Translate a force-time pulse into momentum change

1. Orient: Translate a force-time pulse into momentum change

For fixed mass, Newton’s second law can be written Fnet=Δp/Δt over an interval when Fnet is interpreted as the interval-average force. Multiplying gives impulse J=FavgΔt=Δp. For a varying force, impulse is the signed area under its force–time graph. The vector direction of impulse determines the direction of momentum change, not necessarily final velocity. Conversely, on a momentum-versus-time graph the interval slope Δp/Δt gives average net force; the local slope gives instantaneous net force. A rising, falling or horizontal momentum segment therefore indicates positive, negative or zero net force along the chosen axis, not necessarily positive, negative or zero momentum.

2. Explain and derive

A triangular pulse with peak force Fpeak and duration τ has impulse½Fpeakτ. Its average force is Fpeak/2, not the peak. Equal impulses can arise from different peak forces and durations; matching the area does not require matching the detailed force history.

Work and impulse both involve force but use different horizontal variables. Force-position area is work in joules; force-time area is impulse in N s. An impulse alone, without a mass and initial velocity or without knowing whether it is the net impulse, does not specify net work. For a point object with fixed known mass, known initial velocity and known net impulse, the final velocity is fixed, so the work–energy theorem determines net work without requiring the detailed trajectory. Work by only one selected force can still require additional motion or other-force information. Label the graph axes before assigning meaning to its area.

A rebound can produce a larger momentum change than merely stopping at the same incoming speed. For mass m moving initially at+v and finally at−v, Δp=−2 mv, while stopping gives−mv. Longer duration reduces average force for a fixed impulse, but duration alone cannot predict the peak without a pulse shape. This bounded model is not an injury or contact-material predictor.

3. Work an example

For a triangular+8 N pulse lasting 0.5 s, impulse is½×8×0.5=2 N s. The 2 kg object begins with momentum 2 kg m/s, ends with 4 kg m/s and therefore has final velocity 2 m/s. Reversing the peak to−8 N gives impulse−2 N s and brings it to rest.

4. Predict and investigate

Predict first: If peak force doubles while duration halves, what happens to impulse and the final velocity in the same-mass case?

Compare(8 N,0.5 s) with(4 N,1 s), then reverse the pulse sign. Calculate area, average force and final momentum before checking. Distinguish a zero final velocity from a zero force throughout the event. Sketch a momentum-time line rising from−2 to+2 kg m/s over2 s, then staying flat: the first segment has average net force+2 N and the flat segment zero, even though momentum remains+2 kg m/s. Retain settings and raw evidence in the lesson notebook.

5. Transfer and explain

Try a new practice case

A 2 kg object changes velocity from+3 to−2 m/s. Find its signed impulse.

Your numerical result (N s) Check number

Reveal worked check (practice, not mastery)

This extra practice is not saved and does not award mastery. Record your first answer, explanation and whether you revealed help in the lesson notebook. Use the lesson’s saved question separately.

6. Reflect and retain evidence

How would a rectangular pulse with the same peak and duration differ from the triangular pulse? Which measurement would be needed to distinguish them?

Retain the first answer, signed component sums, scalar energy totals, bounds, assumptions and one changed-case prediction. Public worked checks are practice, not independent mastery evidence. Synthetic readings are not a performed physical experiment.

Reconstruct a momentum graph and its net work

For a 2 kg object initially moving at +1 m/s, a net impulse +4 N s gives final momentum 6 kg m/s, final velocity 3 m/s and net work 9−1=8 J. A constant +2 N force for 2 s and a triangular pulse of peak +4 N over 2 s have the same impulse and hence the same final velocity and net work for this same initial state, despite different histories. Sketch the momentum graph: the constant-force case rises linearly from 2 to 6 kg m/s with slope 2 N. The triangular-force case has a changing local slope; its endpoint change is still 4 kg m/s. Changing only the initial velocity to zero gives final velocity 2 m/s and net work 4 J. Preserve the initial-state premise when comparing claims.

Framework and source notes

Original teaching prose and numerical cases. College Board AP Physics 1 CED, Unit 4 topic mapping: 4.2. Checked 30 October 2026. Two-dimensional cases use component setup and simple separate relations, not a general simultaneous-angle solver. This is a framework cross-reference, not AP authorization.