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A.2 Forces and momentum · Part 1 of 3 · student investigation · 80 minutes

Forces: predict, test, explain

Make four predictions about everyday forces. Then test each one with a simulation on this page: push a heavy box, ride a lift on a scale, drop balls into thick oil, tilt a slope and pull blocks over a pulley. Part 2 takes these ideas into circular motion.

1 Predict · 7
2 Pairs of forces · 10
3 Friction lab · 12
4 Lift lab · 10
5 Oil lab · 12
6 Slope lab · 8
7 Pulley lab · 12
8 Show + reflect · 9

Learning goals: by the end you can…

draw free-body diagrams and name Newton’s third-law pairs.
explain how static friction grows up to μ_sF_N, and why sliding friction is smaller.
find the scale reading in an accelerating lift.
explain terminal speed using weight, buoyancy and drag, and use Stokes’ law.
decide when a block slips on a slope.
solve connected bodies over a pulley.

How to work in this lesson

How to work. Pairs discuss and agree, but each of you writes. Pairs → individual discuss the method, then write your own answer. Individual no discussion until your teacher says so. Your typed answers are saved on this device only.

Everything you need is on this page. Each simulation stays locked until you have made your prediction. Use g = 9.8 m s⁻². Extra help: the forces visual guide and the forces notes.

📘 From the data booklet (A.2)

F = maF_f ≤ μ_sF_NF_f = μ_dF_NF_d = 6πηrvF_b = ρVgF_g = mgF_H = −kxp = mvF = ∆p/∆tJ = F∆t

You do not need to memorize these: they are given in the exam. Your job is to choose the right one. Also in the booklet (maths section): volume of a sphere = ⁴⁄₃πr³.

2. Forces come in pairs

10 minPairs

ATL Thinking: tell apart two ideas that look alike. Communication: use exact words.

A book rests on a table. Two forces act on the book: its weight W (the pull of the Earth) and the normal force F_N (the push of the table).

2.1The two forces on the book are equal and opposite. Are they a Newton’s third-law pair?

Yes: they are equal and opposite No: they act on the same body

2.2What is the third-law partner of the weight of the book?

The normal force of the table on the book The push of the book on the table The gravitational pull of the book on the Earth

Simulation · Find the pairs click a force to see its third-law partner

force chosen—

acts on—

type—

its third-law partner—

2.3What is the third-law partner of F_N (the table pushing up on the book)?

Check your answer

The book pushes down on the table with a contact force of the same size. Same type (contact), different bodies (book and table).

2.4Find the mistake: “A horse pulls a cart. The cart pulls back on the horse with an equal force, so they cancel and the cart can never move.”

What a good answer looks like

The two forces act on different bodies (one on the horse, one on the cart), so they cannot cancel. Whether the cart accelerates depends only on the forces on the cart: the pull of the horse and the friction on the cart.

3. Friction lab: test prediction 1.1

12 minPairs → individual

ATL Research: collect data from a simulation. Thinking: find a rule from data.

A 20 kg box sits on a rough floor. Push it with the slider. Watch the friction force and the graph of friction against push.

Simulation A · Push the box m = 20 kg · μ_s = 0.50 · μ_d = 0.40

push P 0 N

Friction—

Net force—

Acceleration—

Box—

3.1Use the simulation. Fill in the friction force for each push.

Push / N	Friction / N	Moving?
40		no
80		no
120		yes

3.2Find the largest push that does not move the box. Use it to calculate μ_s. Largest push =

3.3Push with 120 N. Calculate the acceleration of the box, then compare with the simulation. a =

m s⁻²

Back check

3.bReverse 3.3: the box (20 kg) accelerates at 2.08 m s⁻² while sliding friction is 78.4 N. What push P gives this? P =

Find the mistake

A student wrote:

“The 20 kg box is at rest and μ_s = 0.50. I push it with 40 N. Friction = μ_smg = 0.50 × 20 × 9.8 = 98 N. Net force = 40 − 98 = −58 N, so the box accelerates backwards at 2.9 m s⁻².”

What is wrong?

The student should use μ_d = 0.40 instead of μ_s. Static friction is only as big as needed: here it is 40 N. μ_smg = 98 N is only its largest possible value. The student should use the mass (20 kg), not the weight, in the friction formula.

3.4Was prediction 1.1 right? Describe the shape of the friction–push graph and explain the “drop” when the box starts to slide.

What a good answer looks like

While the box is at rest, friction is equal to the push (a straight line through the origin, gradient 1): static friction is only as big as it needs to be, so the answer to 1.1 is B, 30 N. It can grow only up to μ_sF_N = 98 N. Once the box slides, friction is the dynamic value μ_dF_N = 78.4 N, which is smaller (μ_d < μ_s) and does not depend on the push.

4. Lift lab: test prediction 1.2

10 minPairs → individual

ATL Thinking: link the direction of the acceleration to the forces. Self-management: check signs.

A 60 kg student stands on a scale in a lift. The scale shows the normal force F_N. Choose a motion and an acceleration.

Simulation B · Ride the lift m = 60 kg

size of the acceleration a 1.5 m s⁻²

Acceleration (up +)—

Scale reading F_N—

Scale shows (in kg)—

4.1Was prediction 1.2 right? Which matters for the scale reading: the direction of the velocity or the direction of the acceleration?

What a good answer looks like

The answer to 1.2 is A: speeding up while going up means the acceleration is upwards, so F_N − mg = ma and F_N > mg. Only the acceleration matters: "moving down and slowing down" gives the same high reading as "moving up and speeding up". At constant velocity the reading equals the weight.

4.2The lift moves up and speeds up at 1.5 m s⁻². Scale reading =

Back check

4.bReverse 4.2: the scale reads 678 N for the 60 kg student. Find the acceleration of the lift from this reading. a =

m s⁻²

4.3The lift moves down and slows down at 2.0 m s⁻². Scale reading =

4.4The cable breaks and the lift falls freely. Scale reading =

Try it for real (optional) · phyphox. The free phyphox app (iOS and Android) has an Elevator experiment. Put the phone flat on the floor of a real lift, start the recording and ride a few floors. Compare the acceleration graph with your answers above: when is a positive, zero and negative?

5. Oil lab: test prediction 1.3

12 minPairs → individual

ATL Research: use a model and real data. Thinking: explain a graph with forces.

A sphere is released from rest in glycerol (density 1260 kg m⁻³, viscosity η = 1.4 Pa s). Three forces act: weight (down), buoyancy (up) and drag (up, growing with speed).

Hinge question

Choose your answer. You will see at once if it is right, and why.

A box slides across a rough floor at constant velocity. You push it horizontally with 50 N. What is the friction force on the box?

A. more than 50 N, because friction must hold the box back B. exactly 50 N C. less than 50 N, because the push must win to keep the box moving D. zero, because the box is moving

Simulation C · Drop a sphere into glycerol played in slow motion

radius r 5.0 mm

Speed now—

Weight—

Buoyancy—

Drag now—

Terminal speed—

5.1Was prediction 1.3 right? Use the simulation: what happens to the terminal speed when you double the radius?

What a good answer looks like

The big ball arrives first (answer B). Doubling r makes the terminal speed 4 times bigger: v_t ∝ r². The weight and buoyancy grow with r³ (volume), but the drag 6πηrv grows only with r, so a bigger sphere must go faster before the drag can balance it.

5.2At terminal speed, which equation is true?

weight = drag weight = buoyancy + drag weight + buoyancy = drag

5.3Choose from the data booklet. For a steel sphere (7800 kg m⁻³) of radius 5.0 mm, calculate the buoyancy force, in millinewtons. F_b =

5.4Calculate the terminal speed of this sphere, then check it with the simulation. v_t =

m s⁻¹

Back check

5.bCheck 5.4 with forces: at v = 0.254 m s⁻¹, the drag on the 5.0 mm steel sphere is 6πηrv =

Sample data · A class measured the terminal speed of steel balls in glycerol (these are sample results for you to analyse).

r / mm	v_t / cm s⁻¹	v_t/r² / cm s⁻¹ mm⁻²
1.0	1.02	1.02
1.5	2.28	1.01
2.0	4.10
2.5	6.32	1.01

5.6The last column is nearly constant. What does this tell you? Then use the 2.0 mm row to find the viscosity η of the glycerol. η =

Pa s

6. Slope lab: test prediction 1.4

8 minIndividual

ATL Thinking: compare two effects that grow together. Self-management: test before you conclude.

Simulation D · Tilt the board μ_s = 0.40

angle θ 15°mass m 5.0 kg

mg sin θ (down the slope)—

F_N = mg cos θ—

Largest friction μ_sF_N—

6.1A 5.0 kg block on a 30° slope: component of the weight down the slope =

6.2Find from the simulation the angle at which the block starts to slip. Then check: tan θ = μ_s. θ =

Back check

6.bCheck 6.2 at the slipping angle 21.8° for the 5.0 kg block: the largest friction μ_smg cos θ =

6.3Was prediction 1.4 right? Explain using the equation.

What a good answer looks like

Same angle (B). Doubling m doubles the force down the slope (mg sin θ) and the largest friction (μ_smg cos θ). In mg sin θ = μ_smg cos θ the mass cancels: tan θ = μ_s.

7. Pulley lab: connected bodies

12 minPairs → individual

ATL Thinking: plan a two-equation solution. Self-management: check with the simulation.

7.1Predict first: a block on a table is pulled over a pulley by a hanging mass m. While they accelerate, the tension in the string is…

A. more than mg B. equal to mg C. less than mg

Simulation E · Pulley set-ups

block on the table M 3.0 kghanging mass m 1.0 kgfriction μ_d on the table 0.00

Acceleration—

Tension T—

Weight of m (mg)—

📘 Choose your equations

F = ma (for each body)F_f = μ_dF_Nweight = mg

Write F = ma for each body separately, then add the equations: the tension cancels.

7.2Smooth table, M = 3.0 kg, m = 1.0 kg. Acceleration a = m s⁻², tension T = N

Back check

7.bCheck 7.2 with the other body: use F = ma for the hanging 1.0 kg mass alone, with T = 7.35 N. a =

m s⁻²

7.3Same masses, but now μ_d = 0.20 on the table. a =

m s⁻²

7.4Two hanging masses, 3.0 kg and 2.0 kg, over a smooth pulley. Tension =

7.5Was prediction 7.1 right? Explain why the tension cannot equal the weight of the hanging mass while it accelerates.

What a good answer looks like

Answer C. The hanging mass accelerates downwards, so the resultant force on it is down: mg − T = ma > 0, so T < mg. If T were equal to mg, the hanging mass would have no resultant force and could not accelerate.

What if? · stretch

7.wA 4.0 kg block sits on a smooth slope at 30° to the horizontal. A string runs from it up over a pulley at the top of the slope to a hanging 3.0 kg mass. Acceleration =

m s⁻²

8. Show it, then reflect

9 minIndividual

ATL Communication + Self-management: choose how to show your understanding, retrieve without notes, plan your next step.

Success criteria: your answer must…

use a free-body diagram for one body;
say when friction is static (≤ μ_sF_N) and when it is dynamic;
link the resultant force to the acceleration (zero for equilibrium and terminal speed);
use one number from a simulation today.

Option A · Written

One paragraph: "Which of my four predictions was wrong, and what did the simulation teach me?"

Option B · Sketch

Free-body diagrams for the sphere in oil at the start and at terminal speed, with arrows to scale.

Option C · Spoken

Explain the lift lab to your partner in 1 minute. Your partner ticks the success criteria.

8.1Write your paragraph or your notes.

What a good answer looks like

Model answer (Option A): I predicted that friction on a box at rest is zero, but the simulation showed friction equal to the push (30 N for a 30 N push). Static friction adjusts to keep the resultant force zero, up to its largest value μ_sF_N = 98 N for the 20 kg box. Above that the box slides and friction drops to μ_dF_N = 78.4 N, so with a 120 N push the resultant force is 41.6 N and a = 2.08 m s⁻².

Exit ticket (no notes)

1A 10 kg box on a floor, μ_s = 0.60. Largest friction before it moves, in N =

2A 50 kg student in a lift that accelerates downwards at 2.0 m s⁻². Scale reading in N =

3At terminal speed, the resultant force on a falling sphere is…

zero equal to the weight equal to the drag

4Masses of 1.0 kg and 1.5 kg hang over a smooth pulley. Acceleration in m s⁻² =

5The third-law partner of the weight of a book is…

the normal force on the book the push of the book on the table the pull of the book on the Earth

6Which idea from today is still the least clear to you?

What a good answer looks like

There is no single right answer. A good answer names one idea exactly, for example: "I'm not sure when friction is μ_sF_N and when it is smaller."

7Feedforward: Next time I will check ________ before I ________.

What a good answer looks like

A good feedforward names an action: "Next time I will draw a free-body diagram for each body before I write F = ma."

TOK · 2 minutes

Theory of knowledge: how do we know? There is no single right answer. Write two or three sentences.

TOKF = μF_N for friction is not a law of nature like F = ma: it is an approximate model that works only for some surfaces. What is the difference between a law and a model, and does it matter?

Points you could make

A law is very general and has been tested in a huge range of situations; a model describes a particular system and works only in a limited range.
Both are accepted because experiments agree with them, and both can be revised.
It matters because we must know the limits of a model before we rely on its predictions.

Go to Part 2: circular motion Read the forces notes