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GCSE/Combined Science/OCR

P2.2Newton’s laws: balanced and unbalanced forces, F = ma, free-body diagrams

Notes

Newton's laws (P2.2)

Gateway A's flagship physics topic. Expect at least one calculation using F = ma, a free-body diagram, and a 6-mark question on safety / stopping distances or momentum.

Newton's three laws

Newton's First Law — inertia

A body continues at rest, or moves at a constant velocity in a straight line, unless acted on by a resultant force.

In other words: if forces are balanced (resultant force = 0), velocity is constant. The body could be stationary or moving at the same speed in the same direction.

Example: a parachutist at terminal velocity. Weight downward = drag upward → resultant = 0 → constant velocity.

Newton's Second Law — F = ma

F (N) = m (kg) × a (m/s²)

The resultant force on a body is proportional to its mass and to its acceleration. Always use resultant force in this equation — not the size of any one force.

Inertial mass: a measure of how difficult it is to change a body's velocity. Higher mass → harder to accelerate.

Newton's Third Law — action–reaction pairs

For every action there is an equal and opposite reaction.

Important: the action and reaction forces act on different objects. So they don't cancel each other out on a single body.

Example: a book on a table. The book pushes down on the table (action); the table pushes up on the book (reaction). The pair are equal in size, opposite in direction, on different bodies.

Resultant force and balanced forces

Forces are vectors — they have direction. To find a resultant on a single line:

  • If forces in the same direction: add.
  • If forces in opposite directions: subtract.

If the resultant = 0, forces are balanced. The object is either stationary or moving at constant velocity.

If the resultant ≠ 0, forces are unbalanced. The object accelerates in the direction of the resultant.

Free-body diagrams

Draw the body as a dot or rectangle. Draw arrows for every force acting on the body — never forces it exerts on others. Label every force with name and (if known) magnitude.

Common labels:

  • Weight (W or mg) — always downward.
  • Normal contact force (N or R) — perpendicular to surface.
  • Friction (F or f) — opposes motion, parallel to surface.
  • Drag (air resistance) — opposes motion through fluid.
  • Thrust / push / tension — applied force.

Worked F = ma examples

Example 1. A 1,200 kg car accelerates at 2.5 m/s². Find the resultant force.

  • F = m × a = 1,200 × 2.5 = 3,000 N.

Example 2. A 70 kg cyclist experiences a forward force of 200 N from pedalling and a 60 N backward drag/friction.

  • Resultant = 200 − 60 = 140 N.
  • a = F / m = 140 / 70 = 2 m/s².

Example 3. A 60 kg lift descends at constant velocity. The cable tension is 588 N. What is the cable tension if the lift then accelerates downward at 1.5 m/s²?

  • Constant velocity: T = mg = 60 × 9.8 ≈ 588 N. ✓
  • Accelerating down at 1.5 m/s²: resultant force on lift is downward, magnitude m × a = 60 × 1.5 = 90 N.
  • Resultant = mg − T (downward), so T = mg − ma = 588 − 90 = 498 N.

Stopping distance

stopping distance = thinking distance + braking distance
  • Thinking distance is the distance travelled during the driver's reaction time. Increased by tiredness, alcohol, drugs, distractions.
  • Braking distance is the distance after the brakes are applied. Increased by greater speed, worn tyres, wet/icy roads, worn brakes.

Doubling the speed roughly quadruples the braking distance (because KE ∝ v²).

Common Gateway-paper mistakes

  1. Using a single force in F = ma instead of the resultant.
  2. Forgetting that constant velocity ≠ zero force — it means zero RESULTANT force.
  3. Mixing up Newton's pair: writing that weight and normal contact force are a Newton-third-law pair (they are NOT — both act on the same body).
  4. Confusing mass (kg) with weight (N): W = mg.
  5. Forgetting to convert g to kg in F = ma calculations.

Try thisQuick check

  • A 0.50 kg ball is thrown with a force of 7.5 N. Acceleration = 7.5 / 0.50 = 15 m/s².
  • A 1,500 kg car decelerates from 30 m/s to 0 in 10 s. a = (0 − 30) / 10 = −3 m/s². F = 1,500 × −3 = −4,500 N (i.e. 4,500 N opposing motion).

AI-generated · claude-opus-4-7 · v3-ocr-combined-science

Practice questions

Try each before peeking at the worked solution.

  1. Question 12 marks

    F = ma calculation

    A 60 kg cyclist accelerates at 1.5 m/s² along a flat road.

    Calculate the resultant force acting on the cyclist.

    [2 marks]

    Ask AI about this

    AI-generated · claude-opus-4-7 · v3-ocr-combined-science

  2. Question 24 marks

    Resultant force from balanced and unbalanced

    A 1,400 kg car experiences a forward force of 4,800 N from the engine and a backward force of 1,300 N from drag and friction.

    (a) Calculate the resultant force on the car. [1]
    (b) Calculate the acceleration of the car. [2]
    (c) State what happens to the resultant force if the car reaches a constant top speed. [1]

    [4 marks]

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    AI-generated · claude-opus-4-7 · v3-ocr-combined-science

  3. Question 33 marks

    Newton's First Law

    Explain, in terms of forces, why a parachutist falling at terminal velocity moves at a constant speed.

    [3 marks]

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    AI-generated · claude-opus-4-7 · v3-ocr-combined-science

  4. Question 45 marks

    Free-body diagram

    A 5 kg book rests on a flat horizontal table.

    (a) Draw a free-body diagram of the book showing all the forces on it. Label each. [3]
    (b) Calculate the size of each force. (g = 9.8 N/kg) [2]

    [5 marks]

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    AI-generated · claude-opus-4-7 · v3-ocr-combined-science

  5. Question 56 marks

    Stopping distance (6-marker)

    A car driver sees a cyclist 70 m ahead and applies the brakes.

    The driver's thinking distance at 20 m/s is 12 m. The braking distance at 20 m/s is 28 m.

    The driver successfully stops 30 m before the cyclist.

    (a) Calculate the total stopping distance. [1]
    (b) State TWO factors that could increase the THINKING distance. [2]
    (c) State THREE factors that could increase the BRAKING distance. [3]

    [6 marks]

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    AI-generated · claude-opus-4-7 · v3-ocr-combined-science

  6. Question 64 marks

    Mass vs weight

    A scientist's mass is 65 kg on Earth.

    (a) Calculate her weight on Earth. (g_Earth = 9.8 N/kg) [1]
    (b) Calculate her weight on the Moon. (g_Moon = 1.6 N/kg) [1]
    (c) State her mass on the Moon. Explain. [2]

    [4 marks]

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    AI-generated · claude-opus-4-7 · v3-ocr-combined-science

Flashcards

P2.2 — Newton’s laws: balanced and unbalanced forces, F = ma, free-body diagrams

12-card SR deck for OCR Combined Science (J250) topic P2.2

12 cards · spaced repetition (SM-2)