OCR A Level Physics A: Units, measurement uncertainty and vectors (H556) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow sections 2.1.1, 2.1.2, 2.2.1 and 2.3.1 of the OCR Advanced GCE in Physics A specification (H556), Version 3.0 (March 2026), for first assessment in 2017. Teaching is in the study guide and the revision notes. Course pages: OCR A Level Physics hub, checklist and free diagnostics. Use g = 9.81 m s⁻² where needed.

Questions

1. The watt is defined as one joule per second. Show that the watt can be written in SI base units as kg m² s⁻³. [2]

2. A student proposes that the period T of a mass m on a spring of spring constant k is T = 2π√(k/m). The unit of k is N m⁻¹. Use base units to show that this equation is not homogeneous. [3]

3. Write each quantity in the unit given, in standard form. [3] (a) 650 nm in m (b) 3.3 MHz in Hz (c) 15 cm³ in m³

4. Estimate the momentum of a cyclist and bicycle moving at a steady speed along a flat road. State your assumptions. [3]

5. An ammeter reads 0.03 A when no current flows. Three readings of one current are 0.52 A, 0.51 A and 0.53 A. (a) Name the type of error and state whether averaging removes it. [2] (b) Find the corrected mean current. [2] (c) State whether the three readings are precise. Give a reason. [1]

6. A resistor has a potential difference of 6.40 ± 0.05 V across it and a current of 0.250 ± 0.005 A through it. Calculate its resistance with its absolute uncertainty. [4]

7. (a) A rod is 85.0 ± 0.5 cm long. A piece of length 32.0 ± 0.5 cm is cut from it. Find the length remaining with its absolute uncertainty. [2] (b) A trolley of mass 0.180 ± 0.002 kg has speed 2.70 ± 0.04 m s⁻¹. Calculate its kinetic energy with an absolute uncertainty. [4]

8. A student measures the resistance R of a wire for different lengths l and plots R against l. The best-fit line has gradient 4.20 Ω m⁻¹. The steepest and shallowest worst lines have gradients 4.55 Ω m⁻¹ and 3.90 Ω m⁻¹. The wire diameter is 0.314 ± 0.003 mm. The resistivity is ρ = gradient × A, where A is the cross-sectional area. (a) Find the absolute and percentage uncertainty in the gradient. [2] (b) Calculate A and its percentage uncertainty. [2] (c) Calculate ρ with its absolute uncertainty. [2] (d) A reference value of 3.50 × 10⁻⁷ Ω m is supplied. Find the percentage difference. [1]

9. (a) From kinetic energy, momentum, displacement, speed and weight, name the vector quantities. [2] (b) A runner jogs once round a 400 m track in 80 s and finishes where she started. State the average speed and the average velocity. [2]

10. Two forces of 9.0 N and 6.0 N act at a point. The angle between their lines of action is 70°. Find the size of the resultant [3] and its angle to the 9.0 N force. [2]

11. A boat’s velocity changes from 6.0 m s⁻¹ due east to 8.0 m s⁻¹ due north. (a) State how the change in velocity is found from the two velocities. [1] (b) Calculate the magnitude of the change in velocity. [2] (c) Give its direction. [1]

12. A 24 kg crate rests on a slope at 22° to the horizontal. A rope pulls it up the slope at 15° above the slope surface, with tension 120 N. Ignore friction. (a) Calculate the components of the weight along and perpendicular to the slope. [2] (b) Calculate the components of the tension along and perpendicular to the slope. [2] (c) Find the resultant force along the slope. [2] (d) Find the normal contact force from the slope. [2]

Answers

1. Joule = newton × metre = (kg m s⁻²)(m) = kg m² s⁻² [1]. Watt = J / s = kg m² s⁻² / s = kg m² s⁻³ [1]. [2] Examiner insight: A “show that” answer tends to gain credit only when the starting unit and each substitution are visible, so write the base form of the newton first.

2. k = N m⁻¹ = kg m s⁻² / m = kg s⁻² [1]. k / m = kg s⁻² / kg = s⁻² [1]. √(s⁻²) = s⁻¹, but T has unit s, so the sides differ: not homogeneous [1]. [3] Examiner insight: Reduce both sides separately and compare them in a final sentence. Stating the conclusion explicitly is what makes the working count.

3. (a) 650 × 10⁻⁹ = 6.5 × 10⁻⁷ m [1] (b) 3.3 × 10⁶ = 3.3 × 10⁶ Hz [1] (c) 15 × (10⁻²)³ = 1.5 × 10⁻⁵ m³ [1] Examiner insight: In part (c) the prefix must be cubed. Writing 15 × 10⁻² is the usual slip. Standard form was asked for, so 0.000015 would not match the instruction.

4. Assume total mass about 85 kg [1] and speed about 6 m s⁻¹ [1]. p = mv = 85 × 6 ≈ 510 kg m s⁻¹ (accept a sensible range, roughly 300 to 900) [1]. [3] Examiner insight: Estimates are judged on whether the assumptions are realistic and the unit is right. A 10 kg cyclist or a speed of 60 m s⁻¹ would be questioned.

5. (a) A systematic (zero) error [1]. Averaging does not remove it, because every reading is shifted equally [1]. (b) Mean = (0.52 + 0.51 + 0.53) / 3 = 0.52 A [1]. Corrected = 0.52 − 0.03 = 0.49 A [1]. (c) Yes: the readings differ by only 0.01 A, so they agree closely [1]. [5] Examiner insight: Subtract the zero error from the mean, not from only one reading. Naming the error as “zero error” without the word systematic may leave the first mark short.

6. R = V / I = 6.40 / 0.250 = 25.6 Ω [1]. % in V = 0.05 / 6.40 × 100 = 0.78 % and % in I = 0.005 / 0.250 × 100 = 2.0 % [1]. Total = 2.78 % [1]. Absolute = 0.0278 × 25.6 = 0.71, so R = 25.6 ± 0.7 Ω [1]. [4] Examiner insight: A quotient needs percentage uncertainties. Adding 0.05 and 0.005 directly is the typical error. Round the uncertainty to one significant figure and match the decimal places of R to it.

7. (a) 85.0 − 32.0 = 53.0 cm [1]. Uncertainties add: 0.5 + 0.5 = 1.0, so 53.0 ± 1.0 cm [1]. (b) E = ½mv² = 0.5 × 0.180 × 2.70² = 0.656 J [1]. % in m = 0.002 / 0.180 × 100 = 1.1 %; % in v = 0.04 / 2.70 × 100 = 1.48 %, doubled to 2.96 % [1]. Total = 4.07 % [1]. Absolute = 0.0407 × 0.656 = 0.027, so E = 0.66 ± 0.03 J [1]. [6] Examiner insight: The factor ½ is exact, so it adds no uncertainty. The squared speed is the step most often left out, and it dominates the total here.

8. (a) Absolute = (4.55 − 3.90) / 2 = 0.325 Ω m⁻¹ [1]. Percentage = 0.325 / 4.20 × 100 = 7.7 % [1]. (b) A = π(0.314 × 10⁻³)² / 4 = 7.74 × 10⁻⁸ m² [1]. % in d = 0.003 / 0.314 × 100 = 0.96 %, so % in A = 2 × 0.96 = 1.9 % [1]. (c) ρ = 4.20 × 7.74 × 10⁻⁸ = 3.25 × 10⁻⁷ Ω m [1]. Total % = 7.7 + 1.9 = 9.6 %, so absolute = 0.31 × 10⁻⁷, giving ρ = (3.25 ± 0.31) × 10⁻⁷ Ω m [1]. (d) |3.25 − 3.50| / 3.50 × 100 = 7.1 % [1]. [7] Examiner insight: Convert the diameter to metres before squaring. The 7.1 % difference is smaller than the 9.6 % uncertainty, which is worth saying if a comparison is requested.

9. (a) Momentum, displacement and weight [2] (one mark if only two are named, or if all three are named with one wrong addition). (b) Average speed = 400 / 80 = 5.0 m s⁻¹ [1]. The runner ends at the start, so displacement is zero and average velocity = 0 [1]. [4] Examiner insight: Weight is a force, so it is a vector. Give the reason in part (b): displacement, not distance, is zero.

10. Place the forces tip to tail, so the interior angle is 180° − 70° = 110° [1]. R² = 9.0² + 6.0² − 2(9.0)(6.0)cos 110° = 117 + 36.9 = 153.9 [1]. R = 12.4 N [1]. Sine rule: sin α / 6.0 = sin 110° / 12.41 [1], so α = 27.0° to the 9.0 N force [1]. [5] Examiner insight: Using cos 70° instead of cos 110° gives a smaller, wrong resultant. A sketch triangle with the interior angle marked helps you catch it. Check by components: 9.0 + 6.0 cos 70° = 11.05 N and 6.0 sin 70° = 5.64 N.

11. (a) Change in velocity = final velocity − initial velocity (add the reverse of the initial velocity) [1]. (b) Δv = 6.0 m s⁻¹ west + 8.0 m s⁻¹ north, which are perpendicular [1]. Magnitude = √(6.0² + 8.0²) = 10 m s⁻¹ [1]. (c) tan⁻¹(6.0 / 8.0) = 36.9°, so 36.9° west of north [1]. [4] Examiner insight: Subtraction means reversing the initial vector. Adding the two velocities gives a vector towards north-east, which is the wrong direction.

12. (a) W = 24 × 9.81 = 235.4 N. Along: 235.4 sin 22° = 88.2 N (down the slope) [1]. Perpendicular: 235.4 cos 22° = 218 N (into the slope) [1]. (b) Along: 120 cos 15° = 116 N (up the slope) [1]. Perpendicular: 120 sin 15° = 31.1 N (away from the slope) [1]. (c) Resultant = 115.9 − 88.2 = 27.7 N up the slope [1]. This is the net force along the slope since friction is ignored [1]. (d) The crate does not move perpendicular to the slope, so R + 31.1 = 218.3 [1], giving R = 187 N [1]. [8] Examiner insight: A labelled sketch showing the angle of the rope to the surface, not to the horizontal, is what stops the sine and cosine being swapped. Keep unrounded values until the final subtraction.

Where marks are usually lost

• Leaving out the unit or giving the wrong power of ten for a squared or cubed prefix.
• Adding absolute uncertainties when quantities are multiplied or divided.
• Forgetting to count a squared quantity twice.
• Giving an absolute uncertainty when a percentage was asked, or the other way round.
• Calling a zero error random.
• Using the interior angle of the triangle incorrectly in the cosine rule.
• Adding vectors when a difference was asked.
• Swapping sin and cos because the angle was measured from the wrong line.
• Using more significant figures for an uncertainty than the data justify.

Next steps

• Revision notes
• Study guide
• OCR A Level Physics hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

OCR, Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), for first assessment in 2017. Sections 2.1.1, 2.1.2, 2.2.1 and 2.3.1.