Connect impulse to signed momentum

Give motion a signed quantity

A 2 kg particle travels right at +2 m/s. Its linear momentum is p = mv = 2 × 2 = +4 kg·m/s. Here p is momentum, m is mass in kilograms and v is signed velocity in metres per second. Positive means rightward in our chosen coordinate system. Momentum is a vector; in this one-dimensional investigation, its sign carries direction.

The same particle traveling left at −2 m/s has momentum −4 kg·m/s. Its kinetic energy is still ½ × 2 × (−2)² = 4 J. Momentum depends on signed velocity; kinetic energy depends on its square. Negative momentum is ordinary leftward motion, not negative energy or an invalid state.

Predict the impulse that would stop the initially rightward particle. The required momentum change is final minus initial: 0 − 4 = −4 kg·m/s. A leftward influence must remove the positive momentum. Reversing the particle to −2 m/s requires a larger change, −4 − (+4) = −8 kg·m/s.

Connect force over time to momentum change

For a constant mass, net force changes momentum. The accumulated effect over a time interval is the net impulse J, with J = Δp = pfinal − pinitial. The symbol J here names impulse; the unit symbol J for joule names energy. Always include the quantity’s units to avoid confusing them.

Impulse has units newton-seconds, N·s. Since one newton is one kg·m/s², multiplying by seconds gives kg·m/s, the same units as momentum. It is not force by itself and is not energy. For a constant net force, impulse equals force multiplied by duration. When force varies, use the signed area under its force-versus-time curve or its time average multiplied by duration.

The lab prescribes two ideal force pulses acting on the same 2 kg particle. They have the same selected impulse and duration. An external actuator supplies the pulse; the particle does not generate its own net impulse. Forces are zero after the pulse. The model omits friction and does not simulate a material collision, deformation or contact sensor.

Background: OpenStax impulse and collisions. The two pulse profiles and numerical comparisons here are original teaching constructions.

Follow three leftward impulses

Keep duration at 1 s. With impulse −4 N·s, final momentum is 4 + (−4) = 0 kg·m/s. Final velocity and kinetic energy are both zero. The particle has just stopped when the pulse ends; with no force afterward, it remains at rest.

With impulse −8 N·s, final momentum is 4 − 8 = −4 kg·m/s. Dividing by mass gives velocity −2 m/s. Kinetic energy is 4 J, equal to its initial value, although momentum changed by −8 kg·m/s. During the pulse the particle slows, momentarily stops, then speeds up leftward. Zero net change in kinetic energy does not mean no force acted or no motion changed.

With impulse −12 N·s, final momentum is −8 kg·m/s, velocity is −4 m/s and kinetic energy is 16 J. A negative impulse has increased kinetic energy overall in this case. The sign of momentum change alone does not determine whether energy rises or falls; the initial momentum matters too.

Save all three cases. Write a comparison that names initial momentum, change and final momentum separately. Calling the −8 N·s impulse the ‘final momentum’ misses the starting +4 kg·m/s.

Read the intermediate evidence

The history table compares both pulses at five times. It reports signed position from the starting point, not total distance traveled. In a reversing case, a particle can move right first and return to its starting position, giving zero displacement despite nonzero travel.

At impulse −8 N·s over 1 s, the rectangular pulse produces momentum +2 kg·m/s after 0.25 s; the triangular pulse produces +3 kg·m/s. The triangle begins with a smaller force magnitude, so less momentum has changed early on. Both reach zero momentum halfway through this particular trial and finish at −4 kg·m/s.

Equal final momentum does not mean the entire histories are identical. Nor does a five-row table measure an experiment: these are exact values from the prescribed force functions. Real evidence would require force/time or velocity measurements, instrument calibration and uncertainty analysis.

Keep the system boundary explicit

Momentum of this particle is not conserved while the actuator supplies a net impulse. The impulse-momentum equation accounts for its change; it does not require the particle’s initial and final momenta to match. A broader closed system that includes the source of the interaction can have conserved total momentum, but the actuator and its support are not modeled here.

Likewise, the particle’s kinetic-energy change is not the impulse. For the fixed mass, calculate energy from the final velocity. Work uses force through displacement, whereas impulse uses force through time. They describe different aspects of the same interaction and have different units.

Transfer to another mass and direction

A 3 kg particle starts at −2 m/s and receives a +9 N·s impulse. Predict initial and final momentum, final velocity and kinetic-energy change. Decide whether it stops or reverses before finishing.

Check the momentum transfer

Initial momentum is −6 kg·m/s. Final momentum is −6 + 9 = +3 kg·m/s, so final velocity is +1 m/s. Initial kinetic energy is 6 J; final kinetic energy is 1.5 J; the change is −4.5 J. The final direction is opposite the initial direction. A continuous positive-force pulse sufficient to produce this impulse passes through zero velocity before the end. The positive impulse does not mean kinetic energy increased.