1D Elastic Collision (Final Velocities)

Equivalent forms

v_2' = \frac{(m_2 - m_1) v_2 + 2 m_1 v_1}{m_1 + m_2}

Two conservation laws solved simultaneously yield a closed-form swap.

Symbol	Name	SI unit	Dimension	Range
$v1_prime$	Final velocity of mass 1output Velocity of m₁ after collision	m/s	LT^-1	-50 – 50
$m1$	Mass 1 Mass of body 1	kg	M	0.1 – 20
$m2$	Mass 2 Mass of body 2	kg	M	0.1 – 20
$v1$	Initial velocity 1 Pre-collision velocity of m₁	m/s	LT^-1	-20 – 20
$v2$	Initial velocity 2 Pre-collision velocity of m₂	m/s	LT^-1	-20 – 20

Unit systems

Symbol	SI	CGS	Imperial
$v1_prime$	m/s	cm/s	ft/s
$m1$	kg	g	slug
$m2$	kg	g	slug
$v1$	m/s	cm/s	ft/s
$v2$	m/s	cm/s	ft/s

Where it holds

Closed system, no external impulse during contact, point/rigid bodies; non-relativistic.

Dimensional analysis

v_{1}' \to [LT^{-1}]

(M\cdot LT^{-1})/M = [LT^{-1}]

Same units.

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Discovery

Christiaan Huygens / John Wallis / Christopher Wren · 1668

Royal Society memoirs of 1668–69 worked out hard-body collision rules — the first quantitative use of momentum conservation in physics.

Try this

Newton's cradle — why does the last ball fly off with exactly the speed of the first?

Two billiard balls of mass m₁ and m₂ collide head-on with initial velocities v₁ and v₂. Find the final velocities assuming a perfectly elastic collision.

Research status: stable

Real-world applications

Billiards / pool shot prediction
Particle physics scattering
Pinball machine simulation
Newton's cradle demonstrations

Common misconceptions

Both velocities reverse — only the relative velocity reverses
Equal masses bounce off equally — they swap velocities
All collisions are elastic — real-world ones rarely are

Experimental verification

Air-track gliders with magnetic bumpers reproduce these results

to \sim 1

% accuracy; Newton's cradle demos for centuries.

Derivation

Conservation of momentum:

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

'.

Conservation of KE:

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

.

Subtract to factor

(v-v')

terms; relative velocity reverses:

v_{1}'-v_{2}' = -(v_{1}-v_{2})

.

Solving the linear system yields the boxed formula.

Limiting cases

m_{1} = m_{2}

⟶

v_{1}' = v_{2}

,

v_{2}' = v_{1}

Velocities swap — like a Newton's cradle.

m_{1}

≫

m_{2}

⟶

v_{1}' \approx v_{1}

;

v_{2}' \approx 2v_{1} - v_{2}

Heavy ball plows through; light ball gets a slingshot.

m_{2} \to \infty

⟶

v_{1}' \to -v_{1}

;

v_{2}' \to 0

Light ball bounces back from an immovable wall.

What if…

What if it's 2D?

Decompose into components along the line of impact; same 1D rules apply along that axis.

What if there's a coefficient of restitution e < 1?

Relative velocity reverses by factor e: $v_{1}'-v_{2}' = -e(v_{1}-v_{2})$ .

What if I want kinetic energy conserved only approximately?

Use e in (0,1) — fully elastic is $e = 1$ , perfectly inelastic is $e = 0$ .

1

2 kg ball hits stationary 5 kg ball

Given ·

m1:: 2
m2:: 5
v1:: 10
v2:: 0

Find · v1_prime

Steps

Numerator: $(m_{1}-m_{2})v_{1} + 2m_{2}v_{2} = -3\cdot 10 + 0 = -30$
Denominator: $m_{1}+m_{2} = 7$
$v_{1}' \approx -4.29\,\mathrm{m/s}$ — light ball bounces back

Result ·

v_{1}' = (2-5)\cdot 10/(2+5) = -30/7 \approx -4.29\,\mathrm{m/s}

2

Equal-mass swap

Given ·

m1:: 1
m2:: 1
v1:: 5
v2:: -3

Find · v1_prime

Steps

$m_{1}=m_{2} so (m_{1}-m_{2})=0$
$v_{1}' = 2m_{2}v_{2}/(m_{1}+m_{2}) = v_{2} = -3$
$v_{2}' = v_{1} = 5$ — they swap

Result ·

v_{1}' = -3\,\mathrm{m/s}

,

v_{2}' = 5\,\mathrm{m/s}

Perfectly Inelastic Collision (1D)→Coefficient of Restitution→Linear Momentum→Kinetic Energy→