Linear Momentum — Physica

Equivalent forms

\vec{p} = m\vec{v}

p_{\text{total}} = \sum_i m_i v_i

The conserved currency of every collision in the universe.

Symbol	Name	SI unit	Dimension	Range
$p$	Momentumoutput Linear momentum of the object	kg·m/s	MLT^-1	0 – 1000
$m$	Mass Mass of the object	kg	M	0.1 – 100
$v$	Velocity Velocity of the object	m/s	LT^-1	0 – 100

Unit systems

Symbol	SI	CGS	Imperial
$p$	kg·m/s	g·cm/s	slug·ft/s
$m$	kg	g	slug
$v$	m/s	cm/s	ft/s

Where it holds

Non-relativistic mechanics: valid for v << 299792458 m/s. For relativistic regimes

use p = \gamma mv

.

Dimensional analysis

p \to [MLT^{-1}]

m\cdot v \to [M]\cdot [LT^{-1}] = [MLT^{-1}]

Both sides match

kg\cdot m/s

.

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Discovery

René Descartes / Isaac Newton · 1644

Descartes introduced 'quantity of motion'; Newton formalized momentum as the natural variable in his second law: F = dp/dt.

Try this

Why does a 10-gram bullet at 400 m/s stop a 70 kg human cold?

Compute the momentum of a 0.01 kg bullet at 400 m/s versus a 70 kg person walking at 1.4 m/s. Compare and see which carries more 'oomph'.

Research status: stable

Real-world applications

Rocket propulsion (recoil from expelled mass)
Car crash analysis and airbag design
Billiards and pool
Particle physics collision experiments

Common misconceptions

Momentum is not the same as kinetic energy — it is a vector, not a scalar
A heavy slow object can have the same momentum as a light fast one
Momentum conservation requires zero external force, not zero internal forces

Experimental verification

Demonstrated in air-track collisions, ballistic pendulum experiments, and modern particle accelerator detectors that measure momentum to parts-per-million precision.

Derivation

Momentum is defined as

p \equiv mv

.

Newton's second law in its original form states

F = dp/dt

; for constant m this reduces to

F = ma

.

In an isolated system,

dp/dt = 0

, so total momentum is conserved.

Limiting cases

v \to 0

⟶

p \to 0

A stationary object has no momentum regardless of mass.

m \to 0

⟶

p \to 0

Massless classical object carries no momentum (photons need relativistic treatment).

v \to c

⟶

p \to \gamma mv

Relativistic correction:

\gamma = 1/\sqrt{1 - v^{2}/c^{2}}

becomes important.

What if…

What if velocity doubles?

Momentum doubles linearly — but kinetic energy quadruples. Momentum is gentler than energy.

What if two objects collide and stick?

Conservation of momentum: $m_{1}v_{1} + m_{2}v_{2} = (m_{1}+m_{2})v_f$ . Kinetic energy is generally NOT conserved (inelastic).

What if v approaches c?

$Use p = \gamma mv$ where $\gamma = 1/\sqrt{1-v^{2}/299792458^{2}}$ . At $v = 0.99c$ , $\gamma \approx 7$ .

1

Bullet vs walking person

Given ·

m:: 0.01
v:: 400

Find · p

Steps

Apply $p = mv$ with $m = 0.01\,\mathrm{kg}$ , $v = 400\,\mathrm{m/s}$
p_bullet $= 4 kg\cdot m/s$
Walker: $p = 70 \times 1.4 = 98 kg\cdot m/s$ — the walker has more momentum, but the bullet delivers it over microseconds → huge force

Result ·

p = mv = 0.01 \times 400 = 4 kg\cdot m/s

2

Truck on the highway

Given ·

m:: 8000
v:: 27

Find · p

Steps

Convert speed if needed $(27\,\mathrm{m/s} \approx 60\,\mathrm{mph})$
$p = mv = 8000 \times 27$
$p = 2.16 \times 10^{5} kg\cdot m/s$

Result ·

p = 8000 \times 27 = 216000 kg\cdot m/s

Newton's Second Law→Impulse-Momentum Theorem→Kinetic Energy→