Cambridge International A-Level Physics · topic 2
Kinematics
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
Work through the linked stages and explain what each result means.
A convincing graph with a hidden timing error
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
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.
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.