Lesson B · Specification 1.11–1.21
This is the lesson where students start writing “because.” Every force has a direction. The resultant is the only force you put into $F=ma$.
The idea I want in your head before any formula
A force is a push or a pull. It is not “energy” and it is not “speed.” Forces change motion or shape. If two people pull a rope equally hard in opposite directions, there are large forces, but nothing accelerates. That is why we spend so long on the word resultant.
1.11, 1.12, 1.16
Specification 1.11 is short and it is examined every year: a force can change an object’s speed, shape or direction. I add a classroom fourth — start or stop motion — which is really a change of speed from or to zero.
| Force | Contact? | Direction I want on a diagram | Everyday example |
|---|---|---|---|
| Gravitational (weight) | Non-contact | Towards the centre of the Earth, from the object’s centre | You standing; a falling apple |
| Electrostatic | Non-contact | Along the line between charges; attract or repel | A rubbed balloon sticking to a wall |
| Magnetic | Non-contact | Towards / away from a pole | Fridge magnet |
| Friction | Contact | Opposes motion (or the tendency to move) | Shoes on a floor; brakes |
| Air resistance / drag | Contact (with air) | Opposite to the velocity | Skydiver; cycling into wind |
| Normal contact | Contact | Perpendicular to the surface, outwards | The desk pushing up on a book |
| Tension | Contact | Along a string / spring, away from the object | Towing a trailer |
| Upthrust | Contact (fluid) | Upwards | A floating boat |
1.12 specifically names gravitational and electrostatic. 1.16 is a sentence you should be able to write in your sleep: friction is a force that opposes motion. If a box is being pulled right, friction is left. If a car is moving right and you brake, friction at the road is left. Friction can also help you walk — it opposes the foot slipping backwards, so it acts forwards. The examiner’s safe line is still “opposes motion / opposes sliding.”
1.13, 1.14
Scalar — magnitude only
Mass, distance, speed, time, energy, temperature, charge. A number and a unit is enough: $60\,\text{kg}$, $12\,\text{s}$.
Vector — magnitude and direction
Displacement, velocity, acceleration, force, weight. $10\,\text{N}$ right is not the same as $10\,\text{N}$ left.
1.14 is one mark if they ask “explain why force is a vector.” You write: “Force has magnitude and direction.” That is the whole answer. Do not write a paragraph.
On a diagram we draw a force as an arrow. The length represents size; the arrowhead is the direction. I want arrows that start on the object (or at its centre for weight). Two arrows the same length in opposite directions mean balanced forces of equal size.
1.15
This paper only asks you to add forces that sit on one line — forwards/backwards or up/down. You do not need Pythagoras or components.
Method I write on the board every time
Example 1 — two forces
$18\,\text{N}$ right, $6\,\text{N}$ left. Resultant $= 18 - 6 = 12\,\text{N}$ right.
Example 2 — three forces
A boat: driving force $500\,\text{N}$ forwards, water resistance $200\,\text{N}$, wind $50\,\text{N}$ backwards. Resultant $= 500 - 200 - 50 = 250\,\text{N}$ forwards.
Example 3 — balanced
A crate pulled at constant velocity: pull $40\,\text{N}$, friction $40\,\text{N}$. Resultant $= 0$. Constant velocity (including remaining at rest) means balanced forces. That is the idea behind Newton’s first law, even though this specification writes it through 1.15 and 1.17.
“No resultant force” is not “no forces”
A book on a table has weight down and a normal contact force up. Both are there. They cancel. Students who write “there are no forces on a book at rest” lose the mark and also wreck terminal-velocity answers later.
1.17, 1.18
1.17: the relationship between unbalanced force, mass and acceleration is
$$F = m \times a$$$F$ is the resultant force in newtons. $m$ is mass in kilograms. $a$ is acceleration in m/s$^2$. Rearrangements you must be fluent with:
$$a = \frac{F}{m} \qquad m = \frac{F}{a}$$Same resultant force: a larger mass gets a smaller acceleration. Same mass: a larger resultant force gets a larger acceleration. That is the whole of the “explain” version.
Worked example — the two-step question they always write
A $900\,\text{kg}$ car has a driving force of $2400\,\text{N}$. Resistive forces total $600\,\text{N}$. Find (a) the resultant force (b) the acceleration.
(a) Do not touch $F=ma$ yet. $F = 2400 - 600 = 1800\,\text{N}$ forwards.
(b) Now $a = 1800 / 900 = 2.0\,\text{m/s}^2$ forwards.
If you put $2400$ into $F=ma$ you have ignored friction. That is the most common 1.17 mistake in my marking.
1.18:
$$W = m \times g$$Mass is the amount of matter, in kg. It is the same on Earth, on the Moon, in deep space. Weight is the gravitational force on that mass, in newtons. $g$ is gravitational field strength, in N/kg. On Earth the paper usually gives $g = 10\,\text{N/kg}$ (sometimes $9.8$). On the Moon $g \approx 1.6\,\text{N/kg}$.
A $60\,\text{kg}$ student:
Bathroom “scales” in kilograms are actually measuring the contact force and converting it. In physics, if they say “a mass of $2.0\,\text{kg}$ is hung on a spring,” the force on the spring is the weight, $20\,\text{N}$, not $2.0\,\text{N}$.
The sentence that scores on “explain the difference”
“Mass is the amount of matter in an object, measured in kg. Weight is the gravitational force on that mass, $W=mg$, measured in N.”
Watch
Cognito on balanced forces, what a non-zero resultant does, and a worked $F=ma$ example. Circular motion and inertia are extra context — focus on the $F=ma$ section for this specification.
1.19, 1.20
I draw this on the board as a timeline, not as a definition to memorise.
Thinking distance at constant speed is just $d = vt$:
$$\text{thinking distance} = \text{speed} \times \text{reaction time}$$At $13\,\text{m/s}$ with a $0.70\,\text{s}$ reaction time, thinking distance $= 9.1\,\text{m}$. Double the speed, thinking distance doubles (if reaction time stays the same).
You do not need kinetic energy on this specification, but the idea helps: a faster car has much more motion to take away, so it travels further while the brakes do that job. Rough classroom numbers (dry road, typical car):
| Speed | Thinking (typical) | Braking (typical dry) | Stopping |
|---|---|---|---|
| $13\,\text{m/s}$ ($47\,\text{km/h}$) | $9\,\text{m}$ | $14\,\text{m}$ | $23\,\text{m}$ |
| $22\,\text{m/s}$ ($80\,\text{km/h}$) | $15\,\text{m}$ | $38\,\text{m}$ | $53\,\text{m}$ |
| $31\,\text{m/s}$ ($110\,\text{km/h}$) | $21\,\text{m}$ | $75\,\text{m}$ | $96\,\text{m}$ |
Do not memorise the table for the exam. Use it to feel that braking distance rises steeply with speed. 1.20 names speed, mass, road condition and reaction time.
| Factor | Thinking? | Braking? | The physics reason I want |
|---|---|---|---|
| Higher speed | Yes | Yes | Further in the same reaction time; more speed to lose under the same braking force |
| Tiredness, alcohol, drugs, distraction, some medicines | Yes | No | Longer reaction time, so $d=vt$ is larger |
| Greater mass / extra passengers / a load | No | Yes | Same braking force gives smaller deceleration ($a=F/m$) |
| Wet, icy, gravel, oily road | No | Yes | Less friction, so a smaller braking force |
| Worn tyres or worn brakes | No | Yes | Smaller frictional / braking force |
How a 2-mark “wet road” question is marked
“Thinking distance stays the same because reaction time and speed are unchanged [1]. Braking distance increases because there is less friction between the tyres and the road [1]. Therefore stopping distance increases.” If you only write “it takes longer to stop,” that is 0.
Watch
1.21
A falling object near the Earth has two important forces: weight down (constant, $W=mg$) and air resistance up (gets larger as speed gets larger, and larger if the area is larger). Terminal velocity is the constant speed reached when those two are equal.
I refuse to let students write “gravity increases as it falls.” Weight does not increase. Speed increases; drag increases.
Snapshot 1 — just released
$v = 0$, so air resistance $R \approx 0$. Resultant $= W$ down. Acceleration is about $g$ ($10\,\text{m/s}^2$).
Snapshot 2 — speeding up
$R$ has grown but $R < W$. Resultant $= W - R$ down, smaller than before. Still accelerating, but $a$ is falling.
Snapshot 3 — terminal velocity
$R = W$. Resultant $= 0$. Acceleration $= 0$. Speed is constant — the maximum for that shape and that weight.
Snapshot 4 — parachute opens (if asked)
Area jumps up, so $R$ suddenly bigger than $W$. Resultant is upwards. The skydiver decelerates. As speed falls, $R$ falls until $R = W$ again — a new, smaller terminal velocity.
The graph starts steep (acceleration $\approx g$), then the gradient gets shallower (acceleration falling), then it becomes horizontal (terminal velocity). If a parachute opens, the line drops — deceleration — and levels off at a lower horizontal line.
Four-mark answer — write four separate points
A raindrop and a crumpled piece of paper reach a much lower terminal velocity than a skydiver because their weight is small compared with the drag they can produce. A streamlined, dense object has a higher terminal velocity.
Watch
Optional second watch
1.
A $2.5\,\text{kg}$ trolley is pulled by $8.0\,\text{N}$. Friction is $3.0\,\text{N}$. Calculate resultant force and acceleration.
2.
A driver’s reaction time increases from $0.60\,\text{s}$ to $1.2\,\text{s}$ at $15\,\text{m/s}$. Calculate the two thinking distances. What happens to braking distance if only reaction time changes?