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This lesson covers AQA Spec 3.4.1.4 (Newton's laws of motion) in full depth. You will master free body diagrams, connected-body problems, pulley systems, and motion on inclined planes.
Spec mapping (AQA 7408 §3.4.1.4–§3.4.1.5): This lesson develops the three laws of motion in their A-Level form: the law of inertia, the second law as F = ma for constant mass (and as F = dp/dt in general), and the third law as paired interaction forces. The strand also covers free-body diagrams, motion on inclined planes, connected bodies via inextensible strings, frictional force F = μN, and the approach to terminal velocity. Newton's laws sit at the centre of the AQA mechanics module: every problem that gives "find the acceleration" or "find the tension" routes through this content. (Refer to the official AQA specification document for exact wording.)
Synoptic links: Newton's laws are the engine that drives the rest of A-Level mechanics. (i) Momentum and impulse (Lesson 4, §3.4.1.6) is a direct restatement of the second law: F = dp/dt is the more general form, and conservation of momentum follows when ΣF_external = 0. (ii) Circular motion (Year 13, §3.6.1.1) applies F = ma with a = v²/r directed centripetally — every circular-motion problem is a Newton's-second-law problem with a particular geometry. (iii) Simple harmonic motion (Year 13, §3.6.1.2) sets F = −kx and applies the second law to give ẍ = −(k/m)x, the defining SHM equation. The third law also reappears in electromagnetic theory (charges exert equal and opposite Coulomb forces) and in fluid statics (the reaction to upthrust acts on the displaced fluid). Treating Newton's laws as "just mechanics" understates how often they recur.
An object remains at rest, or continues to move with constant velocity, unless acted upon by a resultant external force.
Implications: An object moving at constant velocity has zero resultant force. A stationary object has zero resultant force. A change in speed OR direction requires a resultant force.
The rate of change of momentum of an object is directly proportional to the resultant force acting on it, and takes place in the direction of that force.
For constant mass: F = ma
More generally: F = Δp/Δt where p = mv
The unit of force (newton) is defined such that 1 N = 1 kg m s⁻².
When two objects interact, they exert equal and opposite forces on each other. These Newton's Third Law force pairs:
Common Misconception: Weight and normal contact force are NOT a Newton's Third Law pair — they act on the SAME object and are of different types. The Third Law pair of the weight of a book on a table is the gravitational pull of the book on the Earth.
A free body diagram shows all forces acting on a single object. Forces are drawn as arrows from the object's centre, with length proportional to magnitude.
Described diagram — Block on a rough inclined plane at angle θ:
The block sits on a slope. Three forces act:
Choose axes parallel and perpendicular to the slope:
For equilibrium on the slope:
A 5.0 kg block is placed on a smooth slope inclined at 30° to the horizontal. Find the acceleration of the block down the slope.
Solution: The only force component along the slope is mg sin θ (the slope is smooth, so no friction).
F = ma along the slope:
mg sin θ = ma
a = g sin θ = 9.81 × sin 30° = 9.81 × 0.50 = 4.9 m s⁻²
A 5.0 kg block is on a rough slope at 30°. The coefficient of friction is μ = 0.30. Find the acceleration.
Solution:
Normal force: N = mg cos 30° = 5.0 × 9.81 × 0.866 = 42.5 N
Friction force: F = μN = 0.30 × 42.5 = 12.7 N (up the slope)
Net force down the slope = mg sin 30° − F = 5.0 × 9.81 × 0.50 − 12.7 = 24.5 − 12.7 = 11.8 N
Acceleration: a = 11.8 / 5.0 = 2.4 m s⁻²
When two or more objects are connected (e.g., by a string), they move with the same acceleration (assuming an inextensible string). Treat each body separately with its own free body diagram.
Two blocks, A (3.0 kg) and B (5.0 kg), are connected by a light inextensible string on a smooth horizontal surface. A horizontal force of 24 N is applied to block B. Find (a) the acceleration and (b) the tension in the string.
Solution:
(a) Consider the whole system: F = (mₐ + m_b)a
24 = (3.0 + 5.0)a → a = 24/8.0 = 3.0 m s⁻²
(b) Consider block A alone: The only horizontal force on A is the tension T pulling it forward.
T = mₐ × a = 3.0 × 3.0 = 9.0 N
Check with block B: Net force on B = 24 − T = 24 − 9.0 = 15 N. Acceleration of B = 15/5.0 = 3.0 m s⁻² ✓
Described diagram — Atwood machine: Two masses m₁ and m₂ (m₁ > m₂) hang on either side of a frictionless pulley connected by a light inextensible string. m₁ accelerates downward and m₂ accelerates upward, both with the same magnitude of acceleration a. The tension T is the same throughout the string.
For mass m₁ (moving down): m₁g − T = m₁a ... (1)
For mass m₂ (moving up): T − m₂g = m₂a ... (2)
Adding (1) and (2):
m₁g − m₂g = (m₁ + m₂)a
a = (m₁ − m₂)g / (m₁ + m₂)
Substituting back into (2):
T = 2m₁m₂g / (m₁ + m₂)
Masses of 4.0 kg and 6.0 kg are connected over a frictionless pulley. Find the acceleration and tension.
Solution:
a = (m₁ − m₂)g / (m₁ + m₂) = (6.0 − 4.0) × 9.81 / (6.0 + 4.0) = 2.0 × 9.81 / 10.0 = 1.96 m s⁻²
T = 2 × 6.0 × 4.0 × 9.81 / (6.0 + 4.0) = 2 × 6.0 × 4.0 × 9.81 / 10.0 = 470.9 / 10.0 = 47.1 N
Check: For the 6.0 kg mass: 6.0 × 9.81 − 47.1 = 58.86 − 47.1 = 11.76 N. a = 11.76/6.0 = 1.96 m s⁻² ✓
Exam Tip: In pulley problems, always define the positive direction for each mass consistently. For an Atwood machine, if m₁ goes down, define down as positive for m₁ and up as positive for m₂.
Described diagram: A block of mass M sits on a smooth horizontal table. A string passes over a frictionless pulley at the edge of the table and connects to a hanging mass m. The tension T is the same throughout.
For the hanging mass (downward positive): mg − T = ma ... (1)
For the block on the table (horizontal): T = Ma ... (2)
Adding: mg = (M + m)a
a = mg / (M + m)
T = Mmg / (M + m)
A 2.0 kg block on a smooth table is connected via a pulley to a 3.0 kg hanging mass. Find the acceleration and tension.
Solution:
a = mg / (M + m) = 3.0 × 9.81 / (2.0 + 3.0) = 29.43 / 5.0 = 5.89 m s⁻²
T = Mmg / (M + m) = 2.0 × 3.0 × 9.81 / 5.0 = 58.86 / 5.0 = 11.8 N
When an object falls through a fluid, it experiences:
Initially, weight > drag → resultant downward force → acceleration.
As speed increases, drag increases until drag = weight. Then resultant force = 0, so acceleration = 0 (Newton's First Law). The object moves at constant terminal velocity.
For a sphere falling through a viscous fluid at low speeds: drag force F = 6πηrv (Stokes' law), where η is the viscosity, r is the radius, and v is the speed.
At terminal velocity: mg = 6πηrv_t
v_t = mg / (6πηr)
A spherical bead of radius 1.0 mm and density 7800 kg m⁻³ (steel) falls through oil of viscosity η = 0.10 Pa s and density 900 kg m⁻³. Find the terminal velocity, including the upthrust correction. Take g = 9.81 m s⁻².
Solution: At terminal velocity, three forces balance: weight downward, upthrust upward, viscous drag upward.
Volume of bead: V = (4/3) π r³ = (4/3) π × (1.0 × 10⁻³)³ = 4.189 × 10⁻⁹ m³.
Weight: W = ρ_bead × V × g = 7800 × 4.189 × 10⁻⁹ × 9.81 = 3.203 × 10⁻⁴ N.
Upthrust: U = ρ_oil × V × g = 900 × 4.189 × 10⁻⁹ × 9.81 = 3.697 × 10⁻⁵ N.
Net downward force: F_net = W − U = 3.203 × 10⁻⁴ − 3.697 × 10⁻⁵ = 2.833 × 10⁻⁴ N.
At terminal velocity, this equals the Stokes' drag: 6πηrv_t = F_net.
v_t = F_net / (6πηr) = 2.833 × 10⁻⁴ / (6 × π × 0.10 × 1.0 × 10⁻³) = 2.833 × 10⁻⁴ / 1.885 × 10⁻³ = 0.150 m s⁻¹ (3 s.f.)
This experiment (a falling ball in a tall cylinder of oil, with the time-of-fall between two marks measured by stopwatch) is the standard A-Level method for measuring the viscosity of a fluid.
Exam Tip: When asked to explain terminal velocity, always reference Newton's Second Law (F = ma). State that the resultant force decreases as drag increases, so acceleration decreases. When resultant force is zero, acceleration is zero and velocity is constant.
Specimen question modelled on the AQA paper format — not a verbatim past-paper item.
Two masses, A (mass 3.0 kg) and B (mass 5.0 kg), are connected by a light inextensible string passing over a smooth pulley fixed at the top of a smooth inclined plane. Mass A hangs vertically; mass B sits on the slope, which is inclined at 30° to the horizontal. The string is parallel to the slope between B and the pulley.
(a) Draw a free-body diagram for each mass. [2 marks]
(b) By writing Newton's second law for each mass, calculate the acceleration of the system and the tension in the string. Take g = 9.81 m s⁻². [5 marks]
(c) The slope is replaced by a rough slope with coefficient of friction μ = 0.20. Without further calculation, state whether the magnitude of the acceleration increases, decreases, or stays the same, and justify your answer. [2 marks]
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