AP Physics 1 · Topic 4.2

Topic 4.2: Change in Momentum and Impulse

Unit 4: Linear Momentum10-15% of the multiple-choice section

Impulse is the average force on a system multiplied by the time interval it acts, and the impulse of the net external force equals the system's change in momentum. On a graph of net external force against time, impulse is the area under the curve.

AP Physics: Unit 4 (topics 4.2 Change in Momentum and Impulse). AP Physics 1 Unit 4, Topic 4.2. Two learning objectives: 4.2.A, describe the impulse delivered to an object or system, and 4.2.B, describe the relationship between the impulse exerted on an object or system and its change in momentum. Eight essential knowledge statements support them, including 4.2.A.1 (net force equals the rate of change of momentum), 4.2.A.2 (impulse is average force times the time interval), 4.2.A.3 (impulse is a vector along the net force), 4.2.A.4 (impulse is the area under a net external force vs time graph), 4.2.A.5 (net external force is the slope of a momentum vs time graph), 4.2.B.1 (change in momentum is final minus initial), 4.2.B.2 (the impulse-momentum theorem), and 4.2.B.3 (Newton's second law is a direct result of the theorem for constant-mass systems). The boundary statement excludes quantitative analysis of systems whose mass changes with time. The CED's suggested skills for this topic are 1.B, 2.A, 2.D, 3.A, and 3.C. Unit 4 carries 10 to 15 percent of the multiple-choice section.

What Topic 4.2 requires

Topic 4.2 carries two learning objectives and eight essential knowledge statements, five of which come with an equation. It is the equation-heaviest topic in Unit 4.

Learning objective 4.2.A asks you to describe the impulse delivered to an object or system.

4.2.A.1 The rate of change of momentum is equal to the net external force exerted on an object or system: $\vec{F}_{\text{net}} = \frac{\Delta \vec{p}}{\Delta t}$ .
4.2.A.2 Impulse is defined as the product of the average force exerted on a system and the time interval during which that force is exerted on the system: $\vec{J} = \vec{F}_{\text{avg}} \Delta t$ .
4.2.A.3 Impulse is a vector quantity and has the same direction as the net force exerted on the system.
4.2.A.4 The impulse delivered to a system by a net external force is equal to the area under the curve of a graph of the net external force exerted on the system as a function of time.
4.2.A.5 The net external force exerted on a system is equal to the slope of a graph of the momentum of the system as a function of time.

Learning objective 4.2.B asks you to describe the relationship between the impulse exerted on an object or system and the change in momentum of that object or system.

4.2.B.1 Change in momentum is the difference between a system's final momentum and its initial momentum: $\Delta \vec{p} = \vec{p} - \vec{p}_0$ .
4.2.B.2 The impulse-momentum theorem relates the impulse exerted on a system and the system's change in momentum: $\vec{J} = \vec{F}_{\text{avg}} \Delta t = \Delta \vec{p}$ .
4.2.B.3 Newton's second law of motion is a direct result of the impulse-momentum theorem applied to systems with constant mass: $\vec{F}_{\text{net}} = \frac{\Delta \vec{p}}{\Delta t} = m \frac{\Delta \vec{v}}{\Delta t} = m\vec{a}$ .

The boundary statement rules out one whole class of problem: AP Physics 1 does not require students to quantitatively analyze systems in which the mass of the system changes with respect to time. Rockets burning fuel and carts collecting falling sand are the standard examples of what that excludes; the CED does not ask you to compute their motion.

Five suggested skills are listed here: 1.B, create quantitative graphs with appropriate scales and units, including plotting data; 2.A, derive a symbolic expression from known quantities; 2.D, predict new values or factors of change of physical quantities using functional dependence between variables; 3.A, create experimental procedures that are appropriate for a given scientific question; and 3.C, justify or support a claim using evidence. That list tells you the shape of the questions: graphs, derivations, and ratio reasoning rather than plug-and-chug.

Change in momentum comes first (4.2.B.1)

\Delta \vec{p} = \vec{p} - \vec{p}_0

Final minus initial, in that order, and as vectors. The subscript zero for the initial value matches the notation used across the AP Physics 1 equation sheet. In one dimension the subtraction is ordinary arithmetic on signed numbers, which is where the marks go, because the two terms often have opposite signs and the minus sign in front of $\vec{p}_0$ then turns into a plus.

Three things about $\Delta \vec{p}$ that are worth separating from the momentum itself.

It is not the final momentum. A question asking for a change in momentum wants a difference, and an answer equal to $mv$ has skipped the subtraction.
Its magnitude can exceed both momenta involved. An object that reverses direction at the same speed has $\Delta p = mv - (-mv) = 2mv$ , twice either individual magnitude. Reversing a motion always costs more than stopping it, and the impulse-momentum theorem guide works that comparison through with numbers.
It has its own direction. Because it is a vector difference, $\Delta \vec{p}$ can point in a direction neither velocity pointed. In one dimension that just means its sign may differ from the sign of both $p$ and $p_0$ .

For a system rather than a single object, take the total momentum before and the total momentum after, each computed as the vector sum described in Topic 4.1, and subtract those totals. Subtracting object by object and adding afterwards gives the same answer; mixing the two approaches halfway through does not.

Impulse: average force multiplied by time (4.2.A.2)

\vec{J} = \vec{F}_{\text{avg}} \Delta t

Impulse is what a force accomplishes over an interval of time. Two knobs set it: how hard and how long. That makes it the time-based partner of work, which is what a force accomplishes over a distance.

Its unit is the newton second, and a newton second is identical to a kilogram meter per second, so impulse and momentum share a unit. Checking that your impulse answer carries the same unit as the momentum change it is supposed to equal is a fast sanity check.

Essential knowledge 4.2.A.3 fixes the direction: impulse is a vector and points the same way as the net force exerted on the system. It does not point along the velocity, and this is an easy difference to miss. A car braking to a stop moves forward the whole time while both the net force and the impulse point backward.

Two cautions about the word average.

$\vec{F}_{\text{avg}}$ is a time average, not a distance average, and not the peak force. In a real collision the force spikes; the average is the constant force that would deliver the same impulse over the same interval.
Any single force delivers its own impulse, but the impulse that equals the system's momentum change is the impulse of the net external force. If several forces act and only one is given, multiplying that one by $\Delta t$ gives that force's impulse alone.

Essential knowledge 4.2.A.1 states the same relationship as a rate: $\vec{F}_{\text{net}} = \Delta \vec{p} / \Delta t$ . Read it as a definition of what a net force does. A net force is not something that keeps an object moving; it is the rate at which an object's momentum is being changed.

Impulse is the area under a force vs time graph (4.2.A.4)

This is the representation the CED attaches skill 1.B to, and it is the reason skill 1.B (creating quantitative graphs) is attached to this topic. The impulse delivered over any interval equals the area between the net force curve and the time axis over that interval.

The mechanics of reading one:

Split the shape into rectangles and triangles wherever the graph is made of straight segments, find each area, and add them.
Treat area below the time axis as negative. A negative segment subtracts from the impulse and can reduce, cancel, or reverse the momentum change.
For a curved graph, count grid boxes or approximate with trapezoids. AP questions that use curves ask for an estimate and accept a range.
Check the units on the axes before computing. Area on a force-time graph is newtons times seconds, so the answer is in N·s, which is the same as kg·m/s.

Two refinements separate a confident reading from a shaky one. First, the average force is the height of the rectangle whose area matches the actual shape over the same interval: draw that rectangle across a force spike and it will cut through the peak well below the top. Second, 4.2.A.4 says the graph is of the net external force. If a problem plots the applied force while friction also acts, the area under the plotted curve is not the system's momentum change until you account for the other force.

The impulse-momentum theorem guide carries the full procedure and additional graph practice, including the case of a force that reverses partway through.

The slope of a momentum vs time graph is the net force (4.2.A.5)

Essential knowledge 4.2.A.5 runs the graph relationship in the other direction, and it is a statement students often arrive at the exam without. Plot the system's momentum against time and the slope at any instant is the net external force at that instant, because $\vec{F}_{\text{net}} = \Delta \vec{p}/\Delta t$ is exactly a rise over a run.

That gives three readings of a momentum-time graph at a glance:

A horizontal line means zero net external force. The momentum is constant, which is Newton's first law written in momentum language.
A straight, sloped line means a constant net force equal to that slope. Steeper is stronger, and a negative slope means the net force points in the negative direction.
A curved line means a changing net force. The slope of the tangent at a point gives the net force at that moment.

It helps to keep the whole family of graphs straight, because AP questions move between them.

Graph	Its slope gives	The area under it gives
Net external force vs time	no standard quantity in AP Physics 1	impulse, equal to $\Delta \vec{p}$ (4.2.A.4)
Momentum vs time	net external force (4.2.A.5)	no standard quantity in AP Physics 1
Velocity vs time	acceleration	displacement
Position vs time	velocity	no standard quantity in AP Physics 1

For a system of constant mass a momentum-time graph is a velocity-time graph with every value scaled by $m$ , so the two have the same shape. The slopes differ by that same factor of $m$ : one gives acceleration, the other gives net force, which is Newton's second law appearing as a graph property.

The impulse-momentum theorem, and why Newton's second law follows

Set the two definitions equal and the theorem falls out. Essential knowledge 4.2.B.2 states it as one printed line on the AP Physics 1 formula sheet:

\vec{J} = \vec{F}_{\text{avg}} \Delta t = \Delta \vec{p}

The middle expression is what you compute from a force and a duration; the right-hand one is what you compute from masses and velocities. The theorem says they are the same number, so any problem that gives you one gives you the other. That is the entire method: whichever side the question hands you, cross to the other. The impulse-momentum theorem guide sets out the step-by-step procedure and works several problems end to end.

Essential knowledge 4.2.B.3 then reverses the usual teaching order, and the direction matters:

\vec{F}_{\text{net}} = \frac{\Delta \vec{p}}{\Delta t} = m \frac{\Delta \vec{v}}{\Delta t} = m\vec{a}

Newton's second law is a direct result of the impulse-momentum theorem applied to systems with constant mass. The middle step, pulling $m$ out in front of the $\Delta$ , is legal only because the mass does not change, which is precisely the restriction the topic's boundary statement places on the course. The momentum form is the more general statement, and $\vec{F}_{\text{net}} = m\vec{a}$ is the special case you meet first in Unit 2.

The practical payoff is that the impulse form survives situations where $\vec{F}_{\text{net}} = m\vec{a}$ is awkward. During a collision the force varies violently over a few milliseconds, so there is no single acceleration to plug in, but the total impulse over the interval is still well defined and still equals the momentum change. That is why every collision problem in Topic 4.3 can be solved without knowing anything about the force at all.

Fixed change in momentum, adjustable time

Rearranging the theorem gives the relationship that the CED's essential questions for this unit are built around, including why a water balloon breaks on pavement but not when caught carefully, and why cars are designed with crumple zones:

\vec{F}_{\text{avg}} = \frac{\Delta \vec{p}}{\Delta t}

Read it as suggested skill 2.D asks you to, as a functional dependence. Hold $\Delta p$ fixed and the average force is inversely proportional to the stopping time. Double the time and the force halves. Stretch it by a factor of ten and the force falls to one tenth. No safety device reduces the momentum change of a passenger who goes from highway speed to rest; the momentum change is settled by the initial and final velocities. What the device changes is $\Delta t$ .

The same reasoning runs in reverse for a hard, fast interaction. A lump of clay dropped onto concrete stops in a millisecond or two and registers an enormous average force; the same lump dropped onto a cushion has the same $\Delta p$ spread over far longer and barely registers. Keep the landing outcome fixed when you make this comparison, or you have changed $\Delta p$ as well as $\Delta t$ . Catching an egg by drawing your hands back is the classroom version.

Two related comparisons that AP questions like to combine with this one:

Bouncing versus stopping. An object that rebounds has a larger $|\Delta p|$ than one that stops, so for the same contact time it experiences a larger average force. This is why a bouncing ball can knock over an object that an equally fast lump of clay cannot.
Force versus pressure. An airbag also spreads the force over a larger area. That is a separate effect from the timing one, and a full answer to a question about airbags mentions both, attributing the reduced force to the longer $\Delta t$ and the reduced injury to the larger area.

One CED sample activity for this topic sharpens the distinction better than any formula. A pitcher throws a baseball and a catcher catches it. Who exerted more force on the ball? There is no way to know, because neither contact time was given. Who applied the greater impulse? Neither: the magnitudes are the same, since the ball went from rest to speed $v$ and then from speed $v$ back to rest. Who did the greater magnitude of net work? Again the same, for the same reason applied to kinetic energy. Impulse and work are settled by the endpoints; force is not settled until you know the duration.

How Topic 4.2 is tested

The five suggested skills line up with the question types the CED suggests, and momentum questions carry 10 to 15 percent of the 42-question multiple-choice section. Section II has four free-response questions in 95 minutes, in the formats Mathematical Routines, Translation Between Representations, Experimental Design and Analysis, and Qualitative/Quantitative Translation. A calculator is allowed on both sections.

Read an impulse off a force-time graph, then find a velocity (skill 1.B). Often the first half of a longer problem.
Translate between representations (skill 1.B). Given a momentum-time graph, sketch the corresponding force-time graph, or the reverse. A straight momentum line becomes a horizontal force line; a curved one does not.
Derive a symbolic expression (skill 2.A). Show that the average force on an object stopped from speed $v_0$ over time $\Delta t$ is $mv_0/\Delta t$ , with no numbers anywhere.
Predict a factor of change (skill 2.D). If the contact time triples while the momentum change is unchanged, by what factor does the average force change? Answer with the ratio, not a recalculation.
Design an experiment (skill 3.A). The CED's own sample activity connects a spring-loaded lanyard between a cart and a force sensor, with a motion sensor on the other side of the cart, so that force-time and velocity-time data are recorded together and the impulse can be checked against the momentum change. Expect to be asked what to measure, how to linearize it, and what a slope would represent.
Justify a claim (skill 3.C). Explain in words why a longer collision time reduces the force on a passenger, citing the impulse-momentum theorem rather than intuition.

Next, Topic 4.3 applies all of this to a system rather than a single object: if the net external force is zero, the impulse on the system is zero, so its total momentum cannot change. Topic 4.4 then sorts collisions by what happens to the kinetic energy. To check any impulse or collision answer against worked numbers, use the momentum and collision calculator.

Impulse from a three-segment force vs time graph, including a negative area

A 3.5 kg cart starts from rest on a level, low-friction track. A sensor records the net force on the cart along the direction of motion: it rises linearly from 0 to 8.0 N between t = 0 and t = 1.5 s, holds steady at 8.0 N until t = 3.0 s, then switches to a constant -4.0 N from t = 3.0 s to t = 4.0 s. Find (a) the impulse delivered in each segment, (b) the cart's speed at t = 3.0 s and at t = 4.0 s, and (c) the average net force over the full 4.0 s.

Essential knowledge 4.2.A.4: impulse is the signed area under the net-force curve. Take the direction of the initial push as positive.
Segment 1 is a triangle of base 1.5 s and height 8.0 N: $J_1 = \frac{1}{2}(1.5 \, \mathrm{s})(8.0 \, \mathrm{N}) = 6.0 \, \mathrm{N \cdot s}$ .
Segment 2 is a rectangle of width $3.0 - 1.5 = 1.5 \, \mathrm{s}$ and height 8.0 N: $J_2 = (8.0)(1.5) = 12 \, \mathrm{N \cdot s}$ .
Segment 3 lies below the axis, so its area is negative: $J_3 = (-4.0 \, \mathrm{N})(1.0 \, \mathrm{s}) = -4.0 \, \mathrm{N \cdot s}$ .
(b) Through $t = 3.0$ s the impulse is $6.0 + 12 = 18 \, \mathrm{N \cdot s}$ . The cart started at rest, so $\Delta p = p$ and $v = J/m = 18/3.5 = 5.142... = 5.1 \, \mathrm{m/s}$ to two significant figures.
Over the full interval the impulse is $18 - 4.0 = 14 \, \mathrm{N \cdot s}$ , so $v = 14/3.5 = 4.0 \, \mathrm{m/s}$ . The negative segment slowed the cart without reversing it, because the total area is still positive.
(c) $F_{\text{avg}} = J_{\text{total}}/\Delta t = (14 \, \mathrm{N \cdot s})/(4.0 \, \mathrm{s}) = 3.5 \, \mathrm{N}$ . A rectangle 3.5 N tall and 4.0 s wide has the same area as the actual graph, which is what the word average means here.

(a) $+6.0$ , $+12$ , and $-4.0 \, \mathrm{N \cdot s}$ , for a total of $+14 \, \mathrm{N \cdot s}$ . (b) 5.1 m/s at $t = 3.0$ s and 4.0 m/s at $t = 4.0$ s. (c) $F_{\text{avg}} = 3.5 \, \mathrm{N}$ , well below the 8.0 N peak.

Reversing a hockey puck: change in momentum and average force

A 0.17 kg hockey puck slides toward a player at 12 m/s. The player's stick sends it straight back along the same line at 22 m/s, and the stick stays in contact with the puck for 0.015 s. Find (a) the puck's change in momentum, (b) the average force the stick exerts on the puck, and (c) how that compares with the average force needed only to stop the puck in the same contact time, and with the puck's weight.

Set the axis: take the puck's original direction of travel as positive. Then $v_0 = +12 \, \mathrm{m/s}$ and $v = -22 \, \mathrm{m/s}$ . The reversal is carried entirely by that minus sign.
(a) $\Delta p = mv - mv_0 = (0.17)(-22) - (0.17)(+12) = -3.74 - 2.04 = -5.78 \, \mathrm{kg \cdot m/s}$ . To two significant figures, $\Delta p = -5.8 \, \mathrm{kg \cdot m/s}$ : a magnitude of 5.8 kg·m/s directed opposite the puck's original motion.
(b) By the impulse-momentum theorem, $F_{\text{avg}} = \Delta p / \Delta t = (-5.78)/(0.015) = -385.3 \, \mathrm{N}$ , which is $-3.9 \times 10^2 \, \mathrm{N}$ to two significant figures. The minus sign says the force points back toward the player, in the same direction as the impulse, exactly as 4.2.A.3 requires.
(c) To merely stop the puck: $\Delta p = (0.17)(0) - (0.17)(12) = -2.04 \, \mathrm{kg \cdot m/s}$ , so $F_{\text{avg}} = (-2.04)/(0.015) = -136 \, \mathrm{N}$ , or $-1.4 \times 10^2 \, \mathrm{N}$ .
The ratio is $5.78/2.04 = 2.8$ , so reversing the puck takes nearly three times the impulse of stopping it, for the same contact time and therefore nearly three times the average force.
Sanity check on ignoring gravity: the puck's weight is $mg = (0.17)(9.8) = 1.7 \, \mathrm{N}$ , and on ice the normal force cancels it, so the net external force on the puck is only the small ice friction, orders of magnitude below the 390 N stick force. That gap is the condition essential knowledge 4.1.A.3.i uses to license treating this as a collision and dropping the external forces.

(a) $\Delta p = 5.8 \, \mathrm{kg \cdot m/s}$ directed opposite the puck's original motion. (b) $F_{\text{avg}} = 3.9 \times 10^2 \, \mathrm{N}$ in that same direction. (c) Stopping the puck alone would need only $1.4 \times 10^2 \, \mathrm{N}$ , so the reversal costs 2.8 times as much; the 1.7 N weight is negligible over 0.015 s.

Reading the net force off a momentum vs time graph

A 2.5 kg cart moves along a straight track. Its momentum vs time graph is a straight line rising from p = 2.0 kg·m/s at t = 0 to p = 8.0 kg·m/s at t = 4.0 s, then a horizontal line at 8.0 kg·m/s from t = 4.0 s to t = 6.0 s. Find (a) the net external force during each stage, (b) the cart's velocity at t = 0 and at t = 6.0 s, and (c) the impulse delivered between t = 0 and t = 4.0 s, together with the cart's acceleration over that stage.

Essential knowledge 4.2.A.5: the slope of a momentum-time graph is the net external force.
(a) First stage: $F_{\text{net}} = \Delta p/\Delta t = (8.0 - 2.0)/(4.0 - 0) = 6.0/4.0 = 1.5 \, \mathrm{N}$ , constant and positive because the line is straight and rising.
Second stage: the line is horizontal, so the slope and therefore the net external force are zero. The cart coasts at constant velocity, which is Newton's first law seen through momentum.
(b) Momentum and velocity differ only by the constant mass here: $v_0 = p_0/m = 2.0/2.5 = 0.80 \, \mathrm{m/s}$ , and at $t = 6.0$ s, $v = 8.0/2.5 = 3.2 \, \mathrm{m/s}$ . The cart was already moving at $t = 0$ , which the nonzero intercept told you.
(c) Impulse over the first stage is just the momentum change: $J = 8.0 - 2.0 = 6.0 \, \mathrm{N \cdot s}$ . Cross-check against $J = F_{\text{avg}} \Delta t = (1.5)(4.0) = 6.0 \, \mathrm{N \cdot s}$ , which agrees.
Acceleration follows from 4.2.B.3: $a = F_{\text{net}}/m = 1.5/2.5 = 0.60 \, \mathrm{m/s^2}$ . Cross-check from the velocities: $\Delta v = 3.2 - 0.80 = 2.4 \, \mathrm{m/s}$ over 4.0 s gives $0.60 \, \mathrm{m/s^2}$ , the same value.

(a) 1.5 N for the first 4.0 s, then 0 N. (b) 0.80 m/s at $t = 0$ and 3.2 m/s at $t = 6.0$ s. (c) $J = 6.0 \, \mathrm{N \cdot s}$ and $a = 0.60 \, \mathrm{m/s^2}$ , confirmed two independent ways.

Frequently asked questions

What is impulse in AP Physics 1?

Essential knowledge 4.2.A.2 defines impulse as the product of the average force exerted on a system and the time interval during which that force is exerted, written J = F_avg times delta t. It is a vector and points the same way as the net force on the system, not the same way as the velocity. Its unit is the newton second, which is identical to the kilogram meter per second.

What is the difference between impulse and momentum?

Momentum is a property an object has at one instant, equal to mv. Impulse is something delivered to it over an interval, equal to the average net force times the duration. The impulse-momentum theorem links them: the impulse delivered equals the change in momentum, not the momentum itself. They share a unit, which is why the two are easy to confuse and easy to check against each other.

What does the area under a force vs time graph represent?

The impulse delivered over that interval, and therefore the change in momentum, from essential knowledge 4.2.A.4. Break straight-line graphs into rectangles and triangles, count area below the time axis as negative, and estimate curved regions by counting grid boxes. The result is in newton seconds, the same unit as momentum.

What does the slope of a momentum vs time graph represent?

The net external force on the system, from essential knowledge 4.2.A.5. A horizontal line means zero net force and constant momentum, a straight sloped line means a constant net force equal to that slope, and a curved line means the net force is changing. The area under a momentum-time graph has no standard meaning in AP Physics 1, so do not look for one.

Why does increasing the collision time reduce the average force?

Because the change in momentum is fixed by the initial and final velocities, so rearranging the theorem gives F_avg = delta p divided by delta t. With the numerator locked in, the average force is inversely proportional to the stopping time: double the time and the force halves. Airbags, crumple zones, and catching a ball by drawing your hands back all work this way.

Is Newton's second law the same as the impulse-momentum theorem?

They are the same physics stated over different intervals. Essential knowledge 4.2.B.3 puts the momentum form first: F_net = delta p over delta t = m times delta v over delta t = ma, with the last step valid only when the mass is constant. The impulse form is the more general one and it still works during a collision, where the force varies too fast for any single acceleration to be useful.

Does AP Physics 1 test rockets or other changing-mass systems?

Not quantitatively. The Topic 4.2 boundary statement says AP Physics 1 does not require students to quantitatively analyze systems in which the mass of the system changes with respect to time. That is the whole of what the CED commits to, so treat a rocket or a sand-collecting cart as a reasoning scenario rather than a calculation.