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GCSE/Physics/OCR

P2Forces — scalars/vectors, Newton's laws, momentum, force-extension

Notes

P2 Forces

Scalars and vectors

A scalar has magnitude only. A vector has both magnitude and direction.

ScalarsVectors
Speed, distance, mass, energy, time, temperatureVelocity, displacement, force, acceleration, momentum, weight

Vectors are represented by arrows: length shows magnitude, direction shows direction. To add vectors not in line, draw a scale diagram or use a right-angled triangle + Pythagoras.

Resolving vectors (Higher): split a vector into horizontal and vertical components.

  • F_x = F cos θ (horizontal component)
  • F_y = F sin θ (vertical component)

Forces and Newton's laws

Newton's 1st Law: An object remains at rest or moves at constant velocity unless acted on by a resultant force.

Newton's 2nd Law:

F = ma

where F = resultant force (N), m = mass (kg), a = acceleration (m/s²).

Weight: W = mg where g = 10 N/kg (or 9.8 N/kg if specified).

Newton's 3rd Law: When object A exerts a force on object B, object B exerts an equal and opposite force on object A (same type, same line of action, but opposite direction).

Free body diagrams

Draw a dot representing the object. Draw all forces as arrows from the dot, labelled with magnitude and direction. Check whether the resultant is zero (equilibrium) or non-zero (acceleration).

Terminal velocity: when drag = driving force, resultant = 0, acceleration = 0, constant velocity is reached.

Momentum

Momentum (p) = mass × velocity:

p = mv (kg m/s)

Conservation of momentum: in a closed system (no external forces), total momentum before a collision = total momentum after.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Impulse (Higher):

Impulse = FΔt = Δp (change in momentum)

Larger impact time → smaller force for same Δp (crumple zones, air bags).

Newton's 2nd law in terms of momentum (Higher):

F = Δp / Δt = m(v − u) / t = ma

Stopping distances

  • Thinking distance = speed × reaction time (affected by: tiredness, alcohol, drugs, distraction).
  • Braking distance = depends on speed², road condition, tyre condition, brake condition (PAG P2.1 — ruler drop for reaction time).
  • Stopping distance = thinking + braking.

Braking distance ∝ v² because KE = ½mv² — doubling speed quadruples KE, so quadruples braking distance.

Force-extension (Hooke's law)

F = ke

where F = force (N), k = spring constant (N/m), e = extension (m).

This applies up to the limit of proportionality. Beyond the elastic limit, the spring does not return to its original length.

Work done stretching a spring (elastic potential energy):

Eₑ = ½ke²

PAG P2.2: Investigate the relationship between force and extension for a spring. Plot F vs e; gradient = k. Identify the limit of proportionality.

Weight and gravitational field

W = mg

On Earth, g ≈ 10 N/kg. Weight is a force (vector); mass is a scalar.

Common mistakes

  1. Confusing mass and weight: mass is in kg, weight is in Newtons.
  2. Newton's 3rd law pairs: forces must be same type (both contact, both gravitational etc.), on different objects.
  3. Thinking distance vs braking distance: thinking ∝ v; braking ∝ v².
  4. Momentum direction: always assign + and − directions before calculating.
  5. Hooke's law — extension not length: measure extension (change in length), not total length.

AI-generated · claude-opus-4-7 · v3-ocr-physics

Practice questions

Try each before peeking at the worked solution.

  1. Question 16 marks

    Newton's 2nd law — F = ma

    OCR J259/01 — Foundation/Higher

    A car of mass 1200 kg accelerates from rest to 18 m/s in 9.0 s.

    (a) Calculate the acceleration of the car. (2 marks)

    (b) Calculate the resultant force acting on the car. (2 marks)

    (c) In practice, the engine force is 2800 N. Explain why the actual acceleration is less than your calculated value. (2 marks)

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    AI-generated · claude-opus-4-7 · v3-ocr-physics

  2. Question 26 marks

    Conservation of momentum — collision

    OCR J259/02 — Foundation/Higher

    A trolley of mass 0.80 kg moves at 3.0 m/s and collides with a stationary trolley of mass 1.20 kg. After the collision the two trolleys stick together.

    (a) Show that the velocity of the combined trolleys after the collision is 1.2 m/s. (3 marks)

    (b) Calculate the kinetic energy before and after the collision. State whether kinetic energy is conserved. (3 marks)

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    AI-generated · claude-opus-4-7 · v3-ocr-physics

  3. Question 36 marks

    Hooke's law — force-extension graph (PAG)

    OCR J259/01 — Foundation/Higher (PAG P2.2)

    A student hangs masses from a spring and records the extension.

    Force (N)Extension (cm)
    00
    24
    48
    612
    822

    (a) State Hooke's law. (1 mark)

    (b) Calculate the spring constant for the linear region. Give the unit. (3 marks)

    (c) Identify the point at which the spring exceeds the limit of proportionality and explain how you can tell from the data. (2 marks)

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  4. Question 46 marks

    Stopping distances — thinking and braking

    OCR J259/01 — Foundation

    A car travels at 30 m/s. The driver's reaction time is 0.7 s.

    (a) Calculate the thinking distance. (2 marks)

    (b) The braking distance at 30 m/s is 45 m. Calculate the total stopping distance. (1 mark)

    (c) The car then travels at 60 m/s. Explain why the braking distance at 60 m/s is NOT simply double the value at 30 m/s. (3 marks)

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  5. Question 57 marks

    Impulse and force — crumple zones (Higher)

    OCR J259/02 — Higher

    In a crash test, a car of mass 1500 kg decelerates from 13 m/s to 0 m/s.

    (a) Calculate the change in momentum of the car. (2 marks)

    (b) The crash takes 0.15 s. Calculate the average force on the car. (2 marks)

    (c) With a crumple zone, the same crash takes 0.40 s. Calculate the force in this case and explain the safety benefit. (3 marks)

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    AI-generated · claude-opus-4-7 · v3-ocr-physics

Flashcards

P2 — Forces — scalars/vectors, Newton's laws, momentum, force-extension

9-card SR deck for OCR GCSE Physics (J259 Gateway) topic P2

9 cards · spaced repetition (SM-2)