Cambridge IGCSE Physics

1. The gradient is (0.95−0.15)/(1.0−0.2) = 1.0 s m⁻¹, so speed is its reciprocal, 1.0 m s⁻¹.
2. The line is t = x/(1.0 m s⁻¹) − 0.05 s. Its nonzero intercept reveals the common timing offset.
3. Dividing x by t assumes a zero intercept; that assumption fails most strongly over short distances.
4. Choose added trolley mass as the independent variable and interval speed as the dependent variable; keep slope, release position, wheel condition and gate spacing fixed.
5. Use several masses and repeat each condition, recording SI units. Two light gates remove a human start delay; guard the run-out area.
6. The 0.01 s display resolution is finer than the ±0.03 s repeat scatter. Repeats improve precision of a mean but cannot remove the systematic 0.05 s timing error.

Answer (a) Speed = 1.0 m s⁻¹; recorded time is 0.05 s too small. (b) Change mass alone; measure speed over a fixed interval with light gates, repeats, and controlled release.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

1. The slope is 10/27.3 kPa °C⁻¹. Extrapolating 100 kPa down to zero requires a decrease of 273°C.
2. Use absolute temperature T = 273 K and SI pressure 100000 Pa in pV = nRT.
3. n = 100000(0.002)/(8.31×273) ≈0.0882 mol. The mathematical intercept motivates an absolute scale; a real gas condenses before this extrapolation reaches zero.
4. 136.5°C corresponds to 409.5 K, 1.5 times 273 K, so fixed-volume pressure rises by the same factor to 150 kPa.
5. Higher absolute temperature means greater mean molecular kinetic energy; more vigorous momentum transfer to the walls raises pressure at fixed volume.
6. At fixed pressure and fixed particle number, V/T is constant, so the volume increases by 1.5. Dividing by Celsius temperatures would give an invalid proportionality.

Answer (a) Zero-pressure intercept ≈−273°C; amount ≈0.0882 mol. (b) Rigid-container pressure = 150 kPa; at constant pressure the volume would rise to 3.0×10⁻³ m³.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

1. One complete spatial cycle spans 0.80 m, and maximum displacement from equilibrium is 3.0 cm.
2. Frequency is cycles/time = 5/2.0 = 2.5 Hz; T = 1/f = 0.40 s; propagation speed is fλ = 2.0 m s⁻¹.
3. A single spatial profile can move either left or right. A later snapshot or local displacement–time information is needed to establish direction.
4. Superposition adds signed displacements: +3.0+(−2.0) = +1.0 cm upward.
5. Partial cancellation describes displacement at that location and instant. It does not mean the waves permanently destroy each other.
6. The pulses continue in the ideal linear model. Energy is redistributed between kinetic and elastic forms; displacement alone is insufficient to determine all local energy during overlap.

Answer (a) A = 3.0 cm; λ = 0.80 m; T = 0.40 s; f = 2.5 Hz; v = 2.0 m s⁻¹. (b) Resultant displacement is +1.0 cm; in the ideal linear medium both pulses continue through each other.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

1. Outside a magnet, a compass's north-seeking end follows the field direction from north to south. Pointing away from the right end identifies that end as north.
2. The measured gradient is 1.0×10⁻³ T A⁻¹ = μ₀n, so n = 0.001/(4π×10⁻⁷)≈796 m⁻¹.
3. Reversing current reverses field direction. At −1.5 A the linear calibration predicts −1.5 mT; its magnitude is unchanged.
4. Control current rather than supply voltage, because resistance and heating may change. Keep probe orientation and centre position fixed, repeat measurements and subtract the ambient field.
5. Measure a current-up and current-down sequence to detect hysteresis or remanence in the iron core; the air-core proportional model need not describe a saturating core.
6. Field lines represent local directions and relative density. They are not physical wires or trajectories that isolated magnetic poles follow; a complete magnetic field map forms loops.

Answer (a) Right end is north; n≈796 turns m⁻¹; reversed field is −1.5 mT. (b) Vary core material at fixed coil geometry and current; measure signed field at fixed locations and check for residual magnetism.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.