Calculate impulse from force and duration
Use the signed resultant force over its stated interval and connect impulse to momentum change.
Cross-referenced to the 2028–2029 0653 section using original teaching from the related single-science courses. A 0653-specific objective-by-objective and Core/Supplement review is in progress. Written practice is self-checked and does not demonstrate supervised laboratory competence or award an official Cambridge grade.
Before you begin
- Calculate a signed momentum change.
- Convert milliseconds to seconds.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How many seconds is 50 ms?
What you will learn
- Define impulse and use FΔt = Δp.
- State when a constant-force or average-force product is valid.
A force acts for an interval
For a constant resultant force, impulse is force multiplied by the time for which it acts: impulse = FΔt. Its unit is N s, equivalent to kg m/s, and its direction follows the signed force. A horizontal resultant −40 N acting for 50 ms or 0.050 s gives impulse −2.0 N s, equal to momentum change −2.0 kg m/s.
For the original graph, the force is −40 N from 0 to 0.050 s and zero afterwards to 0.100 s. The signed force-time area is −40 × 0.050 = −2.0 N s. The later zero-force interval adds no impulse; using the full 0.100 s with −40 N would incorrectly double the result.
Do not call every collision force constant
A real contact force may vary during a collision. The product Faverage Δt still represents impulse when Faverage is the appropriate time average of the resultant over that interval. A maximum force is not interchangeable with an average force. If no suitable constant force, average or force-time information is supplied, do not invent an impulse from one peak value.
Impulse equals the change in momentum, so a 0.20 kg ball with change −2.0 kg m/s receives impulse −2.0 N s in that horizontal direction. This concerns the resultant on the ball in the chosen direction. Additional relevant forces must be included; a horizontal example avoids adding vertical weight to a horizontal momentum equation.
Sources for this lesson
Pause and explain
May a peak collision force be substituted for the time-average force without other information?
Worked example
A constant horizontal resultant −40 N acts for 50 ms. Find impulse and momentum change.
Show the worked solution
- Convert 50 ms to 0.050 s.
- Impulse = −40 × 0.050 = −2.0 N s.
- Momentum change is −2.0 kg m/s in the same direction.
Answer −2.0 N s; Δp = −2.0 kg m/s.
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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.
Find impulse for −40 N constant resultant acting for 50 ms.
Give me a hint
Use seconds and retain the force sign.
Compare my reasoning
- 50 ms = 0.050 s.
- Impulse = −40 × 0.050.
- Impulse = −2.0 N s, equal to Δp = −2.0 kg m/s.
−2.0 N s.
Look for these in your work
- Convert time.
- Retain direction in the product.
A constant +12 N resultant acts for 0.25 s on a cart of initial momentum −1.0 kg m/s. Find impulse and final momentum.
Give me a hint
Add the impulse to the initial momentum rather than reporting it as the final value.
Compare my reasoning
- Impulse = 12 × 0.25 = +3.0 N s.
- Final momentum = initial momentum + change = −1.0 + 3.0.
- Final momentum = +2.0 kg m/s.
+3.0 N s; final momentum +2.0 kg m/s.
Look for these in your work
- Calculate the impulse.
- Use the supplied initial momentum.
A −40 N force acts for 0.050 s then becomes zero for another 0.050 s. A student uses −40 × 0.100. Explain the error.
Give me a hint
Split the interval by the actual stated force values.
Compare my reasoning
- The first interval contributes −2.0 N s.
- The second interval contributes zero.
- Total impulse remains −2.0 N s; the non-zero force did not act for the whole 0.100 s.
Use only the active 0.050 s for −40 N; total impulse −2.0 N s.
Look for these in your work
- Identify the force's actual duration.
- Include the later zero contribution correctly.
Common mistakes
- Convert ms to s.
- Use the interval during which the force acts.
- An average force is not a peak force.
What can you explain now?
How much impulse does a later zero-force interval add?
Compare with the explanation
Zero.
Force multiplied by that later duration is zero.
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Make it stick
Define impulse, connect its units to momentum change, and explain why a peak force is insufficient.
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.