Learn4PH1Paper 1P · Paper 1 (core content)Forces and motionForces and motion

Forces and motion · Subtopic 3 of 6

Forces and Motion

If the forces on an object are balanced, it stays at rest or keeps moving at constant velocity. An unbalanced (resultant) force makes it accelerate: $F=ma$ . Friction and air resistance oppose motion, which is why a falling object reaches a terminal velocity. Moving in a circle needs a force towards the centre.

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Section 1 of 14

Learning objectives

Find the resultant of forces acting along a line.
Explain that balanced forces mean constant velocity (or rest).
Use $F=ma$ to find force, mass or acceleration.
Describe friction, drag, terminal velocity and the force needed for circular motion.

Pause and think

A car drives along a straight motorway at a steady $30$ m/s with its engine pushing it forward. Is there a resultant force on it?

Section 2 of 14

Core idea

Start hereWhy seatbelts and why not to tailgate

A car doesn't stop the instant you see danger. First you react, travelling at full speed; then the brakes take time to remove the car's motion. Forces, mass and acceleration explain both distances, and why speed matters so much.

The big idea

Resultant force $F=ma$ . Balanced forces: constant velocity or rest. $W=mg$ ( $g=10$ N/kg). Stopping distance = thinking + braking.

1. Resultant force

A car: balanced and unbalanced forces

Driving force minus drag gives the resultant.

a=\dfrac{F}{m}=\dfrac{2000}{1000}=2

m/s².

F=ma,\qquad W=mg

$F$: resultant force (N).
$m$: mass (kg).
$g$: 10 N/kg on Earth.

Worked exampleAccelerating car

1200 kg car; driving force 3000 N; resistance 600 N.

Goal: Acceleration.
Strategy: Resultant first.

$F=3000-600=2400\text{ N}$
Forces in opposite directions subtract.
$a=\frac{2400}{1200}=2\text{ m/s}^2$
$a=F/m$ .

Answer

2

m/s².

✓ CheckWith balanced forces (3000 = 3000)

a

would be 0 ✓.

Common misconceptionMass vs weight A 60 kg person weighs 600 N on Earth and about 100 N on the Moon, but their mass is 60 kg in both places.

2. Stopping distances

Stopping distance = thinking distance + braking distance

Thinking distance grows in proportion to speed; braking distance grows with speed squared. Tiredness, alcohol, wet roads and worn tyres all make it longer.

Braking distance from energy

$Fd=\tfrac12mv^2$
Work done by brakes removes kinetic energy.
$d=\frac{mv^2}{2F}=\frac{v^2}{2a}$
$F=ma$ .

Braking distance

\propto v^2

: double the speed, four times the distance.

Worked exampleAt 20 m/s

Reaction time 0.7 s; deceleration 5 m/s².

Goal: Stopping distance.
Strategy: Thinking at constant speed; braking from the formula.

$20\times0.7=14\text{ m}$
Thinking distance.
$\frac{20^2}{2\times5}=40\text{ m}$
Braking distance.
$14+40=54\text{ m}$
Total.

Answer

54

✓ CheckAt 40 m/s:

28+160=188

m, far more than double ✓.

Thinking distance: tiredness, alcohol, phones.
Braking distance: speed, wet roads, worn tyres/brakes.

Pause and think

Why does braking distance quadruple when speed doubles?

Summary

$F=ma$ with the resultant.
$W=mg$ .
Stopping = thinking + braking.

Resultant force: Overall force after combining all forces.

HistoryNewton's Principia. In

1687

Isaac Newton published the Principia, setting out his three laws of motion and universal gravitation. It was written in Latin and so hard to read that a student is said to have remarked, "There goes the man who has writ a book that neither he nor anybody else understands."

Fun factThe cat's terminal velocity. Cats falling from tall buildings often survive better from higher floors than from medium ones. Once they reach their terminal velocity (about

100

km/h) they relax and spread out like a parachute, increasing air resistance and landing more evenly.

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