Forces, accelerations and Newton's Laws of motion & RP7

SPEC 4.5.6.2 • TOPIC 5 FORCES — PAPER 2 — RP7

Newton's Laws & RP7

Three laws of motion, one equation to recall (F = m a), and the required practical that proves the relationship. The keystone of mechanics.

What You'll Learn

Newton's three laws are the foundation of mechanics — they connect forces to changes in motion. F = m a is the most-used equation in Paper 2.

Lesson Goals

State and apply Newton's First, Second, and Third Laws

Apply F = m a (recall) — directly and in chained calculations

Define inertia and inertial mass HT

Describe RP7: investigating how force and mass affect acceleration ⚗ RP7

Identify when Newton's 1st Law applies (zero resultant force)

Newton's 1st Law

If the resultant force on an object is zero, its velocity doesn't change. Stationary stays stationary; moving stays at the same velocity.

If the resultant force on an object is zero:

A stationary object stays stationary

A moving object continues at the same speed AND in the same direction (constant velocity)

So: the velocity of an object will only change if a resultant force is acting on it.

The big consequence

If you see something moving at constant velocity (constant speed in a straight line), you can immediately conclude the resultant force on it is zero. The driving force balances the resistive forces.

Newton's 1st Law — In Action

Spec point: apply the law to objects with uniform velocity AND objects with changing speed/direction.

Example 1 — vehicle at steady speed

A car travelling at a constant 30 m/s along a straight road has zero resultant force. The forward driving force from the engine equals the backward resistive forces (friction + air resistance). They cancel out.

Example 2 — object on a frictionless surface

An ice puck sliding on smooth ice keeps moving at constant velocity until it hits the boards. With friction nearly zero, the resultant force is approximately zero — so velocity is approximately constant.

Example 3 — change in motion → resultant force exists

If a car speeds up, slows down, OR turns a corner (direction changes), there must be a resultant force. Velocity is a vector — changing direction at constant speed still counts as a change in velocity.

Newton's 2nd Law — F = m a

When a resultant force DOES act, it produces an acceleration in the direction of the force. The 2nd law tells you exactly how big.

The acceleration of an object is:

Proportional to the resultant force (a ∝ F)

Inversely proportional to the mass (a ∝ 1/m)

Resultant force = mass × accelerationF = m × aF — resultant force, Nm — mass, kga — acceleration, m/s²

F is the RESULTANT force

F in F = m a is always the resultant (net) force — the single force you get after combining all forces acting on the object. If a 100 N forward force is opposed by 30 N of friction, F = 70 N, not 100 N. Mark schemes are strict on this.

Symbol recognition

The spec asks you to recognise two symbols used in this topic: ∝ means "is proportional to" (e.g., a ∝ F). ~ (or ≈) means "approximately equal to" — used for estimates.

ESUA Worked Example — Direct F = m a

A box of mass 5.0 kg is pushed across a smooth surface. The box accelerates at 4.0 m/s².

Calculate the resultant force on the box. 3 MARKS

E

Equation: F = m × a

S

Substitute: F = 5.0 × 4.0

U

Unit: newtons (N)

A

Answer: F = 20 N

ESUA — Chained Calculation

A car has a mass of 1500 kg. The car accelerates uniformly from rest to a velocity of 24 m/s in a time of 8.0 s.

Calculate the resultant force acting on the car. 5 MARKS

Step 1

Find a using a = Δv ÷ t:

a = (24 − 0) ÷ 8.0 = 3.0 m/s² (M1, M1)

Step 2

Apply F = m × a:

F = 1500 × 3.0 (M1)

U

Unit: newtons (N)

A

Answer: F = 4500 N (A1, plus M1 for chain reasoning)

Pattern to recognise

Many F=ma questions give you mass, initial/final velocity, and time — NOT acceleration directly. Spot the pattern: a from Δv/t first, then F = m a. Chained calculations are the most common 4–5 mark Paper 2 questions in this topic.

Estimating road-transport forces (spec skill)

The spec asks you to estimate the speed, accelerations, and forces involved in large accelerations for everyday road transport. Use these typical values:

Family car mass: ~1000–1500 kg

Brisk acceleration: ~3 m/s² → resultant force ~3000–5000 N

Hard braking deceleration: ~7–10 m/s² → braking force ~7000–15,000 N on a typical car

Lorry mass: ~10,000+ kg → same acceleration needs ~10× more force

Newton's 3rd Law

Forces come in pairs. When two objects interact, the forces they exert on each other are equal and opposite.

Whenever two objects interact:

The forces they exert on each other are equal in size and opposite in direction.

Forces act on DIFFERENT bodies

The two forces in a Newton's 3rd-Law pair act on different objects. They never cancel out (cancellation requires forces on the same body). To get full marks, your answer must name both bodies: "Object A pushes B; object B pushes A back equally." A one-sided statement scores zero.

Newton's 3rd Law — Examples

InteractionForce 1Force 2 (equal and opposite)

Person standing on floorPerson pushes down on floorFloor pushes up on person

Rocket launchRocket pushes exhaust gas downExhaust gas pushes rocket up

Hitting a wallHand pushes wallWall pushes hand back equally

Inertia & Inertial Mass HT only

These terms appear in HT papers and are tightly defined by the spec. Match the wording carefully.

Inertia

The tendency of objects to continue in their state of rest or of uniform motion. (Spec wording — HT only.)

In plain English: an object "wants" to keep doing what it's doing — to stay still if it's still, or to keep moving at the same velocity if it's moving.

Inertial Mass

Inertial mass is a measure of how difficult it is to change the velocity of an object. It is defined as the ratio of force over acceleration:

m = F ÷ a

This is the m in F = m a — same quantity, defined operationally as the force required to produce a unit acceleration.

Worked link

A larger inertial mass means a smaller acceleration for the same force. If 100 N gives a 50 kg object an acceleration of 2 m/s² (m = 100/2 = 50 kg), the same 100 N applied to a 200 kg object gives only 0.5 m/s² — four times harder to accelerate.

RP7 — Apparatus ⚗ Required Practical 7

Investigate the effect of (1) varying force on acceleration of a constant-mass object and (2) varying mass on acceleration with a constant force.

What each part does

Hanger + masses: provides the accelerating force, F = m_hanger × g

Trolley: the object whose acceleration you're measuring

Pulley: redirects the string's tension into a horizontal force on the trolley

Two light gates: measure the trolley's velocity at two points; acceleration calculated from the change in velocity over the time between gates

Alternative: tickertape attached to the trolley through a timer — gives velocity from dot spacing

RP7 Part 1 — Vary Force, Constant Mass

Test the relationship between resultant force and acceleration, with mass held constant.

VariableDetail

IndependentForce pulling the trolley — varied by changing the mass on the hanger (force = mass on hanger × g)

DependentAcceleration of the trolley — calculated from light-gate (or tickertape) data

ControlsSame trolley and surface (constant friction); total moving mass kept constant ; trolley starts from rest each run

Critical control — TRANSFER masses, don't just add

When you increase the hanger mass, you increase the moving system's total mass — unless you compensate. The trick: take a mass off the trolley and put it on the hanger. The force changes, but the total moving mass stays the same. Without this control, you're varying TWO things at once and can't isolate the F → a relationship.

Method (5 steps)

Place trolley with masses on runway; attach string and hang one mass on hanger. Record this mass

Release trolley from rest. Use light gates (or tickertape) to find the trolley's acceleration

Repeat each force value at least 3 times; take a mean

Transfer a mass from the trolley to the hanger (force increases, total mass unchanged); repeat

Plot acceleration (y) against force (x)

RP7 Part 2 — Vary Mass, Constant Force

Test the relationship between mass and acceleration, with force held constant.

VariableDetail

IndependentTotal mass being accelerated — varied by adding masses ONTO the trolley

DependentAcceleration of the trolley — same measurement method as Part 1

ControlsSame mass on the hanger throughout (constant force); same surface; starts from rest

Method (5 steps)

Fix one mass on the hanger and keep it there for ALL runs in Part 2

Release the trolley from rest with no extra masses on top. Measure acceleration

Add a known mass to the top of the trolley. Measure acceleration again

Repeat for at least 5 different total trolley masses

Plot acceleration (y) against mass (x) — expect a curve. Or plot a against 1/m for a straight-line graph

RP7 — Expected Results

The two graphs confirm the two halves of Newton's 2nd Law.

Conclusion (what to write)

Part 1 result: a is proportional to F (straight line through origin)

Part 2 result: a is inversely proportional to m (curve; or straight line through origin when plotted against 1/m)

Combined: a = F ÷ m, which is F = m a rearranged. Newton's 2nd Law confirmed.

Exam-Style Multi-Part Question

A car has a mass of 1200 kg. The driver applies a constant resultant force of 3600 N forwards.

(a) Calculate the acceleration of the car. 3 MARKS

(b) The car starts from rest. Calculate the velocity of the car after 4.0 s. 3 MARKS

(c) An additional passenger of mass 80 kg gets into the car. The resultant force stays at 3600 N. State and explain how the car's acceleration changes. 3 MARKS

(a)

Equation: F = m a → a = F ÷ m (M1)

Substitute: a = 3600 ÷ 1200 (M1)

Answer: a = 3.0 m/s² (A1)

(b)

Equation: a = Δv ÷ t → Δv = a × t (M1)

Substitute: Δv = 3.0 × 4.0 = 12 m/s (M1)

Answer: v = 0 + 12 = 12 m/s (starts from rest) (A1)

(c)

Mass increases from 1200 kg to 1280 kg (M1). By F = m a (or "a is inversely proportional to m"), the acceleration decreases (M1). New a = 3600 ÷ 1280 ≈ 2.8 m/s² — explicitly comparing to 3.0 m/s² shown earlier. (A1)

Examiner Alerts & Common Errors

Classic exam trap — runway angle question

A common Paper 2 question asks "explain why a trolley on a raised runway accelerates more than on a flat runway". Weak answers say "because of gravity" — true but vague, low marks. Full-mark answer: a component of weight now acts along the runway, which adds to the resultant force; by F = m a, the acceleration increases proportionally.

Two answers to "Explain how raising one end of the RP7 runway changes the trolley's acceleration."

✗ Weak

"The acceleration increases because of gravity."Vague; no mechanism, no equation linkage. 1/3 at best.

✓ Strong

"Raising the runway means a component of the trolley's weight now acts along the slope. This adds to the resultant force on the trolley. By F = m a, since F has increased and m is unchanged, the acceleration increases."Mechanism + equation + comparison. 3/3.

Common errorWhy it costs marks

Uses F as one force, not the resultantF = m a always uses RESULTANT force

Confuses F = m a with W = m gDifferent physics; F = resultant force on object, W = weight (a specific kind of force from gravity)

Newton's 3rd Law stated one-sidedly (only one body named)Mark scheme requires BOTH bodies named

(RP7) Just adds masses to the hanger to vary forceTotal moving mass increases too — not a fair test. Must transfer masses from trolley to hanger

"Inertia is the mass of an object"Inertia is the tendency to resist changes in motion — not the same as mass (HT only)

Checkpoint

You can now

State all three of Newton's laws with correct conditions

Apply F = m a — both directly and in chained calculations with Δv/t

Identify when the resultant force is zero (constant velocity → Newton's 1st)

Name BOTH bodies when describing a Newton's 3rd-Law pair

Define inertia and inertial mass (HT)

Describe RP7 — apparatus, both investigations, the critical mass-transfer control, and expected results

The two habits

1. "F is the resultant force" — say it in your head every time you write F = m a. 2. For chained questions (mass + Δv + t given): get a from Δv/t first, then F from F = m a. Pattern recognition turns 5-mark questions into routine.

Next lesson

5.11 — Forces and braking. Stopping distance = thinking + braking. Reaction times, factors that affect each, and the v² dependence of braking distance.