Newton's Second Law — F = Δp/Δt

Newton's Second Law is the quantitative heart of mechanics. It tells us how much an object accelerates when a given resultant force acts on it. For OCR A-Level Physics A candidates, the Second Law is frequently tested in both its classic form F = ma and its more general form F = Δp/Δt, which is the version stated on the specification and in the data booklet.

This lesson derives both forms carefully, shows when each applies, and works through several exam-style calculations, including variable-mass problems (rockets, rain-collecting carts) that catch out weaker candidates.

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Statement of Newton's Second Law

The resultant force acting on a body is equal to the rate of change of its linear momentum.

Mathematically:

F = Δp/Δt

where:

• F is the resultant force in newtons (N),
• p = mv is the linear momentum in kg m s⁻¹,
• Δt is the time interval in seconds.

The law is a vector equation: force and the change in momentum point in the same direction. The resultant force matters — if more than one force acts, first take their vector sum and then apply the law.

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Deriving F = ma

Writing momentum as p = mv, we have:

F = Δ(mv)/Δt

If the mass m is constant, it comes out of the derivative:

F = m Δv/Δt = m a

because a = Δv/Δt. This is the form most students meet first at GCSE, and it is perfectly valid whenever the mass of the body does not change — that is, in the vast majority of A-Level mechanics problems.

If the mass changes with time (rockets burning fuel, trucks collecting rain, conveyor belts loading sand) you must use the full F = Δp/Δt form. We will come back to that at the end of the lesson.

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Diagram — Second Law Flow

flowchart LR
  A["Resultant force F (N)"] --> B["Rate of change of momentum Δp/Δt"]
  B --> C{"Is mass constant?"}
  C -- Yes --> D["F = m a"]
  C -- No --> E["F = v·(Δm/Δt) + m·(Δv/Δt)"]

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Units — The Definition of the Newton

The newton is defined through the Second Law: 1 N is the resultant force needed to give a 1 kg mass an acceleration of 1 m s⁻². So:

1 N = 1 kg m s⁻²

This is worth writing on the front page of your revision notes. Whenever a calculation gives you units of kg m s⁻², you know it is a force in newtons.

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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. Find the resultant force needed (assume straight line motion).

• Δv = 25 − 10 = 15 m s⁻¹
• a = Δv/Δt = 15 / 5.0 = 3.0 m s⁻²
• F = m a = 1200 × 3.0 = 3600 N

Notice that this is the resultant force — the forward driving force minus air resistance and friction. If air resistance is 800 N, the engine must provide 3600 + 800 = 4400 N.

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Worked Example 2 — Lift Accelerating Upwards

A 70 kg person stands in a lift that accelerates upwards at 1.5 m s⁻². Find the normal contact force the floor exerts on them.

• Weight down: W = mg = 70 × 9.81 = 686.7 N
• Let N = normal force up (unknown).
• Take upwards as positive. Resultant force = N − W = m a.
• N = m(g + a) = 70 × (9.81 + 1.5) = 70 × 11.31 = 791.7 N ≈ 792 N

The person feels "heavier" because the floor pushes up harder than their weight. In a lift accelerating downwards at 1.5 m s⁻², the same calculation gives N = 70 × (9.81 − 1.5) = 582 N, and the person feels lighter. In free fall (a = 9.81 m s⁻² down), N = 0 and the person is weightless.

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Worked Example 3 — Two Blocks on a String

Two blocks of masses m₁ = 3.0 kg and m₂ = 2.0 kg are joined by a light, inextensible string over a frictionless pulley (an Atwood machine). Find their acceleration and the tension in the string.

Treat each mass separately. Let the heavier mass (3.0 kg) descend with acceleration a; by the string constraint the lighter mass rises with the same a.

For m₁ (down positive):

m₁ g − T = m₁ a

For m₂ (up positive):

T − m₂ g = m₂ a