Dynamics: Newton's Second Law

Spec mapping: OCR H556 Module 3.2 — Forces in action (Newton's three laws of motion; net force $F = ma$ ; drag and frictional forces; terminal velocity). Refer to the official OCR H556 specification document for exact wording.

Kinematics tells us how objects move. Dynamics asks why. At the heart of dynamics is Newton's second law — arguably the single most important equation in A-Level Physics, and the conceptual bridge between description and explanation. This lesson covers Module 3.2.1 (Newton's laws) and the early portion of Module 3.2.2 (non-linear motion including drag and terminal velocity), and lays the groundwork for momentum (Module 4.1), energy (Module 3.3) and every later mechanics topic.

Key Definition: Newton's second law in its most general form states that the net (resultant) force on a body equals the rate of change of its momentum: $\vec{F}_{\text{net}} = d\vec{p}/dt$ . For a body of constant mass this simplifies to $\vec{F}_{\text{net}} = m\vec{a}$ . The force and the acceleration are vectors; they always point in the same direction.

Newton's Three Laws — The Statements

OCR examiners regularly test the precise wording of Newton's laws. Memorise these formulations:

First Law (inertia): A body remains at rest or moves with constant velocity in a straight line unless acted upon by a resultant external force. (This defines inertia — the tendency of matter to resist changes in its state of motion. It also defines what we mean by an inertial reference frame: one in which the first law holds.)
Second Law: The rate of change of momentum of a body is directly proportional to the resultant external force applied to it, and occurs in the direction of that force. For constant mass this reduces to $F = ma$ .
Third Law: For every action there is an equal and opposite reaction; forces always come in pairs acting on different bodies.

The first law was, historically, the great conceptual revolution — Aristotle had taught that a body needs a continuous force to keep it moving (because everyday experience, dominated by friction, suggests so). Galileo, paraphrased rather than quoted, identified friction as the culprit and Newton crystallised the law of inertia. The first law is, strictly, a special case of the second law (zero resultant force ⇒ zero acceleration ⇒ velocity constant), but it earns its own statement because it defines the frame in which the second law applies.

Newton's Second Law — Full Mathematical Statement

The most general statement is:

$\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt}$

where $\vec{p} = m\vec{v}$ is the momentum of the body. For a body of constant mass:

$\frac{d\vec{p}}{dt} = \frac{d(m\vec{v})}{dt} = m \frac{d\vec{v}}{dt} = m \vec{a}$

so:

$\boxed{\vec{F}_{\text{net}} = m\vec{a}}$

$\vec{F}_{\text{net}}$ is the resultant (net) force in newtons (N).
$m$ is mass in kilograms (kg).
$\vec{a}$ is acceleration in metres per second squared (m s $^{-2}$ ).

One newton is defined as the resultant force that gives a $1$ kg mass an acceleration of $1$ m s $^{-2}$ . The equation $F = ma$ therefore defines the unit of force.

For systems where mass is changing (rockets burning fuel, raindrops growing as they fall through moist air, conveyor belts loading), $F = ma$ is incorrect — you must use the full $F = dp/dt$ , which becomes $F = ma + v(dm/dt)$ when expanded. For A-Level you will typically be working in the constant-mass regime, but the existence of variable-mass problems is worth remembering as an A* discriminator.

Common Exam Mistake: Substituting the applied force, rather than the resultant force, into $F = ma$ . Always find the vector sum of all forces acting on the body and use that.

Weight and Mass

Mass ( $m$ ) is a scalar measure of the amount of matter in a body — unchanged anywhere in the universe. Its SI unit is kg.

Weight ( $W$ ) is a vector force, the gravitational pull of the local massive body (Earth, Moon, etc.) on the mass:

$\vec{W} = m\vec{g}$

$W$ in newtons, $m$ in kg, $\vec{g}$ in m s $^{-2}$ (equivalently N kg $^{-1}$ ).
$\vec{g}$ is both the free-fall acceleration and the gravitational field strength; the dual interpretation is a direct consequence of Newton's second law.
On Earth $g \approx 9.81$ N kg $^{-1}$ ; on the Moon $g \approx 1.62$ N kg $^{-1}$ ; in deep interplanetary space $g \to 0$ .

A 70 kg astronaut has the same mass on Earth and on the Moon (matter is conserved) but a different weight: 687 N on Earth versus 113 N on the Moon. Apparent weightlessness in orbit (e.g. aboard the International Space Station) is not because $g = 0$ there — $g$ at 400 km altitude is still about $8.69$ m s $^{-2}$ , only 11% less than at sea level — but because the spacecraft and the astronaut are both in continuous free fall around the Earth, so the astronaut feels no normal contact force. Weight and apparent weight are different concepts; the apparent weight is the normal reaction from a support surface, which can vanish in free fall.

Worked Example 1 — Lift Accelerating Upward

A person of mass $65$ kg stands on bathroom scales inside a lift. The lift accelerates upward at $2.0$ m s $^{-2}$ . What reading (in newtons) do the scales show?

Let $R$ be the normal contact force from the scales upward (this is what they measure — the apparent weight). The actual weight $W = mg$ acts downward.

Taking upward positive and applying Newton II to the person:

$R - mg = ma \Rightarrow R = m(g + a) = 65 \times (9.81 + 2.0) = 65 \times 11.81 = 768 \text{ N}$

Compare with the static reading ( $R_{\text{static}} = mg = 637$ N). The person feels heavier because the scales must exert an additional 131 N upward to accelerate them. The "apparent weight gain" — $131$ N, or 21% — is what every aircraft passenger feels as the aircraft pushes upward off the runway.

In a free-falling lift ( $a = -9.81$ m s $^{-2}$ ), $R = m(g + a) = 0$ . The person and the scales are in mutual free fall, with the scales pulling away from the floor at the same rate the person is — apparent weight is zero. This is exactly the sensation of weightlessness astronauts experience in orbit.

Worked Example 2 — Towing a Trailer

A car of mass $1200$ kg tows a trailer of mass $500$ kg. The engine provides a forward driving force of $3600$ N. Resistance forces are $200$ N (on the car) and $150$ N (on the trailer).

(a) Find the acceleration of the combined system.

(b) Find the tension in the tow-bar.

(a) Treat the car-plus-trailer as a single system:

Total mass $= 1200 + 500 = 1700$ kg.
Resultant force $= 3600 - 200 - 150 = 3250$ N (forward).
$a = F_{\text{net}} / m_{\text{total}} = 3250 / 1700 = 1.91$ m s $^{-2}$ .

(b) Now consider the trailer alone:

Forces on trailer: tension $T$ (forward, from the tow-bar) and $150$ N resistance (backward).
Newton II: $T - 150 = 500 \times 1.91 = 956 \Rightarrow T = 1106$ N $\approx 1110$ N (3 s.f.).

Cross-check by considering the car alone:

Forces on car: $3600$ N forward (engine), $200$ N back (resistance), $T$ back (from tow-bar reaction).
Newton II: $3600 - 200 - T = 1200 \times 1.91 = 2294 \Rightarrow T = 1106$ N. ✓

Cross-checking by analysing different free-body subsystems is a habit that catches arithmetic and conceptual errors. Examiners reward it.

Free-Body Diagrams — The Essential Skill

Every dynamics problem should begin with a free-body diagram (FBD): an isolated picture of one body, with every force acting on it drawn as an arrow originating at the body. Forces acting on other bodies are excluded.

For a $1$ kg book on a horizontal table:

Forces on the book only: its weight ( $W = mg$ downward) and the normal contact force ( $N$ upward). The book's gravitational pull on the Earth is not drawn — that is a force on the Earth, not on the book.

Drag and Resistance Forces

Unlike gravity (which is constant near Earth's surface), drag depends on velocity. At low Reynolds numbers (viscous regime, slow speeds, small bodies), drag is approximately linear in velocity:

$F_{\text{drag}} = b v$

At higher Reynolds numbers (turbulent regime, faster speeds, larger bodies), drag is approximately quadratic in velocity:

$F_{\text{drag}} = \tfrac{1}{2} \rho_{\text{fluid}} C_D A v^2$

where $\rho$ is the fluid density, $C_D$ is the dimensionless drag coefficient (depends on shape — about $0.5$ for a sphere, $0.04$ for a streamlined airfoil), and $A$ is the cross-sectional area.

For A-Level work, the key facts are:

Drag increases with speed.
Drag always opposes motion through the fluid (it acts antiparallel to velocity).
Drag reduces the net force, hence the acceleration, of a body moving through a fluid.
When drag balances the driving force (e.g. weight, for a falling body), the net force is zero and the body moves at constant velocity — terminal velocity.

Terminal Velocity

Consider a skydiver dropped from a stationary aircraft. Initially $v = 0$ and the only force is weight $mg$ (downward), so the acceleration is $g$ downward.

As the skydiver speeds up, drag $F_{\text{drag}}(v)$ builds. The net downward force is $mg - F_{\text{drag}}(v)$ , which decreases as $v$ rises. Acceleration falls toward zero.

At terminal velocity $v_T$ , drag exactly balances weight:

$F_{\text{drag}}(v_T) = mg$

From that point onward, the net force is zero and the skydiver falls at constant $v_T$ . For a typical human in a spread-eagle "belly-down" position, $v_T \approx 55$ m s $^{-1}$ (about 120 mph). In a head-down or feet-down position with arms tucked, $v_T$ can reach 90 m s $^{-1}$ — and competitive sky-divers exploit this for vertical speed.

Once the parachute is deployed, $A$ increases by a factor of $\sim 100$ , so $F_{\text{drag}}$ at any given $v$ is $\sim 100\times$ larger. The new terminal velocity is around $5$ – $6$ m s $^{-1}$ , fast enough to feel the impact but slow enough to land safely.

Velocity-time graph for a falling skydiver

flowchart TD
  A[v = 0, t = 0] --> B[Acceleration starts at g, drag negligible]
  B --> C[Drag grows with v; net force shrinks]
  C --> D[v approaches v_T1; gradient approaches zero]
  D --> E[Parachute opens at time t_p]
  E --> F[Drag suddenly large; rapid deceleration]
  F --> G[v approaches new smaller v_T2]
  G --> H[Constant velocity descent to ground]

On a $v$ – $t$ graph, the curve starts with gradient $g$ , asymptotes to $v_{T1}$ , then drops sharply at parachute deployment and asymptotes to $v_{T2}$ . Both asymptotes are horizontal (zero acceleration once drag balances weight).

Worked Example 3 — Non-Uniform Deceleration by Drag

A $0.050$ kg ball is thrown upward at $u = 20$ m s $^{-1}$ through air in which drag is $F_{\text{drag}} = 0.010\,v^2$ (SI units). At the instant of launch, find (a) the weight, (b) the drag force, (c) the resultant force, (d) the deceleration.

(a) $W = mg = 0.050 \times 9.81 = 0.491$ N.

(b) $F_{\text{drag}} = 0.010 \times (20)^2 = 4.00$ N (directed downward, opposing upward velocity).

(d) $a = F_{\text{net}} / m = -4.49 / 0.050 = -89.8$ m s $^{-2}$ .

The initial deceleration is about $9g$ — nearly an order of magnitude greater than gravity alone! At high speeds drag can dominate weight, which is why a bullet's velocity decays rapidly after leaving the barrel and why pebbles tossed at the sea-surface from a great height never accelerate beyond a relatively low terminal velocity.

Note: this is a non-SUVAT problem. As the ball decelerates, $v$ drops, so $F_{\text{drag}}$ drops, so $a$ changes. The motion is uniformly accelerated nowhere along the trajectory. To find the trajectory you must integrate numerically; only at the instant of launch can the values be computed in closed form.

Newton's Third Law — Identifying Action-Reaction Pairs

Newton's third law states: every force has an equal and opposite reaction. The critical caveat: the two forces act on different bodies.

System	Action	Reaction
Book on table	Book pushes down on table (contact force)	Table pushes up on book (normal force)
Earth and ball	Earth pulls ball down (weight)	Ball pulls Earth up (gravitational reaction)
Swimmer in pool	Swimmer pushes water backward	Water pushes swimmer forward
Rocket in flight	Engine pushes exhaust gases backward	Exhaust gases push rocket forward
Walking	Foot pushes ground backward (friction)	Ground pushes foot forward (friction reaction)
Climbing a ladder	Hands pull down on rung	Rung pulls up on hands

Common Exam Mistake: Claiming that the normal force on a book equals its weight because of Newton III. This is wrong. The normal force and the weight are both acting on the book, so they are first-law balanced forces (zero resultant ⇒ no acceleration), not Newton-III partners. The Newton-III partner of the book's weight is the gravitational pull of the book on the Earth, an equal-and-opposite force acting on a different body (the Earth). The Newton-III partner of the normal force on the book is the book's downward push on the table.

A clean way to test "are these two forces a Newton-III pair?" is to ask: (i) do they act on different bodies? (ii) are they equal in magnitude and opposite in direction? (iii) are they of the same physical type (both gravitational, both contact, both electrical, etc.)? Only if all three conditions hold are the forces a genuine Newton-III pair.

Dynamics: Newton's Second Law

Dynamics: Newton's Second Law

Newton's Three Laws — The Statements

Newton's Second Law — Full Mathematical Statement

Weight and Mass

Worked Example 1 — Lift Accelerating Upward

Worked Example 2 — Towing a Trailer

Free-Body Diagrams — The Essential Skill

Drag and Resistance Forces

Terminal Velocity

Velocity-time graph for a falling skydiver

Worked Example 3 — Non-Uniform Deceleration by Drag

Newton's Third Law — Identifying Action-Reaction Pairs

Synoptic Links

More in Physics