Cambridge IGCSE Physics · unit 1
Motion, forces and energy
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
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. 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.
Is a hollow object made from the claimed metal?
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A sealed metal object has mass 156 g. It raises a measuring-cylinder reading from 40 to 100 cm³. Each reading has uncertainty ±1 cm³, and the mass has uncertainty ±1 g. Solid samples of the claimed alloy give the mass–volume data below. Ignore the mass of air in the hollow object. Supplied data (Volume / cm³; Mass / g): (10, 78), (20, 156), (30, 234). (a) Use the reference data to find the alloy density and calculate the sealed object's cavity volume. [3 marks] (b) Estimate a conservative uncertainty in the external volume and in the average density; suggest a method improvement. [3 marks]
Solid alloy samples
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Mass / g
Horizontal axis (x): Volume / cm³
Read the exact data as a table
| Volume / cm³ | Mass / g |
|---|---|
| 10 | 78 |
| 20 | 156 |
| 30 | 234 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Use the reference data to find the alloy density and calculate the sealed object's cavity volume.
Reveal the worked solution
- The mass–volume gradient is 78/10 = 7.8 g cm⁻³, equivalent to 7800 kg m⁻³.
- Metal volume = mass/density = 156/7.8 = 20 cm³. The external volume is 100−40 = 60 cm³.
- Cavity volume = external volume − metal volume = 40 cm³. An average density of 2.6 g cm⁻³ does not by itself disprove the alloy claim.
- For a conservative bound, add the two reading uncertainties: ΔV = 1+1 = 2 cm³.
- The approximate fractional density uncertainty is 1/156 + 2/60 = 0.0397; absolute uncertainty is about 0.10 g cm⁻³.
- Use a narrower calibrated cylinder and read the meniscus at eye level. Ensure complete submersion without trapped external bubbles; repeated readings do not correct a trapped bubble.
Answer (a) Alloy density = 7.8 g cm⁻³; cavity volume = 40 cm³. (b) External volume = 60 ±2 cm³; average density ≈2.6 ±0.10 g cm⁻³.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The journey that ends closer but travels farther
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. Take east as positive. A robot's velocity varies linearly between the supplied points. At t = 0 it is 2 m east of the origin. The graph describes one-dimensional motion; negative velocity means westward motion. Supplied data (Time / s; Velocity east / m s⁻¹): (0, 4), (2, 4), (4, 0), (6, -4), (8, -4). (a) Determine displacement, total distance, final position, average speed and average velocity for the full 8 s. [3 marks] (b) Find the acceleration between 2 and 6 s and explain whether the robot is slowing throughout that interval. [3 marks]
Signed robot velocity
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Velocity east / m s⁻¹
Horizontal axis (x): Time / s
Read the exact data as a table
| Time / s | Velocity east / m s⁻¹ |
|---|---|
| 0 | 4 |
| 2 | 4 |
| 4 | 0 |
| 6 | -4 |
| 8 | -4 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Determine displacement, total distance, final position, average speed and average velocity for the full 8 s.
Reveal the worked solution
- Signed graph areas are +8, +4, −4 and −8 m, so their sum is zero displacement.
- Distance adds the magnitudes, giving 24 m. The final position is the initial 2 m plus zero displacement.
- Average speed = 24/8 = 3 m s⁻¹, whereas average velocity = 0/8 = 0. Distance and speed are scalars; displacement and velocity require direction.
- The gradient is (−4−4)/(6−2) = −2 m s⁻² throughout the interval.
- Before 4 s acceleration opposes velocity, so speed decreases. After 4 s acceleration and velocity both point west, so speed increases.
- A negative acceleration does not alone mean slowing. Compare its direction with the instantaneous velocity.
Answer (a) Displacement = 0 m; distance = 24 m; final position = 2 m east; average speed = 3 m s⁻¹; average velocity = 0. (b) Acceleration = −2 m s⁻²; it slows until 4 s and then speeds up westward.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Reading an accelerating lift's force sensor
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A 60 kg student stands on a scale in a lift. Use g = 10 N kg⁻¹. The graph gives the upward scale force during three separate constant-force phases. Initially the lift is stationary; ignore the short transitions between phases. Supplied data (Phase; Upward normal force / N): (1, 720), (2, 600), (3, 480). (a) Draw or describe the student's free-body diagram and find acceleration in each phase. Identify which forces form a Newton's-third-law pair. [3 marks] (b) Phase 1 lasts 2 s, phase 2 lasts 3 s and phase 3 lasts 2 s. Find total height gained and explain the 600 N reading while the lift moves. [3 marks]
Scale reading during lift phases
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Upward normal force / N
Horizontal axis (x): Phase
Read the exact data as a table
| Phase | Upward normal force / N |
|---|---|
| 1 | 720 |
| 2 | 600 |
| 3 | 480 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Draw or describe the student's free-body diagram and find acceleration in each phase. Identify which forces form a Newton's-third-law pair.
Reveal the worked solution
- The student experiences weight mg = 600 N downward and normal force N upward. There is no additional force called acceleration.
- Taking upward positive, N−600 = 60a gives accelerations +2.0, 0 and −2.0 m s⁻².
- Weight and normal act on the same body and are not a third-law pair. Each has a partner on a different body.
- Phase 1 ends at 4 m s⁻¹ and gains ½(2)(2²) = 4 m. Phase 2 gains (4)(3) = 12 m.
- Phase 3 slows from 4 to 0 m s⁻¹ in 2 s and gains 4 m, so total rise is 20 m.
- Zero resultant means constant velocity, which may be nonzero. Balanced forces do not require the lift to be stationary.
Answer (a) Accelerations: +2.0, 0, −2.0 m s⁻². The scale-on-student and student-on-scale forces form a third-law pair. (b) Height gained = 20 m; a 600 N reading is consistent with upward constant velocity.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.