Impulse vs Force: What Is the Difference?

A force is what acts at an instant. Impulse is force accumulated over a time interval: the average force multiplied by how long it acted. That means the same change in momentum can come from a large force applied briefly or a small force applied for longer, which is the entire physics of an airbag.

AP Physics: Unit 4 (topics 2.2 Forces and Free-Body Diagrams, 4.2 Change in Momentum and Impulse). Force is introduced in AP Physics 1 Unit 2, Force and Translational Dynamics, weighted at 18 to 23 percent of the multiple-choice section over about 22 to 27 class periods; Topic 2.2 carries learning objective 2.2.A, describe a force as an interaction between two objects or systems, with EK 2.2.A.1 and 2.2.A.2. Impulse is introduced in Unit 4, Linear Momentum, weighted at 10 to 15 percent over about 10 to 15 class periods. Topic 4.2, Change in Momentum and Impulse, is where they are connected: learning objective 4.2.A carries EK 4.2.A.1 through 4.2.A.5, covering the rate-of-change statement, the definition of impulse as the average force times the time interval, its vector nature, and the two graph readings, and learning objective 4.2.B carries EK 4.2.B.1 through 4.2.B.3. Suggested skills for Topic 4.2 are 1.B, 2.A, 2.D, 3.A and 3.C. A 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.

One is a rate, the other is a total

Force answers "how hard, right now". Impulse answers "how much, over the whole interval". A stopwatch is the difference between them.

The AP Physics 1 CED sets them up in that order. Essential knowledge 2.2.A.1 says forces are vector quantities that describe the interactions between objects or systems, and 2.2.A.1.i adds that a force exerted on an object is always due to the interaction of that object with another object. That is an instantaneous statement: at any moment, either the interaction is happening or it is not.

EK 4.2.A.2 then defines impulse 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

Every symbol on the right matters. Take away the force and there is no impulse. Take away the time interval and there is no impulse either, which is why an enormous force acting for zero time delivers nothing.

The pair are linked in both directions, and the CED prints both links. EK 4.2.A.1 says the rate of change of momentum is equal to the net external force exerted on an object or system, which reads $\vec{F}_{\text{net}} = \Delta \vec{p}/\Delta t$ . Rearranged, that is the impulse relation. So force is momentum change per unit time, and impulse is momentum change in total. They are the same physics divided by, or multiplied by, a clock.

One consequence worth noticing early: forces add across sources at a single instant, giving a net force. Impulses add across time. You never add two forces that acted at different moments, and you never divide an impulse by anything except a duration.

Impulse vs force, side by side

Question you are asking	Force	Impulse
Symbol	$\vec{F}$	$\vec{J}$
What it describes	An interaction happening now	The accumulated effect of one over an interval
SI unit	$\text{N}$	$\text{N}\cdot\text{s}$ , equivalently $\text{kg}\cdot\text{m/s}$
Vector or scalar	Vector	Vector
Direction	Along the interaction	Along the net force, per EK 4.2.A.3
Needs a duration	No	Yes, always
Defining relation	$\vec{F}_{\text{net}} = \dfrac{\Delta \vec{p}}{\Delta t}$	$\vec{J} = \vec{F}_{\text{avg}}\Delta t$
What it equals	The rate of change of momentum	The change in momentum
On a force against time graph	The height of the curve	The area under the curve
On a momentum against time graph	The slope	The rise between two instants
Doubling the duration	Changes nothing	Doubles it, at fixed average force
Can be huge while the other is tiny	Yes, a very brief impact	Yes, a very gentle very long push
CED essential knowledge	2.2.A.1, 4.2.A.1, 4.2.A.5	4.2.A.2 through 4.2.A.4, 4.2.B.2

The two "can be huge while the other is tiny" cases are the fastest proof that these are different quantities rather than two names for one.

A large force with a negligible impulse: a hammer strikes a nail with several hundred newtons for a few milliseconds. The force is far larger than anything you could produce by hand, and the impulse is a few newton seconds.

A small force with a large impulse: an ion thruster on a spacecraft pushes with a fraction of a newton for months. No single instant of that is impressive, and the accumulated impulse changes the craft's momentum by more than the hammer ever could.

Ranking by force and ranking by impulse are two different orderings of the same list of events, and the first worked example below builds a pair where they disagree outright.

The case that separates them: a bigger force with a smaller impulse

Put two pushes on the same cart, one hard and quick and one gentle and long.

	Shove A	Shove B
Force	$200$ N	$5.0$ N
Duration	$0.050$ s	$4.0$ s
Impulse	$10\ \text{N}\cdot\text{s}$	$20\ \text{N}\cdot\text{s}$
Change in momentum	$10\ \text{kg}\cdot\text{m/s}$	$20\ \text{kg}\cdot\text{m/s}$
Speed gained by a $4.0$ kg cart	$2.5$ m/s	$5.0$ m/s

Shove A applies forty times the force of shove B and produces half the effect. Nothing here is a trick: the durations differ by a factor of eighty, and $80/40 = 2$ .

This is why "which push was stronger" is not a well-posed question about motion. It is a question about force. "Which push changed the motion more" is a question about impulse, and the two have different answers here.

The reverse pairing is just as easy to build. Two shoves with the same impulse can have any force you like, because $\vec{J} = \vec{F}_{\text{avg}}\Delta t$ fixes only their product. A $1000$ N force for $0.010$ s and a $10$ N force for $1.0$ s both deliver $10\ \text{N}\cdot\text{s}$ , and a cart on a frictionless track ends up moving at exactly the same speed either way. It just cares about the total.

The trade-off airbags are built on

In a crash the change in momentum is not negotiable. A car arrives with a certain momentum and leaves with zero, so $\Delta \vec{p}$ is set by the mass and the impact speed before any engineering happens. The impulse the occupant must receive is therefore fixed too.

Rearrange the definition to see what is left to design:

F_{\text{avg}} = \frac{\lvert \Delta p \rvert}{\Delta t}

The numerator is fixed. The only variable an engineer controls is the denominator, and the force falls in exact inverse proportion to it. Stretch the stopping time by a factor of a hundred and the average force drops by a factor of a hundred, which is what the egg in the second worked example does.

Every member of this family is the same calculation:

An airbag and a crumple zone extend the time over which the occupant's momentum goes to zero.
A gymnast bends her knees on landing, converting a millisecond stop into a tenth of a second.
A cricketer or a baseball catcher moves the hands backward with the ball rather than holding them rigid.
Packaging, foam helmet liners and safety netting all buy time and nothing else.

Running it in the other direction is the physics of follow-through. A batter or a golfer wants the largest possible impulse, so they keep the club or bat in contact for as long as they can rather than striking and withdrawing. Same equation, same fixed product, opposite goal.

What none of these do is change the momentum. That is the part students get backwards: an airbag does not absorb momentum, it spreads the same impulse over more time so the force stays survivable.

Reading both quantities off one graph

A graph of net force against time carries both quantities, in two different geometric features, and confusing them is a routine loss of marks.

EK 4.2.A.4 states that 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. So:

The height of the curve at a moment is the force at that moment.
The area between the curve and the time axis over an interval is the impulse for that interval.

Area below the time axis counts as negative, because a force pointing the other way removes momentum.

A tall narrow spike and a low wide plateau can enclose the same area, and that is the airbag picture drawn rather than argued. Two impacts with identical impulses look completely different on the graph, and the difference is exactly what determines whether the object survives.

The partner graph runs the operations backwards. EK 4.2.A.5 says the net external force exerted on a system equals the slope of a graph of the momentum of the system as a function of time. So a momentum against time graph gives you the force as a slope, and the impulse as a rise. Height, slope, area and rise: four features, two quantities, and picking the wrong feature is a wrong answer rather than a rounding error.

You are given	To find the force	To find the impulse
Net force against time	Read the height	Take the area
Momentum against time	Take the slope	Read the rise between two instants

The force-time graph entry has the definition on its own, and impulse vs momentum works through what the momentum graph adds.

Average force is not peak force

The subscript on $\vec{F}_{\text{avg}}$ is not decoration, and ignoring it is a standard way for a correct method to produce a wrong number.

A real impact does not deliver a constant force. Contact begins at zero, rises to a maximum as the materials compress, and falls back to zero as they separate. The impulse is the whole area under that shape. The average force is the constant force that would enclose the same area over the same duration, which is always less than the peak.

For a triangular pulse, the average is exactly half the peak, and the third worked example below runs the numbers: a hammer blow peaking at $900$ N over $0.0080$ s delivers $3.6\ \text{N}\cdot\text{s}$ , and its average force is $450$ N. Using the peak instead would overstate the impulse by a factor of two.

So three different force numbers can describe one impact:

The peak force, which is what breaks things and what a force sensor's maximum reading shows.
The average force, which is what $\vec{J} = \vec{F}_{\text{avg}}\Delta t$ uses and what a question means by "the force exerted during the collision" unless it says otherwise.
The force at some named instant, which is a reading off the graph and is neither of the above.

When a question hands you a curve rather than a number, take the area and then divide by the duration if an average force is wanted. Never multiply the maximum height by the total width.

When force and impulse track each other, and why that lulls you

Hold the duration fixed and the two quantities become proportional. Double the force and the impulse doubles, so any comparison across events of equal length gives the same ranking either way, and nothing forces you to notice which quantity you were reasoning about.

Most textbook problems are exactly like that. A single collision, one stated contact time, and force and impulse rise and fall together throughout.

Two further coincidences are worth naming.

Both are vectors, and they point the same way. EK 4.2.A.3 says impulse is a vector quantity with the same direction as the net force exerted on the system. So direction cannot be used to tell them apart, and a diagram that shows one shows the other.
A constant force makes $\vec{F}_{\text{avg}}$ and $\vec{F}$ the same number. On a frictionless track under a steady push, the average force equals the instantaneous force at every moment, and the distinction quietly disappears until a graph with a curve arrives.

The habit that survives all of this is to check the unit of the number you are about to write down. Newtons answer "how hard", newton seconds answer "how much". If the question asked for the force on the occupant and your line ends in $\text{N}\cdot\text{s}$ , you answered a different question.

Where the confusion costs a mark

Each of these is a scoring event rather than general advice.

Reporting an impulse when the question asked for a force, or the reverse. The units give it away to a reader instantly, and there is no partial credit for the right calculation of the wrong quantity.
Multiplying the peak force by the duration. $\vec{J} = \vec{F}_{\text{avg}}\Delta t$ carries the subscript for a reason. Take the area.
Claiming the harder push always produces the bigger change in motion. Only if the durations match. A $200$ N shove for $0.050$ s changes the momentum half as much as a $5.0$ N shove for $4.0$ s.
Saying an airbag reduces the impulse. It does not. The change in momentum is set by the crash, so the impulse is fixed, and the airbag lengthens $\Delta t$ to reduce the average force. A justification built on "less impulse" is marked wrong even if the final number is right.
Forgetting that a force with no duration delivers nothing. A question that gives a force and no time interval cannot be asking for an impulse, and one that gives an impulse and no time interval cannot be asking for a force.
Reading the height of a force against time graph as the impulse. Height is the force. Area is the impulse. On a momentum against time graph the roles are slope and rise.
Using the net force in the impulse relation while the question asked about one force. $\vec{J} = \vec{F}_{\text{avg}}\Delta t = \Delta \vec{p}$ holds for the net external force and the total momentum change. An individual force delivers its own impulse, which does not equal the object's total momentum change unless that force is the only one acting.
Dividing an impulse by a mass to get a force. It gives a velocity change. Divide by a time to get a force.

What the CED asks, and what the sheet prints

Force is introduced in AP Physics 1 Unit 2, Force and Translational Dynamics, weighted at 18 to 23 percent of the multiple-choice section over about 22 to 27 class periods. Topic 2.2, Forces and Free-Body Diagrams, carries learning objective 2.2.A, describe a force as an interaction between two objects or systems, with EK 2.2.A.1 and 2.2.A.2, and learning objective 2.2.B on free-body diagrams.

Impulse is introduced in Unit 4, Linear Momentum, weighted at 10 to 15 percent over about 10 to 15 class periods. Topic 4.2, Change in Momentum and Impulse, is where the two quantities are formally connected. Learning objective 4.2.A asks you to describe the impulse delivered to an object or system and carries EK 4.2.A.1 through 4.2.A.5: the rate-of-change statement, the definition of impulse, its vector nature, and the two graph readings. Learning objective 4.2.B asks you to describe the relationship between the impulse exerted on an object or a system and the change in momentum, and carries EK 4.2.B.1 through 4.2.B.3, ending with the statement that Newton's second law is a direct result of the impulse-momentum theorem applied to systems with constant mass.

Suggested skills for Topic 4.2 are 1.B, 2.A, 2.D, 3.A and 3.C. A boundary statement adds that AP Physics 1 does not require students to quantitatively analyze systems in which the mass of the system changes with respect to time.

On the AP Physics 1 equation sheet, the two relevant lines are $\vec{F}_{\text{net}} = \frac{\Delta \vec{p}}{\Delta t} = m\frac{\Delta \vec{v}}{\Delta t} = m\vec{a}$ and $\vec{J} = \vec{F}_{\text{avg}}\Delta t = \Delta \vec{p}$ . The symbol key lists $F$ as force, $J$ as impulse and $t$ as time. Read those two lines together and the whole distinction is visible on the page: one divides a momentum change by a time, the other multiplies a force by one.

Going further: the impulse-momentum theorem guide has the three routes to an impulse and the graph method worked in full, how to find net force covers the force side, and impulse vs momentum handles the other confusion in this cluster. The CED framing is on Topic 4.2.

Forty times the force, half the effect

A $4.0$ kg cart sits at rest on a frictionless horizontal track. Shove A applies a constant $200$ N for $0.050$ s. Shove B applies a constant $5.0$ N for $4.0$ s. For each shove find the impulse, the change in momentum and the final speed of the cart, and say which shove is larger by force and which by impulse.

Shove A, impulse. $J = F_{\text{avg}}\Delta t = (200\ \text{N})(0.050\ \text{s}) = 10\ \text{N}\cdot\text{s}$ .
Shove A, momentum and speed. By the impulse-momentum theorem $\Delta p = J = 10\ \text{kg}\cdot\text{m/s}$ , and the cart started at rest so $p = 10\ \text{kg}\cdot\text{m/s}$ . Its speed is $v = p/m = 10/4.0 = 2.5\ \text{m/s}$ .
Shove B, impulse. $J = (5.0\ \text{N})(4.0\ \text{s}) = 20\ \text{N}\cdot\text{s}$ .
Shove B, momentum and speed. $\Delta p = 20\ \text{kg}\cdot\text{m/s}$ , so $v = 20/4.0 = 5.0\ \text{m/s}$ .
Compare. Shove A wins on force by a factor of $200/5.0 = 40$ . Shove B wins on impulse by a factor of $20/10 = 2$ , and therefore on final speed by the same factor of $2$ .
Check the arithmetic of the reversal so it does not feel like a trick. The force ratio is $40$ in A's favour and the duration ratio is $4.0/0.050 = 80$ in B's favour. The impulse ratio is the product of the two, $80/40 = 2$ , in B's favour. Nothing cancelled by accident: the durations simply differed by more than the forces did.
One extra reading. Suppose a question asked which shove the cart's frame had to survive. That is a force question, and the answer is shove A by a factor of forty. Suppose it asked which shove got the cart moving faster. That is an impulse question, and the answer is shove B. Same two events, two opposite verdicts, and the only way to get them both right is to notice which quantity was asked for.

Shove A delivers $10\ \text{N}\cdot\text{s}$ and leaves the cart at $2.5$ m/s. Shove B delivers $20\ \text{N}\cdot\text{s}$ and leaves it at $5.0$ m/s. Shove A is forty times the force and half the impulse, so ranking by force and ranking by effect on motion give opposite answers.

One egg, one impulse, two forces a hundred times apart

A $0.060$ kg egg is dropped and reaches the floor at $5.0$ m/s, where it stops. Find the impulse the floor must deliver. Then find the average force if the egg stops on concrete in $0.0020$ s, and if it stops on a thick pillow in $0.20$ s. Compare each force to the egg's weight. Take downward as positive and use $g = 9.8\ \text{m/s}^2$ .

Declare the axis and keep it: downward is positive, so the egg arrives with positive momentum and the floor's impulse will come out negative.
Momentum on arrival: $p_0 = mv_0 = (0.060\ \text{kg})(+5.0\ \text{m/s}) = +0.30\ \text{kg}\cdot\text{m/s}$ . Final momentum is zero, so $\Delta p = 0 - 0.30 = -0.30\ \text{kg}\cdot\text{m/s}$ and the impulse from the floor is $0.30\ \text{N}\cdot\text{s}$ upward.
That impulse is the same in both landings, and it was fixed before any surface was chosen. It depends only on the egg's mass and its arrival speed. For reference, arriving at $5.0$ m/s corresponds to a drop of $v^2/2g = 25/19.6 = 1.28\ \text{m}$ .
Concrete. $F_{\text{avg}} = \lvert \Delta p \rvert / \Delta t = 0.30/0.0020 = 150\ \text{N}$ upward.
Pillow. $F_{\text{avg}} = 0.30/0.20 = 1.5\ \text{N}$ upward. The stopping time grew by a factor of $100$ and the force fell by a factor of $100$ , exactly as the inverse proportionality demands.
Put both against the egg's weight, $mg = (0.060)(9.8) = 0.588\ \text{N}$ . The concrete delivers about $255$ times the egg's weight; the pillow delivers about $2.6$ times it. The shell breaks in one case and not the other, and the only physical difference between the two landings is the duration.
Note what did not change: the impulse, the change in momentum, the arrival speed, or the energy the egg had on arrival. The pillow did not absorb momentum. It spread the identical impulse over a hundred times as long.

The floor must deliver $0.30\ \text{N}\cdot\text{s}$ upward in both cases. On concrete the average force is $150$ N, about $255$ times the egg's weight; on the pillow it is $1.5$ N, about $2.6$ times its weight. Same impulse, forces a hundredfold apart, because the stopping time differs a hundredfold.

A force-time graph: area, average and peak all differ

A hammer head of mass $0.50$ kg strikes a nail. The force on the hammer rises linearly from $0$ to a peak of $900$ N over $0.0030$ s, then falls linearly back to $0$ over the next $0.0050$ s. Find the impulse delivered to the hammer, the average force, and the hammer's speed just before impact if it comes to rest at the end of the contact. Then find what a student gets by multiplying the peak force by the contact time.

Total contact time: $0.0030 + 0.0050 = 0.0080\ \text{s}$ .
Impulse from the area, per EK 4.2.A.4. The shape is a triangle with base $0.0080$ s and height $900$ N, so $J = \frac{1}{2}(0.0080\ \text{s})(900\ \text{N}) = 3.6\ \text{N}\cdot\text{s}$ . Splitting it into the two right triangles gives the same total: $\frac{1}{2}(0.0030)(900) = 1.35$ and $\frac{1}{2}(0.0050)(900) = 2.25$ , and $1.35 + 2.25 = 3.6\ \text{N}\cdot\text{s}$ .
Average force. $F_{\text{avg}} = J/\Delta t = 3.6/0.0080 = 450\ \text{N}$ . That is exactly half the peak, which is what a triangular pulse always gives, whether or not the peak sits in the middle.
Speed before impact. The hammer ends at rest, so the magnitude of its momentum change equals its initial momentum: $\lvert \Delta p \rvert = 3.6\ \text{kg}\cdot\text{m/s}$ , and $v_0 = 3.6/0.50 = 7.2\ \text{m/s}$ .
The error. Multiplying the peak by the duration gives $(900)(0.0080) = 7.2\ \text{N}\cdot\text{s}$ , exactly twice the true impulse, and it would imply the hammer arrived at $14.4$ m/s. Every downstream number doubles.
Three different force numbers describe this one impact and a question can ask for any of them: the peak, $900$ N; the average, $450$ N; and the force at some named instant, for example at $t = 0.0015$ s, where the rising line is halfway to its peak and the height is $450$ N. Only one of them belongs in $\vec{J} = \vec{F}_{\text{avg}}\Delta t$ .
Sanity check the shape against the physics. The area is a couple of newton seconds and the peak is nearly a kilonewton, which is the signature of a short hard impact: a large force, a small impulse, and a duration measured in milliseconds.

The impulse is $3.6\ \text{N}\cdot\text{s}$ , the average force is $450$ N, half the $900$ N peak, and the hammer arrived at $7.2$ m/s. Using the peak force instead of the average doubles the impulse to $7.2\ \text{N}\cdot\text{s}$ and doubles every quantity derived from it.

Frequently asked questions

What is the difference between impulse and force?

A force describes an interaction happening at an instant, measured in newtons. Impulse is that force accumulated over a time interval, the average force multiplied by the duration, measured in newton seconds. The AP Physics 1 CED defines impulse this way at essential knowledge 4.2.A.2. The practical consequence is that force and impulse can rank two events in opposite orders: a 200 N push lasting 0.050 s delivers 10 newton seconds, while a 5.0 N push lasting 4.0 s delivers 20 newton seconds, so the smaller force produces twice the change in motion. Force answers how hard, impulse answers how much.

Can a smaller force deliver a larger impulse?

Yes, whenever it acts for long enough. Impulse is the product of the average force and the time interval, so only the product is fixed, and either factor can be traded against the other. An ion thruster pushing with a fraction of a newton for months delivers far more impulse than a hammer blow of several hundred newtons lasting milliseconds. This is why a question asking which push was stronger and a question asking which push changed the motion more are different questions with potentially different answers. Only the second one is about impulse, and only impulse equals the change in momentum.

How do you find impulse from a force versus time graph?

Take the area between the curve and the time axis. The AP Physics 1 CED states this at essential knowledge 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. Split the shape into rectangles, triangles and trapezoids and add the areas, counting anything below the time axis as negative. Do not multiply the peak height by the total width, because the definition uses the average force. The height of the curve is the force at that instant, not the impulse.

Do airbags reduce the force or the impulse in a crash?

The force. The impulse is fixed by the crash before any safety device acts, because the occupant arrives with a certain momentum and must end at zero, and that change in momentum is the impulse. Writing the definition as average force equals the size of the momentum change divided by the time interval shows that the only adjustable quantity left is the duration. An airbag, a crumple zone, a helmet liner and bending your knees on landing all lengthen the stopping time, and the average force falls in exact inverse proportion. Saying an airbag absorbs momentum or reduces the impulse is marked as a wrong justification even when the final number is right.

Is the average force the same as the peak force?

No, and confusing them is a routine source of factor-of-two errors. A real impact starts at zero force, rises to a maximum as the materials compress, and falls back to zero. The average force is the constant force that would enclose the same area under the force against time graph over the same duration, so it is always smaller than the peak. For a triangular pulse it is exactly half. A hammer blow peaking at 900 N over 0.0080 s delivers 3.6 newton seconds of impulse and has an average force of 450 N; using 900 N instead would double the impulse and every quantity derived from it.

What are the units of impulse compared with force?

Force is measured in newtons and impulse in newton seconds, which is the same combination of base units as kilogram meters per second. Substituting one newton equals one kilogram meter per second squared into newton seconds gives kilogram meters per second exactly, which it must, because the impulse-momentum theorem sets impulse equal to a change in momentum. The unit is the fastest check on which quantity you have computed. If a question asks for the force on an occupant and your answer carries newton seconds, you calculated the impulse instead, and no credit follows a correct calculation of the wrong quantity.

Is Newton's second law the same as the impulse relation?

They are the same statement written two ways, and the AP Physics 1 CED derives one from the other rather than treating them as separate laws. Essential knowledge 4.2.A.1 says the rate of change of momentum equals the net external force, and EK 4.2.B.3 says Newton's second law of motion is a direct result of the impulse-momentum theorem applied to systems with constant mass, printing the chain from net force equals change in momentum over change in time through to mass times acceleration. The momentum form is the more general one, because pulling the mass out of the change in momentum assumes the mass is constant.