Change in momentum and impulse

Impulse is what a net force does over time: $J = \int F\,dt$, which is just $J = F\Delta t$ when the force holds steady. A constant $60$ N push lasting $0.2$ s delivers $J = (60)(0.2) = 12$ N·s. Multiply newtons by seconds and the unit comes out as kg·m/s: impulse and momentum are measured in the same unit, because one is the change in the other.

Neither ingredient is the impulse by itself. The $60$ N is a force, the $0.2$ s is a time, and only their product is the kick. Doubling either one doubles the impulse; quoting either one alone answers a different question.

And force times distance is something else: that product is work, which belongs to energy, not momentum. Impulse problems never need the distance; they need the time the force acts, or the momentum change it produces.

Three habits cost points: mistaking the ingredients or the work for the impulse (this section and §2), thinking a cushion shrinks the momentum change (§3), and cutting corners on a varying force with peak-times-time (§4).

Integrate Newton's second law, $F = dp/dt$, over an interval and the result is the impulse-momentum theorem: $J = \Delta p$. The impulse is not the final momentum and not a momentum an object carries; it is the change, so the working equation is $p_f = p_i + J$. A $3$ kg cart at $+2$ m/s that receives $+9$ N·s ends with $p_f = 6 + 9 = +15$ kg·m/s.

Because it equals the change, the signs are everything here. A $0.5$ kg ball arrives at a wall at $6$ m/s. If it sticks, the momentum runs from $3$ to $0$: a change of magnitude $mv = 3$ N·s. If it bounces straight back at $6$ m/s, the momentum runs from $+3$ all the way to $-3$: a change of $2mv = 6$ N·s. The bounce is the bigger kick, twice the stick, even though the speed never changed; the velocity reversed, and impulse follows the velocity.

The same logic finishes partial rebounds: a ball down at $6$ m/s that leaves up at $4$ m/s changes by $m(4 - (-6)) = 10m$, not $2m$. Subtracting speeds without signs misses the reversal every time.

The trap is to score equal arrivals as equal impulses, to read the impulse as the momentum the object ends with, or to call a bounce zero change because the speed survived. Write the before and the after with signs, then subtract.

When the motion fixes the momentum change, the impulse is fixed before any choice is made, and the only freedom left is the split: $F = \Delta p / \Delta t$. The same $12$ N·s can be $60$ N for $0.2$ s or $30$ N for $0.4$ s; neither stop removes more momentum than the other.

That trade is how all cushioning works. A crash that must remove a passenger's momentum can do it against a rigid dashboard in $0.05$ s or against an airbag in $0.25$ s: five times the time, one fifth the average force, and exactly the same $\Delta p$ and the same impulse. Airbags, crumple zones, bent knees, and soft landings all stretch the time; none of them shrinks the momentum change.

Watch when the trade applies. It holds when $\Delta p$ is fixed in advance, as in any stop. With the force held fixed instead, more time genuinely means more impulse: that is why a batter follows through. Figure out which quantity the situation fixes before working out the others.

The trap is to credit the gentle stop with a smaller impulse, to say the airbag reduces the momentum change, or to declare the biggest force or the longest time the winner without multiplying. The product is fixed; only its factors move.

When the force varies, the integral is the definition: the impulse is the area under the F-t curve. A triangular pulse peaking at $40$ N over $0.3$ s delivers $\tfrac{1}{2}(40)(0.3) = 6$ N·s: half of the peak-times-time rectangle. Piecewise forces add area by area: $20$ N for $0.5$ s then $40$ N for $0.1$ s is $10 + 4 = 14$ N·s, and the average force is the impulse over the total time, $14/0.6 \approx 23$ N, not the unweighted $30$: each force counts for as long as it acted.

Read the other graph by its slope. Newton's second law in its general form is $F = dp/dt$, so on a p-t graph the slope is the net force: a steady climb from $2$ to $14$ kg·m/s over $4$ s means $3$ N, a flat stretch means zero force at full momentum, and the momentum changes fastest wherever the F-t curve sits highest, not where it is steepest.

Two readings to keep apart: a force returning to zero ends the push without undoing the momentum the area already delivered, and a momentary zero of force means the momentum is holding still right then, not that it is zero.

The trap is peak times total time, averages that ignore duration, p-t values quoted as forces, and the steepness of F-t mistaken for the rate the momentum changes. On F-t the area is the impulse; on p-t the slope is the force.

An impulse question usually comes down to computing the change in momentum, pinning down $\Delta p$ before you split it, or matching the operation to the graph in front of you.

Compute the change. Write $p_i$ and $p_f$ with signs and subtract: $J = \Delta p$, with $p_f = p_i + J$ as the working form. A reversal makes the change large; a stick makes it $mv$; a same-speed bounce makes it $2mv$. No distance is ever needed.

Pin $\Delta p$, then split it. When the motion fixes the momentum change, the impulse is set; $F = \Delta p/\Delta t$ just divides it up. Stretch the time, shrink the force. The trade only exists at fixed $\Delta p$; at fixed force, more time means more impulse.

Match the operation to the graph. On F-t, integrate: the area is the impulse, with the $\tfrac{1}{2}$ on triangles and every force weighted by its duration. On p-t, differentiate: the slope is the force. Never read a value where the question asks for an area or a slope.

Setup. A $0.5$ kg ball hits a wall at $6$ m/s and rebounds straight back at $6$ m/s; contact lasts $0.1$ s. (a) Find the impulse on the ball. (b) Find the average force from the wall. (c) If the contact force instead ramped linearly from zero to a peak and back to zero over the same $0.1$ s, find the peak force that delivers the same impulse.

(a) The change, with signs. Take the rebound direction as positive. The ball arrives with $p_i = (0.5)(-6) = -3$ kg·m/s and leaves with $p_f = +3$, so $J = \Delta p = 3 - (-3) = +6$ N·s. The reversal doubles the stick's $mv = 3$; subtracting speeds without signs would have given zero.

(b) The split. The average force is the impulse over the contact time: $F = \Delta p/\Delta t = 6/0.1 = 60$ N. Had the contact lasted $0.2$ s, the same $6$ N·s would need only $30$ N: the impulse is fixed by the rebound; the time choice only sets the force.

(c) The varying force. A ramp up and back down is a triangle, and its area must still be $6$ N·s: $\tfrac{1}{2}F_{pk}(0.1) = 6$, so $F_{pk} = 120$ N. The peak is exactly double the constant force from (b), because the triangle wastes half its bounding rectangle. Reading $60$ N as the peak, or $F_{pk}\Delta t = 12$ N·s as the impulse, are the two halves of the same shortcut.

The traps to avoid. Do not report the impulse as $3$ N·s (the arrival momentum, not the change), do not let a longer contact suggest a smaller impulse (only the force drops), and do not multiply the peak by the time on the triangle (the area carries the $\tfrac{1}{2}$).