What's on the test
Straight from our study guide. Learn each section top to bottom, then do the practice at the very end.
| Topic | You need to be able to… |
|---|---|
| 1.1 SI units & orders of magnitude | Use SI units properly, scientific notation, metric prefixes, compare to nearest order of magnitude |
| 1.2 Uncertainties & errors | Random vs systematic errors, absolute / fractional / % uncertainty, combine uncertainties (+ − × ÷ powers), uncertainty in gradients & intercepts |
| 1.3 Vectors & scalars | Solve vector problems graphically and algebraically |
| 2 Kinematics | Use the 1st, 2nd and 3rd equations of motion; get velocity from an x–t graph, acceleration and displacement from a v–t graph |
Suggested plan: Mon–Tue = Topic 1 · Wed = Vectors + equations · Thu = graphs + practice · Fri = cheat sheet only.
1.1 SI units & orders of magnitude
Physics uses one agreed set of units so everyone's numbers match. The base SI units you'll use most:
| Quantity | Unit | Symbol |
|---|---|---|
| length | metre | m |
| mass | kilogram | kg |
| time | second | s |
| current | ampere | A |
| temperature | kelvin | K |
| amount | mole | mol |
Everything else is derived from these, e.g. speed = m s−1, acceleration = m s−2.
Prefixes
n = 10−9 · µ = 10−6 · m = 10−3 · c = 10−2 · k = 103 · M = 106 · G = 109
Worked example. Write 0.000 45 m in scientific notation and in µm.
Move the decimal 4 places → 4.5 × 10⁻⁴ m. Since 1 µm = 10⁻⁶ m: 4.5 × 10⁻⁴ ÷ 10⁻⁶ = 450 µm.
Move the decimal 4 places → 4.5 × 10⁻⁴ m. Since 1 µm = 10⁻⁶ m: 4.5 × 10⁻⁴ ÷ 10⁻⁶ = 450 µm.
Order of magnitude = the power of 10 closest to the number. 3 × 10⁴ → order 10⁴. 8 × 10⁴ is closer to 10⁵ → order 10⁵. To compare two values, divide them and find the order of the answer.
Always give final answers with a unit, and keep units in the right format (m s⁻¹ not "mps").
1.2 Errors: precision vs accuracy
Every measurement has errors. Our class example is a cart travelling from A to B — we measure distance and time, and both have errors.
Precise = repeat readings are close together (repeatable), even if wrong.
Accurate = close to the true value.
Accurate = close to the true value.
From the slides:
Ruler: 0.82, 0.85, 0.82 m but true distance is 0.75 m → precise but not accurate (maybe a faulty ruler).
Stopwatch: 1.23, 1.15, 1.37 s, average 1.25 s = true value → accurate but not precise (random fluctuations).
Ruler: 0.82, 0.85, 0.82 m but true distance is 0.75 m → precise but not accurate (maybe a faulty ruler).
Stopwatch: 1.23, 1.15, 1.37 s, average 1.25 s = true value → accurate but not precise (random fluctuations).
Two types of error
| Systematic error | Random error | |
|---|---|---|
| Cause | Faulty equipment or the same mistake every time (e.g. voltmeter reads 1 V when disconnected = zero error) | Random fluctuations (e.g. temperature varying in a liquid, reaction time) |
| Effect | Shifts all readings the same way → affects accuracy | Readings scatter → affects precision |
| Fix | Subtract the zero error / calibrate. Repeating does NOT help | Repeat readings and average (and stir!) |
Rule to remember: small systematic error → accurate. Small random error → precise and reliable.
Absolute, fractional & % uncertainty
The uncertainty tells you the range the real value could be in. Write it as best estimate ± uncertainty.
How big is the reading uncertainty?
- Analogue (ruler, thermometer with a scale): ± half the smallest division.
- Digital (stopwatch, digital meter): ± the smallest division it shows.
- Length needs two readings (left end AND right end), so add both: stick read to ±0.05 cm at each end → 49 ± (0.05 + 0.05) = 49 ± 0.1 cm.
Repeated readings
Uncertainty = (max − min) ÷ 2 (half the range).
Worked example (class slide). Velocities ranged 1.24 to 1.92 m s⁻¹, average 1.50 m s⁻¹.
Half range = (1.92 − 1.24)/2 = 0.340. Add reading error 0.005 → 0.345 → round to 2 s.f. → 1.50 ± 0.35 m s⁻¹.
Half range = (1.92 − 1.24)/2 = 0.340. Add reading error 0.005 → 0.345 → round to 2 s.f. → 1.50 ± 0.35 m s⁻¹.
Uncertainties are given to 1 or 2 s.f., and the value must not have more decimal places than the uncertainty.
The three forms
| Form | Formula | Unit? |
|---|---|---|
| Absolute | Δx | same unit as x |
| Fractional | Δx / x | none |
| Percentage | (Δx / x) × 100% | % |
Worked example. (24.0 ± 0.2) cm → absolute 0.2 cm, fractional 0.2/24.0 = 0.00833, percentage 0.83%.
Backwards: 80.0 cm with 2.5% → Δx = 0.025 × 80.0 = 2.0 → 80.0 ± 2.0 cm.
Backwards: 80.0 cm with 2.5% → Δx = 0.025 × 80.0 = 2.0 → 80.0 ± 2.0 cm.
% uncertainty → tells you about random error/reliability.
% difference = |measured − true| / true × 100% → tells you about accuracy. E.g. 0.82 m vs true 0.75 m → 9.3% off → not accurate.
% difference = |measured − true| / true × 100% → tells you about accuracy. E.g. 0.82 m vs true 0.75 m → 9.3% off → not accurate.
Combining uncertainties
When you calculate something from measured values, the uncertainties carry through. There are just three rules:
1. Add or subtract → add the absolute uncertainties. Δz = Δx + Δy (yes, even for subtraction!)
2. Multiply or divide → add the fractional / % uncertainties. Δz/z = Δx/x + Δy/y
3. Power → multiply the % uncertainty by the power. x² → 2 × %x, √x → ½ × %x. Constants (π, 2, g if given exact) have no uncertainty.
2. Multiply or divide → add the fractional / % uncertainties. Δz/z = Δx/x + Δy/y
3. Power → multiply the % uncertainty by the power. x² → 2 × %x, √x → ½ × %x. Constants (π, 2, g if given exact) have no uncertainty.
Worked example 1 – subtraction. Liquid heated from (21.0 ± 0.5) °C to (76.0 ± 0.5) °C.
Rise = 76.0 − 21.0 = 55.0 °C. Δ = 0.5 + 0.5 = 1.0 → 55.0 ± 1.0 °C (1.8%).
Rise = 76.0 − 21.0 = 55.0 °C. Δ = 0.5 + 0.5 = 1.0 → 55.0 ± 1.0 °C (1.8%).
Worked example 2 – the cart (class slide). d = 0.83 ± 0.02 m, t = 1.25 ± 0.07 s. Find speed.
① Value: s = 0.83/1.25 = 0.664 m s⁻¹
② % uncertainties: 0.02/0.83 = 2.4%, 0.07/1.25 = 5.6% → total 8.0%
③ Absolute: 0.080 × 0.664 = 0.053
④ Answer: 0.66 ± 0.05 m s⁻¹
① Value: s = 0.83/1.25 = 0.664 m s⁻¹
② % uncertainties: 0.02/0.83 = 2.4%, 0.07/1.25 = 5.6% → total 8.0%
③ Absolute: 0.080 × 0.664 = 0.053
④ Answer: 0.66 ± 0.05 m s⁻¹
Worked example 3 – power. KE = ½mv², m = 0.200 ± 0.005 kg, v = 5.0 ± 0.2 m s⁻¹.
KE = 0.5 × 0.200 × 25 = 2.5 J. % = 2.5% + 2 × 4% = 10.5% → ±0.26 → 2.5 ± 0.3 J.
KE = 0.5 × 0.200 × 25 = 2.5 J. % = 2.5% + 2 × 4% = 10.5% → ±0.26 → 2.5 ± 0.3 J.
Worked example 4 – square root. t = √(2h/g), h = 5.00 ± 0.05 m (1%), g = 9.8 exact.
t = √(10/9.8) = 1.01 s. % = ½ × 1% = 0.5% → 1.01 ± 0.01 s.
t = √(10/9.8) = 1.01 s. % = ½ × 1% = 0.5% → 1.01 ± 0.01 s.
Error bars, gradients & intercepts
On a graph, each point gets error bars (± the uncertainty) in x and y. Together they make an error box — the true point could be anywhere inside it.
- Best fit line – goes through all the error boxes.
- Max line – steepest line that still passes through every box.
- Min line – least steep line that still passes through every box.
- No error bars? Judge max/min lines by eye.
Δgradient = (max gradient − min gradient) ÷ 2
or: gradient of best line − gradient of worst acceptable line.
Δintercept = best intercept − worst intercept
or: gradient of best line − gradient of worst acceptable line.
Δintercept = best intercept − worst intercept
Worked example. Max line gradient 2.30, min line 1.90, best 2.10.
Δ = (2.30 − 1.90)/2 = 0.20 → gradient = 2.1 ± 0.2.
Δ = (2.30 − 1.90)/2 = 0.20 → gradient = 2.1 ± 0.2.
1.3 Vectors & scalars
Scalar = size only (magnitude): time, distance, speed, mass.
Vector = size AND direction: displacement, velocity, acceleration, force.
Vector = size AND direction: displacement, velocity, acceleration, force.
A vector is drawn as an arrow – the length shows the size. A 20 m s⁻¹ arrow is twice as long as a 10 m s⁻¹ arrow.
Adding vectors
Graphically: draw to scale, tip-to-tail; the resultant goes from the start of the first to the tip of the last. Measure its length and angle.
Algebraically: same line → add with signs (right/north +, left/south −). At right angles → Pythagoras + tan.
Worked example. Walk 3 m east then 4 m north.
Resultant = √(3² + 4²) = 5 m. Angle: tan θ = 4/3 → θ = 53° north of east.
Distance walked = 7 m (scalar) but displacement = 5 m at 53° N of E (vector).
Resultant = √(3² + 4²) = 5 m. Angle: tan θ = 4/3 → θ = 53° north of east.
Distance walked = 7 m (scalar) but displacement = 5 m at 53° N of E (vector).
Splitting a vector (components). 10 m s⁻¹ at 30° above horizontal: horizontal = 10 cos 30 = 8.7 m s⁻¹, vertical = 10 sin 30 = 5.0 m s⁻¹.
2 Kinematics – describing motion
Reference frame: where you measure from. Sitting at your desk you're still relative to the classroom but moving relative to the Sun. We usually use "standing still on Earth", with +x to the right (and up/north as +).
Distance – total path, always positive (scalar). Displacement Δx = x − x₀ – how far from the start, can be ±.
Drive 60 miles out, 20 back → distance 80, displacement 40.
Speed = distance/time (scalar). Velocity v = Δx/Δt (vector). Acceleration a = Δv/Δt.
Drive 60 miles out, 20 back → distance 80, displacement 40.
Speed = distance/time (scalar). Velocity v = Δx/Δt (vector). Acceleration a = Δv/Δt.
Speeding up or slowing down? If v and a point the same way → speeds up (even if both are negative!). Opposite ways → slows down.
The equations of motion
Only for constant acceleration. Write down what you know (u, v, a, t, s), pick the equation without the one you don't know and don't need.
| # | Equation | Missing |
|---|---|---|
| 1st | v = u + at | s |
| 2nd | s = ut + ½at² | v |
| 3rd | v² = u² + 2as | t |
| avg | s = ½(u + v)t | a |
(Our slides write u as v₀ and s as Δx – same thing.) Free fall: a = g = 9.8 m s⁻² downward. "Dropped" means u = 0. At the top of a throw, v = 0.
1st equation. A car at 24 m s⁻¹ brakes at −5.0 m s⁻². Time to stop?
0 = 24 + (−5.0)t → t = 4.8 s
0 = 24 + (−5.0)t → t = 4.8 s
Average velocity. Motorcycle 0 → 38 m s⁻¹ in 5.0 s.
v̄ = (0 + 38)/2 = 19 m s⁻¹, s = 19 × 5.0 = 95 m
v̄ = (0 + 38)/2 = 19 m s⁻¹, s = 19 × 5.0 = 95 m
2nd equation. Dropped from a 42 m building.
Take down as +: 42 = 0 + ½(9.8)t² → t² = 8.57 → t = 2.9 s
Take down as +: 42 = 0 + ½(9.8)t² → t² = 8.57 → t = 2.9 s
3rd equation (stomp rocket). Max height 15.8 m. Launch speed?
At top v = 0: 0 = u² − 2(9.8)(15.8) → u = 17.6 m s⁻¹. Time up: t = 17.6/9.8 = 1.80 s, so up and down = 3.6 s.
At top v = 0: 0 = u² − 2(9.8)(15.8) → u = 17.6 m s⁻¹. Time up: t = 17.6/9.8 = 1.80 s, so up and down = 3.6 s.
Signs! Choose a + direction at the start and stick to it. Thrown up at 16 m s⁻¹, after 2 s: v = 16 − 9.8×2 = −3.6 m s⁻¹ → it's now moving down.
Motion graphs
| Graph | Gradient means | Area means |
|---|---|---|
| x–t (displacement–time) | velocity | – |
| v–t (velocity–time) | acceleration | displacement |
| a–t | – | change in velocity |
Reading shapes
- x–t horizontal → stationary. Straight sloping → constant velocity. Negative slope → constant velocity backwards. Curve → velocity changing (use a tangent for instantaneous velocity).
- v–t horizontal at 0 → stationary. Horizontal above 0 → constant velocity. Positive slope → accelerating, negative slope → decelerating. Below the axis → moving the opposite way.
Worked example. v–t line from 4 m s⁻¹ to 12 m s⁻¹ in 4 s.
a = gradient = (12 − 4)/4 = 2 m s⁻². s = area of trapezium = ½(4 + 12) × 4 = 32 m.
a = gradient = (12 − 4)/4 = 2 m s⁻². s = area of trapezium = ½(4 + 12) × 4 = 32 m.
One-page cheat sheet
- Analogue ± ½ smallest division · digital ± smallest division · repeats ± ½ range
- + / − → add absolute · × / ÷ → add % · power n → n × %
- Uncertainty to 1–2 s.f.; match decimal places
- Systematic → accuracy (fix: zero/calibrate) · random → precision (fix: repeat)
- Δgradient = (max − min)/2
- v = u + at · s = ut + ½at² · v² = u² + 2as · s = ½(u+v)t · g = 9.8 m s⁻²
- x–t gradient = v · v–t gradient = a · v–t area = s
- Vectors: tip-to-tail, Pythagoras, tan θ = opp/adj
Practice questions
From our class worksheets. Try each one on paper first, then open the worked answer.
- An ammeter has a zero offset error. Does this affect precision, accuracy, both or neither?
Worked answer
Only accuracy (C). A zero error is systematic – it shifts every reading by the same amount, so readings can still be close together. - A ball's fall time from the Eiffel Tower: 7.83 s, 7.56 s, 7.93 s. Give the average and random error.
Worked answer
Average = (7.83+7.56+7.93)/3 = 7.77 s. Half range = (7.93 − 7.56)/2 = 0.185 → 2 s.f. → 7.77 ± 0.19 s (A). - Two lengths (12.4 ± 0.1) cm and (8.3 ± 0.2) cm. Find the total.
Worked answer
12.4 + 8.3 = 20.7 cm; Δ = 0.1 + 0.2 = 0.3 → 20.7 ± 0.3 cm. - d = (100 ± 2) m, t = (20.0 ± 0.5) s. Find v = d/t with uncertainty.
Worked answer
v = 5.0 m s⁻¹. % = 2% + 2.5% = 4.5%. Δv = 0.045 × 5.0 = 0.23 → 5.0 ± 0.2 m s⁻¹. - Radius r = 4.3 ± 0.1 cm. Find (a) the circumference and (b) the area with uncertainties.
Worked answer
% in r = 0.1/4.3 = 2.3%.
(a) C = 2πr = 27.0 cm, same 2.3% → ±0.6 → 27.0 ± 0.6 cm.
(b) A = πr² = 58.1 cm², % = 2 × 2.3 = 4.7% → ±2.7 → 58 ± 3 cm². - F = mv²/r with m = 2.3 ± 0.2 kg, v = 15 ± 1 m s⁻¹, r = 3.4 ± 0.2 m. Find F.
Worked answer
F = 2.3 × 225 / 3.4 = 152 N. % = 8.7% + 2(6.7%) + 5.9% = 27.9%. ΔF = 0.279 × 152 = 42 → F = (1.5 ± 0.4) × 10² N. - Cylinder: height uncertainty 5%, diameter 1%. Uncertainty in volume?
Worked answer
V = π(d/2)²h → 2 × 1% + 5% = 7% (D). - Rectangle L = (2 ± 0.1) cm, W = (3 ± 0.2) cm. Uncertainty in the perimeter?
Worked answer
P = 2L + 2W → Δ = 2(0.1) + 2(0.2) = 0.6 cm (D). - pV = nRT, uncertainties P 2%, V 6%, T 1%. Uncertainty in n?
Worked answer
n = pV/RT → 2 + 6 + 1 = 9% (C). - EPE = ½kx², k ± 2%, x ± 3%. Uncertainty in EPE?
Worked answer
2% + 2 × 3% = 8% (A). - Max gradient line = 4.8, min = 4.2. Best line = 4.5. Quote the gradient.
Worked answer
Δ = (4.8 − 4.2)/2 = 0.3 → 4.5 ± 0.3. - A boat goes 4 m s⁻¹ north across a river flowing 3 m s⁻¹ east. Resultant velocity?
Worked answer
√(4² + 3²) = 5 m s⁻¹; tan θ = 3/4 → θ = 37° east of north. - An object moving at 23 m s⁻¹ reaches 85 m s⁻¹ in 8.3 s. Acceleration?
Worked answer
a = (v − u)/t = (85 − 23)/8.3 = 7.5 m s⁻². - A plane reaches 240 m s⁻¹ in 11 s, travelling 1400 m. Average and initial velocity?
Worked answer
v̄ = 1400/11 = 127 m s⁻¹. v̄ = (u+v)/2 → u = 2(127.3) − 240 = 15 m s⁻¹. - A car brakes at −3.2 m s⁻² and stops in 210 m. Starting speed?
Worked answer
0 = u² + 2(−3.2)(210) → u² = 1344 → u = 36.7 m s⁻¹. - An object is thrown up at 35 m s⁻¹. Maximum height?
Worked answer
At top v = 0: 0 = 35² − 2(9.8)h → h = 1225/19.6 = 62.5 m. - A ball is thrown down at 8 m s⁻¹ off a cliff and lands 6 s later. Cliff height?
Worked answer
Down = +: s = 8(6) + ½(9.8)(6²) = 48 + 176.4 = 224 m. - An arrow is shot straight up and lands 8 s later. Initial velocity and max height?
Worked answer
Up takes 4 s: u = 9.8 × 4 = 39.2 m s⁻¹. h = ½(39.2)(4) = 78.4 m. - A v–t graph goes in a straight line from 0 to 20 m s⁻¹ in 5 s, then stays at 20 m s⁻¹ for 3 s. Find the acceleration and total displacement.
Worked answer
a = 20/5 = 4 m s⁻². Area = ½(5)(20) + 3(20) = 50 + 60 = 110 m.
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