Conservation of Linear Momentum

Equivalent forms

\frac{d\vec{p}_{\text{tot}}}{dt} = \vec{F}_{\text{ext}} = 0

\Delta \vec{p}_1 = -\Delta \vec{p}_2

It is just Newton's third law integrated over time — internal forces cancel in pairs, so the sum survives.

Symbol	Name	SI unit	Dimension	Range
$m_1$	Mass 1 Mass of the first body	kg	M	0.5 – 5
$m_2$	Mass 2 Mass of the second body	kg	M	0.5 – 5
$u_1$	Initial speed 1 Incoming velocity of body 1	m/s	LT^-1	0 – 6
$v_1$	Final speed 1output Outgoing velocity of body 1	m/s	LT^-1	-6 – 6

Unit systems

Symbol	SI	CGS	Imperial
$m_1$	kg	g	lb
$m_2$	kg	g	lb
$u_1$	m/s	cm/s	ft/s
$v_1$	m/s	cm/s	ft/s

Where it holds

Any isolated system (zero net external force) in any inertial frame; relativistically the conserved quantity becomes the four-momentum.

Dimensional analysis

[M][LT^{-1}] = [MLT^{-1}]

[M][LT^{-1}] = [MLT^{-1}]

Both sides are momentum, units

kg\cdot m/s

.

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Discovery

John Wallis, Christopher Wren, Christiaan Huygens · 1668

In 1668 the Royal Society posed the collision problem; Wallis (inelastic), Wren and Huygens (elastic) independently answered it, and Huygens showed total momentum is conserved in any frame. Newton later folded it into his third law: equal-and-opposite forces mean equal-and-opposite momentum changes.

Try this

Two carts crash on a frictionless track. Before you know either final speed, you already know one number won't budge.

Vary the masses, the incoming speed, and the bounciness (restitution e). The two carts' velocities change wildly, but m1u1 + m2u2 stays pinned.

Research status: stable

Real-world applications

Rocket propulsion (exhaust momentum balances vehicle momentum)
Recoil of firearms and the design of recoilless rifles
Vehicle crash analysis and airbag timing
Newton's cradle and billiards

Common misconceptions

Momentum is conserved only in elastic collisions — it is conserved in ALL collisions; only kinetic energy distinguishes them
A bigger mass always has more momentum — momentum is mv, a light fast object can beat a heavy slow one
Momentum is a scalar — it is a vector, so directions (signs) matter

Experimental verification

Air-track gliders with photogate timers reproduce

m_{1}u1 + m_{2}u2 = m_{1}v1 + m_{2}v2

to within a percent; ballistic pendulums and recoil guns confirm it; particle accelerators rely on four-momentum conservation in every event.

Derivation

For two bodies the only forces during contact are the mutual contact forces F12 and F21.

Newton's third law gives

F12 = -F21

at every instant, so

dp1/dt = -dp2/dt

.

Integrating over the interaction,

\Delta p_{1} = -\Delta p_{2}

, i.e.

m_{1}u1 + m_{2}u2 = m_{1}v1 + m_{2}v2

.

Generalizing to N bodies, all internal forces cancel in third-law pairs and

dP_tot/dt = F_ext

; with

F_ext = 0

, P_tot is constant.

Limiting cases

m_1 = m_2

, elastic⟶ Velocities swapBody 1 stops dead and body 2 leaves with u1 — the Newton's-cradle result.

m_1 ≫ m_2⟶ Body 1 barely slowsA truck hitting a ping-pong ball keeps almost all its momentum.

e = 0

(perfectly inelastic)⟶ Common final velocityThey stick:

v = (m_{1}u1 + m_{2}u2)/(m_{1}+m_{2})

; momentum conserved, kinetic energy lost.

What if…

What if there is friction on the track?

Friction is an external force, so momentum is no longer conserved — it bleeds away to the Earth.

What if I switch to the center-of-mass frame?

Total momentum is exactly zero there, before and after — the cleanest frame to analyze any collision.

What if the two stick and then explode apart?

Still conserved: the explosion's internal forces cancel, so the fragments' momenta still sum to the original.

1

Perfectly inelastic catch

Given ·

m 1:: 2
u 1:: 4
m 2:: 3

Find · common final velocity

Steps

Sticking means one final velocity v
Momentum: $m_{1}u1 + m_{2}u2 = (m_{1}+m_{2})v$
$v = 8/5 = 1.6\,\mathrm{m/s}$ ; KE drops from 16 J to 6.4 J

Result ·

v = (2\times 4 + 3\times 0)/(2+3) = 8/5 = 1.6\,\mathrm{m/s}

2

Equal-mass elastic hit

Given ·

m 1:: 2
u 1:: 4
m 2:: 2

Find · v1,

v_{2} (e = 1)

Steps

$Use v_{1} = ((m_{1}-m_{2})u_{1})/(m_{1}+m_{2}) = 0$
$v_{2} = (2\,\mathrm{m}_{1} u_{1})/(m_{1}+m_{2}) = 4\,\mathrm{m/s}$
Total momentum $8 kg\cdot m/s$ before and after

Result ·

v_{1} = 0

,

v_{2} = 4\,\mathrm{m/s}

— velocities swap

Linear Momentum→Impulse-Momentum Theorem→1D Elastic Collision (Final Velocities)→Perfectly Inelastic Collision (1D)→Coefficient of Restitution→