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AS & A-Level Physics 03 — Forces and Momentum

AS & A-Level Physics 03 — Forces and Momentum

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Independent Deckloop revision aligned with the Cambridge International AS & A Level Physics (9702) syllabus, 2025–2027. Not affiliated with or endorsed by Cambridge International Education. Chapter 3 of 26: Forces and momentum. Concepts, worked applications and misconception checks.

Physics EN AS & A-Level
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Newton's Second Law: F = ma

Newton's Second Law of Motion describes the relationship between resultant force, mass, and acceleration for an object of constant mass. The equation F=maF = ma states that the resultant force (FF) acting on an object is equal to the product of its mass (mm) and its acceleration (aa). It is vital to understand that FF represents the vector sum of all forces acting on the object, often called the net force. Since force and acceleration are vector quantities, this equation implies they always point in the same direction. A resultant force causes an object to accelerate, and the magnitude of this acceleration is directly proportional to the force and inversely proportional to the mass. The SI unit of force, the newton (N), is defined from this relationship: one newton is the force required to give a 1 kg mass an acceleration of 1 m s⁻².

Key points

  • The resultant force on an object is the product of its mass and acceleration: Fresultant=maF_{resultant} = ma.
  • FresultantF_{resultant} is the vector sum of all individual forces acting on the object.
  • The acceleration vector (aa) is always in the same direction as the resultant force vector (FresultantF_{resultant}).
  • The SI unit of force is the newton (N), where 1 N=1 kg m s−21 \text{ N} = 1 \text{ kg m s}^{-2}.

Worked example

Question

A model rocket of mass 2.5 kg has an engine that provides a constant upward thrust of 40 N. Assuming air resistance is negligible, calculate its initial upward acceleration. Use g=9.81 m s−2g = 9.81 \text{ m s}^{-2}.

Solution

1. Identify all forces acting on the rocket. There is an upward thrust (T=40 NT = 40 \text{ N}) and a downward gravitational force, its weight (WW).
2. Calculate the weight of the rocket: W=mg=2.5 kg×9.81 m s−2=24.525 NW = mg = 2.5 \text{ kg} \times 9.81 \text{ m s}^{-2} = 24.525 \text{ N}.
3. Determine the resultant force, FresultantF_{resultant}. Taking upwards as the positive direction: Fresultant=T−W=40 N−24.525 N=15.475 NF_{resultant} = T - W = 40 \text{ N} - 24.525 \text{ N} = 15.475 \text{ N}.
4. Apply Newton's Second Law to find the acceleration: a=Fresultant/m=15.475 N/2.5 kg=6.19 m s−2a = F_{resultant} / m = 15.475 \text{ N} / 2.5 \text{ kg} = 6.19 \text{ m s}^{-2}.
5. State the final answer to an appropriate number of significant figures.

The initial upward acceleration is 6.2 m s⁻².

Common pitfalls

  • Using only one of the applied forces for 'F' instead of the resultant (net) force. Always sum all forces vectorially first.
  • Forgetting to include non-obvious forces like weight, friction, or normal contact force when calculating the resultant force.

Prerequisites

  • Understanding of mass is required.
  • Requires definition and calculation of acceleration.
  • Requires addition of vectors (forces) to find the resultant.
Further resources