Newton's Laws of Motion are the foundation of classical mechanics. In Cambridge A-Level Mechanics 1, you will apply all three laws to solve problems involving single particles, particles on slopes, and connected particle systems. Every dynamics problem in M1 comes back to F = ma.
F = ma | N = mgcosθ | Friction: F = μN | Atwood: a = (m₁−m₂)g / (m₁+m₂)
Learning Objectives
State Newton's First, Second, and Third Laws of Motion
Identify Newton's Third Law action-reaction pairs correctly
Set up and solve F = ma equations for a single particle
Apply F = ma with friction: F = μN opposing motion
Resolve forces on inclined planes and apply Newton's Second Law
Analyse two connected particles on a horizontal surface
Solve Atwood machine problems with light inextensible strings
Solve problems involving a particle on a slope connected over a pulley to a hanging particle
Determine when a string goes slack (tension = 0)
Solve lift/elevator problems using apparent weight
Newton's Three Laws
Inertia, F=ma, action-reaction pairs
F=ma Applications
Single particle, net force, friction on horizontal surfaces
Inclined Planes
mgsinθ down slope, N=mgcosθ, friction on slopes
Connected Strings
Two particles, same acceleration, tension transmitted
Pulley Systems
Atwood machine, light inextensible string over smooth pulley
Worked Examples
8 fully worked with mark-scheme steps
Common Mistakes
8 errors that regularly cost marks in exams
Exercises + Practice
70+ self-marking questions across all topics
Learn 1 — Newton's Three Laws of Motion
Newton's First Law — The Law of Inertia
A particle remains at rest, or continues to move in a straight line with constant velocity, unless acted upon by a resultant external force.
ΣF = 0 ⟺ object at rest OR moving at constant velocity
What this means in practice:
• If a book sits still on a table, the resultant force on it is zero (weight balanced by normal reaction).
• If a car cruises at a steady 60 km/h, the driving force equals the resistance — resultant is zero.
• A particle moving at constant velocity is NOT necessarily unforced — the forces just balance.
• The First Law defines what we mean by "no net force" and links to the concept of equilibrium.
Newton's Second Law — F = ma
The resultant force acting on a particle is equal to the product of its mass and its acceleration. The acceleration is in the same direction as the resultant force.
F = ma (resultant force = mass × acceleration)
Key points:
• F is the net (resultant) force — the vector sum of ALL forces acting on the particle.
• m is in kilograms (kg), a is in m s⁻², and F is in Newtons (N). Always check units.
• If multiple forces act: net F = ΣF = F₁ + F₂ + … then apply F = ma.
• The direction of acceleration is always the same as the direction of the net force.
Rearrangements: a = F/m | m = F/a
Example: A 5 kg box is pushed with 30 N. Friction = 10 N opposes motion.
Net force = 30 − 10 = 20 N
Acceleration a = F/m = 20/5 = 4 m s⁻²
Newton's Third Law — Equal and Opposite Reaction Pairs
When body A exerts a force on body B, body B exerts an equal and opposite force on body A. These forces are equal in magnitude, opposite in direction, and act on different objects.
Force of A on B = −(Force of B on A)
Classic examples of action-reaction pairs:
• Earth pulls a book downward (weight) → the book pulls Earth upward with equal force.
• Floor pushes up on a person (normal reaction) → person pushes down on floor with equal force.
• String pulls mass upward (tension) → mass pulls string downward with equal force.
Critical: Action-reaction pairs ALWAYS act on DIFFERENT objects. They can never cancel each other out in one equation of motion.
Distinguishing Action-Reaction Pairs
A common confusion: the weight W of a box on a table and the normal reaction N from the table on the box are NOT Newton's Third Law pairs — they act on the SAME object (the box). They balance because the box is in equilibrium (Newton's First Law).
The true Newton's Third Law pairs here are:
(1) Earth pulls box down with W ↔ box pulls Earth up with W
(2) Table pushes box up with N ↔ box pushes table down with N
Summary Table
Law
Statement
Mathematical Form
First (Inertia)
No net force → constant velocity (or rest)
ΣF = 0 ⟺ a = 0
Second (F=ma)
Net force causes acceleration
ΣF = ma
Third (Pairs)
Forces come in equal-opposite pairs on different objects
F_AB = −F_BA
Always state which object you are applying F = ma to. The equation only works for one object at a time — you must include ALL forces on that one object and no forces on other objects.
Learn 2 — F = ma Applications
Method: Setting Up the Equation of Motion
Step 1: Draw a clear force diagram showing every force on the particle. Step 2: Choose a positive direction (usually the direction of motion or expected acceleration). Step 3: Write ΣF = ma, listing all forces with correct signs. Step 4: Solve for the unknown (acceleration, tension, or applied force).
Single Particle on Horizontal Surface (No Friction)
Forces: Applied force P (→ positive), weight mg (↓), normal reaction N (↑).
Vertical: N − mg = 0 → N = mg (no vertical acceleration).
Horizontal: P = ma → a = P/m.
Example: m = 4 kg, P = 20 N. a = 20/4 = 5 m s⁻².
Friction: F = μN
On a rough surface, friction opposes the motion. The friction force is:
F_friction = μN (at maximum / limiting friction during motion)
μ is the coefficient of (kinetic) friction — a dimensionless number between 0 and 1 typically.
N is the normal reaction — found from the vertical equation first.
On a horizontal surface: N = mg, so F_friction = μmg.
Example: m = 3 kg, P = 18 N, μ = 0.4, g = 10 m s⁻².
N = mg = 30 N. Friction = μN = 0.4 × 30 = 12 N.
Net force = 18 − 12 = 6 N. Acceleration = 6/3 = 2 m s⁻².
Multiple Forces — Net Force = ΣF
When several horizontal forces act, sum them (with signs) to get the net force, then apply F = ma.
Example: Forces 25 N (→), 10 N (←), friction 8 N (←), mass = 7 kg.
Net force = 25 − 10 − 8 = 7 N (→).
a = 7/7 = 1 m s⁻² to the right.
Finding Unknown Applied Force
If acceleration is given and you need to find P:
Rearrange ΣF = ma → P = ma + friction (or P = ma + any resisting forces).
Example: m = 6 kg, a = 3 m s⁻², μ = 0.25, g = 10 m s⁻².
Friction = μmg = 0.25 × 6 × 10 = 15 N.
P − 15 = 6 × 3 = 18. P = 33 N.
Lift (Elevator) Problems
When a particle of mass m is inside a lift with acceleration a:
R − mg = ma (lift accelerating upward, R is normal/apparent weight) R = m(g + a)
Lift accelerating upward: apparent weight R = m(g + a) — heavier than normal.
Lift accelerating downward: R = m(g − a) — lighter than normal.
Lift in free fall (a = g): R = 0 — weightlessness.
Example: 70 kg person in lift accelerating up at 2 m s⁻². g = 10 m s⁻².
R = 70(10 + 2) = 70 × 12 = 840 N. (Normal weight would be 700 N.)
Always resolve vertically first to find N, then use N to find friction, then apply F = ma horizontally. Never substitute mg for N on a surface where there is also a vertical component of applied force.
Learn 3 — Forces on Inclined Planes
Resolving Weight on a Slope at Angle θ
Choose axes parallel and perpendicular to the slope. Weight mg acts vertically downward. Resolve into two components:
Component down the slope = mg sinθ | Component perpendicular into slope = mg cosθ
Perpendicular to slope (no acceleration in this direction):
N − mg cosθ = 0 → N = mg cosθ
Along slope (taking down-slope as positive if particle slides down):
mg sinθ − T − F_friction = ma
Check at extremes: θ = 0° → sin0° = 0 (no sliding force, makes sense on flat). θ = 90° → sin90° = mg (particle in free fall, also makes sense).
Smooth Inclined Plane (No Friction)
Forces on particle: weight mg downward, normal reaction N perpendicular to slope, tension T up the slope (if string attached).
Along slope (up = positive): T − mg sinθ = ma
Perpendicular: N = mg cosθ
Example: m = 3 kg, θ = 30°, T = 0 (free slide), g = 10 m s⁻².
Along slope: −mg sin30° = ma → a = −g sin30° = −10 × 0.5 = −5 m s⁻²
(acceleration is 5 m s⁻² down the slope)
Rough Inclined Plane — Particle Moving Down
Friction acts UP the slope (opposing downward motion).
Along slope (down = positive): mg sinθ − μN = ma
Perpendicular: N = mg cosθ
Substitute: mg sinθ − μmg cosθ = ma → a = g(sinθ − μcosθ)
Example: m = 5 kg, θ = 40°, μ = 0.3, g = 10 m s⁻².
N = 5 × 10 × cos40° = 38.3 N. Friction = 0.3 × 38.3 = 11.5 N.
Net force down slope = 5 × 10 × sin40° − 11.5 = 32.1 − 11.5 = 20.6 N.
a = 20.6/5 = 4.12 m s⁻² down the slope.
Rough Inclined Plane — Particle Moving Up
Friction acts DOWN the slope (opposing upward motion).
Along slope (up = positive): T − mg sinθ − μN = ma
Perpendicular: N = mg cosθ
Net force up = T − mgsinθ − μmgcosθ = ma
The sign of friction changes depending on the direction of motion — always draw it opposing the motion.
Finding the Coefficient of Friction on a Slope
At limiting equilibrium (about to slide, acceleration = 0):
Along slope: mg sinθ = μN = μmg cosθ
Divide both sides by mg cosθ: tanθ = μ μ = tanθ at limiting equilibrium on a slope — a clean, elegant result.
Example: Object in limiting equilibrium on 28° slope. μ = tan28° ≈ 0.532.
Slope with Applied Force and Tension
General equation along slope (taking up-slope as positive, particle moving up):
T + P cosα − mg sinθ − μmg cosθ = ma
where T is tension in a string and P is an applied force at angle α to the slope.
Always identify ALL forces and their directions carefully. Draw the force diagram first.
On inclined plane problems, ALWAYS resolve perpendicular to slope first to find N, then use N = mgcosθ to find friction, then write the along-slope equation. Never skip the perpendicular step.
Learn 4 — Connected Particles (Strings on Horizontal Surface)
The Setup
Two particles A and B are connected by a light inextensible string on a smooth (or rough) horizontal surface. A force P is applied to one particle. They move together.
Key assumptions (light inextensible string):
• String has negligible mass (light) → tension is the same throughout the string.
• String does not stretch (inextensible) → both particles have the same acceleration a.
• The string can only pull (tension T ≥ 0). If T = 0, the string is slack.
Method: Treat Each Particle Separately
Step 1: Draw force diagrams for EACH particle separately. Step 2: Apply F = ma to each particle individually. Step 3: The acceleration a is the same for both particles. Step 4: The tension T appears in both equations (equal magnitude, opposite direction on each particle). Step 5: Solve simultaneously — usually add equations to find a, then substitute back for T.
Example: Smooth Horizontal Surface
Particle A (mass m₁ = 3 kg) connected by string to particle B (mass m₂ = 5 kg). Force P = 24 N applied to B. Smooth surface, g = 10 m s⁻².
For particle A (tension T pulls it forward):
T = m₁a = 3a ... (1)
For particle B (P forward, T backward via Newton's 3rd Law):
P − T = m₂a → 24 − T = 5a ... (2)
Add (1) and (2): 24 = 8a → a = 3 m s⁻²
From (1): T = 3 × 3 = 9 N
Alternative: System Method for Acceleration
For finding a only: treat the whole system as one particle of mass (m₁ + m₂).
Net external force = P (internal tension cancels out).
a = P / (m₁ + m₂) = 24 / (3 + 5) = 24/8 = 3 m s⁻²
Warning: This only gives a. To find T, you MUST use a single-particle equation for one of the particles.
Rough Horizontal Surface
Each particle has its own friction force (based on its own normal reaction = its own weight).
Friction on A = μm₁g; Friction on B = μm₂g.
For A: T − μm₁g = m₁a
For B: P − T − μm₂g = m₂a
Add: P − μ(m₁ + m₂)g = (m₁ + m₂)a → a = [P − μ(m₁+m₂)g] / (m₁+m₂)
When the String Goes Slack
The string becomes slack when the tension T = 0. This happens when the front particle no longer needs to pull the rear particle — e.g. if the applied force is removed or the front particle decelerates faster than the rear one.
If you solve for T and get a negative value, re-examine your setup — a string cannot push, so either the motion is different from assumed or the string is slack.
Always apply F = ma separately to each particle. The mistake of using the total mass to find tension (instead of just acceleration) is one of the most common errors in connected particle problems.
Learn 5 — Connected Particles (Pulleys)
The Atwood Machine
Two particles of masses m₁ and m₂ (m₁ > m₂) are connected by a light inextensible string passing over a smooth fixed pulley. The heavier particle descends, the lighter ascends.
Acceleration: a = (m₁ − m₂)g / (m₁ + m₂) | Tension: T = 2m₁m₂g / (m₁ + m₂)
Key assumptions:
• String is light and inextensible → same tension T throughout, same magnitude of acceleration a.
• Pulley is smooth (frictionless) → tension is the same on both sides.
• One particle goes down, the other goes up — choose a consistent positive direction for each.
Sign convention: Define positive as "direction of motion for each particle." For m₁ descending: down is positive for m₁, up is positive for m₂.
Deriving the Equations
For m₁ (descending, downward = positive):
m₁g − T = m₁a ... (1)
For m₂ (ascending, upward = positive):
T − m₂g = m₂a ... (2)
Add (1) and (2): m₁g − m₂g = (m₁ + m₂)a a = (m₁ − m₂)g / (m₁ + m₂)
From (2): T = m₂(g + a) = m₂g + m₂a. Substituting a: T = 2m₁m₂g / (m₁ + m₂)
Particle on Slope Connected Over Pulley to Hanging Particle
Particle A (mass m₁) on smooth slope at angle θ, connected by string over a pulley at the top to hanging particle B (mass m₂).
For A (along slope, up = positive if B heavier):
T − m₁g sinθ = m₁a ... (1)
For B (downward = positive):
m₂g − T = m₂a ... (2)
If the slope is rough (μ) and A moves up the slope (B descends):
Friction on A acts DOWN the slope (opposing upward motion of A).
Normal reaction on A: N = m₁g cosθ. Friction = μm₁g cosθ.
For A (up slope = positive): T − m₁g sinθ − μm₁g cosθ = m₁a
For B (down = positive): m₂g − T = m₂a
Check 1: If a comes out negative, the assumed direction of motion is wrong — the other particle descends. Check 2: If a = 0, the system is in equilibrium — check limiting friction or tension conditions. Check 3: If T comes out negative, something is wrong — re-examine the setup. Check 4: After finding a and T, substitute back into one original equation to verify.
Never use the total mass (m₁ + m₂) to find tension. The tension is found by applying F = ma to ONE particle alone, using the acceleration already found. Using total mass gives acceleration — it cannot give tension.
Worked Examples
8 fully worked examples covering all key Newton's Laws topics. Study each step carefully.
Example 1 — Single Particle Horizontal (Rough Surface)
A particle of mass 8 kg is pulled along a rough horizontal surface by a horizontal force of 40 N. The coefficient of friction is 0.3. Find the acceleration. (g = 10 m s⁻²)
Step 1 — Normal reaction: No vertical acceleration → N = mg = 8 × 10 = 80 N. B1
Step 2 — Friction: F = μN = 0.3 × 80 = 24 N (opposing motion). M1
Step 3 — Net force: ΣF = 40 − 24 = 16 N. M1
Step 4 — F = ma: 16 = 8a → a = 2 m s⁻²A1
Example 2 — Particle on a Smooth Slope
A particle of mass 5 kg slides down a smooth slope inclined at 35° to the horizontal. Find its acceleration. (g = 9.8 m s⁻²)
Step 1 — Force diagram: Weight 49 N vertically down, Normal N perpendicular to slope. No friction (smooth). B1
Step 2 — Perpendicular to slope: N = mg cos35° = 49 × 0.8192 ≈ 40.1 N. M1
Step 3 — Along slope (down = positive): mg sin35° = ma → a = g sin35° = 9.8 × 0.5736 ≈ 5.62 m s⁻²M1 A1
Example 3 — Rough Slope — Finding μ
A 4 kg block moves at constant speed down a rough slope at 25° to the horizontal. Find the coefficient of friction μ. (g = 10 m s⁻²)
Step 1 — Constant speed → a = 0 → ΣF = 0.B1
Step 2 — Perpendicular: N = mg cos25° = 40 × 0.9063 = 36.25 N. M1 A1
Particles A (4 kg) and B (6 kg) are connected by a string. A force of 30 N is applied to B on a smooth horizontal surface. Find a and T.
Step 1 — System acceleration: Total mass = 10 kg. Net force = 30 N. a = 30/10 = 3 m s⁻²M1 A1
Step 2 — Apply F = ma to particle A alone: T = m_A × a = 4 × 3 = 12 NM1 A1
Check with B: 30 − T = 6 × 3 → 30 − 12 = 18 = 18 ✓ B1
Example 5 — Atwood Machine
Masses m₁ = 7 kg and m₂ = 3 kg hang over a smooth pulley connected by a light inextensible string. Find the acceleration and tension. (g = 10 m s⁻²)
Step 1 — For m₁ (down = positive): m₁g − T = m₁a → 70 − T = 7a ...(1) M1
Step 2 — For m₂ (up = positive): T − m₂g = m₂a → T − 30 = 3a ...(2) M1
Step 3 — Add (1)+(2): 40 = 10a → a = 4 m s⁻²A1
Step 4 — From (2): T = 30 + 3 × 4 = 30 + 12 = 42 NA1
Formula check: a = (7−3)×10/(7+3) = 40/10 = 4 ✓ T = 2×7×3×10/(7+3) = 420/10 = 42 ✓ B1
Example 6 — Particle on Slope with Hanging Mass
Particle A (mass 5 kg) rests on a smooth slope at 30° connected by a string over a pulley to hanging particle B (mass 4 kg). Find a and T. (g = 10 m s⁻²)
Step 1 — For A (up slope = positive, driven by B): T − 5g sin30° = 5a → T − 25 = 5a ...(1) M1
Step 2 — For B (down = positive): 4g − T = 4a → 40 − T = 4a ...(2) M1
Step 3 — Add: 15 = 9a → a = 5/3 ≈ 1.67 m s⁻²A1
Step 4 — From (2): T = 40 − 4 × 5/3 = 40 − 20/3 ≈ 33.3 NA1
Example 7 — Lift Problem
A 60 kg person stands on scales in a lift. The lift accelerates downward at 2.5 m s⁻². Find the reading on the scales. (g = 10 m s⁻²)
Step 1 — Forces on person: Weight W = 600 N downward, Normal R upward (scales reading). B1
Step 2 — F = ma (downward = positive): W − R = ma → 600 − R = 60 × 2.5 = 150 M1
Step 3: R = 600 − 150 = 450 N (scales read 45 kg equivalent — lighter than normal 600 N) A1
Example 8 — Braking Distance on Rough Horizontal Road
A car of mass 1200 kg brakes to rest from 20 m s⁻¹. The braking force is 6000 N and road friction coefficient is 0.1. Find the deceleration and stopping distance. (g = 10 m s⁻²)
Step 1 — Friction force: μmg = 0.1 × 1200 × 10 = 1200 N. Total retarding force = 6000 + 1200 = 7200 N. M1
Step 2 — Deceleration: a = −7200/1200 = −6 m s⁻² (deceleration = 6 m s⁻²) A1
8 mistakes that regularly cost marks in Cambridge M1 Newton's Laws questions. Recognise and avoid each one.
Mistake 1 — Wrong Direction for Friction
WRONG: Drawing friction in the direction of motion "because friction acts along the surface."
CORRECT: Friction OPPOSES the motion or the tendency to move. If the particle moves right, friction acts left. If the particle is about to slide down a slope, friction acts up the slope. Always determine the motion direction first, then draw friction opposing it.
Mistake 2 — Forgetting the Weight Component on a Slope
WRONG: On a slope, writing F = ma as "T = ma" without including the mg sinθ component down the slope.
CORRECT: On a slope, weight has a component along the slope: mg sinθ downward along slope. The equation along the slope is T − mg sinθ = ma (if tension is up the slope). Always decompose weight into its two components before writing F = ma.
Mistake 3 — Using Total Mass to Find Tension in Atwood Machine
WRONG: T = (m₁ + m₂) × a (using combined mass to find tension).
CORRECT: Tension is found by applying F = ma to ONE particle alone. For m₂ ascending: T − m₂g = m₂a → T = m₂(g + a). The total mass method only works for finding the system acceleration — never for tension.
Mistake 4 — Newton's Third Law Pairs Confusion
WRONG: Saying the weight of a book and the normal reaction from the table are Newton's Third Law pairs, and therefore cancel.
CORRECT: Newton's Third Law pairs act on DIFFERENT objects. Weight (Earth on book) pairs with book-on-Earth force. Normal reaction (table on book) pairs with book-on-table force. Weight and normal reaction balance by Newton's First Law (equilibrium), not Third Law.
Mistake 5 — Not Resolving Perpendicular to Slope to Find N
WRONG: Assuming N = mg on a slope and then computing friction as μmg.
CORRECT: On a slope at angle θ, the normal reaction is N = mg cosθ, NOT mg. Friction is then F = μN = μmg cosθ. If you use N = mg, your friction and acceleration will both be wrong.
Mistake 6 — Applying F = ma to the Whole System to Find Tension
WRONG: Net force = P. Total mass = m₁ + m₂. Therefore T = m₁ × (P/(m₁+m₂)) — using system F = ma directly to write a tension equation.
CORRECT: After finding a = P/(m₁+m₂) for the system, apply F = ma to particle A alone: T = m₁a. The key is that internal forces (tension) cancel in the system equation — so you must isolate one particle to find T.
Mistake 7 — Wrong Sign When Particle Goes Slack
WRONG: Continuing to include tension T in the equation after a string goes slack, getting a negative tension and using it.
CORRECT: A string cannot have negative tension — if you get T < 0, the string is slack and T = 0. Re-solve with T = 0 and check the direction of motion. A slack string means the particles move independently.
Mistake 8 — Ignoring Lift Acceleration Direction
WRONG: For a lift accelerating downward, writing R − mg = ma (positive upward), giving R = m(g + a) which is LARGER than mg — but the lift going down should make you feel lighter.
CORRECT: For lift accelerating downward at a m s⁻², take downward as positive: mg − R = ma → R = m(g − a). This gives R < mg (lighter). Always define positive direction clearly and be consistent throughout the equation.
Key Formulas
Newton's Laws
Formula
Meaning
Notes
ΣF = 0
Equilibrium / constant velocity
Newton's First Law
F = ma
Net force = mass × acceleration
Newton's Second Law — F is the resultant
F_AB = −F_BA
Action-reaction pair
Equal magnitude, opposite direction, different objects
Forces and Friction
Formula
Meaning
Notes
W = mg
Weight
g = 9.8 or 10 m s⁻² as specified
N = mg (horizontal surface)
Normal reaction on flat surface
Only if no vertical applied force
N = mg cosθ (slope)
Normal reaction on inclined plane
Found by resolving ⊥ to slope
F = μN
Limiting / kinetic friction
Direction opposes motion
μ = tanθ
Coefficient at limiting equilibrium on slope
Elegant result from sinθ/cosθ
Inclined Plane
Formula
Meaning
Notes
mg sinθ
Weight component down the slope
Causes acceleration / needs balancing
mg cosθ
Weight component ⊥ to slope
Equals N for smooth slope
a = g sinθ
Acceleration on smooth slope
No friction, no string
a = g(sinθ − μcosθ)
Acceleration sliding down rough slope
Friction opposes downward motion
Connected Particles
Formula
Meaning
Notes
a = P/(m₁+m₂)
System acceleration (no friction)
Treat as single particle of total mass
T = m₁a
Tension in string (particle A alone)
Apply F=ma to one particle
a = (m₁−m₂)g/(m₁+m₂)
Atwood machine acceleration
m₁ > m₂; m₁ descends
T = 2m₁m₂g/(m₁+m₂)
Atwood machine tension
Same tension on both sides
Lift Problems
Situation
Equation
Apparent Weight
Lift accelerating upward at a
R − mg = ma
R = m(g + a) — heavier
Lift accelerating downward at a
mg − R = ma
R = m(g − a) — lighter
Lift stationary or constant velocity
R = mg
Normal weight
Free fall (a = g)
R = 0
Weightlessness
Proof Bank
Three proofs to develop deep understanding of Newton's Laws in Mechanics 1.
Proof 1 — Derive the Atwood Machine Formulas
Let masses m₁ > m₂ hang over a smooth pulley connected by a light inextensible string. Let the acceleration of m₁ downward (and m₂ upward) be a, and tension T.
Apply F = ma to m₁ (positive downward):
m₁g − T = m₁a ... (1)
Apply F = ma to m₂ (positive upward):
T − m₂g = m₂a ... (2)
To find T, substitute a into equation (2):
T = m₂g + m₂a = m₂(g + a) = m₂[g + (m₁−m₂)g/(m₁+m₂)]
= m₂g[(m₁+m₂+m₁−m₂)/(m₁+m₂)]
= m₂g[2m₁/(m₁+m₂)] T = 2m₁m₂g / (m₁ + m₂) Q.E.D.
Proof 2 — Prove F = μN from the Definition of Coefficient of Friction
The coefficient of kinetic friction μ is defined as the ratio of the friction force F_friction to the normal reaction N when an object slides on a surface:
μ = F_friction / N (by definition)
Rearranging: F_friction = μN
This is not so much a proof as a derivation from definition. The key physics is:
1. The normal reaction N is found by resolving perpendicular to the surface (Newton's Second Law with zero perpendicular acceleration).
2. On a horizontal surface with no vertical applied force: N = mg.
3. On a slope at angle θ: resolving perpendicular → N − mg cosθ = 0 → N = mg cosθ.
4. Friction = μN = μmg cosθ on the slope.
At limiting equilibrium on a slope:
Along slope: mg sinθ = F_friction = μN = μmg cosθ
Divide both sides by mg cosθ: tanθ = μ Therefore μ = tanθ at limiting equilibrium on a slope. Q.E.D.
Proof 3 — Acceleration of Particle on Slope Connected Over Pulley to Hanging Particle
Let particle A (mass m₁) lie on a smooth slope at angle θ. A light inextensible string connects A over a pulley to hanging particle B (mass m₂). Assume B descends and A moves up the slope.
Apply F = ma to A (up slope = positive):
T − m₁g sinθ = m₁a ... (1)
Apply F = ma to B (downward = positive):
m₂g − T = m₂a ... (2)
Substitute into (2): T = m₂g − m₂a = m₂(g − a) T = m₂g[1 − (m₂ − m₁sinθ)/(m₁+m₂)] = m₁m₂g(1 + sinθ)/(m₁+m₂) Q.E.D.
Note: if a < 0, A actually descends and B ascends — the assumed direction was wrong. Re-solve with friction if the slope is rough.
Force Diagram Builder — Newton's Laws Scenarios
Select a scenario, enter values, and click Calculate to see the F = ma working and results. The canvas shows the force diagram.
Select a scenario, enter values, and click Calculate.
Exercise 1 — Newton's Laws: Identify Forces and State Equations (10 Questions)
Exercise 2 — F = ma Calculations on Horizontal Surface (10 Questions)
Exercise 3 — Forces on Inclined Planes (10 Questions)
Exercise 4 — Connected Particles on Horizontal Surface (10 Questions)
Exercise 5 — Pulley Systems (10 Questions)
Practice — 30 Mixed Questions
Challenge — 15 Harder Questions
Exam Style Questions
8 Cambridge M1-style questions with mark schemes. Attempt each question fully before revealing the mark scheme.
Q1 [3 marks]
A particle of mass 5 kg is pulled along a smooth horizontal surface by a horizontal force P N. The particle accelerates at 3.5 m s⁻². Find P. (g = 10 m s⁻²)
Smooth surface → no friction. Apply F = ma: [M1]
P = 5 × 3.5 = 17.5 N [A1]
Normal reaction N = mg = 50 N (not needed for P but confirms setup) [B1]
Q2 [5 marks]
A 6 kg block is pulled along a rough horizontal surface by a force of 36 N. The coefficient of friction is 0.4. Find the acceleration of the block and the friction force. (g = 10 m s⁻²)
Normal reaction: N = mg = 6 × 10 = 60 N [B1]
Friction: F = μN = 0.4 × 60 = 24 N [M1 A1]
Net force: 36 − 24 = 12 N [M1]
Acceleration: a = 12/6 = 2 m s⁻² [A1]
Q3 [5 marks]
A particle of mass 4 kg slides down a rough slope inclined at 30° to the horizontal. The coefficient of friction is 0.2. Find the acceleration. (g = 10 m s⁻²)
Perpendicular to slope: N = mg cos30° = 40 × (√3/2) ≈ 34.64 N [M1 A1]
Friction (up the slope, opposing downward motion): F = μN = 0.2 × 34.64 ≈ 6.93 N [M1]
Along slope (down = positive): mg sin30° − F = ma → 40 × 0.5 − 6.93 = 4a
20 − 6.93 = 13.07 = 4a → a = 13.07/4 ≈ 3.27 m s⁻² [M1 A1]
Q4 [6 marks]
Particles A (mass 3 kg) and B (mass 5 kg) are connected by a light inextensible string on a smooth horizontal surface. A force of 32 N is applied to B horizontally. Find the acceleration of the system and the tension in the string. (g = 10 m s⁻²)
System (A+B), total mass = 8 kg, net force = 32 N: [M1]
a = 32/8 = 4 m s⁻² [A1]
Apply F = ma to particle A alone: [M1]
T = m_A × a = 3 × 4 = 12 N [A1]
Check with B: 32 − 12 = 5 × 4 → 20 = 20 ✓ [B1 B1]
Q5 [6 marks]
Two particles of masses 8 kg and 3 kg are connected by a light inextensible string over a smooth fixed pulley. Find the acceleration and tension when the system is released from rest. (g = 10 m s⁻²)
For 8 kg (descending, down = positive): 80 − T = 8a ...(1) [M1]
For 3 kg (ascending, up = positive): T − 30 = 3a ...(2) [M1]
Add: 50 = 11a → a = 50/11 ≈ 4.55 m s⁻² [A1]
From (2): T = 30 + 3 × 50/11 = 30 + 150/11 = (330+150)/11 = 480/11 ≈ 43.6 N [M1 A1]
Check formula: T = 2×8×3×10/(8+3) = 480/11 ✓ [A1]
Q6 [7 marks]
A particle A of mass 6 kg rests on a smooth slope inclined at 25° to the horizontal. It is connected by a light inextensible string over a smooth pulley at the top of the slope to a hanging particle B of mass 4 kg. Find the acceleration and the tension in the string. (g = 10 m s⁻²)
For A (up slope = positive, pulled up by B): T − 6g sin25° = 6a
T − 60 × 0.4226 = 6a → T − 25.36 = 6a ...(1) [M1 A1]
For B (down = positive): 4g − T = 4a → 40 − T = 4a ...(2) [M1]
Add (1)+(2): 40 − 25.36 = 10a → 14.64 = 10a → a = 1.46 m s⁻² [A1]
From (2): T = 40 − 4 × 1.46 = 40 − 5.84 = 34.16 N ≈ 34.2 N [M1 A1 A1]
Q7 [4 marks]
A person of mass 65 kg stands in a lift. The lift accelerates upward at 1.5 m s⁻². Find the normal reaction of the lift floor on the person. (g = 10 m s⁻²)
Forces on person: R upward, weight 650 N downward. [B1]
Apply F = ma (upward positive): R − 650 = 65 × 1.5 = 97.5 [M1]
R = 650 + 97.5 = 747.5 N [A1]
This is equivalent to an apparent mass of 74.75 kg (greater than real mass — person feels heavier). [B1]
Q8 [5 marks]
A car of mass 900 kg accelerates from rest to 20 m s⁻¹ in 8 s on a horizontal road. The driving force is constant. Road resistance is 450 N. Find the driving force and the coefficient of friction. (g = 10 m s⁻²)
Acceleration: a = (v−u)/t = 20/8 = 2.5 m s⁻² [B1]
Net force = ma = 900 × 2.5 = 2250 N [M1]
Driving force P − 450 = 2250 → P = 2700 N [A1]
Normal reaction N = mg = 9000 N. Friction = μN = 450 N [M1]
μ = 450/9000 = 0.05 [A1]
Past Paper Questions
5 questions from Cambridge A-Level 9709 Mechanics 1 past papers on Newton's Laws of Motion. Attempt fully before revealing.
Past Paper Q1 — 9709/41/M/J/20 Q3 [6 marks]
A particle of mass 3 kg is pulled up a rough plane inclined at 20° to the horizontal by a force of 25 N acting along the plane. The coefficient of friction between the particle and the plane is 0.35. Find the acceleration of the particle. (g = 10 m s⁻²)
Normal reaction: N = 3 × 10 × cos20° = 30 × 0.9397 = 28.19 N [M1 A1]
Friction (down slope, opposing upward motion): F = 0.35 × 28.19 = 9.87 N [M1]
Along slope (up = positive): 25 − 3g sin20° − F = 3a
25 − 30 × 0.3420 − 9.87 = 3a
25 − 10.26 − 9.87 = 4.87 = 3a [M1]
a = 4.87/3 ≈ 1.62 m s⁻² up the slope [A1 A1]
Past Paper Q2 — 9709/42/O/N/19 Q5 [7 marks]
Particles A (mass 5 kg) and B (mass 3 kg) are connected by a light inextensible string over a smooth pulley. A lies on a rough horizontal table; B hangs freely. The coefficient of friction between A and the table is 0.4. Find the acceleration of the system and the tension. (g = 10 m s⁻²)
Normal reaction on A: N = 5g = 50 N. Friction = 0.4 × 50 = 20 N. [M1 A1]
For A (horizontal, positive in direction of motion): T − 20 = 5a ...(1) [M1]
For B (downward = positive): 3g − T = 3a → 30 − T = 3a ...(2) [M1]
Add: 10 = 8a → a = 1.25 m s⁻² [A1]
From (1): T = 20 + 5 × 1.25 = 20 + 6.25 = 26.25 N [M1 A1]
Past Paper Q3 — 9709/43/M/J/18 Q3 [5 marks]
A particle of mass 2 kg is projected up a smooth slope inclined at 40° to the horizontal with initial speed 8 m s⁻¹. Find the deceleration and the distance travelled before coming to rest. (g = 10 m s⁻²)
Along slope (up = positive): −mg sin40° = ma [M1]
a = −g sin40° = −10 × 0.6428 = −6.43 m s⁻² (deceleration 6.43 m s⁻²) [A1]
Using v² = u² + 2as with v = 0, u = 8, a = −6.428: [M1]
0 = 64 + 2(−6.428)s → s = 64/12.856 ≈ 4.98 m [A1 A1]
Past Paper Q4 — 9709/41/O/N/17 Q4 [7 marks]
Two particles A (4 kg) and B (6 kg) are connected by a light inextensible string over a smooth pulley. A is on a rough slope of 35°; B hangs freely. μ = 0.25 between A and slope. The system is released from rest. Find the acceleration and tension. (g = 10 m s⁻²)
For A: N = 4g cos35° = 40 × 0.8192 = 32.77 N [M1]
Friction on A (down slope if A moves up): F = 0.25 × 32.77 = 8.19 N [M1]
For A (up slope = positive): T − 4g sin35° − 8.19 = 4a
T − 40 × 0.5736 − 8.19 = 4a → T − 22.94 − 8.19 = 4a → T − 31.13 = 4a ...(1) [M1 A1]
For B (down = positive): 60 − T = 6a ...(2) [M1]
Add: 28.87 = 10a → a = 2.89 m s⁻² [A1]
T = 60 − 6 × 2.887 = 60 − 17.32 = 42.68 N ≈ 42.7 N [A1]
Past Paper Q5 — 9709/42/M/J/16 Q5 [8 marks]
A lift of mass 500 kg carries a passenger of mass 70 kg. The lift accelerates upward from rest at 1.2 m s⁻² for 3 s, then travels at constant speed for 5 s, then decelerates to rest at 0.8 m s⁻². Find (i) the tension in the lift cable during acceleration, (ii) the normal reaction on the passenger during deceleration. (g = 10 m s⁻²)
(i) Tension during upward acceleration:
Total mass of lift + passenger = 570 kg. Upward acceleration a = 1.2 m s⁻². [B1]
T − 570g = 570 × 1.2 [M1]
T = 570 × 10 + 570 × 1.2 = 5700 + 684 = 6384 N [A1 A1]
(ii) Normal reaction on passenger during deceleration (a = 0.8 m s⁻² downward):
Lift decelerates → acceleration is 0.8 m s⁻² downward. [B1]
For passenger (taking down as positive): 70g − R = 70 × 0.8 [M1]
700 − R = 56 → R = 700 − 56 = 644 N [A1 A1]