Find average force and contact time from impulse
Rearrange the impulse equation and keep the resultant force distinct from one contact force.
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 momentum changes with reversal.
- Use consistent force and time units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which rearrangement finds average resultant from momentum change and duration?
What you will learn
- Use Faverage = Δp/Δt.
- Find a duration from a stated force and momentum change.
The signed quotient gives an average resultant
A 0.20 kg ball rebounding from +6 to −4 m/s has Δp = −2.0 kg m/s. Over contact time 0.050 s, its average horizontal resultant is −2.0/0.050 = −40 N. The negative result points left for right positive. This is an average over the interval, not an automatically measured peak force.
If a constant −25 N horizontal resultant produces change −2.0 kg m/s, the required duration is Δt = Δp/F = (−2.0)/(−25) = 0.080 s. A physically positive duration follows because the force and its resulting impulse share direction. An inconsistent sign arrangement needs checking rather than an invented negative contact time.
Include other forces in the chosen direction
For a vertical catch, the average contact force and the resultant need not be equal because weight also acts. In the original example, a 0.50 kg object changes from 4 m/s downward to rest in 0.20 s, with upward positive and g = 10 N/kg. Its Δp = 0 − 0.50(−4) = +2.0 kg m/s, so average resultant is +10 N.
Weight is 5 N downward. The contact equation is Fcontact − 5 = 10, giving average contact force 15 N upward. Reporting 10 N as the upward contact force would ignore gravity. Use the relevant forces on the selected object and distinguish a resultant, a contact interaction and a force on a different object.
Pause and explain
In an upward-positive vertical catch, is average upward contact force always equal to average resultant?
Worked example
Find average horizontal resultant for a 0.20 kg ball changing from +6 to −4 m/s in 0.050 s.
Show the worked solution
- Δp = 0.20(−4 − 6) = −2.0 kg m/s.
- Faverage = Δp/Δt = −2.0/0.050.
- Average resultant = −40 N.
Answer −40 N, directed left.
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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.
Find average resultant for Δp = −2.0 kg m/s over 0.050 s.
Give me a hint
Divide change by duration and keep the signed direction.
Compare my reasoning
- Faverage = −2.0/0.050.
- The magnitude is 40 N.
- The signed average resultant is −40 N.
−40 N.
Look for these in your work
- Use momentum change rather than final momentum.
- State the force direction.
A constant −25 N resultant must give Δp = −2.0 kg m/s. Find its duration and express it in milliseconds.
Give me a hint
Solve time = change/force before converting seconds.
Compare my reasoning
- Duration = (−2.0)/(−25) = 0.080 s.
- The time is positive because the signs agree.
- 0.080 s = 80 ms.
0.080 s or 80 ms.
Look for these in your work
- Keep a positive physical duration.
- Convert the time unit correctly.
A 0.50 kg object moving 4 m/s down stops in 0.20 s. With upward positive and g = 10 N/kg, find average resultant and upward contact force.
Give me a hint
Calculate the upward momentum change, then include downward weight in the contact equation.
Compare my reasoning
- Δp = 0 − 0.50(−4) = +2.0 kg m/s, so resultant = +10 N.
- Weight = 0.50 × 10 = 5 N downward.
- Contact − 5 = 10, so average contact = 15 N upward.
Average resultant 10 N upward; average contact force 15 N upward.
Look for these in your work
- Use signed initial downward velocity.
- Account for weight separately.
Common mistakes
- Find momentum change before dividing by time.
- Average is not peak.
- A contact force can differ from the resultant when other relevant forces act.
What can you explain now?
What duration gives impulse −2.0 N s for a constant −25 N resultant?
Compare with the explanation
0.080 s.
Divide the signed impulse by the force; both are negative, giving positive time.
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Make it stick
Derive average force and time from impulse, then explain why a vertical contact force differs from the resultant.
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