United StatesPhysics C: MechanicsSyllabus dot point

How does an impulse change an object's momentum, and how do we compute impulse as the integral of force over time and as the area under a force-time graph?

Topic 4.2 Change in Momentum and Impulse: define impulse as the integral of force over time, relate it to the change in momentum, and interpret the force-time graph and the average force.

A focused answer to AP Physics C: Mechanics Topic 4.2, covering impulse as the time integral of force, the impulse-momentum theorem, impulse as the area under a force-time graph, the role of average force and contact time, and applications to collisions and cushioning, with calculus-based worked examples.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
Impulse as the time integral of force
The impulse-momentum theorem
Average force and contact time
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What this topic is asking

The College Board (Topic 4.2) wants you to define impulse as the time integral of force, to relate it to the change in momentum through the impulse-momentum theorem, and to interpret a force-time graph and the average force. In AP Physics C the integral definition is central: forces during collisions vary sharply in time, and you compute the impulse by integrating $F(t)$ or by taking the area under the force-time curve.

Impulse as the time integral of force

Because impulse integrates force over time (whereas work integrates force over distance), it is the natural quantity for short, intense interactions like collisions, where the force spikes and falls in a fraction of a second. For a constant force the integral is simply $\vec{J} = \vec{F}\,\Delta t$ , but in general the force varies and you must integrate the actual $F(t)$ . This time-integral definition is the AP Physics C generalization, and exam problems routinely supply a time-dependent force to integrate.

The impulse-momentum theorem

Integrating Newton's second law $\vec{F}_{net} = d\vec{p}/dt$ over time gives the impulse-momentum theorem:

\vec{J} = \int \vec{F}_{net}\,dt = \Delta\vec{p} = m\vec{v}_f - m\vec{v}_i.

The impulse delivered by the net force equals the change in the object's momentum. This is the time-domain counterpart of the work-energy theorem (which used force over distance to get the change in kinetic energy). Its power is that you can find the change in velocity from the impulse without knowing the detailed force history, and conversely you can find the impulse, and hence the force, from the velocity change in a collision. Direction matters: a ball that rebounds reverses its velocity, so its momentum change is larger than if it merely stopped.

Average force and contact time

Often you do not know the detailed force but want a representative value. The average force over the interval is defined so that it delivers the same impulse:

\bar{F} = \frac{J}{\Delta t} = \frac{\Delta p}{\Delta t}.

This relationship explains a great deal of safety engineering. For a given momentum change (a car stopping, a person landing), the force is inversely proportional to the time over which the change happens. Extending the contact time, with a crumple zone, an airbag, a bent-knee landing, a padded glove, reduces the peak force in proportion. The momentum change is fixed by the before-and-after velocities; only the time, and hence the force, is under your control.

Impulse from a varying force

A $1.2$ kg object initially moving at $3.0$ m/s in the $+x$ direction is pushed by a force $F(t) = 6.0t$ N (in newtons, $t$ in seconds) along $+x$ from $t = 0$ to $t = 2.0$ s. Find the impulse and the final velocity.

step 1 Integrate the force for the impulse

$J = \int_0^{2.0} 6.0t\,dt = \left[3.0t^2\right]_0^{2.0} = 3.0(2.0)^2 = 12$ N s.

step 2 Apply the impulse-momentum theorem

$J = \Delta p = m(v_f - v_i)$ , so $12 = (1.2)(v_f - 3.0)$ .

step 3 Solve for the final velocity

$v_f - 3.0 = \dfrac{12}{1.2} = 10$ , so $v_f = 13$ m/s.

step 4 Sanity check with average force

The average force is $\bar{F} = J/\Delta t = 12/2.0 = 6.0$ N, the value of $6.0t$ at $t = 1.0$ s (the midpoint of a linear ramp), as expected.

Try this

Q1. A $0.20$ kg ball moving at $5.0$ m/s is brought to rest. Calculate the impulse delivered to it. [2 points]

Cue. $J = \Delta p = m(0 - 5.0) = (0.20)(-5.0) = -1.0$ kg m/s (opposite the motion), magnitude $1.0$ N s.

Q2. A force-time graph is a triangle peaking at $40$ N over a $0.10$ s contact. Calculate the impulse. [2 points]

Cue. Impulse is the triangular area: $\tfrac{1}{2}(0.10)(40) = 2.0$ N s.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2024 (style)5 marksSection II (FRQ, calculus). A

0.50

kg ball moving at

8.0

m/s strikes a wall and rebounds at

6.0

m/s in the opposite direction. The contact lasts

0.020

s. (a) Determine the impulse delivered to the ball. (b) Determine the average force the wall exerts. (c) During contact the force is modelled as

F(t) = F_0\sin(\pi t/\tau)

over

0 \le t \le \tau

with

\tau = 0.020

s; derive the peak force

F_0

in terms of the impulse.

Show worked answer →

A 5-point impulse FRQ ending with a force-model integral.

(a) Impulse (2 points): take the rebound direction as positive. $\Delta p = m(v_f - v_i) = (0.50)(6.0 - (-8.0)) = (0.50)(14) = 7.0$ kg m/s. The impulse equals the change in momentum, $J = 7.0$ N s.
(b) Average force (1 point): $\bar{F} = \dfrac{J}{\Delta t} = \dfrac{7.0}{0.020} = 350$ N.
(c) Peak force (2 points): $J = \int_0^\tau F_0\sin(\pi t/\tau)\,dt = F_0\left[-\dfrac{\tau}{\pi}\cos(\pi t/\tau)\right]_0^\tau = F_0\dfrac{\tau}{\pi}(1 - (-1)) = \dfrac{2F_0\tau}{\pi}$ . So $F_0 = \dfrac{\pi J}{2\tau} = \dfrac{\pi(7.0)}{2(0.020)} = 550$ N.

Markers reward the sign change on rebound and integrating the sinusoidal force to relate the peak force to the impulse.

AP 2021 (style)1 marksSection I (multiple choice). A car is designed with a crumple zone so that a collision lasts longer. For the same change in momentum, a longer collision time results in... (A) a larger average force (B) a smaller average force (C) the same average force (D) zero force. Justify your reasoning.

Show worked answer →

A 1-point conceptual MCQ. The answer is (B).

The impulse $J = \bar{F}\,\Delta t$ equals the change in momentum, which is fixed by the initial and final velocities. For a fixed impulse, increasing the contact time $\Delta t$ decreases the average force $\bar{F} = J/\Delta t$ . This is exactly why crumple zones, airbags and padding reduce the force on occupants: they extend the time over which momentum changes. The trap (A) reverses the relationship.

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Sources & how we know this

AP Physics C: Mechanics Course and Exam Description — College Board (2024)