DP Physics SL A.2 Forces and Momentum

DP Physics SL A.2 Forces and Momentum - Full notes

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DP Physics SL | A.2 Forces and Momentum

DP PHYSICS · THEME A: SPACE, TIME AND MOTION

A.2

Forces and Momentum

Full notes for SL · IB-style

Contents

1	Forces and free-body diagrams
2	Newton's laws of motion
3	Friction, springs, drag and buoyancy
4	Momentum and impulse
5	Conservation of momentum
6	Circular motion
7	Investigating forces and momentum
8	Common misconceptions and links
9	Exam-style question with markscheme
10	Check your understanding
11	Summary and topic map

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and do every Quick check before moving to the next section.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
draw and interpret free-body diagrams and resolve forces
apply Newton's three laws, including lifts and connected bodies
solve problems involving friction, springs, drag and buoyancy
use momentum, impulse and force–time graphs
apply conservation of momentum in one and two dimensions and classify collisions
analyse circular motion, including vertical circles and banked tracks
describe experiments to investigate forces and collisions

There is no additional HL content in this subtopic. SL and HL students study the same material.

Guiding questions

How can we use our knowledge and understanding of the forces acting on a body to predict changes to its motion? How can Newton's laws be modelled mathematically? How can knowledge of forces be used to explain collisions and circular motion?

How this topic is examined

Forces and momentum turn up across the papers: quick multiple-choice items on Newton's laws and momentum, data questions on force–time graphs and trolley experiments, and longer structured questions combining forces, energy and circular motion.

Always start with a free-body diagram and state your positive direction.

Command term	What the examiner wants
State	a brief answer with no explanation
Outline	a brief account or summary
Describe	a detailed account without reasons
Explain	a detailed account including causes, reasons or mechanisms
Calculate	a numerical answer with working and correct units
Determine	obtain the only possible answer, often from a graph or data
Show that	obtain the given result, showing every step (give one more significant figure)
Estimate	an approximate value using reasonable assumptions
Sketch	a graph or diagram showing the general shape and key features
Deduce / Suggest	reach a conclusion from given information; propose a reasonable explanation

How to use these notes

Each section teaches one part of the syllabus in full. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Key point	the one idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
In the real world	where the physics shows up outside the classroom
Try this	something quick you can do yourself to test the idea
Worked example	full solutions with IB-style marks (M1 method, A1 answer, R1 reasoning)
Quick check	short questions to test yourself at the end of each section

1 Forces and free-body diagrams

Every change of motion starts with a force.

Think about it

A book rests on a table. Two forces act on it and it does not move. Which force does the table exert, and why exactly that size?

Mechanics problems become straightforward once you can see every force on an object. This section builds the vocabulary of forces and the single most useful skill in the topic: the free-body diagram.

1.1 What is a force?

A force is a push or a pull that one body exerts on another. Forces are vectors (they have magnitude and direction) measured in newtons (N). A force can change an object's speed, its direction of motion or its shape.

Forces always come from interactions between two bodies. Contact forces need touching surfaces (normal reaction, friction, tension, drag, upthrust); field forces act at a distance (gravitational, electric, magnetic).

Force	Symbol	Direction
weight	W = mg	vertically down, through the centre of mass
normal reaction	N or R	perpendicular to the contact surface
friction	Ff	along the surface, opposing (attempted) sliding
tension	T	along a string, pulling away from the object
drag / air resistance	Fd	opposite to the velocity
upthrust (buoyancy)	Fb	upwards in a fluid
elastic (spring)	F = kx	towards the natural length

WORKED EXAMPLE 1 Weight on two worlds (basic)

Find the weight of a 65 kg astronaut on Earth (g = 9.81 N kg⁻¹) and on the Moon (g = 1.62 N kg⁻¹).

Solution

Earth: W = 65 × 9.81 = 638 NA1

Moon: W = 65 × 1.62 = 105 N; the mass is still 65 kgA1

Common mistake

Treating mass and weight as the same thing. Mass (kg) is the amount of matter and does not change with location; weight (N) is the gravitational force and depends on g.

1.2 Free-body diagrams

A free-body diagram shows one object, usually as a dot or box, with an arrow for every force acting on it. Each arrow starts where the force acts, points in the force's direction and is roughly to scale. Label every arrow.

Only forces acting on the chosen object appear. A common error is to include forces the object exerts on other things, or invented forces such as 'the force of motion'.

•(a) Book on a table: weight down, normal reaction up, equal in size.
•(b) Box pulled at constant velocity: tension balances friction; weight balances normal reaction.
•(c) Skydiver at terminal velocity: drag equals weight.
•(d) Lamp on two cords: the vertical components of T₁ and T₂ together balance the weight.

Figure 1. Free-body diagrams for four common situations. In each case the forces balance.

Key point

Draw the free-body diagram before writing any equation. Most mechanics marks are lost through a missing or invented force.

1.3 Resolving forces into components

A force F at angle θ to the horizontal can be replaced by two perpendicular components: F cos θ horizontally and F sin θ vertically. Together the components have exactly the same effect as F.

On a slope of angle θ, resolve the weight parallel and perpendicular to the slope: mg sin θ down the slope and mg cos θ into the slope. The normal reaction balances mg cos θ when nothing else acts perpendicular to the slope.

Figure 2. Forces on a block resting on a rough slope. The weight is replaced by its two components (blue).

WORKED EXAMPLE 2 Block on a slope (standard)

A 5.0 kg block rests on a 30° slope. Find the components of its weight, the normal reaction and the friction needed to keep it still.

Solution

Parallel: 5.0 × 9.81 × sin 30° = 24.5 NA1

Perpendicular: 5.0 × 9.81 × cos 30° = 42.5 N, so N = 42.5 NA1

Friction balances the parallel component: 24.5 N up the slopeR1

WORKED EXAMPLE 3 Angled pull (standard)

A sledge is pulled by a rope at 25° above the horizontal with a tension of 80 N. Find the horizontal and vertical components.

Solution

Horizontal: 80 cos 25° = 72.5 NA1

Vertical: 80 sin 25° = 33.8 N (this reduces the normal reaction)A1

Quick check

1. Resolve a 50 N force at 40° to the horizontal.

2. Why does pulling a sledge at an angle reduce friction?

Answers: 1. 38.3 N horizontal, 32.1 N vertical 2. the upward component reduces the normal reaction, so friction (μN) is smaller

1.4 Translational equilibrium

An object is in translational equilibrium when the resultant force on it is zero: it is either at rest or moving with constant velocity. Resolve in two perpendicular directions and set each sum to zero.

When exactly three forces act, they can be drawn tip-to-tail to form a closed triangle of forces.

WORKED EXAMPLE 4 Hanging sign (standard)

A 20 N sign hangs from two cables, each at 30° to the horizontal. Find the tension in each cable.

Solution

Vertical: 2T sin 30° = 20M1

T = 20 NA1

WORKED EXAMPLE 5 Unequal angles (exam level)

A 50 N lamp hangs from two cords at 40° and 60° to the horizontal. Find both tensions.

Solution

Horizontal: T₁ cos 40° = T₂ cos 60°M1

Vertical: T₁ sin 40° + T₂ sin 60° = 50M1

T₁ = 25.4 N, T₂ = 38.9 NA1

Quick check

1. What happens to the tension in a washing line as it is pulled more nearly horizontal?

Answers: 1. it increases sharply because sin θ becomes small

1.5 Mass, weight and gravitational field strength

Weight W = mg, where g is the gravitational field strength (N kg⁻¹), numerically equal to the free-fall acceleration (m s⁻²). Near Earth's surface g = 9.81 N kg⁻¹; it is smaller at altitude and on smaller planets. Weight acts through the centre of mass.

Location	g / N kg⁻¹	Weight of a 60 kg person / N
Earth	9.81	589
Moon	1.62	97
Mars	3.71	223
Jupiter (cloud tops)	24.8	1490

Quick check

1. A spring balance reads 12 N for a mass on Mars. Find the mass.

Answers: 1. 12/3.71 = 3.2 kg

2 Newton's laws of motion

Three laws that predict how forces change motion.

Think about it

Why do you lurch forward when a bus brakes suddenly, even though nothing pushed you forward?

2.1 Newton's first law

An object remains at rest, or continues moving at constant velocity, unless acted on by a resultant external force. The tendency to keep moving in the same way is called inertia; mass is a measure of inertia.

A moving object does not need a force to keep moving. It needs a resultant force only to speed up, slow down or change direction.

WORKED EXAMPLE 6 Constant velocity lift (standard)

A lift moves upwards at constant velocity. Compare the tension in the cable with the weight of the lift.

Solution

Constant velocity: resultant force is zeroR1

Tension = weightA1

Common mistake

Saying a force is needed to keep an object moving. On Earth we keep pushing things only because friction and drag oppose motion.

2.2 Newton's second law

The resultant force on an object equals its rate of change of momentum, F = Δp/Δt. For constant mass this becomes F = ma. The acceleration is always in the direction of the resultant force.

Figure 3. Acceleration is proportional to resultant force; the gradient is 1/m.

WORKED EXAMPLE 7 Car (standard)

A 1200 kg car has a driving force of 3000 N and total resistance 600 N. Calculate its acceleration.

Solution

Resultant = 3000 − 600 = 2400 NM1

a = 2400/1200 = 2.0 m s⁻²A1

WORKED EXAMPLE 8 Braking (standard)

A 900 kg car slows from 20 m s⁻¹ to rest in 4.0 s. Find the average braking force.

Solution

a = −20/4.0 = −5.0 m s⁻²M1

F = 900 × 5.0 = 4500 N opposite to the motionA1

Quick check

1. A 2.0 kg trolley accelerates at 1.5 m s⁻² against 1.0 N of friction. Find the applied force.

Answers: 1. F − 1.0 = 3.0: F = 4.0 N

2.3 Apparent weight in lifts

A person in a lift feels the normal reaction from the floor, not their weight. Apply F = ma with up as positive: N − mg = ma. When the lift accelerates upwards, N > mg and you feel heavier; accelerating downwards, N < mg and you feel lighter; in free fall N = 0 (weightlessness).

Figure 4. The normal reaction in a lift depends on the acceleration, not on the direction of the velocity.

WORKED EXAMPLE 9 Lift (exam level)

A 70 kg person stands on scales in a lift accelerating upwards at 1.5 m s⁻². Find the scale reading in newtons.

Solution

N − mg = maM1

N = 70(9.81 + 1.5) = 792 NA1

WORKED EXAMPLE 10 Slowing down (exam level)

The lift is now moving upwards but slowing at 1.5 m s⁻². Find the reading.

Solution

Acceleration is downwards: N = 70(9.81 − 1.5)M1

= 582 NA1

2.4 Newton's third law

If body A exerts a force on body B, then B exerts a force on A that is equal in size, opposite in direction and of the same type. Third-law partners act on different bodies, so they never cancel each other.

Force	Third-law partner
Earth's pull on a book (its weight)	the book's gravitational pull on Earth
table's push on the book (N)	the book's push on the table
foot pushes the ground backwards	ground pushes the foot forwards (lets you walk)
rocket pushes exhaust backwards	exhaust pushes the rocket forwards

Figure 5. Earth pulls the apple down; the apple pulls Earth up with an equal force.

Common mistake

Calling weight and normal reaction a third-law pair. They act on the same object and are different types of force.

2.5 Connected bodies

For objects joined by light, inextensible strings, the whole system shares the same acceleration. First treat the system as one object to find a, then apply F = ma to one object alone to find the tension.

Figure 6. A block on a smooth table connected over a pulley to a hanging mass.

WORKED EXAMPLE 11 Pulley (exam level)

In the figure, the table is smooth. Find the acceleration and the tension in the string.

Solution

System: 2.0 × 9.81 = (3.0 + 2.0)aM1

a = 3.92 m s⁻²A1

Table block: T = 3.0 × 3.92 = 11.8 N (less than 19.6 N)A1

Quick check

1. Why is the tension less than the weight of the hanging mass?

Answers: 1. the hanging mass accelerates downwards, so the resultant on it is downwards

3 Friction, springs, drag and buoyancy

The contact forces in detail.

3.1 Static and dynamic friction

Static friction acts when surfaces are not sliding and adjusts to match the applied force, up to a maximum μsN. Once sliding starts, dynamic (kinetic) friction acts with a roughly constant value μdN. Usually μd < μs, so it is harder to start something sliding than to keep it moving.

Figure 7. Friction on a 40 kg crate as the applied force increases.

WORKED EXAMPLE 12 Coefficients (standard)

A 40 kg crate needs 160 N to start moving and 120 N to keep it moving at constant velocity. Find μs and μd.

Solution

N = 40 × 9.81 = 392 NM1

μs = 160/392 = 0.41; μd = 120/392 = 0.31A1 A1

3.2 Friction on slopes

On a slope, the normal reaction is mg cos θ, so maximum friction is μmg cos θ. A block is on the point of slipping when mg sin θ = μs mg cos θ, giving tan θ = μs. Measuring this angle is a quick way to find μs.

WORKED EXAMPLE 13 Sliding block (exam level)

A block slides down a 35° slope with μd = 0.40. Find its acceleration.

Solution

a = g(sin 35° − 0.40 cos 35°)M1

= 9.81(0.574 − 0.328) = 2.4 m s⁻²A1

Try this

Place a coin on a book and slowly tilt the book. Measure the angle at which the coin starts to slide; tan θ gives μs for coin on paper.

3.3 Springs and Hooke's law

Within the limit of proportionality, the extension of a spring is proportional to the force: F = kx, where k is the spring constant (N m⁻¹). Beyond this limit the graph curves; beyond the elastic limit the spring is permanently stretched.

Figure 8. Force–extension graph for a spring. The gradient of the straight part is k.

WORKED EXAMPLE 14 Spring constant (standard)

A 0.50 kg mass extends a spring by 4.0 cm. Calculate k.

Solution

k = mg/x = 4.905/0.040M1

= 123 N m⁻¹A1

WORKED EXAMPLE 15 Series and parallel (exam level)

Two identical springs (k = 120 N m⁻¹) support a 6.0 N load (a) side by side; (b) one below the other. Find the extension in each case.

Solution

(a) parallel: keff = 240 N m⁻¹; x = 2.5 cmA1

(b) series: keff = 60 N m⁻¹; x = 10 cmA1

3.4 Drag and terminal velocity

Drag acts opposite to velocity and increases with speed. For a small sphere moving slowly through a fluid, Stokes' law gives Fd = 6πηrv. A falling object accelerates until drag (plus upthrust) equals weight; it then falls at constant terminal velocity.

Figure 9. Velocity of a falling object approaching terminal velocity. The gradient (acceleration) falls to zero.

WORKED EXAMPLE 16 Terminal velocity of a sphere (exam level)

A steel ball (r = 1.0 mm, mass 3.3 × 10⁻⁵ kg) falls through glycerol (η = 1.4 Pa s). Ignoring upthrust, estimate its terminal velocity.

Solution

mg = 6πηrvM1

v = 3.24 × 10⁻⁴/(6π × 1.4 × 1.0 × 10⁻³) = 0.012 m s⁻¹A1

3.5 Buoyancy

An object in a fluid experiences an upthrust equal to the weight of fluid displaced (Archimedes' principle): Fb = ρVg. An object floats when upthrust equals its weight, which happens if its average density is less than the fluid's.

WORKED EXAMPLE 17 Submerged block (standard)

A 2.0 × 10⁻³ m³ block is fully submerged in water (ρ = 1000 kg m⁻³). Find the upthrust.

Solution

Fb = 1000 × 2.0 × 10⁻³ × 9.81M1

= 19.6 NA1

WORKED EXAMPLE 18 Floating (exam level)

A wooden block of density 600 kg m⁻³ floats in water. What fraction of its volume is submerged?

Solution

Weight = upthrust: 600Vg = 1000Vsub gM1

Vsub/V = 0.60A1

In the real world

A steel ship floats because its hull encloses a large volume of air, so its average density is less than that of water.

4 Momentum and impulse

Mass in motion, and how forces change it.

Think about it

Why does a cricketer move their hands backwards when catching a fast ball?

4.1 Linear momentum

Linear momentum p = mv is a vector with units kg m s⁻¹ (or N s). Newton's second law in its general form says the resultant force equals the rate of change of momentum.

WORKED EXAMPLE 19 Momentum (basic)

Find the momentum of a 0.16 kg cricket ball at 35 m s⁻¹ and of a 1200 kg car at 2.0 m s⁻¹.

Solution

Ball: 5.6 kg m s⁻¹A1

Car: 2400 kg m s⁻¹A1

4.2 Impulse and force–time graphs

Impulse J = FΔt = Δp. When the force varies, the impulse equals the area under the force–time graph.

Figure 10. Force–time graph for a kick. The shaded area is the impulse.

WORKED EXAMPLE 20 Kick (standard)

Using the graph, find the impulse on a 0.45 kg ball initially at rest and its final speed.

Solution

Area = ½(1)(800) + 2(800) + ½(1)(800) = 2400 N ms = 2.4 N sM1 A1

v = 2.4/0.45 = 5.3 m s⁻¹A1

WORKED EXAMPLE 21 Bat and ball (exam level)

A 0.16 kg ball at 30 m s⁻¹ is hit straight back at 40 m s⁻¹. Contact lasts 2.0 ms. Find the average force.

Solution

Δp = 0.16(40 − (−30)) = 11.2 kg m s⁻¹M1

F = 11.2/0.0020 = 5600 NA1

Common mistake

Forgetting the change of direction. A ball reversing from −30 to +40 m s⁻¹ has Δv = 70 m s⁻¹, not 10 m s⁻¹.

4.3 Increasing the collision time

For the same change in momentum, a longer collision time gives a smaller average force. This is the principle behind crumple zones, airbags, seatbelts, helmets, crash mats and bending your knees on landing.

Figure 11. Two collisions with the same impulse (equal areas): the longer one has a much smaller peak force.

In the real world

Car crumple zones extend the collision time from a few milliseconds to about 0.1 s, reducing the average force on occupants by a factor of ten or more.

Quick check

1. A 70 kg person moving at 15 m s⁻¹ is stopped by an airbag in 0.15 s. Find the average force.

Answers: 1. Δp = 1050 kg m s⁻¹; F = 7000 N

4.4 Rockets and jet propulsion

A rocket pushes exhaust gas backwards at high speed; by Newton's third law the gas pushes the rocket forwards. The thrust equals the rate of change of momentum of the exhaust: thrust = (Δm/Δt) × exhaust speed. Rockets work in a vacuum because they do not need air to push against.

Figure 12. A rocket: the force on the exhaust and the thrust on the rocket are a third-law pair.

WORKED EXAMPLE 22 Thrust (exam level)

A rocket ejects 250 kg of gas per second at 3000 m s⁻¹. Its mass is 50 000 kg. Find the thrust and the initial acceleration on vertical launch.

Solution

Thrust = 250 × 3000 = 7.5 × 10⁵ NA1

a = (7.5 × 10⁵ − 4.9 × 10⁵)/50 000 = 5.2 m s⁻²M1 A1

4.5 Force from a stream of mass

When a stream of material hits a surface, the force equals the rate of change of momentum: F = (Δm/Δt)Δv. This applies to water jets, rocket exhaust, wind on sails and helicopter downdraft.

WORKED EXAMPLE 23 Hose (exam level)

Water leaves a hose at 2.0 kg s⁻¹ at 15 m s⁻¹ and stops on hitting a wall. Find the force on the wall.

Solution

F = 2.0 × 15M1

= 30 NA1

WORKED EXAMPLE 24 Helicopter (exam level)

A 1500 kg helicopter hovers by pushing air down at 20 m s⁻¹. Find the mass of air pushed down per second.

Solution

Force needed = 1500 × 9.81 = 14 700 NM1

Δm/Δt = 14 700/20 = 740 kg s⁻¹A1

5 Conservation of momentum

Collisions and explosions.

5.1 The principle of conservation of momentum

In a closed system (no resultant external force), the total momentum is constant. During a collision the bodies exert equal and opposite forces on each other for the same time (Newton's third law), so their momentum changes are equal and opposite.

Key point

Total momentum before = total momentum after, as long as no external resultant force acts. Momentum is a vector: choose a positive direction.

5.2 Collisions

Draw 'before' and 'after' diagrams, choose a positive direction, write total momentum before = total momentum after, and solve.

Figure 13. A 2.0 kg trolley collides with a stationary 1.0 kg trolley and they stick together.

WORKED EXAMPLE 25 Sticking together (standard)

Find the common velocity after the collision shown.

Solution

2.0 × 3.0 + 0 = 3.0vM1

v = 2.0 m s⁻¹A1

WORKED EXAMPLE 26 Head-on (exam level)

A 3.0 kg ball at 4.0 m s⁻¹ hits a 2.0 kg ball moving at 1.0 m s⁻¹ towards it. Afterwards the 3.0 kg ball is at rest. Find the 2.0 kg ball's velocity.

Solution

3.0(4.0) + 2.0(−1.0) = 0 + 2.0vM1

v = 5.0 m s⁻¹ in the original direction of the 3.0 kg ballA1

5.3 Explosions and recoil

If a system starts at rest, its total momentum is zero and stays zero. The parts fly apart with equal and opposite momenta: the lighter part moves faster.

Figure 14. A rifle and bullet: total momentum is zero before and after firing.

WORKED EXAMPLE 27 Recoil (standard)

A 4.0 kg rifle fires a 10 g bullet at 400 m s⁻¹. Find the recoil velocity.

Solution

0 = 0.010 × 400 + 4.0vM1

v = −1.0 m s⁻¹ (backwards)A1

5.4 Elastic and inelastic collisions

Momentum is conserved in all collisions in a closed system, but kinetic energy is conserved only in elastic collisions. In inelastic collisions some kinetic energy becomes internal energy and sound; in a perfectly inelastic collision the bodies stick together. In explosions kinetic energy increases (from chemical or elastic energy).

Type	Momentum	Kinetic energy
elastic	conserved	conserved
inelastic	conserved	decreases
perfectly inelastic (stick)	conserved	maximum possible decrease
explosion	conserved (zero)	increases

WORKED EXAMPLE 28 Energy lost (exam level)

For the trolleys in the collision above, calculate the kinetic energy lost.

Solution

Before: ½(2.0)(3.0)² = 9.0 J; after: ½(3.0)(2.0)² = 6.0 JM1

Loss = 3.0 J (33%)A1

WORKED EXAMPLE 29 Is it elastic? (exam level)

A 3.0 kg ball at 4.0 m s⁻¹ hits a stationary 2.0 kg ball. Afterwards the 3.0 kg ball moves at 0.80 m s⁻¹. Find the other velocity and decide whether the collision is elastic.

Solution

12 = 2.4 + 2.0v: v = 4.8 m s⁻¹A1

KE before 24 J; after 0.96 + 23.04 = 24 J: elasticR1

5.5 Collisions in two dimensions

Momentum is conserved separately in each perpendicular direction. Resolve every momentum into x and y components, apply conservation to each, then recombine.

Figure 15. Momentum vectors for a glancing collision: the final momenta add to the initial momentum.

WORKED EXAMPLE 30 Glancing collision (exam level)

A 1.0 kg puck at 4.0 m s⁻¹ strikes an identical stationary puck. Afterwards the first moves at 2.0 m s⁻¹ at 60° to its original line. Find the second puck's velocity.

Solution

x: 4.0 = 2.0 cos 60° + vx: vx = 3.0M1

y: 0 = 2.0 sin 60° + vy: vy = −1.73M1

v = 3.46 m s⁻¹ at 30° below the original lineA1

5.6 Momentum–time graphs

The gradient of a momentum–time graph is the resultant force. During a collision, the two bodies' graphs have equal and opposite gradients, and their sum is a horizontal line.

Figure 16. Momentum of two trolleys during a collision lasting 2 ms. The total stays constant.

Quick check

1. Use the graph to find the average force on trolley B during the collision.

Answers: 1. 2 kg m s⁻¹/0.002 s = 1000 N

6 Circular motion

Moving in a circle means accelerating.

Think about it

A car travels round a roundabout at a steady 30 km h⁻¹. Is it accelerating?

6.1 Describing circular motion

For motion in a circle of radius r: period T is the time for one revolution; frequency f = 1/T; angular velocity ω = 2π/T = 2πf (rad s⁻¹); linear speed v = ωr = 2πr/T. Angles are measured in radians: 2π rad = 360°.

WORKED EXAMPLE 31 Wheel (standard)

A wheel of radius 0.30 m rotates 10 times per second. Find ω and the speed of the rim.

Solution

ω = 2π × 10 = 62.8 rad s⁻¹A1

v = 62.8 × 0.30 = 18.8 m s⁻¹A1

6.2 Centripetal acceleration

At constant speed the direction of the velocity keeps changing, so the object accelerates. The change in velocity Δv points towards the centre, so the acceleration is centripetal: a = v²/r = ω²r.

Figure 17. Velocity vectors at two points: the change Δv points towards the centre.

WORKED EXAMPLE 32 Earth's orbit (exam level)

Earth orbits the Sun at radius 1.5 × 10¹¹ m in 3.16 × 10⁷ s. Find its centripetal acceleration.

Solution

a = 4π²r/T² = 4π²(1.5 × 10¹¹)/(3.16 × 10⁷)²M1

= 5.9 × 10⁻³ m s⁻²A1

6.3 Centripetal force

A resultant force F = mv²/r = mω²r towards the centre is needed. It is not a new force: it is provided by a real force such as tension, friction, gravity or a component of the normal reaction. If it is removed, the object moves off along a tangent.

Situation	Force providing the centripetal force
ball on a string (horizontal)	tension
car on a flat bend	friction between tyres and road
satellite in orbit	gravitational force
charge in a magnetic field	magnetic force qvB
car on a banked track	horizontal component of the normal reaction

WORKED EXAMPLE 33 Flat bend (exam level)

A 1000 kg car rounds a flat bend of radius 50 m. The maximum friction is 6000 N. Find its maximum speed.

Solution

6000 = 1000v²/50M1

v = 17.3 m s⁻¹A1

Common mistake

Adding 'centripetal force' as an extra arrow on a free-body diagram. The centripetal force is the resultant of the forces already drawn.

6.4 Vertical circles

In a vertical circle, the resultant towards the centre must still equal mv²/r, but weight helps at the top and opposes at the bottom. The tension is therefore greatest at the bottom. At the top, the string stays taut only if v ≥ √(gr).

Figure 18. Forces on a ball whirled in a vertical circle, at the top and at the bottom.

WORKED EXAMPLE 34 Top and bottom (exam level)

A 0.20 kg ball on a 0.50 m string moves at 4.0 m s⁻¹ at the top and 6.0 m s⁻¹ at the bottom. Find the tension at each point.

Solution

Top: T + mg = mv²/r: T = 0.20(32) − 1.96 = 4.4 NM1 A1

Bottom: T − mg = mv²/r: T = 0.20(72) + 1.96 = 16.4 NA1

WORKED EXAMPLE 35 Hump-backed bridge (exam level)

A 900 kg car goes over a bridge of radius 30 m at 12 m s⁻¹. Find the normal reaction at the top, and the speed at which the car would just leave the road.

Solution

mg − N = mv²/r: N = 900(9.81 − 4.8) = 4.5 kNM1 A1

N = 0 when v = √(gr) = 17.2 m s⁻¹A1

6.5 The conical pendulum

A mass on a string moving in a horizontal circle forms a cone. The vertical component of tension balances the weight (T cos θ = mg) and the horizontal component provides the centripetal force (T sin θ = mv²/r). Dividing gives tan θ = v²/(rg).

Figure 19. A conical pendulum: the string sweeps out a cone.

WORKED EXAMPLE 36 Conical pendulum (exam level)

A 0.10 kg bob moves in a horizontal circle of radius 0.30 m with the string at 25° to the vertical. Find the tension and the speed.

Solution

T = mg/cos 25° = 1.08 NM1 A1

v = √(rg tan 25°) = 1.17 m s⁻¹A1

6.6 Banked tracks

On a banked track, the horizontal component of the normal reaction provides the centripetal force even without friction. For no sideways friction, tan θ = v²/(rg).

Figure 20. A car on a banked track: N has a component towards the centre of the bend.

WORKED EXAMPLE 37 Banking angle (exam level)

Find the banking angle for a bend of radius 80 m designed for 20 m s⁻¹ with no sideways friction.

Solution

tan θ = 400/(80 × 9.81) = 0.51M1

θ = 27°A1

In the real world

Velodrome tracks are banked at up to about 45° at the bends so cyclists can corner at high speed without relying on tyre friction.

7 Investigating forces and momentum

Practical skills and data analysis.

7.1 Verifying F = ma

Accelerate a trolley with hanging masses, keeping the total mass constant by moving masses from the trolley to the hanger. Measure acceleration with light gates or a motion sensor. Plot a against F: a straight line through the origin confirms a ∝ F; the gradient is 1/m.

Figure 21. Data from a trolley experiment. The best-fit line does not pass through the origin.

WORKED EXAMPLE 38 Interpreting (exam level)

Using the graph, find the total mass of the system and suggest why the line does not pass through the origin.

Solution

Gradient = 0.80/0.50 = 1.6 kg⁻¹: m = 0.63 kgM1 A1

Positive a-intercept: a systematic error, e.g. the runway is tilted too steeply (over-compensating for friction)R1

7.2 Measuring a coefficient of friction

Increase the angle of an inclined plane until a block just starts to slide: μs = tan θ. Alternatively, pull the block with a force sensor at constant velocity on a horizontal surface: μd = F/mg.

WORKED EXAMPLE 39 Uncertainty (exam level)

The slipping angle is 28 ± 2°. Find μs and its uncertainty.

Solution

tan 28° = 0.53; tan 26° = 0.49; tan 30° = 0.58M1

μs = 0.53 ± 0.05A1

7.3 Investigating terminal velocity

Drop ball bearings into a tall tube of glycerol or oil with rubber bands marked at equal intervals. Time the ball between marks: when times between successive marks are equal, terminal velocity has been reached. Repeat for balls of different radii to test Stokes' law (v ∝ r²).

Figure 22. Terminal velocity against radius squared for steel balls in glycerol.

WORKED EXAMPLE 40 Testing Stokes' law (standard)

Explain why plotting v against r² tests Stokes' law.

Solution

At terminal velocity mg = 6πηrv, with m ∝ r³R1

So v ∝ r²: a straight line through the originR1

7.4 Investigating collisions

Two trolleys on a friction-compensated runway, with light gates measuring speeds before and after, test momentum conservation. Velcro gives perfectly inelastic collisions; magnets give nearly elastic ones. Compare total momentum and kinetic energy before and after.

Try this

Use a phone slow-motion video and a metre rule to measure the speeds of two toy cars before and after they collide. Check whether total momentum is conserved within your uncertainty.

Quick check

1. Why must the runway be friction-compensated?

Answers: 1. so friction does not act as an external resultant force on the system

8 Common misconceptions and links

Students often think…	Actually…
a moving object needs a force to keep moving	constant velocity needs no resultant force
weight and normal reaction are a third-law pair	they act on the same body; partners act on different bodies
centripetal force is an extra force	it is the resultant of real forces
momentum is lost in inelastic collisions	momentum is conserved; kinetic energy is not
heavier objects fall faster	acceleration is g for all (no drag)
in a lift, the scale shows your weight	it shows the normal reaction

How A.2 connects to the rest of the course

Topic	Connection
A.1 Kinematics	acceleration, motion graphs
A.3 Work, energy and power	energy in collisions
A.4 Rigid body mechanics (HL)	rotational analogue of Newton's laws
D.1 Gravitational fields	orbits: gravity provides centripetal force
D.3 Motion in EM fields	circular paths of charges

9 Exam-style question with markscheme

Try this under exam conditions (about 20 minutes) before reading the markscheme.

Figure 23. Force–time graph for a ball being kicked.

A 0.45 kg football at rest is kicked. The graph shows the force on the ball.

(a) Determine the impulse on the ball. [2]

(b) Calculate the speed of the ball after the kick. [2]

(c) The ball is later whirled in a horizontal circle of radius 0.80 m on a string at 6.0 m s⁻¹ (a training device). Calculate the tension, ignoring gravity. [2]

(d) State and explain Newton's third law as applied to the kick. [2]

(e) The ball hits a wall and rebounds. Explain why momentum is conserved but kinetic energy is not. [2]

Markscheme

Part	Answer	Marks
(a)	area = ½(1)(800) + 2(800) + ½(1)(800) N ms	M1
	2.4 N s	A1
(b)	v = 2.4/0.45	M1
	5.3 m s⁻¹	A1
(c)	T = 0.45 × 36/0.80	M1
	20 N	A1
(d)	ball exerts an equal and opposite force on the foot	A1
	forces act on different bodies, same type, same time	R1
(e)	no external horizontal force on ball–wall–Earth system	R1
	kinetic energy converted to internal energy and sound	R1

Examiner's comments. In (a), many answers forgot to convert milliseconds. In (d), the best answers named both bodies and the type of force.

10 Check your understanding

Section A: multiple choice

1. The unit of impulse is

A. N s B. N m C. kg m s⁻² D. J s

2. A resultant force of 12 N acts on a 4.0 kg mass. The acceleration is

A. 48 m s⁻² B. 0.33 m s⁻² C. 3.0 m s⁻² D. 16 m s⁻²

3. In a perfectly inelastic collision

A. kinetic energy is conserved B. momentum is conserved and the objects stick together C. momentum is lost D. objects bounce apart

4. The area under a force–time graph is

A. work done B. acceleration C. power D. change in momentum

5. For an object moving in a circle at constant speed, the acceleration is

A. zero B. towards the centre C. tangential D. outwards

6. The coefficient of dynamic friction is usually

A. less than the static coefficient B. greater than the static coefficient C. equal to 1 D. independent of the surfaces

7. A 70 kg person in a lift accelerating downwards at 2 m s⁻² has a normal reaction of about

A. 690 N B. 830 N C. 140 N D. 550 N

8. Third-law force pairs

A. act on the same body B. are always contact forces C. act on different bodies D. cancel each other

Section B: structured

9. A 1500 kg car accelerates from rest to 20 m s⁻¹ in 8.0 s against a resistance of 500 N. Find the driving force. [3]

10. A 3.0 kg ball at 4.0 m s⁻¹ collides head-on with a 2.0 kg ball at rest. The 3.0 kg ball moves on at 0.80 m s⁻¹. Find the velocity of the 2.0 kg ball and determine whether the collision is elastic. [5]

11. Explain how an airbag reduces injury in a crash. [3]

12. A car of mass 900 kg travels over a hump-backed bridge of radius 30 m at 12 m s⁻¹. Find the normal reaction at the top. [3]

Answers

1. A 2. C 3. B 4. D 5. B 6. A 7. D 8. C

9. a = 2.5 m s⁻²; F = 1500(2.5) + 500 = 4250 N.

10. 12 = 2.4 + 2v: v = 4.8 m s⁻¹; KE before 24 J, after 0.96 + 23.0 = 24 J: elastic.

11. Increases stopping time; same Δp, smaller average force; spreads force over larger area.

12. mg − N = mv²/r: N = 900(9.81 − 4.8) = 4.5 kN.

11 Summary and topic map

Idea	Remember
Free-body diagrams	all forces on one object; resolve at angles
Newton I	no resultant force: constant velocity
Newton II	F = ma = Δp/Δt
Newton III	equal, opposite, different bodies
Contact forces	F ≤ μsN; F = kx; 6πηrv; ρVg
Momentum	conserved in closed systems; impulse = area under F–t
Circular motion	a = v²/r = ω²r; F = mv²/r towards centre

Before the exam, make sure you…

•draw a free-body diagram for every force problem
•use components for angled forces and 2D collisions
•remember kinetic energy is not conserved in inelastic collisions
•identify the real force providing centripetal force

Forces and Momentum on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 24. Topic map for Forces and Momentum.

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