A-Level · course A-Level
Cambridge International A-Level Physics
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
A convincing graph with a hidden timing error
A trolley travels at constant speed. A single observer measures elapsed time after each distance from the start line. The stopwatch resolves 0.01 s, the distance uncertainty is ±0.01 m, and repeat readings scatter by ±0.03 s. The time–distance data are supplied below. The observer starts the watch after the trolley passes the start line. Supplied data (Distance / m; Recorded time / s): (0.2, 0.15), (0.4, 0.35), (0.6, 0.55), (0.8, 0.75), (1, 0.95). (a) Find speed from the graph gradient and the timing offset. Explain why distance divided by each recorded time gives inconsistent speeds. [3 marks] (b) Design a fair test of the claim that added mass changes speed, and distinguish precision from accuracy in this evidence. [3 marks]
Elapsed time against distance
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Recorded time / s
Horizontal axis (x): Distance / m
Read the exact data as a table
| Distance / m | Recorded time / s |
|---|---|
| 0.2 | 0.15 |
| 0.4 | 0.35 |
| 0.6 | 0.55 |
| 0.8 | 0.75 |
| 1 | 0.95 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Find speed from the graph gradient and the timing offset. Explain why distance divided by each recorded time gives inconsistent speeds.
Reveal the worked solution
- 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.
- The line is t = x/(1.0 m s⁻¹) − 0.05 s. Its nonzero intercept reveals the common timing offset.
- Dividing x by t assumes a zero intercept; that assumption fails most strongly over short distances.
- 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.
- Use several masses and repeat each condition, recording SI units. Two light gates remove a human start delay; guard the run-out area.
- 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.
A satellite raised to twice the orbital radius
A spherical planet has radius R = 6.0×10⁶ m and surface gravitational field g₀ = 10 N kg⁻¹. A 100 kg satellite follows a circular orbit at radius 2R measured from the planet's centre. Ignore rotation and atmosphere. Use GM = g₀R² and the supplied inverse-square field data. Supplied model: Use g = GM/r², circular acceleration v²/r and orbital period T = 2πr/v. Supplied data (Radius / R; Field magnitude / N kg⁻¹): (1, 10), (2, 2.5), (3, 1.111111), (4, 0.625). (a) Find the satellite's weight, orbital speed and orbital period. Explain what supplies the centripetal resultant. [3 marks] (b) The radius is increased to 4R. Predict ratios of field, speed and period, and explain why astronauts can feel weightless despite nonzero gravity. [3 marks]
Planetary gravitational field
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Field magnitude / N kg⁻¹
Horizontal axis (x): Radius / R
Read the exact data as a table
| Radius / R | Field magnitude / N kg⁻¹ |
|---|---|
| 1 | 10 |
| 2 | 2.5 |
| 3 | 1.111111 |
| 4 | 0.625 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Find the satellite's weight, orbital speed and orbital period. Explain what supplies the centripetal resultant.
Reveal the worked solution
- At 2R the field is g₀/4 = 2.5 N kg⁻¹, so the gravitational force is mg = 250 N. The satellite's mass remains 100 kg.
- Gravity alone supplies mv²/r: v = √(gr) = √(2.5×1.2×10⁷)≈5477 m s⁻¹.
- Period is circumference/speed, T = 2πr/v≈1.38×10⁴ s. 'Centripetal force' names the inward resultant; it is not an additional interaction alongside gravity.
- For a circular orbit g∝r⁻², v∝r⁻¹/² and T∝r³/². Doubling r therefore gives the stated ratios.
- Astronauts and spacecraft accelerate together in free fall. Little support force acts on the astronauts, so their apparent weight can be near zero.
- The gravitational field is not zero in orbit, and constant speed still requires acceleration because velocity direction changes.
Answer (a) Weight = 250 N; orbital speed ≈5.48 km s⁻¹; period ≈1.38×10⁴ s. (b) Field becomes 1/4, speed 1/√2 and period 2√2 times the initial orbital values.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.