AP · course AP
AP 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.
The journey that ends closer but travels farther
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.
Heating ice while heat escapes
A 0.10 kg sample of ice begins at 0°C. A heater supplies 100 W, while a separate calibration shows a constant 20 W loss to the surroundings during this experiment. The sample-temperature observations are supplied below. Use latent heat of fusion 3.36×10⁵ J kg⁻¹ and water specific heat capacity 4200 J kg⁻¹ K⁻¹. Ignore vessel heat capacity. Supplied data (Time / s; Temperature / °C): (0, 0), (210, 0), (420, 0), (525, 20). (a) Predict the melting duration and then the time to reach 20°C after melting. Check both against the data. [3 marks] (b) Explain the constant-temperature interval with a particle model and evaluate a calculation that ignores losses. [3 marks]
Observed heating curve
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Temperature / °C
Horizontal axis (x): Time / s
Read the exact data as a table
| Time / s | Temperature / °C |
|---|---|
| 0 | 0 |
| 210 | 0 |
| 420 | 0 |
| 525 | 20 |
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): Predict the melting duration and then the time to reach 20°C after melting. Check both against the data.
Reveal the worked solution
- Net heating power is 100−20 = 80 W, since the sample does not receive the full heater input.
- Melting needs mL = 0.10(3.36×10⁵) = 33600 J; duration is 33600/80 = 420 s.
- Warming liquid by 20 K needs mcΔT = 0.10(4200)(20) = 8400 J; duration is 8400/80 = 105 s. Both match the supplied curve.
- During melting, transferred energy changes intermolecular potential energy rather than increasing mean kinetic energy, so temperature need not rise.
- The internal energy still increases through the phase change; temperature is not a measure of the total internal energy.
- Ignoring losses gives 33600/100 = 336 s, 84 s too short. Insulation and a lid can reduce unwanted transfers; repeats alone do not remove systematic heat loss.
Answer (a) Melting takes 420 s; subsequent warming takes 105 s. (b) Temperature stays constant while internal energy increases through a change in particle arrangement; ignoring losses underestimates the required time.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Calculus extension: a reversal hidden inside the interval
A 2.0 kg cart has one-dimensional position x(t) = t³−6t²+9t metres for 0≤t≤4 s, with t in seconds. The supplied position samples are evidence points on this cubic, not straight-line motion between them. No other translational kinetic-energy stores are included. Supplied data (Time / s; Position / m): (0, 0), (1, 4), (2, 2), (3, 0), (4, 4). (a) Differentiate to find velocity, acceleration and net force. Locate reversals and compute total distance and displacement. [3 marks] (b) Calculate net impulse from 0 to 4 s and check net work from 0 to 2 s with kinetic energy. Explain why zero impulse need not mean no force. [3 marks]
Selected position samples on a cubic trajectory
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Position / m
Horizontal axis (x): Time / s
Read the exact data as a table
| Time / s | Position / m |
|---|---|
| 0 | 0 |
| 1 | 4 |
| 2 | 2 |
| 3 | 0 |
| 4 | 4 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Differentiate to find velocity, acceleration and net force. Locate reversals and compute total distance and displacement.
Reveal the worked solution
- Differentiate x(t): v = dx/dt = 3(t−1)(t−3), and a = dv/dt = 6t−12. Newton's law gives F = ma = 12t−24 N.
- Velocity changes sign at 1 and 3 s. Position values at 0,1,3,4 s are 0,4,0,4 m, so path length is 4+4+4 = 12 m.
- Net displacement is x(4)−x(0) = 4 m. The intermediate t = 2 s sample does not create an additional reversal; smooth cubic behaviour must be used.
- Integrating force gives J = ∫₀⁴(12t−24)dt = [6t²−24t]₀⁴ = 0 N s, matching m[v(4)−v(0)] = 2(9−9).
- At t = 0, v = 9 m s⁻¹ and K = 81 J. At t = 2, v = −3 m s⁻¹ and K = 9 J, so net work is −72 J.
- A direct work integral is ∫F(t)v(t)dt, not ∫F(t)dt. Positive and negative force impulses can cancel even though the force is nonzero during most of the interval.
Answer (a) v = 3t²−12t+9; a = 6t−12; F = 12t−24 N; reversals at 1 and 3 s; distance = 12 m, displacement = 4 m. (b) Net impulse = 0; net work from 0 to 2 s = −72 J.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Finding the location where a field vanishes
Two fixed point charges lie on the x-axis: +4.0 nC at x = 0 and +1.0 nC at x = 0.30 m. Use k = 9.0×10⁹ N m² C⁻² and V = kQ/r with zero potential at infinity. The supplied graph gives the field magnitude of the +4.0 nC charge alone. Supplied model: For point charges use field magnitude E = k|Q|/r² and force F = qE, adding field vectors with directions. Supplied data (Distance from left charge / m; Field magnitude / N C⁻¹): (0.1, 3600), (0.2, 900), (0.3, 400). (a) Locate the zero-resultant-field point between the charges and find the electric potential there. [3 marks] (b) Find force on a −2.0 nC test charge at x = 0.10 m, and describe useful field-line features without treating them as literal objects. [3 marks]
Field of the left charge alone
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Field magnitude / N C⁻¹
Horizontal axis (x): Distance from left charge / m
Read the exact data as a table
| Distance from left charge / m | Field magnitude / N C⁻¹ |
|---|---|
| 0.1 | 3600 |
| 0.2 | 900 |
| 0.3 | 400 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Locate the zero-resultant-field point between the charges and find the electric potential there.
Reveal the worked solution
- Between two positive charges their fields oppose. Equality requires 4/x² = 1/(0.30−x)², giving x/(0.30−x) = 2.
- Solve x = 0.20 m, with distances 0.20 and 0.10 m. Field directions, not just magnitudes, must be checked.
- Potential adds as a scalar: V = 9.0×10⁹[(4.0×10⁻⁹)/0.20+(1.0×10⁻⁹)/0.10] = 270 V. Zero field does not require zero potential.
- At x = 0.10 m the left charge contributes +3600 N C⁻¹, while the right contributes −9.0/(0.20²) = −225 N C⁻¹.
- Net field is +3375 N C⁻¹. F = qE = (−2.0×10⁻⁹)(3375) = −6.75×10⁻⁶ N; a negative charge feels force opposite the field.
- Lines point away from positive charges, never cross and are denser where field is stronger in a consistent drawing. They represent a continuous vector field, not discrete channels carrying test charges.
Answer (a) Zero field at x = 0.20 m; potential there = +270 V. (b) Resultant field = 3375 N C⁻¹ east; force = 6.75×10⁻⁶ N west.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.