Physics · Mechanics

Conservation of momentum

A selected system conserves total momentum when external impulse over the chosen interval is zero or negligible.

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Formula

m_{1} v_{1} + m_{2} v_{2} = m_{1} v_{1}^{'} + m_{2} v_{2}^{'}

m in kg · v in m/s · momenta in kg·m/s

Variables & units

Variables & units
Symbol	Meaning	Unit
m₁, m₂	Positive masses of the two colliding bodies	kg
v₁, v₂	Signed pre-collision velocities in the same reference frame	m/s
v₁', v₂'	Signed post-collision velocities; a prime is not a derivative	m/s

LaTeX

m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'

QUANTA STUDYquanta-study.de

Diagram in two rows: at the top the cart m₁ = 2·m₂, drawn twice as wide, moves with v₁ toward the resting cart m₂, at the bottom m₁ continues with v₁' = v₁/3 and m₂ moves off with v₂' = 4/3·v₁. — Elastic collision with m₁ = 2·m₂, the velocity arrows are to scale: the total momentum is the same before and after, m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'.

Perfectly inelastic collision calculator

Find the common velocity after a collision in which both bodies stick together and no external force acts along the line of impact.

First mass m₁

Mass of the first body, e.g. 200 g for a glider on an air track.

Second mass m₂

Mass of the second body, e.g. 100 g.

First initial velocity v₁

Velocity of the first body before the collision, positive in your chosen direction, e.g. 0.6 m/s.

Second initial velocity v₂

Velocity of the second body before the collision; negative if it comes the other way, e.g. −0.3 m/s.

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

Result: Common final velocity v_e

v_e = 0.3 m/s

With m₁ = 200 g, m₂ = 100 g, v₁ = 0.6 m/s and v₂ = -0.3 m/s, v_e = 0.3 m/s.

Working · rearrange for vₑ, substitute, evaluate

1. Rearrange for v_e
$v_f=\frac{m_1v_1+m_2v_2}{m_1+m_2}$
2. Substitute values (in SI units)
- m₁ = 200 g = 0.2 kg
- m₂ = 100 g = 0.1 kg
$v_f=\frac{0.2\cdot 0.6+0.1\cdot (-0.3)}{0.2+0.1}=0.3$
3. Result
v_e = 0.3 m/s

Momentum is conserved. Kinetic energy is generally not conserved in this collision model.

Quick tips

Total momentum divided by total mass: v = (m₁ · v₁ + m₂ · v₂) ÷ (m₁ + m₂). The result always lies between v₁ and v₂.
Equal masses heading towards each other at equal speeds: afterwards both stand still. Kinetic energy is lost, momentum is not.

Assumptions and limits

Two positive constant masses stick together after collision. No external impulse along the selected axis; all velocities are signed in the same inertial frame. Not a general elastic or vector solution.

Derivation & background

Summing Newton’s force balances cancels classical internal force pairs. The remainder is $\frac{d\vec P}{dt}=\sum\vec F_{\mathrm{external}}$ , hence $\Delta\vec P=\vec J_{\mathrm{external}}$ . With negligible external impulse, momentum is conserved for elastic and inelastic collisions. Kinetic energy is conserved only for elastic collisions; sticking is the perfectly inelastic case.

Exam blueprint

Validity range

Fixed material system, common inertial frame and zero or negligible external impulse in the selected direction. Kinetic energy needs a separate balance.

Derivation steps

Internal force pairs cancel; only external impulse changes total momentum.

1Select a fixed collection of bodies, a common inertial frame and the direction to balance.
2Write dp⃗_i/dt = F⃗_i,external + Σ_jF⃗_i←_j for each body.
3Sum over all bodies. Classical internal force pairs are equal and opposite, cancelling in this sum.
4Hence $\frac{d\vec P}{dt}=\sum\vec F_{\mathrm{external}}$ ; integration gives P⃗after − P⃗before = J⃗external.
5Only for zero or sufficiently small J⃗external use P⃗after = P⃗before. For two bodies in 1D: $m_{1}v_{1} + m_{2}v_{2} = m_{1}v_{1}$ ′ + m₂v₂′.
6For sticking, add v₁′ = v₂′ = v′. For an elastic collision, additionally use conservation of kinetic energy instead.

Rearrangements

Final velocity in a perfectly inelastic collision

v' = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}

Both bodies move on together after the collision.

Recoil velocity

v_2' = -\frac{m_1 v_1'}{m_2}

From total $p = 0$ : what flies forward pushes the rest backward.

Task variant

A cart (3 kg, 4 m/s) couples to a cart at rest (1 kg). Find v'.

$v^{\prime}=\frac{m_1v_1+m_2v_2}{m_1+m_2}=\frac{3\cdot4+1\cdot0}{3+1}\,\mathrm{m\,s^{-1}}=3\,\mathrm{m\,s^{-1}}$ .

An initially stationary cannon (200 kg) fires a ball (2 kg) at $\frac{100 m}{s}$ relative to the ground. External horizontal impulse is negligible. What is its recoil speed?

$0=2\cdot100\,\mathrm{kg\,m\,s^{-1}}+200\,\mathrm{kg}\cdot v_2^{\prime}$ .
Hence $v_2^{\prime}=-\frac{200}{200}\,\mathrm{m\,s^{-1}}=-1\,\mathrm{m\,s^{-1}}$ , or $\frac{1 m}{s}$ backwards.

Common mistakes

Dropping the signs for opposite velocities.

Define one direction as positive; oncoming motion gets a negative v.

Additionally demanding conservation of kinetic energy in an inelastic collision.

In an inelastic collision kinetic energy converts into other forms. Total energy remains conserved in the inclusive isolated system; momentum conservation additionally requires negligible external impulse.

Not adding the masses after coupling.

After a perfectly inelastic collision the total mass m₁+m₂ moves together.

Applying momentum conservation although external forces dominate.

Check external impulse. Short duration alone does not guarantee conservation; a significant external contribution requires the complete momentum balance.

Exam context

Exam classics: coupling carts, the ballistic pendulum, recoil, usually combined in two stages with an energy argument.

Formula cluster

Conservation laws

Momentum balance, energy accounting and an appropriate collision condition complement one another; they do not automatically determine duration or peak force.

p = m·v

Momentum

Ekin = ½mv²

Kinetic Energy

ΣF_ext = F_res = ma

Newton's Second Law

Worked example

Inelastic collision: a cart (m₁ = 2 kg, v₁ = 3 m/s) hits a cart at rest $(m_{2} = 1 kg)$ and couples to it: $v'=\frac{2\cdot3+1\cdot0}{2+1}=2\,\mathrm{m/s}$ .

Applications

Accident reconstruction, billiards and ball sports, the rocket equation, particle physics (scattering experiments)

How do you calculate with Conservation of Momentum?

Here is how to work through a typical Conservation of Momentum (p_vor = p_nach) task step by step:

1
Task
A cart (3 kg, 4 m/s) couples to a cart at rest (1 kg). Find v'.
Solution path
$v^{\prime}=\frac{m_1v_1+m_2v_2}{m_1+m_2}=\frac{3\cdot4+1\cdot0}{3+1}\,\mathrm{m\,s^{-1}}=3\,\mathrm{m\,s^{-1}}$ .
2
Task
An initially stationary cannon (200 kg) fires a ball (2 kg) at $\frac{100 m}{s}$ relative to the ground. External horizontal impulse is negligible. What is its recoil speed?
Solution path
1. $0=2\cdot100\,\mathrm{kg\,m\,s^{-1}}+200\,\mathrm{kg}\cdot v_2^{\prime}$ .
2. Hence $v_2^{\prime}=-\frac{200}{200}\,\mathrm{m\,s^{-1}}=-1\,\mathrm{m\,s^{-1}}$ , or $\frac{1 m}{s}$ backwards.

Understand and apply the equation

Two trolleys collide: one slows down while the other speeds up. Their combined momentum can nevertheless remain unchanged. The key is the system boundary. Only after identifying the included bodies and external impulses can you apply conservation.

When does conservation of momentum apply?

The total momentum of a fixed collection of classical bodies has the same value before and after a process when external impulse over that interval vanishes. Measure all velocities in the same inertial frame. A single component is conserved when external impulse in that direction is zero.[1]

First consider both trolleys together and imagine a boundary around them. Their mutual contact forces are internal. Track friction or a push from outside is external. If you instead select only one trolley, contact from the other trolley is external too. “Internal” and “external” describe the selected balance rather than different kinds of force.[1]

A fixed selection of matter is closed when no mass crosses its boundary. This alone does not ensure momentum conservation: the environment can still impart impulse. If external impulse vanishes over an interval, initial and final momenta match. If net external force is zero at every instant, total momentum remains constant throughout the process.[1]

\begin{aligned}&\vec P=\sum_i m_i\vec v_i\\&\vec P_{\mathrm{after}}-\vec P_{\mathrm{before}}=\int_{t_1}^{t_2}\vec F_{\mathrm{ext}}\,\mathrm dt\end{aligned}

All velocities use the same inertial frame. Zero external impulse gives P⃗after = P⃗before.[1]

A cart pair on a horizontal track may approximately conserve horizontal momentum despite gravity and support forces. What matters is their horizontal component and horizontal friction impulse. Components can be balanced separately. A brief collision does not automatically justify ignoring every external force; explain why its integral is small.[1]

Why internal forces disappear from the balance

The trolleys may accelerate each other strongly during a collision. Newton’s third law still gives equal and opposite contact forces at each instant: $\vec F_{1\leftarrow2}=-\vec F_{2\leftarrow1}$ . This does not make force on either trolley zero. It means the internal contributions cancel pairwise when the two equations of motion are added. For each individual trolley, Newton’s second law relates net force to the rate of change of its momentum; the internal terms cancel only after the two balances are added.[6][8]

\begin{aligned}\frac{\mathrm d\vec p_1}{\mathrm dt}&=\vec F_{1,\mathrm{ext}}+\vec F_{1\leftarrow2}\\\frac{\mathrm d\vec p_2}{\mathrm dt}&=\vec F_{2,\mathrm{ext}}+\vec F_{2\leftarrow1}\\\frac{\mathrm d(\vec p_1+\vec p_2)}{\mathrm dt}&=\vec F_{1,\mathrm{ext}}+\vec F_{2,\mathrm{ext}}\end{aligned}

For more bodies, sum all momenta and external forces in the same way.[6]

\begin{aligned}&\frac{\mathrm d\vec P}{\mathrm dt}=\sum_i\vec F_{i,\mathrm{ext}}\\&\Delta\vec P=\vec J_{\mathrm{ext}}\end{aligned}

Integration gives the change across the collision; $\vec J_{\mathrm{ext}}=0$ is the conservation condition.[6]

For two bodies in one dimension this becomes $m_1v_1+m_2v_2=m_1v_1^{\prime}+m_2v_2^{\prime}$ . A prime denotes the post-collision state, not a derivative. If both final velocities are unknown, this is just one equation. Add a suitable condition, such as sticking together or, for an elastic collision, conservation of kinetic energy.[2]

Elastic, inelastic or perfectly inelastic?

v_{\mathrm{joined}}=\frac{m_1v_1+m_2v_2}{m_1+m_2}

This rearrangement requires both bodies to move together afterwards and negligible external impulse.

Choose a positive axis direction and assign negative velocities to motion in the opposite direction. Masses are positive. After sticking, common velocity lies between initial velocities because it is a mass-weighted average. A result outside that range signals a sign or arithmetic error in this model.

An elastic collision also preserves total kinetic energy. An inelastic collision changes it, with some energy becoming internal energy, deformation or sound. Perfectly inelastic means sticking together. Total energy of a sufficiently inclusive isolated system remains conserved even when its kinetic energy decreases.[2]

Why is one momentum equation insufficient for every collision?

If both final velocities are unknown, the one-dimensional momentum balance allows many possible outcomes. Sticking adds equal final velocities as a condition. An elastic collision instead adds conservation of kinetic energy. The collision type must be specified or justified by a model; it does not follow simply from the word “collision”.[2]

For a concrete comparison, a 1 kg trolley at +4 m/s approaches a stationary 1 kg trolley. Sticking gives both a final velocity of +2 m/s. In an ideal elastic one-dimensional collision, the first trolley stops and the second moves at +4 m/s. Both outcomes conserve the same total momentum; only the second also conserves kinetic energy.[2]

\begin{aligned}P_{\mathrm{before}}&=4\,\mathrm{kg\,m\,s^{-1}}\\P_{\mathrm{stick}}&=(1+1)(2)\,\mathrm{kg\,m\,s^{-1}}=4\,\mathrm{kg\,m\,s^{-1}}\\P_{\mathrm{elastic}}&=(1)(0)+(1)(4)=4\,\mathrm{kg\,m\,s^{-1}}\\K_{\mathrm{before}}&=8\,\mathrm J\\K_{\mathrm{stick}}&=4\,\mathrm J,\qquad K_{\mathrm{elastic}}=8\,\mathrm J\end{aligned}

The same initial data give different collision outcomes when the additional model condition changes.[2]

How do you calculate common velocity and impulse?

Sticking with opposite initial directions

Given: m₁ = 2 kg, v₁ = +3 m/s; m₂ = 1 kg, v₂ = −1 m/s. They stick together; external horizontal impulse is negligible.
Find: Find their common velocity and the change in kinetic energy.

Approach: With negligible external horizontal impulse use $P=m_1v_1+m_2v_2=(m_1+m_2)v^{\prime}$ , and calculate kinetic energies separately. Add signed initial momenta.
$P_{\mathrm{before}}=2\cdot3+1\cdot(-1)=5\,\mathrm{kg\,m/s}$
Divide this momentum by combined mass.
$v^{\prime}=\frac5{2+1}=\frac53\,\mathrm{m/s}$
Add both initial kinetic energies.
$E_{\mathrm{before}}=\frac12\cdot2\cdot3^2+\frac12\cdot1\cdot(-1)^2=9.50\,\mathrm J$
Find final kinetic energy and the difference.
$\begin{aligned}E_{\mathrm{after}}&=\frac12\cdot3\left(\frac53\right)^2=\frac{25}6\,\mathrm J\\{} \Delta E_{\mathrm{kin}}&=-\frac{16}3\,\mathrm J\end{aligned}$

Result: Both move at +1.667 m/s. Kinetic energy decreases by 5.333 J.

Check: $3\cdot\frac53\,\mathrm{kg\,m\,s^{-1}}=5\,\mathrm{kg\,m\,s^{-1}}$ . This confirms total momentum. Final velocity lies between −1 and +3 m/s. The kinetic-energy decrease is conversion, not disappearance of total energy.

Equal opposite impulses, unequal velocity changes

Given: m₁ = 4 kg, v₁ = +2 m/s; m₂ = 2 kg, v₂ = −1 m/s. They stick; external horizontal forces are negligible. Contact time is 0.050 s.
Find: Calculate final velocity, both impulses and mean contact forces.

Approach: First use $v^{\prime}=\frac{m_1v_1+m_2v_2}{m_1+m_2}$ ; then each cart has $J_i=m_i(v^{\prime}-v_i)$ and $\overline F_i=\frac{J_i}{\Delta t}$ . Conserve total momentum and divide by total mass.
$v^{\prime}=\frac{4\cdot2+2\cdot(-1)}6=1\,\mathrm{m/s}$
Use each body’s mass to find its momentum change.
$\begin{aligned}J_1&=4(1-2)=-4\,\mathrm{N\,s}\\{} J_2&=2(1-(-1))=+4\,\mathrm{N\,s}\end{aligned}$
Calculate mean contact forces from $\overline F_i=\frac{J_i}{\Delta t}$ .
$\begin{aligned}\overline F_{1,x}&=-80\,\mathrm N\\{} \overline F_{2,x}&=+80\,\mathrm N\end{aligned}$

Result: Final velocity is +1 m/s. Contact impulses are −4 and +4 N·s; mean forces are −80 and +80 N.

Check: $J_1+J_2=0$ ; total momentum is 6 kg·m/s both before and after. Yet $\Delta v_1=-1\,\mathrm{m\,s^{-1}}$ and $\Delta v_2=+2\,\mathrm{m\,s^{-1}}$ differ because the masses differ.

How do you balance momentum in multiple directions?

\begin{aligned}&P_x=\sum_i m_iv_{i,x}\\&P_y=\sum_i m_iv_{i,y}\\&\vec P=M\vec v_{\mathrm{CM}}\end{aligned}

Each conserved component requires zero external impulse in that particular direction.[3]

The collision diagram shows an elastic special case: $m_1=2m_2$ , v₂ = 0, $v_1^{\prime}=\frac{v_1}{3}$ and $v_2^{\prime}=\frac{4v_1}{3}$ . The unequal arrows represent velocities, not forces. Substitution confirms equal total momentum and kinetic energy before and after. Sticking carts would instead have equal final velocities.

For two equal masses initially moving at (2, 0) and (0, 2) m/s, sticking gives $\vec v^{\prime}=(1,1)\,\mathrm{m\,s^{-1}}$ . Speed is √2 m/s, not 2 m/s. Solve components before finding magnitude. With no net external force, the centre of mass continues at constant velocity even while individual parts accelerate strongly.[3]

How do you explain recoil for fixed masses and rockets?

If two initially stationary bodies push apart, $m_1v_1+m_2v_2=0$ . A 6 kg body moving at +2 m/s requires a 12 kg partner to move at −1 m/s when external impulse is zero. New kinetic energy may come from a released spring or stored internal energy; momentum conservation does not specify its source.

A rocket together with its expelled gas can form a closed system. The rocket alone is open: gas carries momentum across its boundary. A two-fixed-mass equation cannot simply be reused with time-dependent rocket mass. Gravity and drag also contribute external forces.[4]

When is an external impulse small enough?

External friction is often neglected during a short collision. Justify this with an estimate: multiply a known bound on external force by collision duration. Compare that possible external impulse with the momentum changes you want to resolve. When large initial momenta almost cancel, even a small external contribution can matter for the remaining total momentum.[1]

Estimate the accuracy of a trolley balance

Given: Two sticking trolleys have combined mass 4.0 kg and initial momentum +6.0 kg·m/s. Collision duration is 0.020 s; total external horizontal force magnitude is at most 2.0 N.
Find: Estimate how far final velocity can differ from the result neglecting external impulse.

Approach: bound external impulse first, then divide by combined mass.
$\begin{aligned}|J_{\mathrm{ext}}|&\le F_{\mathrm{max}}\Delta t\\v_f&=\frac{P_i+J_{\mathrm{ext}}}{M}\\|v_f-v_0|&\le\frac{F_{\mathrm{max}}\Delta t}{M}\end{aligned}$
Calculate the idealised value and maximum deviation.
$\begin{aligned}v_0&=\frac{6.0}{4.0}\,\mathrm{m\,s^{-1}}=1.50\,\mathrm{m\,s^{-1}}\\|J_{\mathrm{ext}}|&\le0.040\,\mathrm{N\,s}\\|v_f-v_0|&\le0.010\,\mathrm{m\,s^{-1}}\end{aligned}$

Result: Actual final velocity lies between +1.49 and +1.51 m/s.

Check: Maximum possible momentum change 0.040 N·s is about 0.67% of initial momentum. External-force direction is unknown, so an interval is stated.

What conservation alone cannot determine

Momentum conservation alone does not determine collision duration, peak force, deformation or both final velocities. A vehicle alone does not conserve momentum when hitting an anchored obstacle. For a system containing vehicle, obstacle and Earth, their contact forces may be internal; the chosen boundary decides.

The derivation here uses classical bodies and pairwise internal forces. In electromagnetic systems, fields can also carry momentum, so particle momenta alone may be insufficient. Total momentum remains a key relativistic balance, but $\vec p=m\vec v$ must be replaced by the appropriate relativistic expression.[5][7]

Five problems with solutions

Calculate · Foundation

A 1.5 kg cart moving at +4 m/s hits a stationary 2.5 kg cart. They stick with zero external horizontal impulse. Find v′.

Show hint

Divide initial momentum by combined mass.

Learning goal: Find common velocity after a sticking collision.

Show solution and calculation

$P=6\,\mathrm{kg\,m\,s^{-1}},\quad M=4\,\mathrm{kg}$ .
Hence $v^{\prime}=\frac PM=+1.5\,\mathrm{m\,s^{-1}}$ .

v^{\prime}=+1.5\,\mathrm{m/s}

Calculate · Application

Two 0.5 kg bodies move at +2 and −2 m/s and stick. Find v′ and the converted kinetic energy.

Show hint

Add momentum and energy separately.

Learning goal: Check momentum balance and converted kinetic energy separately.

Show solution and calculation

$P=0$ , so $v^{\prime}=0$ .
Initial energy is $\frac12\cdot0.5\cdot4\,\mathrm J+\frac12\cdot0.5\cdot4\,\mathrm J=2\,\mathrm J$ ; final kinetic energy is zero.
Thus 2 J of kinetic energy is converted.

\begin{aligned}v^{\prime}&=0\\{} \Delta E_{\mathrm{kin}}&=-2\,\mathrm J\end{aligned}

Transfer · Application

A system initially has P_x = 6 kg·m/s. A constant external force F_x = −5 N acts for 0.20 s. Find final momentum.

Show hint

Use the external-impulse balance rather than $P_{\mathrm{after}}=P_{\mathrm{before}}$ .

Learning goal: Update total momentum using a specified external impulse.

Show solution and calculation

$J_{\mathrm{external}}=-1\,\mathrm{N\,s}$ ; $P_{x,\mathrm{after}}=(6-1)\,\mathrm{kg\,m\,s^{-1}}=5\,\mathrm{kg\,m\,s^{-1}}$ .
Total momentum is not conserved over this interval.

P_{x,\mathrm{after}}=5\,\mathrm{kg\,m/s}

Interpret · Application

On an ideal horizontal track, a compressed spring pushes apart two initially stationary carts of masses 1 kg and 3 kg. Afterwards the light cart moves at +6 m/s. The spring is massless and external horizontal forces vanish. Find the heavy cart’s velocity. Why is momentum not conserved for the light cart alone, but conserved for the system of both carts and spring?

Show hint

Spring force is external to an individual cart but internal to the combined system. Balance the initially zero total momentum.

Learning goal: Choose the system boundary for a momentum balance.

Show solution and calculation

Approach: choose both carts plus the massless spring as the system and form the horizontal momentum balance.
The heavy cart acquires −2 m/s.
The light cart alone receives a spring impulse, reaching +6 kg·m/s.

Check: the heavy cart has −6 kg·m/s. Stored spring energy enables the motion; momentum conservation alone does not determine that energy.

\begin{aligned}P_{\mathrm{initial}}&=0\\0&=m_1v_1+m_2v_2\\v_2&=-\frac{m_1v_1}{m_2}=-\frac{(1\,\mathrm{kg})(6\,\mathrm{m/s})}{3\,\mathrm{kg}}=-2\,\mathrm{m/s}\end{aligned}

[1][3]

Error analysis · Application

A 2 kg cart moving at +3 m/s hits a stationary 2 kg cart; they stick. External impulse is negligible. A solution uses kinetic-energy conservation to obtain common velocity $\frac3{\sqrt2}\,\mathrm{m/s}$ . Correct the approach and use total momentum and kinetic-energy loss to show why that result is impossible.

Show hint

In a perfectly inelastic collision, total momentum is conserved, not kinetic energy. Check the proposed final momentum.

Learning goal: Recognize invalid kinetic-energy conservation in a sticking collision.

Show solution and calculation

Approach: determine common velocity from conserved total momentum.
It is +1.5 m/s.
Kinetic energy decreases from 9 J to 4.5 J; 4.5 J transfers into other energy forms.

Check: the incorrect proposal gives about 8.49 kg·m/s instead of the initial 6 kg·m/s, requiring an external impulse that the model excludes.

\begin{aligned}v_f&=\frac{m_1v_1+m_2v_2}{m_1+m_2}=1.5\,\mathrm{m/s}\\E_i&=\frac12(2)(3^2)\,\mathrm J=9\,\mathrm J\\E_f&=\frac12(4)(1.5^2)\,\mathrm J=4.5\,\mathrm J\\P_{\mathrm{wrong}}&=4\frac3{\sqrt2}\,\mathrm{kg\,m/s}\approx8.49\,\mathrm{kg\,m/s}\end{aligned}

[1][2]

What to take away

Choose the system boundary first and account for external impulse. Add signed, directed momenta in the same frame. Sticking and elastic rebound impose different additional conditions. Conservation of momentum determines neither collision duration nor peak force and does not replace a complete energy balance.

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

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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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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.

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Change history

October 1, 2026Clarified validity conditions directly; exposed existing collision, recoil and component answers and linked momentum, Newton’s law and energy in context.
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. Conservation of Momentum. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/impulserhaltung

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Frequently asked questions about Conservation of Momentum

How do you apply conservation of momentum to collisions?+

First select the system, a common inertial frame and a positive direction. Check that external impulse in this direction is negligible during the selected interval. Then add signed momenta, Pbefore = m₁v₁ + m₂v₂, and set Pafter = Pbefore. In a perfectly inelastic collision the bodies stick: v′ = (m₁v₁ + m₂v₂)/(m₁ + m₂). A 2 kg cart moving at $+\frac{3 m}{s}$ and coupling to a stationary 1 kg cart gives $+\frac{2 m}{s}$ . For two separate unknown final velocities, momentum alone is insufficient. Add kinetic-energy conservation for an elastic collision; other collisions need a suitable additional condition.

What is the difference between elastic and inelastic collisions?+

Both collision types permit momentum conservation when external impulse on the selected system is zero or negligible. An elastic collision additionally has equal total kinetic energy before and after. In an inelastic collision this energy changes through exchange with other forms, such as internal energy or deformation. Sticking identifies the perfectly inelastic case; not every inelastic collision produces sticking. Brief contact does not automatically make a collision elastic. The energy balance is what decides. Total energy of a sufficiently inclusive isolated system remains conserved in both types even when its final kinetic energy is smaller. Check momentum and energy balances separately.

Why is momentum conserved but kinetic energy not always?+

For a fixed collection of classical bodies, mutual internal forces are equal and opposite. Summing their force balances cancels these contributions. Total momentum therefore changes only through external impulse. If this is zero, P⃗ is conserved; merely retaining the same matter is not enough. Kinetic energy obeys a different balance and may become internal energy, sound or deformation during a collision. Total energy does not disappear. An elastic collision has equal kinetic energy of the selected bodies before and after. This extra property cannot be inferred from momentum conservation alone; it must be part of the problem’s assumptions or collision model.

How do you calculate recoil, for example of a cannon or rocket?+

For two initially stationary bodies and zero external impulse, separation gives $0 = m_{1}v_{1} + m_{2}v_{2}$ . Thus $v_{2} = -(\frac{m_{1}}{m_{2}})v_{1}$ , with both velocities measured in the same original frame. For a freely recoiling 200 kg cannon and a 2 kg projectile at $+\frac{100 m}{s}$ in the ground frame, cannon velocity is $-\frac{1 m}{s}$ . A projectile speed relative to the cannon must first be converted. Anchoring or friction transfers additional external impulse. In a rocket, gas continuously crosses the boundary of the rocket alone; use a momentum-flow balance or the closed rocket-plus-gas system. The simple two-mass formula does not replace the rocket equation.

When must you not apply conservation of momentum?+

Do not set Pafter = Pbefore when external impulse on the selected system is significant over the interval. Use the more general balance ΔP⃗ = ∫F⃗external dt. A vehicle alone does not retain momentum when striking an anchored wall; contact forces transfer momentum to the wall and Earth. Enlarging the system may make those forces internal. Short collision duration alone does not guarantee negligible external influence. A component may also remain conserved during a long process if its external impulse is zero. System boundary, direction and the force–time integral decide, not a blanket “collisions yes, extended processes no” rule. Open mass flows additionally need a momentum-flow balance.

Quanta exam set

Curated exam set for "Conservation of Momentum":

Question (front)

Which formula describes Conservation of Momentum?

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Question (front)

How do you rearrange Conservation of Momentum for Final velocity in a perfectly inelastic collision?

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Question (front)

Which common mistake happens with Conservation of Momentum?

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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, §9.3 Conservation of Linear Momentum

Related formulas

\Delta\vec p=\int_{t_1}^{t_2}\vec F_{\mathrm{ext}}dt, \qquad \Delta\vec L=\int_{t_1}^{t_2}\vec\tau_{\mathrm{ext}}dt

Linear and angular impulse–momentum balance

More formulas: Mechanics

F_G = G \cdot \frac{m_1 \cdot m_2}{r^2}

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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Conservation of momentum

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Perfectly inelastic collision calculator

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How do you calculate with Conservation of Momentum?

When does conservation of momentum apply?

Why internal forces disappear from the balance

Elastic, inelastic or perfectly inelastic?

Why is one momentum equation insufficient for every collision?

How do you calculate common velocity and impulse?

Sticking with opposite initial directions

Equal opposite impulses, unequal velocity changes

How do you balance momentum in multiple directions?

How do you explain recoil for fixed masses and rockets?

When is an external impulse small enough?

Estimate the accuracy of a trolley balance

What conservation alone cannot determine

Five problems with solutions

A 1.5 kg cart moving at +4 m/s hits a stationary 2.5 kg cart. They stick with zero external horizontal impulse. Find v′.

Two 0.5 kg bodies move at +2 and −2 m/s and stick. Find v′ and the converted kinetic energy.

A system initially has Px = 6 kg·m/s. A constant external force Fx = −5 N acts for 0.20 s. Find final momentum.

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More formulas: Mechanics

Retain Conservation of Momentum for exams

A system initially has P_x = 6 kg·m/s. A constant external force F_x = −5 N acts for 0.20 s. Find final momentum.