MYP Physics · topic MYP-04.02
Impulse and force–time evidence
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 safer catch with the same momentum change
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. 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.