Physics · Mechanics

Momentum: a vector quantity of motion

Q: What is the difference between momentum and kinetic energy?

Momentum p⃗ = mv⃗ is a vector; kinetic energy $K = \frac{1}{2}mv^{2}$ is a scalar. At fixed mass, momentum magnitude grows linearly with speed, while energy grows quadratically. Equal masses moving in opposite directions can have zero total momentum and positive total kinetic energy. In a collision, total momentum of the selected system is conserved only when its external impulse vanishes or is negligible. Kinetic energy is additionally conserved only in an elastic collision. If equal masses with equal opposite velocities stick, they stop in this frame. Their former kinetic energy is converted into forms such as internal energy and deformation rather than destroyed.

Q: What is impulse and what does it have to do with momentum?

Impulse is the time integral of net force: J⃗ = ∫F⃗net $dt = \Delta{}p$⃗. F⃗Δt expresses the same relation for constant force or time-averaged net force. In a force–time graph, each component is a signed area. For a fixed momentum change, longer interaction time reduces mean force but does not determine its peak. With $\Delta{}p = 10 N\cdot{}s$, a duration of 0.05 s gives a mean 200 N, whereas 0.5 s gives a mean 20 N. This explains part of the effect of longer stopping processes. Load and protective performance also require the actual force history, distance and absorbed energy.

Q: Why is momentum a vector and why does that matter?

Because velocity has a direction, momentum $p = m\cdot{}v$ inherits it. In practice this means: in every problem you first fix a positive direction and give motions against it a negative sign. A car with $p = +\frac{20\,000 kg\cdot{}m}{s}$ and an oncoming one with $p = -\frac{15\,000 kg\cdot{}m}{s}$ together have $+\frac{5\,000 kg\cdot{}m}{s}$, not 35,000. The vector also matters for reversals: if a ball $(m = 0.5 kg)$ hits a wall at $\frac{10 m}{s}$ and bounces back at $\frac{8 m}{s}$, then $\Delta{}p = 0.5\cdot{}(-8 - 10) = -\frac{9 kg\cdot{}m}{s}$; the magnitudes add. Anyone who merely subtracts magnitudes here significantly underestimates the momentum change and hence the force.

Classical momentum is mass times directional velocity in the selected frame; one-dimensional components carry signs.

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Formula

Calculate

p = m \cdot v

p in kg·m/s = [N·s] · m in kilograms [kg] · v in m/s

Variables & units

Variables & units
Symbol	Meaning	Unit
$p$	Momentum vector; signed component in one dimension	kg·m/s
$m$	Mass of the selected body	kg
$v$	Directional velocity in the specified frame	m/s

LaTeX

p = m \cdot v

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Original plot of directed momentum components: horizontal axis vx/v₀, vertical axis px/(m·v₀), with positive v₀. A line through the origin runs from the negative to the positive quadrant. — For the same mass in a fixed reference frame, px = m·vx. Both components change sign together. Normalised points (−1,−1), (0,0), (1,1) lie on a straight line.

Momentum calculator

Find momentum, mass or velocity of a body along one fixed direction, with signs for direction.

Find

Mass m

Mass of the body, e.g. 450 g for a football.

Velocity v

Velocity with sign, positive in your chosen direction and negative against it, e.g. 72 km/h.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for p, substitute, evaluate

1. Rearrange for p
$p=mv$
2. Substitute values (in SI units)
- m = 450 g = 0.45 kg
- v = 72 km/h = 20 m/s
$p=0.45\cdot 20=9$
3. Result
p = 9 kg·m/s

Result: Momentum p (signed)

p = 9 kg·m/s

With m = 450 g and v = 72 km/h, p = 9 kg·m/s.

Quick tips

Formula triangle: p on top, m and v below. Cover the quantity you need and the rest is the calculation, e.g. v = p ÷ m.
Unit check: kg · m/s is the same as N · s. By hand, divide grams by 1000 and km/h by 3.6 first: 450 g is 0.45 kg, 72 km/h is 20 m/s.

Assumptions and limits

Classical momentum p = mv along a fixed axis, m > 0. Body momentum is not impulse over a time interval, although their dimensions agree.

Derivation & background

For constant mass in an inertial frame, $\vec F_{\mathrm{net}}=\frac{d\vec p}{dt}$ . Impulse is the integral $\int\vec F_{\mathrm{net}}\,dt=\Delta\vec p$ . F⃗Δt applies for constant or time-averaged net force. Longer contact time reduces mean force for the same momentum change, but does not determine peak force or the complete load history.

Exam blueprint

Validity range

p⃗ = mv⃗ applies classically at speeds small compared with c. Impulse balance uses a fixed selection of matter in an inertial frame; open systems need a momentum-flow balance.

Derivation steps

The classical momentum definition connects directional motion with impulse.

1Select the body, inertial frame and positive axis directions.
2Define classical momentum by p⃗ = mv⃗; components satisfy p_x = mv_x and similarly for other axes.
3For constant mass, differentiation gives dp⃗/dt = m dv⃗/dt = m a⃗.
4Newton’s second law equates m a⃗ with net force; the definition alone does not prove this law.
5Integrate over the interaction: Δp⃗ = ∫F⃗net dt. Initial and final momentum must use the same frame.
6For constant or time-averaged net force, Δp⃗ = F⃗meanΔt; mean force is not peak force.

Rearrangements

Velocity from momentum

v = \frac{p}{m}

For the same momentum the lighter body is faster.

Impulse

\overline{\vec F}_{\mathrm{net}}\,\Delta t=\Delta\vec p

Uses mean net force over $\Delta{}t >0$ ; for unchanged momentum change, a longer duration lowers its magnitude but does not determine peak force.

Task variant

A body has $p =\frac{ 15 kg\cdot{}m}{s}$ at $m = 3 kg$ . Find v.

$v=\frac pm=\frac{15}{3}\,\mathrm{m\,s^{-1}}=5\,\mathrm{m\,s^{-1}}$ .

A ball changes its momentum by $\frac{10 kg\cdot{}m}{s}$ in 0.05 s. What average force acts?

$\overline F=\frac{\Delta p}{\Delta t}=\frac{10}{0.05}\,\mathrm N=200\,\mathrm N$ .

Common mistakes

Confusing momentum with kinetic energy.

$p = m\cdot{}v$ is linear in v and a vector; $E_{\mathrm{kin}} = \frac{1}{2}mv^{2}$ is quadratic and scalar.

Ignoring signs for opposite motions.

Fix one direction as positive and insert the opposite direction as negative.

Giving the unit as newtons.

Momentum has $\frac{kg\cdot{}m}{s} (= N\cdot{}s)$ ; the newton is the unit of force.

Exam context

The building block for collision problems and impulse considerations (impact forces, airbags), often combined with momentum conservation.

Formula cluster

Momentum and collisions

Momentum definition, the conservation law and Newton laws form one unit.

p_vor = p_nach

Conservation of Momentum

ΣF_ext = F_res = ma

Newton's Second Law

Ekin = ½mv²

Kinetic Energy

Worked example

A football $(m = 0.43 kg)$ flies at $v =\frac{ 25 m}{s}$ : $p = 0.43 \times{} 25 \approx{}\frac{ 10.8 kg\cdot{}m}{s}$ .

Applications

Crash test evaluation, airbag design, rocket propulsion (recoil), ball sports analysis

How do you calculate with Momentum?

Here is how to work through a typical Momentum (p = m·v) task step by step:

1
Task
A body has $p =\frac{ 15 kg\cdot{}m}{s}$ at $m = 3 kg$ . Find v.
Solution path
$v=\frac pm=\frac{15}{3}\,\mathrm{m\,s^{-1}}=5\,\mathrm{m\,s^{-1}}$ .
2
Task
A ball changes its momentum by $\frac{10 kg\cdot{}m}{s}$ in 0.05 s. What average force acts?
Solution path
$\overline F=\frac{\Delta p}{\Delta t}=\frac{10}{0.05}\,\mathrm N=200\,\mathrm N$ .

Understand and apply the equation

A heavy trolley and a light ball can move equally fast yet respond very differently to an impulse. Momentum combines mass, velocity and direction. It describes what changes when an object slows down, turns or rebounds.

Calculating momentum: symbol, magnitude and direction

The symbol for momentum is p; as a vector we write $\vec p$ . Momentum first answers a question about a state: what directed quantity of motion does this object have now? In the classical model we define $\vec p=m\vec v$ . At equal velocity, twice the mass gives twice the momentum; at equal mass, twice the velocity doubles momentum. For positive mass, momentum and velocity point in the same direction. Every value refers to a specified reference frame.[1][3]

\begin{aligned}&\vec p=m\vec v\\&p_x=mv_x\\&p_y=mv_y\\&p_z=mv_z\end{aligned}

Magnitude is $|\vec p|=m|\vec v|$ . In one dimension, p_x is instead a signed component.

If positive x points right, rightward motion gives p_x > 0 and leftward motion gives p_x < 0. Momentum can change even at unchanged speed, such as during reversal. Uniform circular motion likewise has a changing momentum vector.

For an extended body, add the momenta of its parts. Total momentum is $\vec P=M\vec v_{\mathrm{CM}}$ , where $M$ is total mass and the velocity is that of the centre of mass. Rotation about a stationary centre can involve nonzero constituent momenta whose vector sum vanishes. Rotation also has angular momentum, a separate quantity from the linear momentum studied here.[3]

Momentum change: compare initial and final states

Momentum is not a force. A force changes momentum over a time interval. To describe a change, you therefore need two states and subtract initial momentum from final momentum. Choose the axis before calculating: a negative sign denotes the opposite direction, not a negative amount of motion.[2]

\begin{aligned}\Delta\vec p&=\vec p_2-\vec p_1\\\Delta\vec p&=m(\vec v_2-\vec v_1)\quad(m=\mathrm{const.})\end{aligned}

The second line requires the same object with unchanged mass.[2]

A ball returning at the same speed has the same kinetic energy but reversed momentum. Stopping removes its initial momentum; rebounding also creates momentum in the opposite direction. Reversal therefore produces a larger momentum change than merely stopping from the same initial velocity.[1][4]

What is the unit of momentum? Components and rearrangements

The SI unit of momentum is $\mathrm{kg\,m\,s^{-1}}$ , equivalent to $\mathrm{N\,s}$ : kilogram metre per second, or newton second. $\mathrm N$ alone is a force unit. $\mathrm{N\,m}$ instead measures work or energy, or torque, rather than linear momentum.[1][2]

\begin{aligned}&1\,\mathrm{kg\,m\,s^{-1}}=1\,\mathrm{N\,s}\\&\vec v=\frac{\vec p}{m}\\&|\vec p|=\sqrt{p_x^2+p_y^2+p_z^2}\end{aligned}

Solving for velocity requires m > 0. A newton second measures momentum; a newton metre is a different unit.

Convert grams to kilograms and km/h to m/s. For 120 g and 36 km/h, m = 0.120 kg and speed is 10 m/s, giving |p⃗| = 1.20 kg·m/s. Specify direction separately. Using $m=\frac{p_x}{v_x}$ requires v_x ≠ 0 and components describing the same state.

At fixed mass, each momentum component is linear in the corresponding velocity component. A normalised plot of $\frac{p_x}{mv_0}$ against $\frac{v_x}{v_0}$ therefore has slope one. Negative values belong to the same relationship provided the frame remains unchanged.

What is the difference between momentum and impulse?

Momentum describes a state of motion; impulse describes the momentum change caused by net force over a time interval. For a constant-mass body in an inertial frame, differentiating the definition gives $\frac{\mathrm d\vec p}{\mathrm dt}=m\frac{\mathrm d\vec v}{\mathrm dt}=m\vec a$ . Newton’s second law equates this with net force. The momentum definition alone does not prove the force law; it exposes its connection with a change of motion.[2]

\begin{aligned}\frac{\mathrm d\vec p}{\mathrm dt}&=m\frac{\mathrm d\vec v}{\mathrm dt}\\&=m\vec a=\vec F_{\mathrm{net}}\end{aligned}

For constant mass, this writes Newton’s force law in momentum form.[2]

\begin{aligned}&\vec F_{\mathrm{net}}=\frac{\mathrm d\vec p}{\mathrm dt}\\&\vec J=\int_{t_1}^{t_2}\vec F_{\mathrm{net}}(t)\,\mathrm dt=\vec p_2-\vec p_1\end{aligned}

J⃗ denotes impulse here, not the joule unit. In a force–time graph, its component is the signed area.[2]

Use $\vec F\Delta t$ only for constant force or time-averaged net force. For fixed $\Delta\vec p$ , a longer interaction reduces the magnitude of that mean force. It does not determine peak force: different curves can have the same area. Assessing protective systems also involves distance, load history and absorbed energy.[2]

Same momentum change, different force histories

A force–time graph reveals what mean force hides. Its horizontal axis is time; its vertical axis is a signed component of net force. Areas above the axis count positively and those below negatively. A high brief pulse and a lower longer pulse can have the same area and hence the same momentum change.[2]

Compare rectangular and triangular pulses

Given: A body receives either constant rightward force 50 N for 0.040 s or a triangular pulse of the same duration, rising linearly from zero to 100 N and falling back to zero. Mass remains constant.
Find: Compare impulse, mean force and peak force.

Approach: impulse is the area under the net-force component in each case.
$J_{\mathrm{rect}}=F\Delta t,\qquad J_{\mathrm{tri}}=\frac12F_{\mathrm{peak}}\Delta t$
Evaluate the rectangular area; include the factor one half for the triangle.
$\begin{aligned}J_{\mathrm{rect}}&=(50\,\mathrm N)(0.040\,\mathrm s)=2.0\,\mathrm{N\,s}\\J_{\mathrm{tri}}&=\frac12(100\,\mathrm N)(0.040\,\mathrm s)=2.0\,\mathrm{N\,s}\\\overline F&=\frac{2.0\,\mathrm{N\,s}}{0.040\,\mathrm s}=50\,\mathrm N\end{aligned}$

Result: Both histories deliver +2.0 N·s and the same mean force of +50 N. Their peaks differ: 50 N and 100 N.

Check: Check: the symmetric triangle’s mean force is half its peak. Equal momentum change therefore does not determine peak force.

Equal impulse, different peaks

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Both histories deliver 2 N·s rightward between 0 and 0.04 s. The areas under the curves are equal: 50 N times 0.04 s for the rectangle and one half times 100 N times 0.04 s for the triangle. Both average forces are 50 N.[2]

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Two examples with directions

A momentum vector in a plane

Given: Mass m = 0.36 kg; velocity v⃗ = (3, 4) m/s in a fixed Cartesian frame.
Find: Calculate both momentum components, magnitude and direction.

Approach: In the classical model use $\vec p=m\vec v$ , with components and vector magnitude measured in the same frame. Multiply each component by the same mass.
$\vec p=(0.36\cdot3,\;0.36\cdot4)=(1.08,\;1.44)\,\mathrm{kg\,m/s}$
Find magnitude from the sum of squared components.
$|\vec p|=\sqrt{1.08^2+1.44^2}=1.80\,\mathrm{kg\,m/s}$
Both components are positive; find the angle from the positive x axis.
$\theta=\arctan(\frac{1.44}{1.08})\approx53.13^\circ$

Result: p⃗ = (1.08, 1.44) kg·m/s; magnitude 1.80 kg·m/s, directed approximately 53.1° above positive x.

Check: Speed is $5\,\mathrm{m\,s^{-1}}$ . $m|\vec v|=0.36\cdot5\,\mathrm{kg\,m\,s^{-1}}=1.80\,\mathrm{kg\,m\,s^{-1}}$ confirms the magnitude. Momentum and velocity have the same angle.

A ball reverses direction

Given: m = 0.25 kg; v₁ = −8 m/s, v₂ = +6 m/s; contact lasts 0.020 s. Values use the same frame and axis.
Find: Find momentum change and mean net force during contact.

Approach: Use $J_x=\Delta p_x=m(v_2-v_1)$ ; mean net force is $\overline F_x=\frac{J_x}{\Delta t}$ . Calculate signed initial and final momenta.
$\begin{aligned}p_1&=-2.00\,\mathrm{kg\,m/s}\\{} p_2&=+1.50\,\mathrm{kg\,m/s}\end{aligned}$
Subtract initial momentum from final momentum.
$\Delta p=1.50-(-2.00)=+3.50\,\mathrm{N\,s}$
Divide by contact time.
$\overline F_{\mathrm{net},x}=\frac{3.50}{0.020}=+175\,\mathrm N$

Result: Impulse is +3.50 N·s; mean net force is 175 N rightward.

Check: $\Delta v=+14\,\mathrm{m\,s^{-1}}$ and $m\Delta v=3.50\,\mathrm{kg\,m\,s^{-1}}$ confirm the result. Subtracting speeds as 6 − 8 misrepresents reversal. A wall’s contact force equals net force only when other x forces are negligible.

Why momentum is not energy

E_{\mathrm{kin}}=\frac{|\vec p|^2}{2m},\qquad m>0

This relation is classical and uses the same body and reference frame.[4]

Two 1 kg bodies moving at +4 and −4 m/s have zero total momentum. Each has 8 J of kinetic energy, for a total of 16 J. Adding momentum magnitudes would give the wrong total momentum. Energy is a scalar; opposite directions do not cancel its sum.

At equal momentum magnitude, halving mass doubles kinetic energy. Equal momentum therefore means neither equal energy nor equal stopping force. Stopping force also depends on duration and the history of momentum change.

Reference frames and system boundaries

\begin{aligned}&\vec v\,^{\prime}=\vec v-\vec u\\&\vec p\,^{\prime}=\vec p-m\vec u\end{aligned}

u⃗ is the second inertial frame’s constant velocity relative to the first.[9]

A 2 kg parcel moves at +5 m/s relative to the ground, so $p=+10\,\mathrm{kg\,m\,s^{-1}}$ . In a cart frame moving at +3 m/s, $v^{\prime}=+2\,\mathrm{m\,s^{-1}}$ and $p^{\prime}=+4\,\mathrm{kg\,m\,s^{-1}}$ . Both are correct in their frames. Under a constant frame shift, Δp remains the same for the same constant-mass body.

An individual ball does not conserve momentum during wall contact. Conservation of momentum requires a selected system and a balance of its external impulse. With ball, wall and Earth together, their contact forces are internal. Earth’s large mass may acquire the missing momentum through an extremely small velocity change.[5]

Where $p=mv$ needs extension

For a rocket or material flow, mass inside a selected boundary varies. Applying F = ma blindly, or differentiating a changing selection of matter, misses momentum carried by entering or leaving mass. Use a closed total system or add the momentum flux balance.[6]

\begin{aligned}\vec p&=\gamma m\vec v\\{} \gamma&=\frac1{\sqrt{1-\frac{v^2}{c^2}}}\end{aligned}

For a massive particle at relativistic speeds, this expression is required; m denotes rest mass.[7]

When speed is small compared with c, γ is close to one and the classical definition is appropriate. Substituting m = 0 into $p=mv$ does not establish zero momentum for light. Photons carry momentum, related to energy by $p=\frac Ec$ .[7][8]

Five problems with solutions

Calculate · Foundation

A 150 g mass moves at −12 m/s. Find p_x and its magnitude.

Show hint

Convert mass first and retain the sign.

Learning goal: Find signed momentum component and magnitude separately.

Show solution and calculation

$m=0.150\,\mathrm{kg}$ ; $p_x=mv_x=-1.80\,\mathrm{kg\,m\,s^{-1}}$ ; magnitude is 1.80 kg·m/s.

p_x=-1.80\,\mathrm{kg\,m/s}

Calculate · Application

A 3 kg body has p⃗ = (−6, 9) kg·m/s. Find its velocity.

Show hint

Divide each component by the same mass using $v_i=\frac{p_i}{m}$ .

Learning goal: Reconstruct velocity components from a momentum vector.

Show solution and calculation

$\vec v=\frac{\vec p}{m}=(-2,3)\,\mathrm{m\,s^{-1}}$ .
Speed is $\begin{aligned}|\vec v|&=\sqrt{(-2)^2+3^2}\,\mathrm{m\,s^{-1}}\\{}&=\sqrt{13}\,\mathrm{m\,s^{-1}}\approx3.606\,\mathrm{m\,s^{-1}}\text{.}\end{aligned}$

\vec v=(-2,3)\,\mathrm{m/s}

Transfer · Application

A constant net force of 15 N acts rightward for 0.40 s on a 2 kg body initially moving at −1 m/s. Find final velocity.

Show hint

Add $J=F\Delta t$ to signed initial momentum.

Learning goal: Combine a directed impulse with existing initial velocity.

Show solution and calculation

$p_1=-2\,\mathrm{kg\,m\,s^{-1}}$ , $J=+6\,\mathrm{N\,s}$ , $p_2=p_1+J=+4\,\mathrm{kg\,m\,s^{-1}}$ .
Hence $v_2=\frac{p_2}{m}=+2\,\mathrm{m\,s^{-1}}$ .

v_2=+2\,\mathrm{m/s}

Interpret · Application

A classical 2 kg body moves at +5 m/s in frame S. A second inertial frame S′ moves at +3 m/s relative to S in the same direction. Calculate momentum in both frames. A diagram therefore shows different arrow lengths. Does that difference violate physics?

Show hint

Transform velocity first using the Galilean transformation. Do not compare momenta from different frames without checking their references.

Learning goal: Interpret momentum as a directed quantity that depends on inertial frame.

Show solution and calculation

Approach: multiply the velocity in each frame by the same mass.
Momentum is +10 kg·m/s in S and +4 kg·m/s in S′.
This difference is expected: momentum depends on frame.

Check: the difference equals mass times relative frame velocity, here 6 kg·m/s. A conservation balance must be formed entirely within one fixed inertial frame.

\begin{aligned}v'&=v-u=5-3=2\,\mathrm{m/s}\\p&=mv=+10\,\mathrm{kg\,m/s}\\p'&=mv'=+4\,\mathrm{kg\,m/s}\\p-p'&=mu=6\,\mathrm{kg\,m/s}\end{aligned}

[1][9]

Error analysis · Application

Two separate bodies of 0.40 kg each have velocities +3 m/s and −3 m/s in the same inertial frame. A solution claims: “Total momentum is zero, so both are at rest and total kinetic energy is zero.” Check both conclusions using individual momenta and energies. No collision occurs.

Show hint

Momentum adds with signs. Kinetic energy contains squared velocity and is positive for each moving body.

Learning goal: Distinguish zero total momentum from absence of motion or kinetic energy.

Show solution and calculation

Approach: add directed momenta first, then scalar energies.
Momenta +1.2 and −1.2 kg·m/s do total zero.
Each body nevertheless has 1.8 J, totaling 3.6 J.
Both conclusions are false.

Check: a stationary centre of mass and internal motion can coexist without contradiction.

\begin{aligned}p_1&=+1.2\,\mathrm{kg\,m\,s^{-1}}\\{}p_2&=-1.2\,\mathrm{kg\,m\,s^{-1}}\\{}P&=p_1+p_2=0\\{}E_{\mathrm{tot}}&=2\left(\frac12\cdot0.40\cdot3^2\right)\,\mathrm J=3.6\,\mathrm J\end{aligned}

[1][4][3]

What to take away

First specify the object, frame and positive direction. Calculate momentum component by component and changes as final minus initial. Force–time area gives impulse, not work. Equal total momenta can describe different motions and energies; mean force does not determine peak force.

Sources for this explanation

Quanta · Updated October 1, 2026

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This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

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October 1, 2026Explained the momentum symbol and SI unit directly and distinguished impulse from a state of motion; linked Newton’s law, kinetic energy and momentum conservation within existing explanatory paragraphs.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Revised the introduction and learning progression; added explanatory intermediate steps, an application example and a summary. Retained existing examples and all five exercises.

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Quanta. Momentum. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/impuls

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Frequently asked questions about Momentum

How do you calculate the momentum of a body?+

Multiply the mass in kilograms by the velocity in metres per second: $p = m\cdot{}v$ , with the unit $\frac{kg\cdot{}m}{s}$ . A football of 0.43 kg at $v =\frac{ 25 m}{s}$ has $p = 0.43 \times{} 25 \approx{}\frac{ 10.8 kg\cdot{}m}{s}$ . Important: momentum is a vector; it points in the direction of motion, and in problems with oncoming traffic the two directions need opposite signs. Unlike kinetic energy, the velocity enters only linearly: a ball twice as fast has twice the momentum but four times the energy. Convert $\frac{km}{h}$ to $\frac{m}{s}$ before substituting (divide by 3.6), otherwise the unit is wrong.

What is the difference between momentum and kinetic energy?+

Momentum p⃗ = mv⃗ is a vector; kinetic energy $K = \frac{1}{2}mv^{2}$ is a scalar. At fixed mass, momentum magnitude grows linearly with speed, while energy grows quadratically. Equal masses moving in opposite directions can have zero total momentum and positive total kinetic energy. In a collision, total momentum of the selected system is conserved only when its external impulse vanishes or is negligible. Kinetic energy is additionally conserved only in an elastic collision. If equal masses with equal opposite velocities stick, they stop in this frame. Their former kinetic energy is converted into forms such as internal energy and deformation rather than destroyed.

What is impulse and what does it have to do with momentum?+

Impulse is the time integral of net force: J⃗ = ∫F⃗net $dt = \Delta{}p$ ⃗. F⃗Δt expresses the same relation for constant force or time-averaged net force. In a force–time graph, each component is a signed area. For a fixed momentum change, longer interaction time reduces mean force but does not determine its peak. With $\Delta{}p = 10 N\cdot{}s$ , a duration of 0.05 s gives a mean 200 N, whereas 0.5 s gives a mean 20 N. This explains part of the effect of longer stopping processes. Load and protective performance also require the actual force history, distance and absorbed energy.

Why is momentum a vector and why does that matter?+

Because velocity has a direction, momentum $p = m\cdot{}v$ inherits it. In practice this means: in every problem you first fix a positive direction and give motions against it a negative sign. A car with $p = +\frac{20\,000 kg\cdot{}m}{s}$ and an oncoming one with $p = -\frac{15\,000 kg\cdot{}m}{s}$ together have $+\frac{5\,000 kg\cdot{}m}{s}$ , not 35,000. The vector also matters for reversals: if a ball $(m = 0.5 kg)$ hits a wall at $\frac{10 m}{s}$ and bounces back at $\frac{8 m}{s}$ , then $\Delta{}p = 0.5\cdot{}(-8 - 10) = -\frac{9 kg\cdot{}m}{s}$ ; the magnitudes add. Anyone who merely subtracts magnitudes here significantly underestimates the momentum change and hence the force.

What is the unit of momentum and how do you check it?+

The SI momentum unit is kilogram metre per second. It equals the newton second because $1\,\mathrm N=1\,\mathrm{kg\,m\,s^{-2}}$ . The form $\mathrm{kg\,m\,s^{-1}}$ fits $\vec p=m\vec v$ , while $\mathrm{N\,s}$ fits the impulse balance. Units help distinguish quantities: joules describe energy and newtons describe force. Other consistent units are valid too. Before comparing numerical momenta, convert them to the same unit and use the same reference frame. Grams or kilometres per hour cause errors if their numbers are inserted as kilograms or metres per second without conversion. For example, converting 100 g and $\frac{36 km}{h}$ gives $p=1\,\mathrm{kg\,m\,s^{-1}}$ .

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Vol. 1, §9.1 Linear Momentum

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\mathbf r_{S}=\frac{\sum_i m_i\mathbf r_i}{\sum_i m_i}

Center of Mass of a System

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Newton's Law of Universal Gravitation

E_{pot} = m \cdot g \cdot h

Potential Energy (Gravitational)

F = k \cdot x

Hooke's Law (Spring Force)

s = \frac{1}{2} a t^2 + v_0 t

Uniformly Accelerated Motion (Position–Time Equation)

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Symbol

Meaning

Unit

p

Momentum vector; signed component in one dimension

kg·m/s

m

Mass of the selected body

v

Directional velocity in the specified frame

m/s

Derivation & background

Exam blueprint

Validity range

p⃗ = mv⃗ applies classically at speeds small compared with c. Impulse balance uses a fixed selection of matter in an inertial frame; open systems need a momentum-flow balance.

Derivation steps

The classical momentum definition connects directional motion with impulse.

1Select the body, inertial frame and positive axis directions.
2Define classical momentum by p⃗ = mv⃗; components satisfy p_x = mv_x and similarly for other axes.
3For constant mass, differentiation gives dp⃗/dt = m dv⃗/dt = m a⃗.
4Newton’s second law equates m a⃗ with net force; the definition alone does not prove this law.
5Integrate over the interaction: Δp⃗ = ∫F⃗net dt. Initial and final momentum must use the same frame.
6For constant or time-averaged net force, Δp⃗ = F⃗meanΔt; mean force is not peak force.

How do you calculate with Momentum?

Here is how to work through a typical Momentum (p = m·v) task step by step:

Task

A body has $p =\frac{ 15 kg\cdot{}m}{s}$ at $m = 3 kg$ . Find v.

Solution path

$v=\frac pm=\frac{15}{3}\,\mathrm{m\,s^{-1}}=5\,\mathrm{m\,s^{-1}}$ .

Task

A ball changes its momentum by $\frac{10 kg\cdot{}m}{s}$ in 0.05 s. What average force acts?

Solution path

$\overline F=\frac{\Delta p}{\Delta t}=\frac{10}{0.05}\,\mathrm N=200\,\mathrm N$ .

Calculating momentum: symbol, magnitude and direction

\begin{aligned}&\vec p=m\vec v\\&p_x=mv_x\\&p_y=mv_y\\&p_z=mv_z\end{aligned}

Magnitude is $|\vec p|=m|\vec v|$ . In one dimension, p_x is instead a signed component.

Momentum change: compare initial and final states

\begin{aligned}\Delta\vec p&=\vec p_2-\vec p_1\\\Delta\vec p&=m(\vec v_2-\vec v_1)\quad(m=\mathrm{const.})\end{aligned}

The second line requires the same object with unchanged mass.[2]

What is the unit of momentum? Components and rearrangements

\begin{aligned}&1\,\mathrm{kg\,m\,s^{-1}}=1\,\mathrm{N\,s}\\&\vec v=\frac{\vec p}{m}\\&|\vec p|=\sqrt{p_x^2+p_y^2+p_z^2}\end{aligned}

Solving for velocity requires m > 0. A newton second measures momentum; a newton metre is a different unit.

What is the difference between momentum and impulse?

\begin{aligned}\frac{\mathrm d\vec p}{\mathrm dt}&=m\frac{\mathrm d\vec v}{\mathrm dt}\\&=m\vec a=\vec F_{\mathrm{net}}\end{aligned}

For constant mass, this writes Newton’s force law in momentum form.[2]

\begin{aligned}&\vec F_{\mathrm{net}}=\frac{\mathrm d\vec p}{\mathrm dt}\\&\vec J=\int_{t_1}^{t_2}\vec F_{\mathrm{net}}(t)\,\mathrm dt=\vec p_2-\vec p_1\end{aligned}

J⃗ denotes impulse here, not the joule unit. In a force–time graph, its component is the signed area.[2]

Same momentum change, different force histories

Compare rectangular and triangular pulses

Given: A body receives either constant rightward force 50 N for 0.040 s or a triangular pulse of the same duration, rising linearly from zero to 100 N and falling back to zero. Mass remains constant.
Find: Compare impulse, mean force and peak force.

Approach: impulse is the area under the net-force component in each case.
$J_{\mathrm{rect}}=F\Delta t,\qquad J_{\mathrm{tri}}=\frac12F_{\mathrm{peak}}\Delta t$
Evaluate the rectangular area; include the factor one half for the triangle.
$\begin{aligned}J_{\mathrm{rect}}&=(50\,\mathrm N)(0.040\,\mathrm s)=2.0\,\mathrm{N\,s}\\J_{\mathrm{tri}}&=\frac12(100\,\mathrm N)(0.040\,\mathrm s)=2.0\,\mathrm{N\,s}\\\overline F&=\frac{2.0\,\mathrm{N\,s}}{0.040\,\mathrm s}=50\,\mathrm N\end{aligned}$

Result: Both histories deliver +2.0 N·s and the same mean force of +50 N. Their peaks differ: 50 N and 100 N.

Check: Check: the symmetric triangle’s mean force is half its peak. Equal momentum change therefore does not determine peak force.

Equal impulse, different peaks

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Two examples with directions

A momentum vector in a plane

Given: Mass m = 0.36 kg; velocity v⃗ = (3, 4) m/s in a fixed Cartesian frame.
Find: Calculate both momentum components, magnitude and direction.

Approach: In the classical model use $\vec p=m\vec v$ , with components and vector magnitude measured in the same frame. Multiply each component by the same mass.
$\vec p=(0.36\cdot3,\;0.36\cdot4)=(1.08,\;1.44)\,\mathrm{kg\,m/s}$
Find magnitude from the sum of squared components.
$|\vec p|=\sqrt{1.08^2+1.44^2}=1.80\,\mathrm{kg\,m/s}$
Both components are positive; find the angle from the positive x axis.
$\theta=\arctan(\frac{1.44}{1.08})\approx53.13^\circ$

Result: p⃗ = (1.08, 1.44) kg·m/s; magnitude 1.80 kg·m/s, directed approximately 53.1° above positive x.

Check: Speed is $5\,\mathrm{m\,s^{-1}}$ . $m|\vec v|=0.36\cdot5\,\mathrm{kg\,m\,s^{-1}}=1.80\,\mathrm{kg\,m\,s^{-1}}$ confirms the magnitude. Momentum and velocity have the same angle.

A ball reverses direction

Given: m = 0.25 kg; v₁ = −8 m/s, v₂ = +6 m/s; contact lasts 0.020 s. Values use the same frame and axis.
Find: Find momentum change and mean net force during contact.

Approach: Use $J_x=\Delta p_x=m(v_2-v_1)$ ; mean net force is $\overline F_x=\frac{J_x}{\Delta t}$ . Calculate signed initial and final momenta.
$\begin{aligned}p_1&=-2.00\,\mathrm{kg\,m/s}\\{} p_2&=+1.50\,\mathrm{kg\,m/s}\end{aligned}$
Subtract initial momentum from final momentum.
$\Delta p=1.50-(-2.00)=+3.50\,\mathrm{N\,s}$
Divide by contact time.
$\overline F_{\mathrm{net},x}=\frac{3.50}{0.020}=+175\,\mathrm N$

Result: Impulse is +3.50 N·s; mean net force is 175 N rightward.

Why momentum is not energy

E_{\mathrm{kin}}=\frac{|\vec p|^2}{2m},\qquad m>0

This relation is classical and uses the same body and reference frame.[4]

Reference frames and system boundaries

\begin{aligned}&\vec v\,^{\prime}=\vec v-\vec u\\&\vec p\,^{\prime}=\vec p-m\vec u\end{aligned}

u⃗ is the second inertial frame’s constant velocity relative to the first.[9]

Where $p=mv$ needs extension

\begin{aligned}\vec p&=\gamma m\vec v\\{} \gamma&=\frac1{\sqrt{1-\frac{v^2}{c^2}}}\end{aligned}

For a massive particle at relativistic speeds, this expression is required; m denotes rest mass.[7]

Five problems with solutions

Calculate · Foundation

A 150 g mass moves at −12 m/s. Find p_x and its magnitude.

Show hint

Convert mass first and retain the sign.

Learning goal: Find signed momentum component and magnitude separately.

Show solution and calculation

$m=0.150\,\mathrm{kg}$ ; $p_x=mv_x=-1.80\,\mathrm{kg\,m\,s^{-1}}$ ; magnitude is 1.80 kg·m/s.

p_x=-1.80\,\mathrm{kg\,m/s}

Calculate · Application

A 3 kg body has p⃗ = (−6, 9) kg·m/s. Find its velocity.

Show hint

Divide each component by the same mass using $v_i=\frac{p_i}{m}$ .

Learning goal: Reconstruct velocity components from a momentum vector.

Show solution and calculation

$\vec v=\frac{\vec p}{m}=(-2,3)\,\mathrm{m\,s^{-1}}$ .
Speed is $\begin{aligned}|\vec v|&=\sqrt{(-2)^2+3^2}\,\mathrm{m\,s^{-1}}\\{}&=\sqrt{13}\,\mathrm{m\,s^{-1}}\approx3.606\,\mathrm{m\,s^{-1}}\text{.}\end{aligned}$

\vec v=(-2,3)\,\mathrm{m/s}

Transfer · Application

A constant net force of 15 N acts rightward for 0.40 s on a 2 kg body initially moving at −1 m/s. Find final velocity.

Show hint

Add $J=F\Delta t$ to signed initial momentum.

Learning goal: Combine a directed impulse with existing initial velocity.

Show solution and calculation

$p_1=-2\,\mathrm{kg\,m\,s^{-1}}$ , $J=+6\,\mathrm{N\,s}$ , $p_2=p_1+J=+4\,\mathrm{kg\,m\,s^{-1}}$ .
Hence $v_2=\frac{p_2}{m}=+2\,\mathrm{m\,s^{-1}}$ .

v_2=+2\,\mathrm{m/s}

Interpret · Application

A classical 2 kg body moves at +5 m/s in frame S. A second inertial frame S′ moves at +3 m/s relative to S in the same direction. Calculate momentum in both frames. A diagram therefore shows different arrow lengths. Does that difference violate physics?

Show hint

Transform velocity first using the Galilean transformation. Do not compare momenta from different frames without checking their references.

Learning goal: Interpret momentum as a directed quantity that depends on inertial frame.

Show solution and calculation

Approach: multiply the velocity in each frame by the same mass.
Momentum is +10 kg·m/s in S and +4 kg·m/s in S′.
This difference is expected: momentum depends on frame.

Check: the difference equals mass times relative frame velocity, here 6 kg·m/s. A conservation balance must be formed entirely within one fixed inertial frame.

\begin{aligned}v'&=v-u=5-3=2\,\mathrm{m/s}\\p&=mv=+10\,\mathrm{kg\,m/s}\\p'&=mv'=+4\,\mathrm{kg\,m/s}\\p-p'&=mu=6\,\mathrm{kg\,m/s}\end{aligned}

[1][9]

Error analysis · Application

Two separate bodies of 0.40 kg each have velocities +3 m/s and −3 m/s in the same inertial frame. A solution claims: “Total momentum is zero, so both are at rest and total kinetic energy is zero.” Check both conclusions using individual momenta and energies. No collision occurs.

Show hint

Momentum adds with signs. Kinetic energy contains squared velocity and is positive for each moving body.

Learning goal: Distinguish zero total momentum from absence of motion or kinetic energy.

Show solution and calculation

Approach: add directed momenta first, then scalar energies.
Momenta +1.2 and −1.2 kg·m/s do total zero.
Each body nevertheless has 1.8 J, totaling 3.6 J.
Both conclusions are false.

Check: a stationary centre of mass and internal motion can coexist without contradiction.

\begin{aligned}p_1&=+1.2\,\mathrm{kg\,m\,s^{-1}}\\{}p_2&=-1.2\,\mathrm{kg\,m\,s^{-1}}\\{}P&=p_1+p_2=0\\{}E_{\mathrm{tot}}&=2\left(\frac12\cdot0.40\cdot3^2\right)\,\mathrm J=3.6\,\mathrm J\end{aligned}

[1][4][3]

What to take away

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

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An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 1, 2026Explained the momentum symbol and SI unit directly and distinguished impulse from a state of motion; linked Newton’s law, kinetic energy and momentum conservation within existing explanatory paragraphs.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Revised the introduction and learning progression; added explanatory intermediate steps, an application example and a summary. Retained existing examples and all five exercises.

Cite this reference page

Quanta. Momentum. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/impuls

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Frequently asked questions about Momentum

How do you calculate the momentum of a body?+

What is the difference between momentum and kinetic energy?+

What is impulse and what does it have to do with momentum?+

Why is momentum a vector and why does that matter?+

What is the unit of momentum and how do you check it?+

Momentum: a vector quantity of motion

Variables & units

Momentum calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Momentum and collisions

Worked example

Applications

How do you calculate with Momentum?

Calculating momentum: symbol, magnitude and direction

Momentum change: compare initial and final states

What is the unit of momentum? Components and rearrangements

What is the difference between momentum and impulse?

Same momentum change, different force histories

Compare rectangular and triangular pulses

Equal impulse, different peaks

Two examples with directions

A momentum vector in a plane

A ball reverses direction

Why momentum is not energy

Reference frames and system boundaries

Where p=mvp=mvp=mv needs extension

Five problems with solutions

A 150 g mass moves at −12 m/s. Find px and its magnitude.

A 3 kg body has p⃗ = (−6, 9) kg·m/s. Find its velocity.

A constant net force of 15 N acts rightward for 0.40 s on a 2 kg body initially moving at −1 m/s. Find final velocity.

A classical 2 kg body moves at +5 m/s in frame S. A second inertial frame S′ moves at +3 m/s relative to S in the same direction. Calculate momentum in both frames. A diagram therefore shows different arrow lengths. Does that difference violate physics?

Two separate bodies of 0.40 kg each have velocities +3 m/s and −3 m/s in the same inertial frame. A solution claims: “Total momentum is zero, so both are at rest and total kinetic energy is zero.” Check both conclusions using individual momenta and energies. No collision occurs.

What to take away

Sources for this explanation

Frequently asked questions about Momentum

Subject references and further reading

Related formulas

More formulas: Mechanics

Retain Momentum for exams

Momentum: a vector quantity of motion

Variables & units

Momentum calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Momentum and collisions

Worked example

Applications

How do you calculate with Momentum?

Calculating momentum: symbol, magnitude and direction

Momentum change: compare initial and final states

What is the unit of momentum? Components and rearrangements

What is the difference between momentum and impulse?

Same momentum change, different force histories

Compare rectangular and triangular pulses

Equal impulse, different peaks

Two examples with directions

A momentum vector in a plane

A ball reverses direction

Why momentum is not energy

Reference frames and system boundaries

Where p=mvp=mvp=mv needs extension

Five problems with solutions

A 150 g mass moves at −12 m/s. Find px and its magnitude.

A 3 kg body has p⃗ = (−6, 9) kg·m/s. Find its velocity.

A constant net force of 15 N acts rightward for 0.40 s on a 2 kg body initially moving at −1 m/s. Find final velocity.

A classical 2 kg body moves at +5 m/s in frame S. A second inertial frame S′ moves at +3 m/s relative to S in the same direction. Calculate momentum in both frames. A diagram therefore shows different arrow lengths. Does that difference violate physics?

Two separate bodies of 0.40 kg each have velocities +3 m/s and −3 m/s in the same inertial frame. A solution claims: “Total momentum is zero, so both are at rest and total kinetic energy is zero.” Check both conclusions using individual momenta and energies. No collision occurs.

What to take away

Sources for this explanation

Frequently asked questions about Momentum

Where $p=mv$ needs extension

A 150 g mass moves at −12 m/s. Find p_x and its magnitude.

Where $p=mv$ needs extension

A 150 g mass moves at −12 m/s. Find p_x and its magnitude.