Perfectly Inelastic Collision (1D)

Equivalent forms

v_f = (m_1 v_1 + m_2 v_2)/(m_1+m_2)

The final velocity is simply the center-of-mass velocity — a one-line consequence of momentum conservation.

Symbol	Name	SI unit	Dimension	Range
$v_f$	Final common velocityoutput Shared velocity of stuck bodies after collision	m/s	LT^-1	-50 – 50
$m1$	Mass 1 Mass of body 1	kg	M	0.01 – 20
$m2$	Mass 2 Mass of body 2	kg	M	0.01 – 20
$v1$	Velocity 1 Pre-collision velocity of m₁	m/s	LT^-1	-500 – 500
$v2$	Velocity 2 Pre-collision velocity of m₂	m/s	LT^-1	-50 – 50

Unit systems

Symbol	SI	CGS	Imperial
$v_f$	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, non-relativistic. Energy lost goes to heat/deformation/sound.

Dimensional analysis

v_f \to [LT^{-1}]

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

m/s on both sides.

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Discovery

Christiaan Huygens · 1656

Huygens' De Motu Corporum (posthumous 1703) framed collision problems via the conservation of momentum; inelastic outcomes followed once kinetic-energy loss was understood as heat.

Try this

A bullet embeds in a wooden block — how fast does the block move afterward?

A 0.02 kg bullet at 300 m/s hits a 2 kg block at rest and sticks. Find the common final velocity.

Research status: stable

Real-world applications

Ballistic pendulum (measuring bullet speed)
Car crash crumple-zone analysis
Catching a thrown ball
Asteroid impact crater models

Common misconceptions

Kinetic energy is conserved — it is NOT in inelastic collisions
Final velocity equals the average of $v_{1} and v_{2}$ — only when masses are equal
Inelastic $= no$ momentum conservation — momentum is always conserved (closed system)

Experimental verification

Ballistic pendulum: a bullet embeds in a swinging block; measure the swing height to deduce muzzle velocity. Standard since the 18th century.

Derivation

Momentum conservation:

m_{1}v_{1} + m_{2}v_{2} = (m_{1}+m_{2})v_f \Rightarrow v_f = (m_{1}v_{1}+m_{2}v_{2})/(m_{1}+m_{2})

.

Kinetic energy after

is \tfrac{1}{2}(m_{1}+m_{2})v_f^{2}

; the difference goes into deformation/heat.

Limiting cases

m_{1}

≪

m_{2}

,

v_{2} = 0

⟶

v_f \approx (m_{1}/m_{2}) v_{1}

Tiny projectile barely budges a heavy target.

m_{1} = m_{2}

⟶

v_f = (v_{1}+v_{2})/2

Equal masses → mean velocity.

v_{1} = v_{2}

⟶

v_f = v_{1}

No relative motion

\to no

real collision.

What if…

What if the collision is partially inelastic?

Use coefficient of restitution e ∈ (0,1); fully sticking is $e = 0$ .

Where does the lost kinetic energy go?

Heat, sound, plastic deformation, structural damage.

What if the bodies later separate?

By definition that's not perfectly inelastic anymore — you need a more general restitution model.

1

Bullet into block

Given ·

m1:: 0.02
m2:: 2
v1:: 300
v2:: 0

Find · v_f

Steps

Numerator: $6 kg\cdot m/s$
Denominator: 2.02 kg
$v_f \approx 2.97\,\mathrm{m/s}$ — KE drops from $900\,\mathrm{J} to \approx 8.9\,\mathrm{J}$ (most goes to heat/damage)

Result ·

v_f = (0.02\cdot 300 + 0)/(2.02) \approx 2.97\,\mathrm{m/s}

2

Two equal carts

Given ·

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

Find · v_f

Steps

Net momentum: $5 - 3 = 2 kg\cdot m/s$
Combined mass: 2 kg
$v_f = 1\,\mathrm{m/s}$ — KE drops from 17 J to 1 J

Result ·

v_f = 1\,\mathrm{m/s}

1D Elastic Collision (Final Velocities)→Coefficient of Restitution→Linear Momentum→Kinetic Energy→