MYP Physics · unit 20
Independent assessment practice
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.
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 braking model tested at two speeds
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A 2.0 kg trolley reaches a level rough strip after descending from rest through height 1.8 m on a smooth ramp. Use g = 10 m s⁻². On the strip the opposing force depends on distance as shown. It remains 8 N beyond x = 4 m. Neglect rotational energy. Supplied data (Distance along strip / m; Resisting force / N): (0, 4), (4, 8), (6, 8). (a) Find entry speed and stopping distance, using the force–distance graph rather than a constant deceleration formula. [3 marks] (b) If the entry speed doubles, determine the new stopping distance and explain the energy scaling. [3 marks]
Resisting force on the rough strip
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Resisting force / N
Horizontal axis (x): Distance along strip / m
Read the exact data as a table
| Distance along strip / m | Resisting force / N |
|---|---|
| 0 | 4 |
| 4 | 8 |
| 6 | 8 |
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): Find entry speed and stopping distance, using the force–distance graph rather than a constant deceleration formula.
Reveal the worked solution
- The trolley gains mgh = 36 J, so ½(2)v² = 36 and v = 6.0 m s⁻¹.
- Work against resistance over the first 4 m is trapezium area ½(4+8)(4) = 24 J, leaving 12 J of kinetic energy.
- At 8 N it travels another 12/8 = 1.5 m, stopping at 5.5 m. Resistance is not constant on the first section, so neither is deceleration.
- Kinetic energy varies as v², so doubling speed multiplies it by four, giving 144 J.
- The first 4 m still dissipates 24 J. The remaining 120 J requires 120/8 = 15 m beyond that point.
- Total distance is 19 m. It is not four times 5.5 m because the spatial resistance profile has a lower-force entry section.
Answer (a) Entry speed = 6.0 m s⁻¹; stopping distance = 5.5 m. (b) Kinetic energy becomes 144 J and stopping distance becomes 19 m.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Reconstructing a travelling wave and an overlapping pulse
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. At one instant a transverse wave has the supplied displacement–position profile. Adjacent points are joined smoothly as a sinusoid. A fixed sensor observes five complete cycles in 2.0 s. Later an upward 3.0 cm pulse and a downward 2.0 cm pulse overlap at the sensor in a linear medium. Supplied data (Position / m; Displacement / cm): (0, 0), (0.025, 0.5853), (0.05, 1.1481), (0.075, 1.6667), (0.1, 2.1213), (0.125, 2.4944), (0.15, 2.7716), (0.175, 2.9424), (0.2, 3), (0.225, 2.9424), (0.25, 2.7716), (0.275, 2.4944), (0.3, 2.1213), (0.325, 1.6667), (0.35, 1.1481), (0.375, 0.5853), (0.4, 0), (0.425, -0.5853), (0.45, -1.1481), (0.475, -1.6667), (0.5, -2.1213), (0.525, -2.4944), (0.55, -2.7716), (0.575, -2.9424), (0.6, -3), (0.625, -2.9424), (0.65, -2.7716), (0.675, -2.4944), (0.7, -2.1213), (0.725, -1.6667), (0.75, -1.1481), (0.775, -0.5853), (0.8, 0). (a) Find amplitude, wavelength, period, frequency and propagation speed. Explain why the snapshot does not determine the direction of travel. [3 marks] (b) Find the displacement during pulse overlap and explain what happens to the pulses and their transported energy afterward. [3 marks]
Wave snapshot
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Displacement / cm
Horizontal axis (x): Position / m
Read the exact data as a table
| Position / m | Displacement / cm |
|---|---|
| 0 | 0 |
| 0.025 | 0.5853 |
| 0.05 | 1.1481 |
| 0.075 | 1.6667 |
| 0.1 | 2.1213 |
| 0.125 | 2.4944 |
| 0.15 | 2.7716 |
| 0.175 | 2.9424 |
| 0.2 | 3 |
| 0.225 | 2.9424 |
| 0.25 | 2.7716 |
| 0.275 | 2.4944 |
| 0.3 | 2.1213 |
| 0.325 | 1.6667 |
| 0.35 | 1.1481 |
| 0.375 | 0.5853 |
| 0.4 | 0 |
| 0.425 | -0.5853 |
| 0.45 | -1.1481 |
| 0.475 | -1.6667 |
| 0.5 | -2.1213 |
| 0.525 | -2.4944 |
| 0.55 | -2.7716 |
| 0.575 | -2.9424 |
| 0.6 | -3 |
| 0.625 | -2.9424 |
| 0.65 | -2.7716 |
| 0.675 | -2.4944 |
| 0.7 | -2.1213 |
| 0.725 | -1.6667 |
| 0.75 | -1.1481 |
| 0.775 | -0.5853 |
| 0.8 | 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): Find amplitude, wavelength, period, frequency and propagation speed. Explain why the snapshot does not determine the direction of travel.
Reveal the worked solution
- One complete spatial cycle spans 0.80 m, and maximum displacement from equilibrium is 3.0 cm.
- 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.
- A single spatial profile can move either left or right. A later snapshot or local displacement–time information is needed to establish direction.
- Superposition adds signed displacements: +3.0+(−2.0) = +1.0 cm upward.
- Partial cancellation describes displacement at that location and instant. It does not mean the waves permanently destroy each other.
- 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.
A battery whose terminal voltage depends on the load
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A battery supplies an external load. Its measured terminal-voltage versus total-current data are supplied. A 6.0 Ω resistor and a 3.0 Ω resistor are connected in parallel across its terminals. Model the battery as a fixed emf in series with internal resistance. Supplied data (Current / A; Terminal voltage / V): (0, 12), (1, 11), (2, 10), (3, 9). (a) Infer emf and internal resistance, then find total current, terminal voltage and both branch currents. [3 marks] (b) Calculate useful load power and battery internal heating, then predict the load voltage if the 3 Ω branch is disconnected. [3 marks]
Battery load test
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Terminal voltage / V
Horizontal axis (x): Current / A
Read the exact data as a table
| Current / A | Terminal voltage / V |
|---|---|
| 0 | 12 |
| 1 | 11 |
| 2 | 10 |
| 3 | 9 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Infer emf and internal resistance, then find total current, terminal voltage and both branch currents.
Reveal the worked solution
- V = ε−Ir has intercept ε = 12 V and gradient −r = −1.0 V A⁻¹, so internal resistance is 1.0 Ω.
- The parallel load is (1/6+1/3)⁻¹ = 2.0 Ω. Total current is ε/(Rload+r) = 12/3 = 4.0 A, giving terminal voltage 12−4 = 8.0 V.
- Both branches have 8.0 V across them, so I6 = 8/6 = 4/3 A and I3 = 8/3 A. Their sum checks the junction current rule.
- Power delivered externally is VI = 8(4) = 32 W. Internal heating is I²r = 16 W, and εI = 48 W checks energy conservation.
- With only the 6 Ω branch, total current is 12/(6+1) = 12/7 A.
- Terminal voltage is I(6) = 72/7 V, greater than before because the internal voltage drop is smaller. The emf is unchanged in this model.
Answer (a) emf = 12 V; r = 1.0 Ω; total current = 4.0 A; terminal voltage = 8.0 V; branch currents = 4/3 A and 8/3 A. (b) Load power = 32 W; internal heating = 16 W; disconnected-case terminal voltage = 72/7 ≈10.3 V.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.