Conservation of Momentum — Elastic & Inelastic Collisions

Here is the one-sentence verdict: in every collision of an isolated system, total momentum is conserved, but kinetic energy is conserved only if the collision is elastic. Get that distinction right and most collision problems fall into place.

Elastic, Inelastic, Perfectly Inelastic At A Glance

Momentum is a vector. For an object of constant mass in introductory mechanics,

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

so direction matters: in one dimension, rightward is often positive and leftward negative. For a two-object system in one dimension, the momentum statement is

$$m_1 v_{1,i} + m_2 v_{2,i} = m_1 v_{1,f} + m_2 v_{2,f}$$

valid when the system is isolated well enough over the collision interval. The three collision types differ only in what else is conserved:

Collision type	Total momentum	Total kinetic energy	Objects after impact
Elastic	Conserved	Conserved	Separate, generally different velocities
Inelastic	Conserved	Not conserved	Separate, some KE lost to heat, sound, deformation
Perfectly inelastic	Conserved	Not conserved	Stick together, one shared velocity

For an elastic collision, the extra kinetic-energy equation is

$$\frac{1}{2}m_1 v_{1,i}^2 + \frac{1}{2}m_2 v_{2,i}^2 = \frac{1}{2}m_1 v_{1,f}^2 + \frac{1}{2}m_2 v_{2,f}^2$$

When To Use Which

Momentum is not conserved for a single object on its own; it is conserved for the total system when the net external impulse is negligible. During a short collision the forces between the colliding objects are internal, so they rearrange momentum between the objects without changing the total. So always start from momentum conservation. Add the kinetic-energy equation only when the problem states or implies the collision is elastic. If the objects stick together, treat it as perfectly inelastic and use a single final velocity.

Worked Example: A Perfectly Inelastic Collision

A $2\ \mathrm{kg}$ cart moves right at $4\ \mathrm{m/s}$ and collides with a $1\ \mathrm{kg}$ cart at rest. The carts stick together. Find their final velocity.

This is perfectly inelastic, so momentum is conserved and both carts share one final velocity $v_f$. Before the collision,

$$p_i = (2)(4) + (1)(0) = 8\ \mathrm{kg \cdot m/s}$$

After, the combined mass is $3\ \mathrm{kg}$, so

$$p_f = (3)v_f$$

Setting initial equal to final,

$$8 = 3v_f$$ $$v_f = \frac{8}{3}\ \mathrm{m/s}$$

about $2.67\ \mathrm{m/s}$ to the right.

Now compare kinetic energy before and after:

$$K_i = \frac{1}{2}(2)(4^2) = 16\ \mathrm{J}$$ $$K_f = \frac{1}{2}(3)\left(\frac{8}{3}\right)^2 = \frac{32}{3}\ \mathrm{J} \approx 10.67\ \mathrm{J}$$

Momentum stays the same, but kinetic energy drops from $16\ \mathrm{J}$ to about $10.67\ \mathrm{J}$. That missing kinetic energy is transformed into thermal energy, sound, or internal deformation. It does not violate momentum conservation, because the two laws have different conditions and apply to different quantities.

Try It With A Twist

Keep the same masses, but let the second cart move left before the collision. Recompute the total initial momentum using a negative sign for the leftward cart, then solve for the shared final velocity again. The sign bookkeeping is the whole point: opposite-moving objects partially cancel instead of adding.

Exam Traps And Confusion Points

Treating momentum like a plain number. Momentum has direction. In one dimension, a sign convention is essential; in two or three dimensions, conserve momentum component by component.

Forgetting to define the system. If you track only one object in a collision, its momentum usually changes. The conservation law applies to the total momentum of the isolated system.

Using kinetic-energy conservation for every collision. That is valid only for elastic collisions. For inelastic ones, use momentum conservation first.

Ignoring external impulse. If outside forces matter over the time interval, total momentum for your chosen system may not stay constant. The isolation condition is part of the rule, not optional.

Where Conservation Of Momentum Is Used

It appears in collision analysis, recoil problems, explosions, particle interactions, and lab-cart experiments. The same principle explains why a firearm recoils, why billiard balls exchange motion, and why two objects that stick together move more slowly than the faster object did before impact.