Newton's Second Law — F = Δp/Δt

Spec mapping: OCR H556 Module 3.2 — Forces in action (Newton's three laws of motion; resultant force; mass and weight; the general form $F = dp/dt$ and the constant-mass special case $F = ma$ ). Refer to the official OCR H556 specification document for exact wording.

Newton's Second Law is the quantitative heart of mechanics. Where Newton 1 told us that a resultant force changes the motion, Newton 2 tells us how much. It is the bridge between forces (cause) and accelerations (effect), the equation that converts a free-body diagram into a numerical prediction, and the law that — applied to an isolated system — gives the conservation of momentum that dominates lessons 4-10.

For OCR A-Level Physics A candidates the law is tested in both its classic constant-mass form $F = ma$ and its more general rate-of-change-of-momentum form $\vec{F} = d\vec{p}/dt$ , which is the version stated on the H556 specification. This treatment goes well beyond the " $F = ma$ plug-in" of GCSE: we derive the general form, identify when the special case fails (variable-mass systems), and connect Newton 2 cleanly to the momentum chapter that follows.

Statement of Newton's Second Law

The resultant force acting on a body equals the rate of change of its linear momentum. The change of momentum takes place in the direction of the resultant force.

In equation form (general):

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

where $\vec{p} = m\vec{v}$ is the linear momentum and $\vec{F}_{\text{net}}$ is the resultant (vector sum) external force. The law is a vector equation in two senses: (i) it determines both the magnitude and direction of $d\vec{p}/dt$ , and (ii) it is obeyed component-by-component in 2-D and 3-D ( $F_x = dp_x/dt$ etc.).

This is the form printed in the OCR data booklet and the one mark schemes expect when an extended-prose question asks "state Newton's second law".

Deriving F = ma from F = dp/dt

Starting from $\vec{p} = m\vec{v}$ :

$\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt} = \frac{d(m\vec{v})}{dt}$

Apply the product rule:

$\frac{d(m\vec{v})}{dt} = m \frac{d\vec{v}}{dt} + \vec{v} \frac{dm}{dt}$

If the mass is constant ( $dm/dt = 0$ ):

$\vec{F}_{\text{net}} = m \frac{d\vec{v}}{dt} = m\vec{a}$

This is the form most students meet first at GCSE; it is the special case of Newton 2 valid whenever the body's mass does not change. For 99% of A-Level mechanics problems (cars, balls, blocks, lifts) the constant-mass form is exact.

If the mass changes with time (rockets burning fuel, conveyor belts being loaded, raindrops growing as they fall through moist air) the $\vec{v}\,dm/dt$ term cannot be neglected — see "Variable-mass systems" below. The fact that $F = ma$ is the special case, not the full law, is a Top-band discriminator: candidates who write only $F = ma$ when asked to "state Newton 2" lose a wording mark.

Diagram — When to use which form

flowchart TD
  A[Newton 2: F_net = dp/dt] --> B{Is mass constant?}
  B -- "Yes (typical A-Level)" --> C[F_net = m·a — use this]
  B -- "No (rocket, rain, conveyor)" --> D[F_net = m·dv/dt + v·dm/dt]
  C --> E[Solve for a; combine with SUVAT or F-t graph]
  D --> F[Thrust term v·dm/dt is non-zero]

Units and the Definition of the Newton

The newton is defined through Newton 2: $1$ N is the resultant force that gives a $1$ kg mass an acceleration of $1$ m s $^{-2}$ . So:

$1 \text{ N} = 1 \text{ kg m s}^{-2}$

Equivalently, momentum has units of $\text{kg m s}^{-1} = \text{N s}$ , foreshadowing the impulse-momentum theorem $J = F\Delta t = \Delta p$ (lesson 5).

Whenever a derived quantity comes out with units of kg m s $^{-2}$ in a calculation, you know it is a force in newtons — a useful dimensional sanity-check during multi-step problems.

Vector Form and Components

Newton 2 is a vector equation, valid component-by-component:

$\sum F_x = m a_x \qquad \sum F_y = m a_y \qquad \sum F_z = m a_z$

The magnitude of the acceleration is $|\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2}$ and its direction is given by the ratios of the components. In 2-D the direction makes angle $\theta = \arctan(a_y/a_x)$ with the $x$ -axis.

This is the workhorse for incline, projectile and circular-motion problems.

Mass and Weight

Mass ( $m$ ) is a scalar measure of inertia — a body's resistance to acceleration, and (by the equivalence principle) also a measure of its gravitational coupling. SI unit kg. Mass is invariant: a $70$ kg astronaut has $70$ kg of mass on Earth, on the Moon, in deep space.

Weight ( $W$ ) is the gravitational force a massive body exerts on $m$ :

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

SI unit N. The vector $\vec{g}$ has dual interpretation: free-fall acceleration and gravitational field strength, with identical numerical value because of the equivalence principle. At Earth's surface $g \approx 9.81$ N kg $^{-1}$ ( $= 9.81$ m s $^{-2}$ ); on the Moon $g \approx 1.62$ N kg $^{-1}$ ; in deep interplanetary space $g \to 0$ .

A $70$ kg astronaut weighs $687$ N on Earth, $113$ N on the Moon, and effectively nothing in deep space — but always has $70$ kg of mass.

Apparent weight ≠ weight. Apparent weight is the normal contact force exerted on you by the supporting surface; it equals your true weight only in equilibrium. In an accelerating lift apparent weight is $m(g+a)$ where $a$ is upward acceleration; in a free-falling lift apparent weight is zero (you feel "weightless" even though your true weight $mg$ is unchanged).

Worked Example 1 — Constant-Mass Car

A $1200$ kg car accelerates from $10$ m s $^{-1}$ to $25$ m s $^{-1}$ in $5.0$ s along a straight road. Find the resultant force on the car. If air resistance and rolling friction together oppose motion with $800$ N, find the forward driving force from the tyres.

$\Delta v = 25 - 10 = 15$ m s $^{-1}$ .
$a = \Delta v/\Delta t = 15/5.0 = 3.0$ m s $^{-2}$ .
$F_{\text{net}} = ma = 1200 \times 3.0 = 3600$ N (forward).

This is the resultant force — the vector sum of every horizontal force on the car. The forward driving force from the tyres is:

$F_{\text{drive}} = F_{\text{net}} + F_{\text{resist}} = 3600 + 800 = 4400 \text{ N}$

Common Top-band-vs-mid-band discriminator: the candidate who writes $F = ma = 3600$ N and stops gets the resultant right but doesn't name it as the resultant or recognise that the engine has more work to do. Always specify which force you are calculating.

Worked Example 2 — Lift Accelerating Upwards

A $70$ kg person stands on bathroom scales in a lift accelerating upwards at $1.5$ m s $^{-2}$ . What do the scales read?

The scales measure the normal contact force $N$ on the person. Let upward be positive.

Weight down: $W = mg = 70 \times 9.81 = 686.7$ N.
Newton 2 vertical: $N - W = ma$ .
$N = m(g + a) = 70 \times (9.81 + 1.5) = 70 \times 11.31 = 792$ N (3 s.f.).

The person feels "heavier" by 105 N (about 15%) because the floor must push up harder than the weight to accelerate them upward. For a lift accelerating downward at $1.5$ m s $^{-2}$ the same calculation gives $N = m(g - a) = 70 \times 8.31 = 582$ N (15% lighter). For a free-falling lift ( $a = -g$ ): $N = m(g - g) = 0$ — apparent weight zero, exactly the sensation astronauts feel in orbit.

Worked Example 3 — Atwood Machine

Two blocks of masses $m_1 = 3.0$ kg (heavier) and $m_2 = 2.0$ kg (lighter) are joined by a light, inextensible string passing over a frictionless pulley. Find their common acceleration and the string tension.

Apply Newton 2 to each block separately. Let $a$ be the downward acceleration of $m_1$ (equal to the upward acceleration of $m_2$ by the string constraint).

For $m_1$ (downward positive): $\quad m_1 g - T = m_1 a$

For $m_2$ (upward positive): $\quad T - m_2 g = m_2 a$

Add the two equations:

$(m_1 - m_2)g = (m_1 + m_2)a \quad \Rightarrow \quad a = \frac{(m_1 - m_2)g}{m_1 + m_2} = \frac{1.0 \times 9.81}{5.0} = 1.96 \text{ m s}^{-2}$

Substitute into the $m_2$ equation:

$T = m_2(g + a) = 2.0 \times (9.81 + 1.96) = 23.5 \text{ N}$

Cross-check using the $m_1$ equation: $T = m_1(g - a) = 3.0 \times (9.81 - 1.96) = 3.0 \times 7.85 = 23.5$ N. ✓

Whenever two bodies are connected by an inextensible string, Newton 2 must be applied to each body separately, and the equations solved simultaneously. The constraint $a_1 = a_2$ (in magnitude) closes the system.

Worked Example 4 — Block on a Frictionless Incline

A $5.0$ kg block slides down a frictionless incline at $25°$ to the horizontal. Find (a) its acceleration and (b) the normal contact force.

Resolve gravity into components along ( $\parallel$ ) and perpendicular ( $\perp$ ) to the slope.

(a) Along the slope (downslope positive): $\quad mg \sin\theta = ma \Rightarrow a = g \sin\theta = 9.81 \sin 25° = 4.14$ m s $^{-2}$ .

(b) Perpendicular to slope (no perpendicular motion ⇒ equilibrium): $\quad N - mg\cos\theta = 0 \Rightarrow N = mg\cos\theta = 5.0 \times 9.81 \cos 25° = 44.4$ N.

The mass cancels in (a) — every frictionless incline at $25°$ gives the same acceleration regardless of mass, because gravity scales with mass. This is the same equivalence-principle insight that drives Galilean free-fall: all bodies accelerate at $g$ in vacuum.

Only the component of gravity along the slope drives acceleration; the perpendicular component is balanced by $N$ .

Newton 2 + SUVAT — The Canonical Dynamics Workflow

Newton 2 gives you the acceleration; SUVAT then gives you the kinematics. The standard Paper-1 dynamics question runs:

Draw a free-body diagram for each body.
Apply Newton 2 along the relevant axes to find $a$ .
Feed $a$ into SUVAT ( $v = u + at$ , $s = ut + \tfrac{1}{2}at^2$ , $v^2 = u^2 + 2as$ , etc.) to find time, displacement, final velocity.

Example workflow: for the incline block above, if released from rest and the incline is $8.0$ m long, the block reaches the bottom with velocity $v = \sqrt{2as} = \sqrt{2 \times 4.14 \times 8.0} = 8.14$ m s $^{-1}$ , in time $t = v/a = 8.14/4.14 = 1.97$ s. Newton 2 → $a$ ; SUVAT → trajectory.

Note: SUVAT requires constant acceleration. If $F_{\text{net}}$ depends on $v$ (e.g. drag, terminal velocity, lesson 1 worked example 5), then $a$ is non-constant and SUVAT is invalid — the problem must be solved numerically or with calculus.

Variable-Mass Systems — The Full F = dp/dt

When mass changes with time, apply the product rule to $\vec{p} = m\vec{v}$ :

$\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt} = m\frac{d\vec{v}}{dt} + \vec{v}\frac{dm}{dt}$

$m\,d\vec{v}/dt$ is the familiar $ma$ term — force needed to accelerate the existing mass.
$\vec{v}\,dm/dt$ is the thrust term — force needed to bring incoming mass (or eject outgoing mass) up to the body's velocity, or to recoil against ejected mass.

For A-Level, the qualitative existence of this term is a Top-band discriminator. Quantitative variable-mass problems are unusual at A-Level but not unheard of.

Worked Example 5 — Rain-Collecting Cart

A $20$ kg flat-bed cart rolls along frictionless rails at $\vec{v} = 3.0$ m s $^{-1}$ . Rain falls vertically and lands on the cart at $dm/dt = 0.50$ kg s $^{-1}$ . Find the forward horizontal force needed to keep the cart at constant velocity.

The rain has zero horizontal velocity when it lands; to bring it up to the cart's $3.0$ m s $^{-1}$ , the cart must push it forward. Apply $\vec{F} = d\vec{p}/dt$ to the cart-plus-rain system, horizontally:

Since the cart's velocity is constant, $m\,dv/dt = 0$ . The whole force goes into accelerating incoming rain:

$F = v \frac{dm}{dt} = 3.0 \times 0.50 = 1.5 \text{ N forward}$

If no force is applied, the system's momentum is conserved (a closed system), but velocity drops as mass grows: $v_{\text{new}} = p / m_{\text{new}} < v_{\text{old}}$ (covered formally in lesson 6).

Worked Example 6 — Rocket Thrust (Qualitative)

A rocket ejects burnt fuel rearwards at high exhaust speed $v_{\text{ex}}$ relative to the rocket. By Newton 2 + 3 applied to the rocket-plus-exhaust system, the forward thrust on the rocket is:

$F_{\text{thrust}} = v_{\text{ex}} \frac{dm}{dt}$

where $dm/dt$ is the rate of mass ejection (positive). Typical chemical rocket: $v_{\text{ex}} \sim 3000$ m s $^{-1}$ , $dm/dt \sim 2000$ kg s $^{-1}$ at lift-off, giving $F_{\text{thrust}} \sim 6 \times 10^6$ N — enough to lift a Saturn V (mass $\sim 3 \times 10^6$ kg) against gravity. As fuel burns, the rocket's mass shrinks, so for constant thrust the acceleration grows — accounting for the late-stage rapid acceleration of any rocket flight. We revisit rockets quantitatively in lesson 10 (explosions and recoil).

Newton 2 in Words — The Full-Mark Statement

When an extended-prose question asks "state Newton's second law", the canonical wording is:

The resultant force acting on a body is directly proportional to the rate of change of its linear momentum. The change in momentum takes place in the direction of the resultant force.

Either "rate of change of momentum" or "force equals mass times acceleration" earns credit at A-Level, but the rate-of-change form is the higher-quality answer because it generalises to variable-mass systems. The OCR data booklet contains both $\vec{F} = d\vec{p}/dt$ and $\vec{F} = m\vec{a}$ .

Common wording faults:

"Force equals mass times acceleration" alone — earns credit, but is the constant-mass special case.
"Force causes acceleration" — too vague; loses the direct-proportionality and the rate-of-change-of-momentum framing.
Omitting "resultant force" — drops the same mark that mid-band Newton-1 answers drop.
Omitting the direction clause — Newton 2 is a vector equation, and OCR mark schemes credit the "direction of resultant force" phrase.

Newton's Second Law — F = Δp/Δt

Newton's Second Law — F = Δp/Δt

Statement of Newton's Second Law

Deriving F = ma from F = dp/dt

Diagram — When to use which form

Units and the Definition of the Newton

Vector Form and Components

Mass and Weight

Worked Example 1 — Constant-Mass Car

Worked Example 2 — Lift Accelerating Upwards

Worked Example 3 — Atwood Machine

Worked Example 4 — Block on a Frictionless Incline

Newton 2 + SUVAT — The Canonical Dynamics Workflow

Variable-Mass Systems — The Full F = dp/dt

Worked Example 5 — Rain-Collecting Cart

Worked Example 6 — Rocket Thrust (Qualitative)

Newton 2 in Words — The Full-Mark Statement

Synoptic Links

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