Cambridge International A-Level Physics · topic 3
Dynamics
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.
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.
A collision with an external impulse
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.
A safer catch with the same momentum change
A 0.20 kg ball arrives horizontally at 15 m s⁻¹ and is brought to rest. Design A produces the triangular stopping-force magnitude shown below. Design B gives the same impulse with a triangular pulse lasting 0.060 s. Ignore gravity during the short horizontal catch. Supplied data (Time / s; Force magnitude / N): (0, 0), (0.01, 300), (0.02, 0). (a) Check design A's impulse and find its average force. Determine the peak force needed for design B. [3 marks] (b) Evaluate the claim that extending the catch reduces the impulse and the energy absorbed. [3 marks]
Stopping-force magnitude: design A
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Force magnitude / N
Horizontal axis (x): Time / s
Read the exact data as a table
| Time / s | Force magnitude / N |
|---|---|
| 0 | 0 |
| 0.01 | 300 |
| 0.02 | 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): Check design A's impulse and find its average force. Determine the peak force needed for design B.
Reveal the worked solution
- The required momentum-change magnitude is mΔv = 0.20(15) = 3.0 kg m s⁻¹, opposite the initial motion.
- The triangle area is ½(0.020)(300) = 3.0 N s, consistent with stopping. Average force is 3.0/0.020 = 150 N.
- For design B, ½(0.060)Fpeak = 3.0, giving 100 N. The average is 50 N, half the peak for a triangular pulse.
- The ball has the same initial and final momentum, so its impulse cannot change merely by spreading the catch over time.
- Its kinetic-energy reduction is ½(0.20)(15²) = 22.5 J in both cases. The longer catch can distribute that transfer over a larger distance.
- Lower peak force can reduce injury or damage, but actual safety also depends on pressure, contact area, material deformation and the available stopping distance.
Answer (a) Impulse magnitude = 3.0 N s; A average = 150 N; B peak = 100 N. (b) Both designs require the same 3.0 N s impulse and remove 22.5 J of ball kinetic energy.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.