AP Physics · course AP C:M
AP Physics C: Mechanics
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.
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.
A moving receiver below a launch platform
A ball leaves a platform horizontally at 6.0 m s⁻¹, 5.0 m above the ground. A receiver starts 4.0 m horizontally from the point directly below the launch and moves away at constant speed 2.0 m s⁻¹. Use g = 10 m s⁻² and neglect air resistance; model both objects as points. Supplied data (Time / s; Receiver horizontal position / m): (0, 4), (0.5, 5), (1, 6), (1.5, 7). (a) Determine whether the ball meets the receiver at ground level and calculate the impact velocity components. [3 marks] (b) The ball is instead launched horizontally at 7.0 m s⁻¹. Find the receiver's required starting position and explain the role of the two independent components. [3 marks]
Receiver position after launch
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Receiver horizontal position / m
Horizontal axis (x): Time / s
Read the exact data as a table
| Time / s | Receiver horizontal position / m |
|---|---|
| 0 | 4 |
| 0.5 | 5 |
| 1 | 6 |
| 1.5 | 7 |
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 whether the ball meets the receiver at ground level and calculate the impact velocity components.
Reveal the worked solution
- Vertical motion gives 5.0 = ½(10)t², hence t = 1.0 s.
- The ball reaches x = 6.0t = 6.0 m; the receiver reaches x = 4.0+2.0t = 6.0 m, so they coincide.
- Horizontal velocity stays 6.0 m s⁻¹; vertical velocity is −gt = −10 m s⁻¹. Speed is √136 ≈11.7 m s⁻¹, directed about 59° below horizontal.
- Flight time remains 1.0 s because initial vertical velocity and vertical drop are unchanged.
- The ball reaches x = 7.0 m, while the receiver travels 2.0 m during the fall. Its initial x must therefore be 5.0 m.
- Horizontal speed changes range without changing fall time under this model; a finite receiver or air resistance would require a more detailed tolerance model.
Answer (a) Both reach x = 6.0 m at 1.0 s; impact velocity is (6.0, −10) m s⁻¹. (b) Required starting position = 5.0 m from the launch vertical.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Reading an accelerating lift's force sensor
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.
Why a parachute opens before the new terminal speed
A 2.0 kg falling package has weight 20 N. Downward is positive. Before a parachute opens, drag follows the supplied measured speed–force data. After opening, drag at every speed is four times the previous value. Use the evidence to infer the drag model over this range. Supplied data (Speed / m s⁻¹; Upward drag / N): (0, 0), (2, 2), (4, 8), (6, 18), (8, 32). (a) Infer the drag relationship, find both terminal speeds, and determine the acceleration immediately after opening at 6 m s⁻¹. [3 marks] (b) Explain why the package does not instantaneously jump to its new terminal speed, and evaluate the model as it slows. [3 marks]
Drag before the parachute opens
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Upward drag / N
Horizontal axis (x): Speed / m s⁻¹
Read the exact data as a table
| Speed / m s⁻¹ | Upward drag / N |
|---|---|
| 0 | 0 |
| 2 | 2 |
| 4 | 8 |
| 6 | 18 |
| 8 | 32 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Infer the drag relationship, find both terminal speeds, and determine the acceleration immediately after opening at 6 m s⁻¹.
Reveal the worked solution
- Each nonzero data pair gives D/v² = 0.50 N s² m⁻², supporting quadratic drag over the measured range.
- Before opening, terminal motion requires 0.50v² = 20, giving √40 m s⁻¹. After opening D = 2.0v², giving √10 m s⁻¹.
- At 6 m s⁻¹ just after opening, drag is 72 N upward. Net force is 20−72 = −52 N, so a = −52/2 = −26 m s⁻².
- The package initially still moves downward; the upward acceleration changes its velocity over time rather than resetting it.
- As it slows, drag falls and the upward resultant becomes smaller, approaching zero at the new terminal speed.
- The fourfold-drag model assumes the canopy is fully inflated. Inflation time, air density and oscillation can invalidate it during the opening transient.
Answer (a) D = 0.50v²; terminal speeds ≈6.32 and 3.16 m s⁻¹; acceleration = −26 m s⁻² immediately after opening. (b) Finite upward resultant reduces downward speed continuously until drag balances weight.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.