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Mechanics · Subtopic 2 of 8

Newton's Laws and Momentum

An object keeps its velocity unless a resultant force acts (first law). The resultant force equals the rate of change of momentum, F=ΔpΔtF=\dfrac{\Delta p}{\Delta t}, which is F=maF=ma for constant mass (second law). Forces come in equal and opposite pairs on different objects (third law). Drag grows with speed, so falling objects reach a terminal velocity.

Cambridge 9702 Paper 1Cambridge 9702 Paper 2Edexcel YPH11 WPH11AQA 7408 Paper 1≈ 40 minRevision 5 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • State and apply Newton's three laws.
  • Use F=maF=ma and F=ΔpΔtF=\dfrac{\Delta p}{\Delta t} with free-body diagrams.
  • Distinguish mass and weight; solve lift problems.
  • Explain terminal velocity in terms of drag.
Pause and think

A horse pulls a cart. By Newton's third law, the cart pulls back on the horse with an equal force. So how does anything ever move?

Section 2 of 14

Core idea

Start hereThe force of a hose

Point a hose at a wall and you can feel the wall being pushed. Nothing solid touches it, just a stream of water losing its momentum. Newton's second law is really about the rate of change of momentum.

The big idea

F=ΔpΔtF=\frac{\Delta p}{\Delta t} (gives F=maF=ma for constant mass). Area under FF–tt = impulse. Third-law pairs: equal, opposite, same type, on different bodies.

1. The laws

  • 1st: constant velocity unless a resultant force acts.
  • 2nd: F=d(mv)dtF=\frac{d(mv)}{dt}.
  • 3rd: A on B equals and opposes B on A, same type.
From F=ΔpΔtF=\frac{\Delta p}{\Delta t} to F=maF=ma
  1. F=Δ(mv)ΔtF=\frac{\Delta(mv)}{\Delta t}
    General form.
  2. =mΔvΔt=m\frac{\Delta v}{\Delta t}
    If mm is constant.
  3. =ma=ma
    Definition of aa.
If instead vv is constant and mass flows at rate ΔmΔt\frac{\Delta m}{\Delta t}: F=vΔmΔtF=v\frac{\Delta m}{\Delta t}.

2. Impulse

Force–time graph: the area is the change in momentum
246810122004006008000time (ms)force (N)impulse = Δp
A kick: the force rises and falls over 10 ms. The area (≈ 5.1 N s) is the momentum given to the ball.
Worked exampleA kick

Impulse (area) 5.1 N s on a 0.45 kg ball at rest; contact 10 ms.

Goal
Speed; average force.
Strategy
Impulse = Δp\Delta p.
  1. v=5.10.45=11.3 m s−1v=\frac{5.1}{0.45}=11.3\text{ m s}^{-1}
    Δp=mv\Delta p=mv.
  2. Fˉ=5.10.010=510 N\bar F=\frac{5.1}{0.010}=510\text{ N}
    Average force.
Answer11.311.3 m s−1^{-1}; 510510 N.
✓ CheckPeak force on the graph exceeds the average ✓.
Worked exampleHose on a wall

0.5 kg s−1^{-1} at 12 m s−1^{-1}, stopped dead.

Goal
Force.
Strategy
F=vΔmΔtF=v\frac{\Delta m}{\Delta t}.
  1. ΔpΔt=12×0.5\frac{\Delta p}{\Delta t}=12\times0.5
    Momentum lost per second.
  2. =6 N=6\text{ N}
    Force on the water; equal and opposite on the wall.
Answer66 N.
✓ CheckIf the water bounced back, the force would double ✓.

3. Third-law pairs

Newton's third law pairs act on different bodies
ABpull on A by Bpull on B by A
Equal size, opposite direction, same type (here both electrostatic or both contact), acting on different objects. Weight and normal reaction on one book are not a third-law pair.
Common misconceptionWeight and reaction A book's weight and the table's normal force both act on the book and are different types: not a third-law pair.
Pause and think

What is the third-law partner of the book's weight?

Summary

  • F=Δp/ΔtF=\Delta p/\Delta t.
  • Impulse = area under FF–tt.
  • Third-law pairs on different bodies.
Key words
Impulse
Change in momentum.
History"If I have seen further…" Newton built on Galileo's work on inertia and Descartes' ideas of momentum. His Principia (16871687) stated the three laws so clearly that they ran engineering for over 200200 years, and still do for anything slower than light and bigger than an atom.
Fun factFelix Baumgartner's jump. In 20122012 Felix Baumgartner jumped from 3939 km up. In the thin air his terminal velocity was so high that he broke the sound barrier, reaching about 13601360 km/h, before the thicker air lower down slowed him.
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