Impulse vs Momentum: What Is the Difference?

One is a state, one is a change

Momentum is what a system has. Impulse is what happens to it. That is the whole distinction, and everything below is a consequence of it.

The AP Physics 1 CED defines them in two separate topics and the wording does the work for you. Essential knowledge 4.1.A.1 says linear momentum is defined by the equation $\vec{p} = m\vec{v}$. Read that as a snapshot: give it a mass and a velocity at one instant and it returns a number. EK 4.2.A.2 says 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, $\vec{J} = \vec{F}_{\text{avg}}\Delta t$. Read that as a process: it needs a force and it needs a stopwatch.

Then EK 4.2.B.2 ties them together. The impulse-momentum theorem relates the impulse exerted on a system and the system's change in momentum:

Look at what that equation does and does not say. Impulse equals the CHANGE in momentum, not the momentum. The $\Delta$ is not decoration. EK 4.2.B.1 spells out what it means: change in momentum is the difference between a system's final momentum and its initial momentum, $\Delta \vec{p} = \vec{p} - \vec{p}_0$. So a system with momentum $20\ \text{kg}\cdot\text{m/s}$ has received an impulse of $20\ \text{N}\cdot\text{s}$ only if it started at rest. Start it at $15\ \text{kg}\cdot\text{m/s}$ and the impulse was $5\ \text{N}\cdot\text{s}$.

Both are vectors, and they do not have to point the same way. EK 4.1.A.2 says momentum has the same direction as the velocity. EK 4.2.A.3 says impulse has the same direction as the net force exerted on the system. A car braking has momentum forward and is receiving an impulse backward. Nothing is wrong with that picture; it is what slowing down looks like.

Impulse vs momentum, side by side

The row worth staring at is the one about zeros, because it is the fastest way to prove the two are different quantities rather than two names for one.

An object at rest has zero momentum and can be receiving a large impulse. Push a stationary crate for two seconds; at the instant you start, $\vec{p} = 0$ and the impulse is already accumulating.

An object coasting at constant velocity has large momentum and is receiving zero net impulse. A loaded truck at highway speed carries an enormous momentum. If the net force on it is zero, $\Delta \vec{p} = 0$, and the net impulse over any interval you like is zero.

Two quantities that can each be zero while the other is large are not the same quantity. The rest of this page is about the situations where a problem tries to make you treat them as one.

The case that separates them: bounce or stick

Send the same object at the same wall at the same speed twice. Change nothing except whether it bounces. The momentum on arrival is identical in both trials. The impulse is not.

Take the initial direction of motion as positive and hold that convention for the rest of this section.

Those are the numbers for a $0.145$ kg ball arriving at $40.0$ m/s, worked in full below. The bouncing trial takes $1.75$ times the impulse of the sticking trial, from an identical starting state, because the wall has to do two jobs instead of one: first remove all the incoming momentum, then supply momentum in the opposite direction.

This is why a momentum figure alone never tells you the impulse. You need two states, before and after. It is also why the CED's Unit 4 sample instructional activities point a teacher at exactly this comparison, asking students to use momentum bar charts to explain why a dart bouncing off a cart makes the cart move faster than the dart sticking to the cart, passing through the cart, or stopping and dropping after colliding with the cart.

The reverse case is just as sharp. Two objects can receive the same impulse and end up with completely different momenta, because impulse is added to whatever momentum was already there. Worked example three below has a pair of carts that share one impulse magnitude between four different momentum values.

Two graphs, and each one answers a different question

The CED gives momentum and impulse a graphical reading each, and the two readings are not interchangeable.

EK 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. Force on the vertical axis, time on the horizontal, and impulse is the area. Notice that area needs both axes. A tall thin spike and a short wide plateau can enclose the same area, which is the whole engineering point of a crumple zone: hold the impulse fixed and stretch $\Delta t$, and $F_{\text{avg}}$ falls.

EK 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. Momentum on the vertical axis, time on the horizontal. Here the momentum is the height of the curve, and the force is the steepness.

Put those side by side and the state-against-change split shows up as geometry.

The trap is reading the wrong feature. On a momentum against time graph, a student asked for the impulse over an interval sometimes reports the final height. The height is the final momentum. The impulse is the change in height. Those agree only when the graph starts at zero, which brings us to the coincidence section below.

Newton's second law is this relationship rearranged

Students often meet $\vec{F}_{\text{net}} = m\vec{a}$ years before they meet impulse, which makes impulse feel like an extra topic. The CED puts the dependency the other way round.

EK 4.2.B.3 states that Newton's second law of motion is a direct result of the impulse-momentum theorem applied to systems with constant mass, and prints the chain:

Read it left to right. EK 4.2.A.1 supplies the first equality: the rate of change of momentum is equal to the net external force exerted on an object or system. The second equality pulls a constant $m$ out of $\Delta(m\vec{v})$. The third is just the definition of acceleration. So $m\vec{a}$ is a special case, valid when the mass does not change, and the momentum form is the general statement.

That matters for a boundary you should know about. A boundary statement under Topic 4.2 says AP Physics 1 does not require students to quantitatively analyze systems in which the mass of the system changes with respect to time. So you will not be asked to run the rocket calculation. You should still know why the momentum form is the one the CED derives the other from, because a qualitative question can ask which form survives when the mass changes.

One consequence is worth writing on your formula sheet in your own words: $F_{\text{avg}} = \Delta p / \Delta t$. Impulse fixes the product of force and time, so the force and the time trade off against each other. Airbags, crumple zones, catching a ball with your hands moving backward, and a gymnast bending her knees on landing are all the same calculation: $\Delta p$ is set by the physics of the situation, and lengthening $\Delta t$ is the only variable you control.

The units look interchangeable, and they are not read the same way

$1\ \text{N}\cdot\text{s}$ and $1\ \text{kg}\cdot\text{m/s}$ are the same combination of base units. Substitute $1\ \text{N} = 1\ \text{kg}\cdot\text{m/s}^2$ into the first and you get the second exactly. They have to match, because the impulse-momentum theorem sets them equal.

So dimensional analysis cannot catch this confusion for you. What the two unit labels record is which definition you used.

• Writing $\text{kg}\cdot\text{m/s}$ says you multiplied a mass by a velocity. That is a state.
• Writing $\text{N}\cdot\text{s}$ says you multiplied a force by a time. That is a transfer.

The symbol key on the AP Physics 1 Table of Information lists $J$ as impulse and $p$ as momentum, and, like every entry in that key, gives no unit for either. Since both labels denote the same unit, either one is dimensionally correct for either quantity. Use whichever matches the route you took to the number: it makes your work readable, and it stops you from computing a mass times a velocity when the question asked for a force times a time.

When the two numbers coincide, and why that lulls you

Impulse and final momentum are numerically equal in exactly one situation: the system started at rest.

If $\vec{p}_0 = 0$, then $\Delta \vec{p} = \vec{p} - 0 = \vec{p}$, and the impulse equals the final momentum. That is not a coincidence about impulse; it is what subtracting zero does.

The problem is how many textbook problems open from rest. A ball dropped from rest, a cart released from rest, a puck struck while stationary, an explosion starting from a system at rest: all of them let you write impulse and final momentum as the same number and never notice you conflated two ideas. Then a question arrives where the object was already moving, and the habit produces an answer that is too large by exactly the initial momentum.

Two more near coincidences are worth naming so you do not mistake them for the same thing.

• In an isolated two-object interaction, the impulses on the two objects are equal in magnitude and opposite in direction, but their momenta are not. EK 4.3.A.3.i states it directly: the impulse exerted by one object on a second object is equal and opposite to the impulse exerted by the second object on the first, and it notes this is a direct result of Newton's third law. Worked example three puts numbers on it.
• Total momentum can be unchanged while both objects receive large impulses. EK 4.3.A.3 says that in the absence of net external forces, any change to the momentum of an object within a system must be balanced by an equivalent and opposite change of momentum elsewhere within the system. The system total holds still; the parts do not.

The reflex worth building is small. When a question says impulse, write down two momenta and subtract. On a problem that starts from rest that costs you one extra line, and everywhere else it saves the answer.

Where the confusion costs a mark

Each of these is a scoring event, not general advice.

• Reporting the final momentum when the question asked for the impulse. The single most expensive version of this error, and it is invisible when the object started at rest, which is why it survives practice and appears on the exam.
• Dropping the sign on a rebound. A ball arriving at $+40$ m/s and leaving at $30$ m/s in the other direction has $\Delta p = m(-30) - m(+40)$, not $m(30 - 40)$. Declare a positive direction before the first line and the subtraction takes care of itself.
• Calling the impulse zero because the speed did not change. A ball bouncing off a wall at the same speed it arrived has $\lvert \vec{p} \rvert$ unchanged and $\Delta \vec{p}$ equal to twice the incoming momentum in magnitude. Momentum is a vector, so a reversal is the largest change available, not the smallest.
• Using the peak force instead of the average force. $\vec{J} = \vec{F}_{\text{avg}}\Delta t$ carries the subscript for a reason. If a problem gives a force against time curve, take the area, not the maximum height times the width.
• Writing that a heavy slow object and a light fast object with the same momentum received the same impulse. They received the same impulse only if both started from rest. Equal momentum is a statement about their present states.
• Saying momentum is not conserved because an impulse acted. An impulse on one part of a system does not change the system total unless it came from outside the system. EK 4.3.B.1 states that momentum is conserved in all interactions, and EK 4.3.B.2 gives the condition under which a chosen system's total momentum is constant: the net external force on that system is zero.
• Treating $\vec{F}_{\text{net}} = m\vec{a}$ as the primary law and the impulse relation as a shortcut. EK 4.2.B.3 derives the second law from the impulse relation, not the other way round.

What the CED asks, and what the sheet prints

This comparison sits across the first two topics of AP Physics 1 Unit 4, Linear Momentum. Unit 4 is weighted at 10 to 15 percent of the multiple-choice section and estimated at about 10 to 15 class periods.

Topic 4.1 has one learning objective, 4.1.A: describe the linear momentum of an object or system. Three pieces of essential knowledge sit under it: 4.1.A.1 gives $\vec{p} = m\vec{v}$, 4.1.A.2 says momentum is a vector with the same direction as the velocity, and 4.1.A.3 says momentum can be used to analyze collisions and explosions. A boundary statement adds that unless otherwise stated, the general term momentum will refer specifically to linear momentum. Suggested skills for Topic 4.1 are 1.C, 2.B, 2.C and 3.B.

Topic 4.2 has two learning objectives. 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 definition, its vector nature, and the two graph readings. 4.2.B asks you to describe the relationship between the impulse exerted on an object or a system and the change in momentum of the object or system, and 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. Note 2.A, derive a symbolic expression, and 3.C, justify or support a claim using evidence: this topic is examined as reasoning at least as often as arithmetic.

On the AP Physics 1 equation sheet, impulse appears in exactly one equation, $\vec{J} = \vec{F}_{\text{avg}}\Delta t = \Delta \vec{p}$, and momentum appears in four: that one, plus $\vec{p} = m\vec{v}$, plus $\vec{F}_{\text{net}} = \frac{\Delta \vec{p}}{\Delta t} = m\frac{\Delta \vec{v}}{\Delta t} = m\vec{a}$, plus $\vec{v}_{\text{cm}} = \frac{\sum \vec{p}_i}{\sum m_i} = \frac{\sum m_i \vec{v}_i}{\sum m_i}$. The sheet does not print $\Delta \vec{p} = \vec{p} - \vec{p}_0$, even though the CED lists it as the relevant equation under EK 4.2.B.1. That subtraction is the one piece of this topic you supply from memory, and it is the piece the whole distinction rests on.

For the solving routine rather than the distinction, the impulse-momentum theorem guide has the three ways to find an impulse and the force-time graph method. Conservation of momentum covers what happens when you widen the system so the impulses become internal, and what is conserved in a collision covers why momentum survives an impact that kinetic energy does not. The CED framing for each topic is on Topic 4.1 and Topic 4.2.