Physics & Mechanics

Conservation of Momentum Calculator

Calculate final velocities after a simple 1D collision. Proves that the total momentum before the collision equals total momentum after.

kg
m/s
kg
m/s
Total Initial Momentum
20
Final Velocity (Perfectly Inelastic)0.8 m/s

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The Universe's Accounting System

One of the most profound and unbreakable laws of the universe is the Law of Conservation of Momentum.

It states that in a closed, isolated system (meaning no external forces like friction or air resistance are interfering), the total momentum of all objects involved must remain completely constant, regardless of what they do to each other. They can crash, bounce, stick together, or even explode—the total momentum before the event will perfectly equal the total momentum after the event.

Momentum ($p$) is simply an object's mass multiplied by its velocity ($p = mv$).

Perfectly Inelastic Collisions

This calculator models a specific, common scenario: a perfectly inelastic collision in one dimension.

In this scenario, two objects crash into each other and, instead of bouncing off, they mechanically deform, lock together, and continue moving as a single, combined, heavier mass. Examples include two train cars coupling together, a bullet embedding into a block of wood, or two football players tackling each other.

Because the two objects stick together, they share a single, identical final velocity.

The Formula

m1v1+m2v2=(m1+m2)vf\begin{aligned} m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f \end{aligned}

Where:
m1,m2m_1, m_2=
Masses of the two objects
v1,v2v_1, v_2=
Initial velocities before impact
vfv_f=
Final combined velocity after they stick together

Example Calculation

Imagine a heavy freight train car (Mass 1 = $10,000 , \text{kg}$) rolling down the track at $5 , \text{m/s}$. It crashes into and couples with a stationary car (Mass 2 = $15,000 , \text{kg}$, Velocity 2 = $0 , \text{m/s}$). How fast do the two linked cars move together after the crash?

  1. Total Initial Momentum: $(10000 \cdot 5) + (15000 \cdot 0) = 50,000 + 0 = \mathbf{50,000 , \text{kg}\cdot\text{m/s}}$.
  2. Combined Mass: $10000 + 15000 = 25,000 , \text{kg}$.
  3. Final Velocity: $v_f = \frac{50000}{25000} = \mathbf{2.0 , \text{m/s}}$.

To conserve the $50,000 , \text{kg}\cdot\text{m/s}$ of total momentum, the newly combined, much heavier $25,000\text{kg}$ object must roll at exactly $2.0 , \text{m/s}$.

Frequently Asked Questions

No. This is a critical distinction. In a perfectly inelastic collision, momentum is perfectly conserved, but massive amounts of kinetic energy are 'lost'. Where does it go? It is transformed into the heat, sound, and mechanical deformation required to crush the objects together.

Because velocity is a vector, direction matters immensely. If Car 1 is moving right at 5 m/s, its velocity is +5. If Car 2 is moving left to crash into it, its velocity must be entered as a negative number (e.g., -5). The total momentum will be the mathematical sum of the positive and negative momentums.

Yes, an explosion is simply a reverse inelastic collision. A stationary bomb has a total momentum of exactly 0. When it explodes, fragments fly in all directions. If you measured the mass and velocity vectors of every single tiny fragment and added them all together, they would perfectly cancel out to exactly 0.