AQA GCSE Physics Physics Paper 2 (Higher), November 2021: Question 7

✅ Correct Answer

friction (accept tyre friction / grip)

💡 Key Knowledge

Force A acts between the rear tyre and the road surface. For a bicycle to accelerate forwards, there must be a forward traction force provided by friction acting on the driving wheel.

❌ Common Errors

Students often incorrectly write "air resistance", "gravity", or "push". Remember that driving force forwards on wheels is always due to friction.

✅ Correct Answer

Total distance = 162 m

📐 Step-by-Step Calculation

1. Identify region: Between Y (t = 30 s, v = 5.4 m/s) and Z (t = 70 s, v = 0 m/s).
2. Split shape: Break the area under the graph into a rectangle and a triangle.
3. Rectangle area (t = 30 to 50 s): Width (20 s) × Height (5.4 m/s) = 108 m
4. Triangle area (t = 50 to 70 s): 0.5 × Base (20 s) × Height (5.4 m/s) = 54 m
5. Total distance: 108 + 54 = 162 m

🧠 Exam Technique

Always state your breakdown clearly (area of rectangle + area of triangle). Examiners award method marks even if arithmetic slips, provided the geometric approach is sound.

✅ Correct Answer

1. The force on the pedal causes a moment about the pedal axle.
2. This creates a tension force in the chain, which causes a moment about the rear axle.

💡 Key Knowledge

A moment is the turning effect of a force, calculated as Force × perpendicular distance from the pivot . Here, energy is transferred mechanically through rotation, linking the pedal axle via the chain to the rear wheel axle (Gear B).

❌ Common Errors

Vague answers like "it turns the wheel" fail to gain credit. You must explicitly reference the chain transmitting the force/tension to create a secondary moment at the rear axle.

✅ Correct Answer

Acceleration = 0.16 m/s²

📐 Step-by-Step Calculation

1. Select equation: v² - u² = 2as (where v = 2.4 m/s, u = 0, s = 18 m).
2. Substitute values: 2.4² - 0² = 2 × a × 18
3. Simplify: 5.76 = 36a
4. Rearrange & Solve: a = 5.76 / 36 = 0.16 m/s²
5. Alternative method: Use kinematic equations to find time first (t = 15 s), then use a = Δv / t. Both score full marks.

❌ Common Errors

Forgetting to square the velocity term ( v² ) is a major trap. Always double-check that you are using the correct kinematic formula when time is not provided.

✅ Correct Answer

Magnitude: 212 N to 218 N (or 214 N)
Direction: 20° to 22° from the horizontal (or 68°–70° from vertical / valid bearing 290°–292°)

🧠 Exam Technique & Construction

1. Choose a scale: e.g., 1 cm = 20 N (so 200 N = 10 cm, 75 N = 3.75 cm).
2. Draw components: Draw a horizontal arrow representing 200 N, and a vertical arrow representing 75 N head-to-tail.
3. Complete triangle/parallelogram: Draw the resultant vector from the start point to the end point with an arrowhead pointing top-left.
4. Measure: Measure length with a ruler to convert back via your scale, and measure the angle using a protractor.

❌ Common Errors

Failing to include directional arrowheads on the resultant line, or measuring the angle from the wrong axis (e.g., using vertical instead of horizontal without stating it clearly).