the tension has to be. A washing line pulled almost straight can snap under a small load.

For any equilibrium problem: resolve, then write “forces up = forces down” and “forces left = forces right”.

A resultant force

Newton’s second law.

\(F = ma\)

\(F\) is the resultant force
The acceleration is in the direction of the resultant force
In this form the mass must be constant

A 1200 kg car · driving force 3000 N Resistive forces 600 N Resultant \(F\) = driving force − resistive forces \(F = 3000 - 600 = 2400 \,\mathrm{N}\) \(a = \dfrac{F}{m} = \dfrac{2400}{1200} = \class{res}{2.0 \,\mathrm{m} \,\mathrm{s}^{-2}}\)

A.2 FORCES AND MOMENTUM · SL + HL

Forces come in pairs

Newton’s third law.

If A exerts a force on B, then B exerts an equal and opposite force on A
The two forces act on different bodies
They are the same type of force

A BOOK ON A TABLETWO PAIRS

A.2 FORCES AND MOMENTUM · SL + HL

Why the pair never cancels

Forces cancel only when they act on the same body. The two forces of a pair act on two different bodies, so each appears on a different free-body diagram. A horse can pull a cart forwards even though the cart pulls back equally hard on the horse: the cart’s motion depends only on the forces on the cart.

Finding the partner of a force

Swap the two bodies in its description. The partner of “the force of the Earth on the book” is “the force of the book on the Earth”. It is still gravitational, and it is the same size, but the Earth is so massive that its acceleration is far too small to notice.

Calling the weight of the book and the normal force on it a third-law pair. They are equal here, but both act on the book and they are different types of force. They balance because of the first law, not the third.

The story · London, 1687

Background story · not assessed

Three laws in one book.

Painted portrait of Isaac Newton as a middle-aged man with long gray hair, in a dark gown — **Isaac Newton** in 1689, painted by Godfrey Kneller

Printed title page in Latin: Philosophiae Naturalis Principia Mathematica, by Is. Newton, with the imprimatur of S. Pepys dated 5 July 1686 and the year 1687 in Roman numerals — The **title page** of the first edition, dated MDCLXXXVII

The Principia opens with three “axioms, or laws of motion”
Licensed by Samuel Pepys, president of the Royal Society, on 5 July 1686, and printed in 1687
Edmond Halley saw it through the press at his own financial risk
Newton’s second law speaks of a change of “motion”: what we now call momentum

The same three laws, in the same order, over 300 years later.

Portrait: Godfrey Kneller, public domain. Title page: Wikimedia Commons, public domain.

A.2 FORCES AND MOMENTUM · SL + HL

Worked example

A 70 kg person in a lift.

Up is positive · \(m = 70 \,\mathrm{kg}\) \(F_\mathrm{N} - mg = ma\) so \(F_\mathrm{N} = m(g + a)\) Accelerating upwards: \(a = +1.5 \,\mathrm{m} \,\mathrm{s}^{-2}\) \(F_\mathrm{N} = 70(9.8 + 1.5) = \class{res}{791 \,\mathrm{N}}\) Accelerating downwards: \(a = -1.5 \,\mathrm{m} \,\mathrm{s}^{-2}\) \(F_\mathrm{N} = 70(9.8 - 1.5) = \class{res}{581 \,\mathrm{N}}\)

ACCELERATING UPWARDS

A.2 FORCES AND MOMENTUM · SL + HL

What you feel is the normal force

You have no sense of your weight itself: what you notice is the floor pushing up on your feet. When the lift accelerates upwards the floor pushes harder than your weight and you feel heavier. When it accelerates downwards the floor pushes less and you feel lighter. Bathroom scales read the normal force too.

Steady speed

With \(a = 0\) the normal force is 686 N, the same as at rest, whether the lift is going up or down. It is the acceleration that matters, not the direction of travel. A lift slowing down on its way up has a downward acceleration, so the person feels lighter.

If the cable broke, the lift and the person would both fall at 9.8 m s⁻². Then \(F_\mathrm{N} = 70(9.8 - 9.8) = 0\): the person would feel weightless, although their weight is still 686 N.

Quick check

Newton’s laws.

01A parcel slides at constant velocity. Resultant force on it?

Zero

02Forces of 6.0 N and 8.0 N act at right angles. Size of the resultant?

\(\sqrt{6.0^{2} + 8.0^{2}} = 10 \,\mathrm{N}\)

03A resultant force of 45 N acts on a 15 kg trolley. Acceleration?

\(\frac{45}{15} = 3.0 \,\mathrm{m} \,\mathrm{s}^{-2}\)

04The Earth pulls an apple down with 1.0 N. Its third-law partner?

The apple pulls the Earth up with 1.0 N

05Why are a book’s weight and the normal force on it not a third-law pair?

Both act on the same body

06A lift going up slows down. Normal force: more or less than the weight?

Less: the acceleration is downward

PRESS → TO REVEAL EACH ANSWER

Part 02 · SL + HL

The forces.

Six contact forces and three field forces

A.2 FORCES AND MOMENTUM · SL + HL

A field force and a contact force

Weight and the normal force.

\(F_\mathrm{g} = mg\)

Weight: the gravitational force on a body. Straight down
\(g = 9.8 \,\mathrm{N} \,\mathrm{kg}^{-1}\) near the Earth’s surface

Normal force \(F_\mathrm{N}\): the part of a contact force at right angles to the surface
A surface can only push, never pull

\(F_\mathrm{N}\) is not always equal to \(mg\). Find it from the other forces.

A.2 FORCES AND MOMENTUM · SL + HL

How a table knows how hard to push

A surface behaves like a very stiff spring. A book placed on a table squashes it by an amount far too small to see, and the squashed surface pushes back. The heavier the load, the more it is squashed and the harder it pushes: the normal force adjusts itself to whatever is needed to stop the body sinking in.

Three cases where it is not \(mg\)

In a lift accelerating upwards it is more than \(mg\). On a slope it is \(mg \cos \theta\), less than \(mg\). If you press down on a box with an extra 20 N, it is \(mg + 20\). In each case, write the second law at right angles to the surface and solve for \(F_\mathrm{N}\).

Mass is the amount of matter, in kg, and is the same everywhere. Weight is a force, in N, and depends on where the body is: \(g\) on the Moon is about a sixth of its value here.

\(\mathrm{N} \,\mathrm{kg}^{-1}\) and \(\mathrm{m} \,\mathrm{s}^{-2}\) are the same unit, so \(g\) is both the gravitational field strength and the acceleration of free fall.

Resolving the weight

A body on a slope.

Resolve the weight along the slope and at right angles to it
Along: \(mg \sin \theta\), down the slope
At right angles: \(F_\mathrm{N} = mg \cos \theta\)

FORCES ON A BLOCK ON A SLOPE

A.2 FORCES AND MOMENTUM · SL + HL

Contact force 2

Friction.

Acts along the surface, against sliding
Static, not sliding: \(F_\mathrm{f} \le \mu_\mathrm{s} F_\mathrm{N}\)
Dynamic, sliding: \(F_\mathrm{f} = \mu_\mathrm{d} F_\mathrm{N}\)
\(\mu_\mathrm{d}\) is less than \(\mu_\mathrm{s}\)

PUSHING HARDER ON A BODY AT REST

A.2 FORCES AND MOMENTUM · SL + HL

Why static friction is an inequality

Static friction is only as large as it needs to be to stop the body sliding. Push a heavy crate with 5 N and friction is 5 N; push with 12 N and it is 12 N. \(\mu_\mathrm{s} F_\mathrm{N}\) is the most it can reach, not its value. Only at the moment the body is about to slip does \(F_\mathrm{f} = \mu_\mathrm{s} F_\mathrm{N}\).

What the coefficients depend on

Each is a number with no unit that belongs to a pair of surfaces: rubber on dry road is about 0.8, steel on ice about 0.03. In this model friction does not depend on the area of contact or on the speed of sliding.

Friction is what lets you walk. Your foot pushes backwards on the ground, and static friction from the ground pushes you forwards.

Writing \(F_\mathrm{f} = \mu mg\) for every problem. The equation uses the normal force, which equals \(mg\) only on a level surface with no other vertical forces.

Worked example

Pushing a 5.0 kg crate.

Level floor · \(\mu_\mathrm{s} = 0.40\) · \(\mu_\mathrm{d} = 0.30\) \(F_\mathrm{N} = mg = 5.0 \times 9.8 = 49 \,\mathrm{N}\) Most static friction: \(\mu_\mathrm{s} F_\mathrm{N} = 0.40 \times 49 = \class{res}{19.6 \,\mathrm{N}}\) A 15 N push: it stays at rest. Friction \(= \class{res}{15 \,\mathrm{N}}\) A 25 N push: it slides \(F_\mathrm{f} = \mu_\mathrm{d} F_\mathrm{N} = 0.30 \times 49 = 14.7 \,\mathrm{N}\) \(a = \dfrac{F - F_\mathrm{f}}{m} = \dfrac{25 - 14.7}{5.0} = \class{res}{2.1 \,\mathrm{m} \,\mathrm{s}^{-2}}\)

FORCES ON THE CRATE

A.2 FORCES AND MOMENTUM · SL + HL

Worked example

Sliding down a 20° slope.

\(\mu_\mathrm{d} = 0.25\) · \(\theta = 20^{\circ}\) · find the acceleration \(F_\mathrm{N} = mg \cos \theta\) \(F_\mathrm{f} = \mu_\mathrm{d} F_\mathrm{N} = \mu_\mathrm{d} \, mg \cos \theta\) \(ma = mg \sin \theta - \mu_\mathrm{d} \, mg \cos \theta\) The mass cancels \(a = g(\sin \theta - \mu_\mathrm{d} \cos \theta)\) \(a = 9.8(\sin 20^{\circ} - 0.25 \cos 20^{\circ}) = \class{res}{1.0 \,\mathrm{m} \,\mathrm{s}^{-2}}\)

FORCES ON THE BLOCKNOT TO SCALE

A.2 FORCES AND MOMENTUM · SL + HL

Why the mass does not matter

A heavier block is pulled down the slope harder, presses on it harder and so meets more friction, and is harder to accelerate. All three effects are proportional to the mass, so it divides out. A full sledge and an empty one slide down the same slope with the same acceleration.

The angle at which it would just start to slip

A block at rest is about to slide when the component of its weight down the slope equals the largest static friction: \(mg \sin \theta = \mu_\mathrm{s} mg \cos \theta\), so \(\tan \theta = \mu_\mathrm{s}\). Tilting a surface until a block slips is a quick way to measure \(\mu_\mathrm{s}\).

Without friction the acceleration would be \(9.8 \sin 20^{\circ} = 3.4 \,\mathrm{m} \,\mathrm{s}^{-2}\). Friction has removed more than two thirds of it.

Contact force 3

Tension.

The pull of a stretched string or rope, along its length
A light string has the same tension all the way along

\(m_{1} = 3.0 \,\mathrm{kg}\) · \(m_{2} = 2.0 \,\mathrm{kg}\) \(a = \dfrac{(m_{1} - m_{2})g}{m_{1} + m_{2}} = \dfrac{1.0 \times 9.8}{5.0} = \class{res}{1.96 \,\mathrm{m} \,\mathrm{s}^{-2}}\) \(T - m_{2}g = m_{2}a\) \(T = m_{2}(g + a) = 2.0(9.8 + 1.96) = \class{res}{23.5 \,\mathrm{N}}\)

TWO MASSES OVER A PULLEY

A.2 FORCES AND MOMENTUM · SL + HL

Where the first line comes from

The string keeps the two masses moving together, so they share one acceleration. Treat them as a single 5.0 kg system: the 29.4 N weight drives it one way round the pulley and the 19.6 N weight opposes it. The tension pulls equally on both and is internal to the system, so it does not appear.

Why the tension lies between the two weights

The 2.0 kg mass accelerates upwards, so the string must pull up on it with more than its weight of 19.6 N. The 3.0 kg mass accelerates downwards, so the string must pull up on it with less than its weight of 29.4 N. Check with that mass: \(29.4 - 23.5 = 5.9 \,\mathrm{N}\), and \(3.0 \times 1.96 = 5.9 \,\mathrm{N}\).

A string can only pull. On a free-body diagram, tension always points away from the body, along the string.

The pulley is taken to be light and frictionless too. A heavy pulley needs a difference in tension to make it turn faster (A.4). “Light” means the string’s own mass can be ignored. A heavy rope hanging down has a larger tension at the top than at the bottom, because the top holds up the rope beneath it.

Contact force 4

Hooke’s law.

\(F_\mathrm{H} = -kx\)

\(x\): extension or compression
\(k\): spring constant, in N m⁻¹
The minus sign: the force is opposite to \(x\)

STRETCHING A SPRING

A.2 FORCES AND MOMENTUM · SL + HL

A worked case

A 0.30 kg mass hung from a spring stretches it by 6.0 cm. At rest the spring force equals the weight: \(k \times 0.060 = 0.30 \times 9.8\), so \(k = \frac{2.94}{0.060} = 49 \,\mathrm{N} \,\mathrm{m}^{-1}\). A 0.50 kg mass would stretch the same spring by \(\frac{4.9}{49} = 0.10 \,\mathrm{m}\).

Why it is called a restoring force

Stretch the spring and it pulls back; squash it and it pushes out. Either way the force points towards the natural length, against the displacement, and that is what the minus sign says. In a calculation of size only, use \(F = kx\).

A large \(k\) means a stiff spring: a car’s suspension spring is about 20 000 N m⁻¹, the spring in a ballpoint pen about 200 N m⁻¹.

The law holds only up to the limit of proportionality. Beyond it the graph curves, and the spring may not return to its natural length.

Contact force 5

Buoyancy.

\(F_\mathrm{b} = \rho V g\)

The upward force of a fluid on a body in it
\(\rho\): density of the fluid
\(V\): volume of fluid displaced

It equals the weight of the fluid displaced.

A BODY HELD UNDER THE SURFACE

A.2 FORCES AND MOMENTUM · SL + HL

Where the force comes from

Pressure in a fluid increases with depth. The bottom of a body is deeper than its top, so the fluid pushes up on the bottom harder than it pushes down on the top. The difference is the buoyancy force.

A worked case

A ball of mass 1.2 kg and volume \(2.0 \times 10^{-3} \,\mathrm{m}^{3}\) is held under water of density 1000 kg m⁻³. Buoyancy: \(1000 \times 2.0 \times 10^{-3} \times 9.8 = 19.6 \,\mathrm{N}\). Weight: \(1.2 \times 9.8 = 11.8 \,\mathrm{N}\). On release the resultant is 7.8 N upwards, so it rises.

Floating

The ball comes to rest at the surface when the buoyancy has fallen to equal its weight: it then displaces only \(1.2 \times 10^{-3} \,\mathrm{m}^{3}\) of water, so 60% of it is under the surface. A body floats if its average density is less than the fluid’s.

Using the density or the mass of the body in \(\rho V g\). The equation is about the fluid that has been pushed aside.

Contact force 6

Viscous drag.

A SPHERE FALLING THROUGH A FLUIDPRESS → TO PLAY

\(F_\mathrm{d} = 6 \pi \eta r v\)

On a small sphere of radius \(r\) moving at speed \(v\)
\(\eta\): viscosity of the fluid, in Pa s
Always against the velocity

Terminal velocity: \(F_\mathrm{g} = F_\mathrm{b} + F_\mathrm{d}\)

A.2 FORCES AND MOMENTUM · SL + HL

Viscosity is how much a fluid resists flowing

Layers of fluid drag on each other as they slide past. Water has a viscosity of about 0.001 Pa s; honey is several thousand times more viscous, which is why a spoon sinks through it so slowly.

How this explains terminal velocity

The weight and the buoyancy do not change as the sphere falls, but the drag is proportional to the speed. At the start there is no drag and the sphere accelerates. As it speeds up the drag grows, the resultant force shrinks, and the acceleration falls. When the drag has grown enough to make the three forces balance, the speed stops changing. This is the force behind the speed–time curve met in A.1.

The equation is for a small sphere moving slowly enough for the fluid to flow smoothly round it. It does not describe a skydiver or a car.

The story · Cambridge, 1851

Background story · not assessed

Stokes and the pendulum.

George Gabriel Stokes (1819–1903)

A swinging pendulum is slowed by the air it moves through. By how much?
1851: Stokes’s paper on the friction of fluids and the motion of pendulums
In it, the drag on a slowly moving sphere: \(F_\mathrm{d} = 6 \pi \eta r v\)
He held Newton’s old chair at Cambridge, as Lucasian Professor, for 54 years

A question about clocks gave the law for every small sphere in a fluid.

Photograph: Accademia delle Scienze di Torino, public domain.

A.2 FORCES AND MOMENTUM · SL + HL

Worked example

A steel ball falling through oil.

\(r = 1.0 \,\mathrm{mm}\) · \(\rho_\mathrm{s} = 7800\), \(\rho_\mathrm{o} = 900 \,\mathrm{kg} \,\mathrm{m}^{-3}\) · \(\eta = 0.50 \,\mathrm{Pa} \,\mathrm{s}\) \(V = \tfrac{4}{3} \pi r^{3} = 4.19 \times 10^{-9} \,\mathrm{m}^{3}\) \(F_\mathrm{g} = \rho_\mathrm{s} V g = 7800 \times V \times 9.8 = 3.20 \times 10^{-4} \,\mathrm{N}\) \(F_\mathrm{b} = \rho_\mathrm{o} V g = 900 \times V \times 9.8 = 0.37 \times 10^{-4} \,\mathrm{N}\)

At terminal velocity \(F_\mathrm{d} = F_\mathrm{g} - F_\mathrm{b} = 2.83 \times 10^{-4} \,\mathrm{N}\) \(v = \dfrac{F_\mathrm{d}}{6 \pi \eta r} = \dfrac{2.83 \times 10^{-4}}{6 \pi \times 0.50 \times 1.0 \times 10^{-3}}\) \(v = \class{res}{0.030 \,\mathrm{m} \,\mathrm{s}^{-1}}\)

A.2 FORCES AND MOMENTUM · SL + HL

Two densities, two jobs

The steel’s density gives the mass of the ball and so its weight. The oil’s density gives the buoyancy. Both use the same volume because the ball is completely under the surface, so it displaces its own volume of oil.

How the radius changes the answer

Weight and buoyancy depend on the volume, so on \(r^{3}\); the drag depends on \(r\). Dividing one by the other, the terminal velocity depends on \(r^{2}\): a ball of twice the radius falls four times as fast. This is why fine drizzle drifts down while large raindrops fall quickly.

Convert the radius to meters before anything else. A radius left in millimeters makes the volume wrong by a factor of a thousand million.

Timing a ball between two marks on a tall tube of liquid, once it has reached terminal velocity, is a standard way to measure viscosity.

Forces at a distance

Field forces.

Force	Acts on	Direction	Studied in
Gravitational \(F_\mathrm{g}\)	Anything with mass	Always attracts	Here, then D.1
Electric \(F_\mathrm{e}\)	Anything with charge	Attracts or repels	D.2
Magnetic \(F_\mathrm{m}\)	Moving charges and currents	At right angles to the motion	D.3

On a free-body diagram they are drawn like any other force, and labeled \(F_\mathrm{g}\), \(F_\mathrm{e}\) or \(F_\mathrm{m}\).

A.2 FORCES AND MOMENTUM · SL + HL

What a field is

A body with mass or charge changes the space around it, and a second body placed in that space feels a force without being touched. The region where this happens is the field. The idea is developed properly in theme D; in this topic you only need to recognize these forces, name them and include them in a resultant.

An example with two of them

A small charged oil drop can be held at rest between two charged plates. Its free-body diagram has two arrows: \(F_\mathrm{g}\) downwards and \(F_\mathrm{e}\) upwards, equal in length. Newton’s first law applies exactly as it does to a book on a table.

The Charges in Fields simulation ↗ shows the paths that electric and magnetic forces give to moving charges.

Quick check

The forces.

01A 12 kg box rests on a level floor. Normal force?

\(12 \times 9.8 = 118 \,\mathrm{N}\)

02The box slides, with \(\mu_\mathrm{d} = 0.20\). Friction force?

\(0.20 \times 118 = 24 \,\mathrm{N}\)

03A spring with \(k = 80 \,\mathrm{N} \,\mathrm{m}^{-1}\) is stretched by 0.15 m. Size of the force?

\(80 \times 0.15 = 12 \,\mathrm{N}\)

04What is the normal force on a block of mass \(m\) on a slope at angle \(\theta\)?

\(mg \cos \theta\)

05In \(F_\mathrm{b} = \rho V g\), which density and which volume?

Density of the fluid, volume of fluid displaced

06A sphere falls through a liquid. Its speed doubles. What happens to the viscous drag?

It doubles

PRESS → TO REVEAL EACH ANSWER

Part 03 · SL + HL

Momentum and impulse.

What a force changes, and how much

A.2 FORCES AND MOMENTUM · SL + HL

Mass in motion

Linear momentum.

\(p = mv\)

A vector, in the direction of the velocity
Unit: kg m s⁻¹, the same as N s

A 1500 kg car at 20 m s⁻¹ \(p = mv = 1500 \times 20 = \class{res}{3.0 \times 10^{4} \,\mathrm{kg} \,\mathrm{m} \,\mathrm{s}^{-1}}\) A 0.050 kg ball at 40 m s⁻¹, the other way \(p = mv = 0.050 \times (-40) = \class{res}{-2.0 \,\mathrm{kg} \,\mathrm{m} \,\mathrm{s}^{-1}}\)

A.2 FORCES AND MOMENTUM · SL + HL

What momentum tells you

It measures how hard a moving body is to stop. A 0.16 kg cricket ball at 40 m s⁻¹ and a 1600 kg car creeping at 0.40 m s⁻¹ have the same kinetic energy, 128 J. But the ball’s momentum is 6.4 kg m s⁻¹ and the car’s is 640 kg m s⁻¹: stopping the car in the same time takes a force a hundred times as large.

Why the sign matters

Two identical trolleys moving towards each other at the same speed have momenta of \(+mv\) and \(-mv\). Their total momentum is zero, although both are moving. Every momentum calculation starts by choosing a positive direction.

“Linear” separates this from angular momentum, the momentum of rotation, which HL students meet in A.4.

The full second law

Force changes momentum.

\(F = \dfrac{\Delta p}{\Delta t}\)resultant force = rate of change of momentum

Constant mass

\(F = ma\)

\(\Delta p = m \Delta v\), and \(\dfrac{\Delta v}{\Delta t}\) is the acceleration

A stream of mass

Use \(F = \dfrac{\Delta p}{\Delta t}\)

The exhaust of a rocket, a jet of water, a conveyor belt being loaded

A.2 FORCES AND MOMENTUM · SL + HL

This is the law as Newton stated it

\(F = ma\) is the special case for a body whose mass stays the same. When mass streams in or out, apply \(F = \frac{\Delta p}{\Delta t}\) to the stream itself: for a rocket, the momentum given to the exhaust each second is the thrust. Then, at any instant, the rocket’s acceleration is the resultant force divided by its mass at that moment. As fuel burns, the same thrust produces a larger and larger acceleration.

Using it for a stream of matter

When matter arrives or leaves steadily, look at the momentum that changes in one second. A hose that delivers 12 kg of water each second at 15 m s⁻¹ onto a wall, where the water stops, destroys \(12 \times 15 = 180 \,\mathrm{kg} \,\mathrm{m} \,\mathrm{s}^{-1}\) of momentum every second. The wall therefore pushes on the water with 180 N, and by the third law the water pushes on the wall with 180 N.

On a graph of momentum against time, the gradient is the resultant force.

Force × time

Impulse.

\(J = F \Delta t = \Delta p\)

\(F\): the average resultant force
\(\Delta t\): the time of contact
Impulse = change in momentum

FORCE AGAINST TIME IN A COLLISION

A.2 FORCES AND MOMENTUM · SL + HL

Why the area

A constant force gives a rectangle on this graph, and its area, height × width, is \(F \Delta t\). A real collision force rises and falls, so cut the time into thin strips: each strip’s area is the small change in momentum in that short time, and together they make the whole area. The average force is the height of the rectangle that has the same area over the same time.

Why a longer stop is a gentler stop

A passenger in a crash loses a fixed amount of momentum, so the impulse is fixed. A seat belt and an airbag stretch the stop from a few thousandths of a second to a few tenths, so the average force is reduced to a few percent of its original value. Crumple zones, crash helmets, bending your knees on landing and drawing your hands back to catch a ball all work in the same way.

For a curved graph, find the area by counting squares, as for a velocity–time graph in A.1. One square is worth its height in N times its width in s.

Worked example

A 0.15 kg ball rebounds from a wall.

Towards the wall is positive \(\Delta p = m(v - u)\) \(\Delta p = 0.15(-8.0 - 12) = \class{res}{-3.0 \,\mathrm{kg} \,\mathrm{m} \,\mathrm{s}^{-1}}\) In contact for 0.020 s \(F = \dfrac{\Delta p}{\Delta t} = \dfrac{-3.0}{0.020} = \class{res}{-150 \,\mathrm{N}}\)

BEFORE AND AFTER

A.2 FORCES AND MOMENTUM · SL + HL

Why the change is 20 m s⁻¹ and not 4

The wall first has to take away all 12 m s⁻¹ of the ball’s velocity and then give it 8.0 m s⁻¹ in the opposite direction. The velocities are in opposite directions, so the change is \(-8.0 - 12 = -20 \,\mathrm{m} \,\mathrm{s}^{-1}\). A ball that bounces back needs a larger impulse than one that hits at the same speed and sticks.

Reading the minus signs

Both answers are negative: the impulse and the force on the ball point away from the wall, as the diagram shows. By the third law the ball pushes on the wall with 150 N towards the wall.

Subtracting the speeds: \(0.15(12 - 8.0) = 0.60\). That treats the two velocities as if they were in the same direction and gives an answer five times too small.

Worked example · changing mass

The thrust of a rocket.

Burns 2.5 kg of fuel each second Exhaust leaves at 2000 m s⁻¹ relative to the rocket In 1.0 s the gas gains \(\Delta p = m_\mathrm{gas} \, v = 2.5 \times 2000\) \(F = \dfrac{\Delta p}{\Delta t} = \dfrac{5000}{1.0} = \class{res}{5000 \,\mathrm{N}}\)

The rocket pushes the gas backwards with 5000 N
The gas pushes the rocket forwards with 5000 N: the thrust

A.2 FORCES AND MOMENTUM · SL + HL

A rocket does not push on the air

In this gas calculation, take backwards as positive. The gas starts at rest relative to the rocket and leaves at \(+2000 \,\mathrm{m} \,\mathrm{s}^{-1}\), so \(\Delta p_\mathrm{gas} = m_\mathrm{gas}(v - 0) > 0\). The rocket receives the opposite impulse. It pushes on its own exhaust, and the exhaust pushes back. That is why a rocket works in the vacuum of space, where there is nothing else to push against. Seen as a whole, the rocket and its gas keep the same total momentum: the gas gains momentum backwards and the rocket gains the same amount forwards.

Will it lift off?

A thrust of 5000 N can raise a rocket only if it weighs less than that: a mass below \(\frac{5000}{9.8} = 510 \,\mathrm{kg}\). For a 400 kg rocket the resultant is \(5000 - 3920 = 1080 \,\mathrm{N}\) upwards and the first acceleration is \(\frac{1080}{400} = 2.7 \,\mathrm{m} \,\mathrm{s}^{-2}\). As fuel burns, the weight falls and the acceleration rises.

The general pattern for a steady stream: force = mass per second × change in velocity.

The story · Auburn, Massachusetts, 16 March 1926

Background story · not assessed

The first liquid-fueled rocket.

Black and white photograph of Robert Goddard in a long coat and cap, standing in a snowy field beside a tall metal frame that holds a slender rocket — **Robert Goddard** with the rocket in its launch frame, March 1926

1920: a newspaper editorial scoffs that a rocket cannot work in space, with no air to push against
Goddard’s rocket burns gasoline in liquid oxygen
A flight of 2.5 s: it rises 12.5 m and lands 56 m away, in a cabbage field

A rocket pushes on its own exhaust. It needs nothing else.

Photograph: Esther C. Goddard, NASA, public domain.

A.2 FORCES AND MOMENTUM · SL + HL

Part 04 · SL + HL

Collisions and explosions.

Momentum is conserved. Kinetic energy may not be

A.2 FORCES AND MOMENTUM · SL + HL

The principle

Conservation of momentum.

TWO BODIES COLLIDEPRESS → TO PLAY

The total momentum of a system stays constant…
…unless a resultant external force acts on it

\(m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}\)

A.2 FORCES AND MOMENTUM · SL + HL

It follows from the third law

While the bodies touch, each pushes on the other with an equal and opposite force, for exactly the same time. So they receive equal and opposite impulses: whatever momentum one gains, the other loses. The total cannot change.

Internal and external forces

The forces the two bodies exert on each other are internal to the system and always cancel in this way. Friction from the track or a push from outside is external and does change the total. Collisions are so brief that external forces usually have no time to matter, which is why the principle works so well for them.

Draw a “before” and an “after” picture, mark the positive direction, and give every velocity a sign before writing the equation.

Try it in the Collision Lab ↗.

Worked example

Two cars lock together.

A MOVING CAR HITS A PARKED CARPRESS → TO PLAY

Momentum before = momentum after \(m_{1}u_{1} = (m_{1} + m_{2})v\) \(v = \dfrac{m_{1}u_{1}}{m_{1} + m_{2}}\) \(= \dfrac{800 \times 15}{2000} = \class{res}{6.0 \,\mathrm{m} \,\mathrm{s}^{-1}}\) Kinetic energy: \(E_\mathrm{k} = \tfrac{1}{2}mv^{2}\) Before: \(\tfrac{1}{2} \times 800 \times 15^{2} = 90 \,\mathrm{kJ}\) After: \(\tfrac{1}{2} \times 2000 \times 6.0^{2} = \class{res}{36 \,\mathrm{kJ}}\)

A.2 FORCES AND MOMENTUM · SL + HL

Energy in collisions

Elastic or inelastic?

	Momentum	Total energy	Kinetic energy
Elastic collision	Conserved	Conserved	Conserved
Inelastic collision	Conserved	Conserved	Decreases
Explosion	Conserved	Conserved	Increases

To decide: work out the total kinetic energy before and after, and compare.

A.2 FORCES AND MOMENTUM · SL + HL

Elastic collisions are rare at everyday sizes

Any collision you can hear or that leaves a dent has transferred some kinetic energy elsewhere. Steel balls and snooker balls come close. Collisions between the molecules of a gas are perfectly elastic, which is one of the assumptions of the ideal gas model in B.3.

How kinetic energy can change when momentum cannot

Momentum is a vector, so gains in one direction cancel losses in the other. Kinetic energy is a scalar that is never negative, so nothing cancels: it can only be shared out differently or transferred to another form. In an explosion it comes from a store, such as chemical energy or a compressed spring.

A collision in which the bodies stick together loses the most kinetic energy that conservation of momentum allows.

Worked example

Is this collision elastic?

TWO TROLLEYS ON A TRACKPRESS → TO PLAY

Momentum before = momentum after \(m_{1}u_{1} = m_{1}v_{1} + m_{2}v_{2}\) \(v_{2} = \dfrac{m_{1}(u_{1} - v_{1})}{m_{2}}\) \(= \dfrac{2.0(3.0 - 1.0)}{1.0} = \class{res}{4.0 \,\mathrm{m} \,\mathrm{s}^{-1}}\) Kinetic energy: \(E_\mathrm{k} = \tfrac{1}{2}mv^{2}\) Before: \(\tfrac{1}{2} \times 2.0 \times 3.0^{2} = 9.0 \,\mathrm{J}\) After: \(1.0 + 8.0 = 9.0 \,\mathrm{J}\) Elastic

A.2 FORCES AND MOMENTUM · SL + HL

A collision in reverse

Explosions.

TWO SKATERS PUSH APARTPRESS → TO PLAY

Total momentum before = 0 \(0 = m_{1}v_{1} + m_{2}v_{2}\) \(v_{1} = -\dfrac{m_{2}v_{2}}{m_{1}}\) \(= -\dfrac{40 \times 3.0}{60} = \class{res}{-2.0 \,\mathrm{m} \,\mathrm{s}^{-1}}\) Kinetic energy: \(E_\mathrm{k} = \tfrac{1}{2}mv^{2}\) Before: \(0\) After: \(120 + 180 = \class{res}{300 \,\mathrm{J}}\)

A.2 FORCES AND MOMENTUM · SL + HL

What counts as an explosion

Any event in which bodies that were together fly apart because of forces between them: skaters pushing off, a gun firing a bullet, a nucleus emitting an alpha particle, a firework bursting. The forces are internal, so the total momentum stays what it was. From rest that is zero, so the momenta of the pieces must cancel.

Who gets the energy

The two skaters leave with equal sizes of momentum, 120 kg m s⁻¹ each, so the lighter one moves faster, and has more of the kinetic energy: 180 J against 120 J. This is why a bullet carries almost all the energy of a shot and the heavy rifle very little. The 300 J came from chemical energy in the skaters’ muscles.

Equal and opposite momenta mean the speeds are in the inverse ratio of the masses: the masses are 3 : 2, so the speeds are 2 : 3.

Higher level only

Collisions in two dimensions.

A GLANCING COLLISION, SEEN FROM ABOVEPRESS → TO PLAY

Momentum is a vector
It is conserved along \(x\) and along \(y\) separately
Resolve every momentum, then write one equation for each direction

A.2 FORCES AND MOMENTUM · ADDITIONAL HL

Why each direction is separate

The force between the two pucks has an \(x\) part and a \(y\) part. Each part acts equally and oppositely on the two pucks, so the argument from the third law works for each direction on its own. It is the same independence of perpendicular directions that you used for projectiles in A.1.

Reading the diagram before calculating

Before the collision there is no momentum along \(y\) at all. Afterwards the 2.0 kg puck moves along \(+y\), so the 3.0 kg puck must have exactly as much momentum along \(-y\). And since the 2.0 kg puck now has no momentum along \(x\), the 3.0 kg puck must carry all of the original \(x\) momentum.

Choose the \(x\) direction along the velocity of the incoming body. Then the momentum before the collision has only one component.

The pucks do not meet head-on: the first strikes the second off-center. The force on each acts along the line joining their centers at the moment of contact, so the puck that was at rest leaves along that line. A head-on collision would keep both on the original line.

Higher level only · worked example

Finding the second puck’s velocity.

Along \(x\) \(m_{1}u_{1} = m_{2}v_{2x}\) \(v_{2x} = \dfrac{m_{1}u_{1}}{m_{2}} = \dfrac{2.0 \times 5.0}{3.0} = \class{res}{3.33 \,\mathrm{m} \,\mathrm{s}^{-1}}\) Along \(y\) \(0 = m_{1}v_{1} + m_{2}v_{2y}\) \(v_{2y} = -\dfrac{m_{1}v_{1}}{m_{2}} = -\dfrac{2.0 \times 2.0}{3.0} = \class{res}{-1.33 \,\mathrm{m} \,\mathrm{s}^{-1}}\)

Put the components together \(v_{2} = \sqrt{v_{2x}^{2} + v_{2y}^{2}}\) \(= \sqrt{3.33^{2} + 1.33^{2}} = \class{res}{3.6 \,\mathrm{m} \,\mathrm{s}^{-1}}\) \(\tan \theta = \dfrac{|v_{2y}|}{v_{2x}} = \dfrac{1.33}{3.33}\) \(\theta = \class{res}{22^{\circ}}\) below the \(x\) direction

A.2 FORCES AND MOMENTUM · ADDITIONAL HL

Is it elastic?

Before: \(\frac{1}{2} \times 2.0 \times 5.0^{2} = 25 \,\mathrm{J}\). After: \(\frac{1}{2} \times 2.0 \times 2.0^{2} + \frac{1}{2} \times 3.0 \times 3.6^{2} = 4.0 + 19.4 = 23.4 \,\mathrm{J}\). Kinetic energy has decreased, so the collision is inelastic. The energy test is done with speeds, not components, because kinetic energy is a scalar.

A check with a triangle

The two momenta after the collision must add, as vectors, to the momentum before it. Here they are 4.0 kg m s⁻¹ along \(y\) and \(3.0 \times 3.6 = 10.8 \,\mathrm{kg} \,\mathrm{m} \,\mathrm{s}^{-1}\) at 22° below \(x\). Drawn tip to tail they close a right-angled triangle with the original 10 kg m s⁻¹: \(\sqrt{10.8^{2} - 4.0^{2}} = 10\).

Adding the speeds or the sizes of the momenta: \(10 = 4.0 + 3.0v\). Momenta in different directions add as vectors, never as plain numbers.

Quick check

Momentum and collisions.

01Momentum of a 0.40 kg ball moving at 15 m s⁻¹?

\(0.40 \times 15 = 6.0 \,\mathrm{kg} \,\mathrm{m} \,\mathrm{s}^{-1}\)

02A resultant force of 30 N acts for 0.50 s. Change in momentum?

\(30 \times 0.50 = 15 \,\mathrm{N} \,\mathrm{s}\)

03What does the area under a force–time graph give?

The impulse: the change in momentum

04A 3.0 kg trolley at 4.0 m s⁻¹ hits and sticks to a 1.0 kg trolley at rest. Speed afterwards?

\(\frac{3.0 \times 4.0}{4.0} = 3.0 \,\mathrm{m} \,\mathrm{s}^{-1}\)

05What is conserved in an elastic collision but not in an inelastic one?

Kinetic energy

06A stationary body explodes into two pieces. What is their total momentum?

Zero

PRESS → TO REVEAL EACH ANSWER

Part 05 · SL + HL

Circular motion.

Constant speed, but always accelerating

A.2 FORCES AND MOMENTUM · SL + HL

Describing motion in a circle

Angular velocity.

\(v = \dfrac{2 \pi r}{T} = \omega r\)

\(T\): the period, the time for one revolution
\(\omega\): the angle turned through each second, in rad s⁻¹

\(r = 2.0 \,\mathrm{m}\) · \(T = 5.0 \,\mathrm{s}\) \(\omega = \dfrac{2 \pi}{T} = \dfrac{2 \pi}{5.0} = \class{res}{1.26 \,\mathrm{rad} \,\mathrm{s}^{-1}}\) \(v = \omega r = 1.26 \times 2.0 = \class{res}{2.5 \,\mathrm{m} \,\mathrm{s}^{-1}}\)

MOVING ROUND A CIRCLE

A.2 FORCES AND MOMENTUM · SL + HL

Why radians

An angle in radians is the arc length divided by the radius. So a body that turns through \(\Delta \theta\) moves a distance \(r \Delta \theta\) along the arc, with no conversion factor. Divide both sides by the time taken and you have \(v = \omega r\). One full turn is an arc of \(2 \pi r\), so it is \(2 \pi\) rad, and turning through that in one period gives \(\omega = \frac{2 \pi}{T}\).

One angular velocity, many speeds

Every point on a spinning roundabout turns through the same angle each second, so all share one \(\omega\). But a child at the rim travels round a bigger circle in the same time than a child near the middle, and so moves faster: \(v\) is proportional to \(r\).

Given a rate of rotation in revolutions per minute, divide by 60 for revolutions per second, then multiply by \(2 \pi\) for rad s⁻¹.

Changing direction is accelerating

Centripetal acceleration.

\(a = \dfrac{v^{2}}{r} = \omega^{2} r = \dfrac{4 \pi^{2} r}{T^{2}}\)

The speed is constant, but the direction of the velocity keeps changing
The acceleration points towards the center
It is always at right angles to the velocity

VELOCITY AND ACCELERATIONCONSTANT SPEED

A.2 FORCES AND MOMENTUM · SL + HL

Why towards the center

Acceleration is the change in velocity divided by the time. Take the velocity arrows at two points close together on the circle: they are the same length but turned through a small angle. The arrow you must add to the first to get the second points inwards, along the radius. An acceleration at right angles to the velocity changes its direction and leaves its size alone.

A worked case

For the body on the last slide, \(a = \omega^{2} r = 1.26^{2} \times 2.0 = 3.2 \,\mathrm{m} \,\mathrm{s}^{-2}\), about a third of \(g\). The three forms of the equation are the same thing with \(v = \omega r\) or \(\omega = \frac{2 \pi}{T}\) put in: use whichever matches the data you have.

Doubling the speed on the same circle makes the acceleration four times as large. Doubling the radius at the same speed halves it.

What makes it turn?

Centripetal force.

\(F = \dfrac{mv^{2}}{r} = m \omega^{2} r\)

The resultant force towards the center
It is not an extra force: something has to provide it

Motion	Provided by
A ball whirled on a string	Tension
A car on a level bend	Friction
A satellite in orbit	Gravitational force
An electron in a magnetic field	Magnetic force

A.2 FORCES AND MOMENTUM · SL + HL

What happens if it disappears

Cut the string and the ball does not fly outwards. With no resultant force it obeys the first law and carries on in a straight line along the tangent, in the direction it was moving at that instant.

The “force” that throws you outwards

In a car turning left you seem to be flung to the right. Nothing is pushing you that way. Your body is trying to carry on in a straight line while the car turns underneath you, until the door or the seat belt supplies the inward force that makes you turn too. There is no outward force to draw.

Why the speed stays the same

The force is at right angles to the velocity, so it has no component along the direction of motion and does no work. With no work done the kinetic energy cannot change.

Drawing “centripetal force” as one more arrow on a free-body diagram. Draw the real forces: tension, friction, weight, normal force. Their resultant is the centripetal force.

Worked example · horizontal circle

How fast can a car take a bend?

Level road · radius 50 m · \(\mu_\mathrm{s} = 0.80\) Friction provides the centripetal force Most friction: \(\mu_\mathrm{s} F_\mathrm{N} = \mu_\mathrm{s} mg\) \(\dfrac{mv^{2}}{r} = \mu_\mathrm{s} mg\)

The mass cancels \(v = \sqrt{\mu_\mathrm{s} g r}\) \(v = \sqrt{0.80 \times 9.8 \times 50}\) \(v = \class{res}{20 \,\mathrm{m} \,\mathrm{s}^{-1}}\)

A.2 FORCES AND MOMENTUM · SL + HL

Why static friction, when the car is moving

A rolling tire does not slide over the road: the patch touching the ground is at rest on it for an instant. The sideways grip is therefore static friction, which can be anything up to \(\mu_\mathrm{s} F_\mathrm{N}\). The car skids when the bend demands more than that.

Rain and tight bends

On a wet road \(\mu_\mathrm{s}\) may fall to 0.40. The greatest speed becomes \(\sqrt{0.40 \times 9.8 \times 50} = 14 \,\mathrm{m} \,\mathrm{s}^{-1}\): halving the grip cuts the safe speed by a factor of \(\sqrt{2}\), not 2. Halving the radius does the same.

On a banked track the surface is tilted, so the normal force leans towards the center of the bend and its horizontal component supplies some or all of the centripetal force. Less friction is needed. You may be asked to explain this, but not to calculate with it.

Horizontal circle

A ball whirled on a string.

Only two forces act: tension and weight
Vertical: \(T \cos \theta = mg\)
Towards the center: \(T \sin \theta = \dfrac{mv^{2}}{r}\)

Divide: \(\tan \theta = \dfrac{v^{2}}{rg}\)

A CONICAL PENDULUM

A.2 FORCES AND MOMENTUM · SL + HL

Why the string cannot be horizontal

The ball has no vertical acceleration, so something must balance its weight, and the only upward force available is the vertical part of the tension. A horizontal string has none. However fast you whirl the ball, the string always slopes down a little.

A worked case

A ball moves at 2.0 m s⁻¹ in a horizontal circle of radius 0.60 m. \(\tan \theta = \frac{2.0^{2}}{0.60 \times 9.8} = 0.68\), so the string is at 34° to the vertical. The mass has canceled: a heavy ball and a light one at the same speed and radius hang at the same angle.

Resolve vertically and towards the center, not along the string. The acceleration is horizontal, so those are the directions in which the second law is simplest.

Vertical circle · non-uniform circular motion

Top and bottom of a vertical circle.

The weight helps at the top and opposes at the bottom
Top: \(T_\mathrm{top} + mg = \dfrac{mv_\mathrm{top}^{2}}{r}\)
Bottom: \(T_\mathrm{bottom} - mg = \dfrac{mv_\mathrm{bottom}^{2}}{r}\)

Non-uniform circular motion: the speed is greatest at the bottom.

A BALL ON A STRINGSPEED NOT CONSTANT

A.2 FORCES AND MOMENTUM · SL + HL

Uniform or non-uniform

In uniform circular motion the speed stays the same, as on the earlier slides. Here the ball moves freely under gravity, so it slows as it climbs and speeds up as it falls: \(v_\mathrm{top}\) is less than \(v_\mathrm{bottom}\). The equations on the slide still hold at the top and at the bottom, each with the speed the ball has there.

Building each equation

In both places, take “towards the center” as positive and add up the real forces. At the top the center is below the ball, so tension and weight both point towards it. At the bottom the center is above, so the tension points towards it and the weight away. The resultant towards the center must equal \(\frac{mv^{2}}{r}\).

The slowest it can go over the top

As the speed at the top is reduced, less force is needed and the tension falls. A string cannot push, so the least possible tension is zero. Then the weight alone is the centripetal force: \(mg = \frac{mv^{2}}{r}\), so \(v = \sqrt{gr}\). Any slower and the string goes slack before the top.

Only the top and the bottom are examined. At other points the weight has a component along the path as well, which is what changes the speed.

The same two equations describe a car on a hump-backed bridge or a rider on a loop: replace the tension by the normal force, with its direction.

Worked example · non-uniform circular motion

A 0.25 kg ball on a 0.80 m string.

Weight \(mg = 0.25 \times 9.8 = 2.45 \,\mathrm{N}\) Bottom · \(v_\mathrm{bottom} = 7.0 \,\mathrm{m} \,\mathrm{s}^{-1}\) \(T_\mathrm{bottom} = \dfrac{mv_\mathrm{bottom}^{2}}{r} + mg\) \(T_\mathrm{bottom} = \dfrac{0.25 \times 7.0^{2}}{0.80} + 2.45 = \class{res}{17.8 \,\mathrm{N}}\)

Top · \(v_\mathrm{top} = 4.2 \,\mathrm{m} \,\mathrm{s}^{-1}\) \(T_\mathrm{top} = \dfrac{mv_\mathrm{top}^{2}}{r} - mg\) \(T_\mathrm{top} = \dfrac{0.25 \times 4.2^{2}}{0.80} - 2.45 = \class{res}{3.1 \,\mathrm{N}}\) Least: \(v_\mathrm{top} = \sqrt{gr} = \sqrt{9.8 \times 0.80} = \class{res}{2.8 \,\mathrm{m} \,\mathrm{s}^{-1}}\)

A.2 FORCES AND MOMENTUM · SL + HL

Why the string is most likely to break at the bottom

There the tension has two jobs: it must hold up the weight and also supply the whole centripetal force, at the point where the ball is moving fastest. At the top the weight supplies part of the centripetal force for free, and the ball is at its slowest.

Where the two speeds come from

They are not independent. Between the bottom and the top the ball rises by one diameter, 1.6 m, and kinetic energy is transferred to gravitational potential energy. For each kilogram: \(\frac{1}{2}v^{2} = \frac{1}{2} \times 7.0^{2} - 9.8 \times 1.6 = 8.8\), so \(v = 4.2 \,\mathrm{m} \,\mathrm{s}^{-1}\). This is conservation of energy, from A.3. The equations of motion cannot be used here, because the acceleration is not constant.

Using \(T = \frac{mv^{2}}{r}\) alone in a vertical circle. That is right only for a horizontal circle on a smooth table, where the weight plays no part in the plane of the motion.

Quick check

Circular motion.

01A body moves round a circle at constant speed. Direction of its acceleration?

Towards the center

02A wheel turns once every 0.50 s. Angular velocity?

\(\frac{2 \pi}{0.50} = 12.6 \,\mathrm{rad} \,\mathrm{s}^{-1}\)

03A car moves at 10 m s⁻¹ round a circle of radius 25 m. Its acceleration?

\(\frac{10^{2}}{25} = 4.0 \,\mathrm{m} \,\mathrm{s}^{-2}\)

04Its mass is 900 kg. Centripetal force, and what provides it?

\(900 \times 4.0 = 3600 \,\mathrm{N}\), from friction

05Why does a centripetal force not change the speed?

It is at right angles to the velocity

06A ball on a string moves in a vertical circle. Where is the tension greatest?

At the bottom

PRESS → TO REVEAL EACH ANSWER

Summary

Forces and momentum on one slide.

Newton’s laws

Find the resultant

No resultant force: constant velocity
\(F = ma\). Forces come in pairs on different bodies

The forces

Name every one

\(F_\mathrm{f} \le \mu_\mathrm{s} F_\mathrm{N}\) at rest, \(\mu_\mathrm{d} F_\mathrm{N}\) sliding
\(F_\mathrm{H} = -kx\), \(F_\mathrm{b} = \rho V g\), \(F_\mathrm{d} = 6 \pi \eta r v\)

Momentum

Impulse changes it

\(p = mv\), a vector
\(F \Delta t = \Delta p\): the area under a force–time graph

Collisions

Momentum is conserved

Kinetic energy only if elastic
Explosion from rest: total momentum zero

Collisions · HL

Two dimensions

Conserve momentum along \(x\) and along \(y\) separately

Circular motion

Resultant to the center

\(a = \dfrac{v^{2}}{r} = \omega^{2} r\), and \(v = \omega r\)
Top and bottom of a vertical circle differ

A.2 FORCES AND MOMENTUM · SL + HL

Summary

Key words.

Resultant forceThe vector sum of all the forces

Translational equilibriumNo resultant force

Normal forceAt right angles to the surface

Coefficient of frictionRatio of friction to normal force

BuoyancyWeight of the fluid displaced

Viscous dragA fluid’s force against motion

MomentumMass × velocity. A vector

ImpulseForce × time: change in momentum

Elastic collisionKinetic energy is conserved

Angular velocityAngle turned through each second

Centripetal accelerationTowards the center of the circle

Centripetal forceThe resultant towards the center

A.2 FORCES AND MOMENTUM · SL + HL