MYP · course MYP
MYP 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
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.
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.
A collision with an external impulse
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A 2.0 kg cart moving east at 3.0 m s⁻¹ collides and sticks to a 1.0 kg cart moving west at 1.0 m s⁻¹. During the 0.20 s interaction the combined system also experiences the external force shown below. Take east as positive. Supplied data (Time / s; Force east / N): (0, 0), (0.1, -10), (0.2, 0). (a) Calculate the final velocity, explaining why momentum is not conserved for the two-cart system alone. [3 marks] (b) Compare initial and final kinetic energies. Explain why their difference cannot all be assigned to deformation from the data given. [3 marks]
External force during collision
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Force east / N
Horizontal axis (x): Time / s
Read the exact data as a table
| Time / s | Force east / N |
|---|---|
| 0 | 0 |
| 0.1 | -10 |
| 0.2 | 0 |
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): Calculate the final velocity, explaining why momentum is not conserved for the two-cart system alone.
Reveal the worked solution
- Initial signed momentum is 2(3)+1(−1) = 5 kg m s⁻¹.
- External impulse is the triangular signed area, ½(0.20)(−10) = −1.0 N s; final momentum is therefore 4.0 kg m s⁻¹.
- The stuck mass is 3.0 kg, so v = 4/3 m s⁻¹ east. Internal collision forces cancel in the system sum, but the stated external impulse does not.
- Initial kinetic energy is ½(2)(3²)+½(1)(1²) = 9.5 J; final is ½(3)(4/3)² = 8/3 J.
- The reduction is 9.5−8/3 = 41/6 J. The collision may transfer energy to sound, internal energy and deformation.
- The external force may also do work, which requires displacement information. Impulse alone does not specify work, so the energy loss cannot be attributed entirely to collision deformation.
Answer (a) Final velocity = 4/3 m s⁻¹ east. (b) Kinetic energy falls from 9.5 J to 8/3 J, a reduction of 41/6 ≈6.83 J.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.