Compare stopping times and average forces
Keep the same momentum change while comparing duration, and avoid treating average force as a guaranteed peak.
Exam scope and assessment notes
Extended only (Supplement) · Physics 0625, 2026–2028, section 1.6. Define and use signed momentum p = mv, impulse FΔt = Δ(mv), conservation in simple one-dimensional systems, and resultant force as momentum change per unit time. F is the relevant resultant, with a constant force or stated average over the interval. Conservation requires negligible net external impulse on the chosen system; collision outcomes need enough supplied information. Written practice is self-checked. Supervised practical experience, cumulative assessment and human subject-teacher review remain pending.
Before you begin
- Calculate stopping momentum change.
- Use average resultant = change / duration.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is stopping Δp for 0.50 kg initially at +10 m/s?
What you will learn
- Explain longer stopping time for the same change in momentum.
- Distinguish a mean-force comparison from an unsupported peak or safety claim.
Hold the momentum change fixed
A 0.50 kg ball stops from +10 m/s, so Δp = 0 − 0.50 × 10 = −5.0 kg m/s. If stopping takes 0.020 s, average resultant is −5.0/0.020 = −250 N. If the same initial and final states take 0.100 s, it is −50 N. The impulse is −5.0 N s in both cases, but the longer interval reduces the average magnitude by a factor of five.
Moving the hands back during a controlled catch, cushioning or a deforming structure can increase the time over which momentum changes. Explain the comparison using the same object's initial and final states and the relevant resultant. A different rebound or initial speed changes Δp and cannot be ranked by time alone.
An average is a limited conclusion
The −250 N and −50 N values are averages over their stated intervals. Without the force-time profiles, those values do not establish the exact peak forces or certify a real protective design. A force concentrated into a short part of a long interval can still have a large peak. The diagram compares averages rather than measured contact histories.
If the ball instead rebounds at −4 m/s from the same +10 m/s, Δp = 0.50(−4 − 10) = −7.0 kg m/s. Even at 0.100 s its average resultant is −70 N, different from the −50 N stopping case. Specify both momentum change and duration before explaining a force reduction; do not turn a useful physical trend into an unconditional safety guarantee.
Pause and explain
For the same Δp, stopping time becomes five times longer. What happens to average force magnitude?
Worked example
A 0.50 kg ball stops from +10 m/s in either 0.020 s or 0.100 s. Compare average resultants and impulses.
Show the worked solution
- Both have Δp = −5.0 kg m/s and impulse −5.0 N s.
- Average forces are −5.0/0.020 = −250 N and −5.0/0.100 = −50 N.
- Five times longer duration gives one fifth the average force magnitude for the same change.
Answer −250 N and −50 N; both impulses −5.0 N s.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Compare average resultants when 0.50 kg stops from +10 m/s in 0.020 s and 0.100 s.
Give me a hint
Use the same signed change in both time quotients.
Compare my reasoning
- Δp = −5.0 kg m/s.
- First average = −5.0/0.020 = −250 N.
- Second average = −5.0/0.100 = −50 N.
−250 N and −50 N.
Look for these in your work
- Hold initial and final states fixed.
- Compare average magnitudes after keeping their signs.
The same 0.50 kg ball rebounds from +10 to −4 m/s in 0.100 s. Find Δp and average resultant, then compare with stopping over that duration.
Give me a hint
The rebound has a different final velocity despite the same interval.
Compare my reasoning
- Δp = 0.50(−4 − 10) = −7.0 kg m/s.
- Average resultant = −7.0/0.100 = −70 N.
- Stopping instead gives −50 N, so the rebound's average magnitude is greater.
Δp = −7.0 kg m/s; average −70 N, compared with stopping −50 N.
Look for these in your work
- Use the reversed final sign.
- Do not assume equal impulse just because times match.
Two stopping arrangements have the same Δp, and the longer one has lower average force. Does that alone prove its peak is lower or its design is certified safe? Explain.
Give me a hint
Identify what the interval averages do not reveal.
Compare my reasoning
- The longer duration reduces the average for the stated fixed change.
- The force profiles may differ, so their peaks are not supplied by those averages.
- A real protection assessment needs relevant measured behaviour and appropriate review beyond this model.
No; a lower average alone does not establish the peak or certify a protective design.
Look for these in your work
- Retain the valid average trend.
- State the missing time-profile and review evidence.
Common mistakes
- Keep the same momentum change for a time-only comparison.
- Rebounding changes the final momentum.
- An average does not establish a measured peak or certified protection.
What can you explain now?
Would a rebound at −4 m/s have the same Δp as stopping from +10 m/s for 0.50 kg?
Compare with the explanation
No; rebound Δp is −7.0 kg m/s, compared with stopping −5.0.
The reversed final velocity supplies an additional negative final momentum.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Explain a longer-time stopping comparison, then give a rebound or varying-force case that limits the conclusion.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.