Compare force pulses with the same impulse

Compare equal areas before comparing heights

The particle begins with momentum +4 kg·m/s. Select impulse −4 N·s and duration 1 s. Both displayed pulses remove the same momentum and stop the particle. The rectangular pulse uses a constant −4 N force over the interval. The symmetric triangular pulse starts at zero, reaches −8 N halfway through, then returns to zero.

Their signed areas are equal. For the rectangle, area is height multiplied by width: −4 N × 1 s = −4 N·s. For the triangle, area is half the base multiplied by the signed peak: ½ × 1 s × (−8 N) = −4 N·s. The larger peak compensates for the smaller force magnitude during the rest of the triangular pulse.

The graph holds both axes fixed across trials: time 0–3 s and force −48 to +48 N. Force below the zero line contributes negative impulse. Solid and dashed line styles identify the profiles without requiring color. Read the time width as well as the force height; a narrow tall pulse and a broad low pulse can have the same area.

Average force is not necessarily peak force

For a known impulse J and positive duration T, Faverage = J/T. Average force is signed. Peak magnitude means the largest absolute force during the pulse, so it is nonnegative. The rectangle has peak magnitude equal to the absolute average. For this symmetric triangle, peak magnitude is twice the absolute average.

At −4 N·s over 1 s, average force is −4 N, rectangular peak magnitude is 4 N and triangular peak magnitude is 8 N. At the same impulse over 2 s, average force becomes −2 N, rectangular peak becomes 2 N and triangular peak becomes 4 N. Final momentum remains zero because the area has not changed.

Now compare duration 0.5 s. Average force is −8 N; rectangular and triangular peaks are 8 N and 16 N. Capture the three durations and explain the inverse relationship. This statement holds impulse fixed. Keeping force fixed while doubling duration would instead double impulse and change final momentum.

What equal impulse determines

For the same initial momentum and mass, equal net impulse fixes equal final momentum and velocity. It also fixes the same final kinetic energy for this particle because that energy can be calculated from mass and final velocity. The force history is still needed to describe how the particle moves in between.

At −4 N·s over 1 s, after one quarter of the pulse the rectangular history has momentum 3 kg·m/s. The triangular history has momentum 3.5 kg·m/s. By halfway both have momentum 2 kg·m/s, but their positions differ: approximately 0.750 m and 0.833 m. A matching momentum at one instant does not erase the earlier difference in motion.

Both profiles happen to produce the same final displacement here because each is symmetric in time about the midpoint and has the same impulse and duration. This is an additional property of these profiles, not a general consequence of equal impulse. Moving most of the same impulse toward the beginning or end of the interval would change the final displacement.

Changing duration also changes the journey. With a −4 N·s stop, the final displacement is 0.5 m for duration 0.5 s, 1 m for 1 s and 2 m for 2 s for either displayed profile. Equal final velocity does not imply equal stopping position or time.

Separate the model from a contact claim

A longer interaction duration often motivates reducing average force for a specified momentum change. But a statement about average force is not automatically a bound on peak force. An arbitrary pulse could contain a sharp high spike and low forces for most of a long interval. The peak reductions in this lab follow from the chosen rectangular and triangular shapes remaining the same while their duration changes.

A real impact adds further questions: does the object bounce, how does the contact deform, how is force distributed, and does an instrument resolve the fastest variations? A bounce changes final velocity and therefore changes the required impulse. This widget cannot certify protective equipment or estimate injury; it only compares prescribed net forces on an ideal particle.

The OpenStax impulse discussion supplies the force-time-area and average-force relationship. Use the explicit graph shape whenever inferring a peak.

Optional mathematical account of the history

Explore cumulative impulse

Let s = t/T be the fraction of pulse duration elapsed, from 0 to 1. For the rectangle, the fraction of total impulse delivered is s. For the symmetric triangle it is 2s² through the first half, and 1 − 2(1 − s)² through the second half. Multiply that fraction by signed total impulse and add initial momentum. These formulas come from areas of the relevant triangle pieces, not from a numerical integration step. Dividing momentum by mass gives velocity; integrating velocity gives the displayed positions.

Test a profile not selected in the controls

A positive triangular pulse rises from zero to +12 N and returns to zero over 0.5 s. A second pulse is rectangular and lasts 1.5 s. Predict the triangular impulse, the constant force needed for equal impulse in the rectangle, and whether their final velocities must match.

Check the pulse-profile transfer

The triangle supplies ½ × 0.5 × 12 = +3 N·s. The rectangle needs +3/1.5 = +2 N. Their final momenta match only if their initial momenta match; with equal masses, their final velocities also match. With the same 2 kg particle initially at +2 m/s, both finish at +3.5 m/s. Their durations and intermediate trajectories differ. Equal impulse alone does not promise equal final displacement.