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Physics

Conservation of Linear Momentum

The total momentum of an isolated system remains constant.

\sum p_{ini t ia l} = \sum p_{f ina l}

Study tools

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Source & review

Difficulty

Intermediate

Read time

6 min

Prerequisites

Basic Physics

Review statusNeeds review

Source

FormuLab initial formula library

Initial content draft pending verification against authoritative course or textbook sources.

Definition

Understanding the core concept

Imagine throwing a ball; the ball goes forward, and you feel a slight push backward. This is because momentum, a measure of an object's mass and velocity, can't just appear or disappear. In any closed system where no external forces act, the total 'oomph' (momentum) before an event, like a collision or explosion, must equal the total 'oomph' after the event.

Variables & Units

Understanding each component

Symbol	Meaning	Units
$\sum p_{ini t ia l}$	Total initial momentum of the system	$kg \cdot m/s$
$\sum p_{f ina l}$	Total final momentum of the system	$kg \cdot m/s$
$p$	Momentum (where $\mathbf{p} = m\mathbf{v}$)	$kg \cdot m/s$
$m$	Mass of an object	$kg$
$v$	Velocity of an object	$m/s$

Real-World Applications

Where this formula is used in practice

Collisions and Impacts

Used to analyze the velocities of objects before and after a collision, from billiard balls to car crashes.

Rocket Propulsion

Explains how rockets generate thrust by expelling high-velocity exhaust gases, conserving the total momentum of the rocket-fuel system.

Recoil of Firearms

Calculates the backward velocity of a gun after firing a projectile, ensuring momentum is conserved.

Astronomy

Helps understand orbital mechanics and predict the motion of celestial bodies interacting through gravitational forces.

Worked Example

Step-by-step calculation with real numbers

Problem

A 2 kg ball moving at 5 m/s collides head-on with a stationary 3 kg ball. After the collision, the 2 kg ball moves backward at 1 m/s. What is the final velocity of the 3 kg ball?

Given

m_{1} = 2 kg

v_{1 i} = 5 m/s

m_{2} = 3 kg

v_{2 i} = 0 m/s

v_{1 f} = - 1 m/s

Solution

m_{1} v_{1 i} + m_{2} v_{2 i} = m_{1} v_{1 f} + m_{2} v_{2 f}

(2 kg) (5 m/s) + (3 kg) (0 m/s) = (2 kg) (- 1 m/s) + (3 kg) v_{2 f}

10 kg \cdot m/s + 0 = - 2 kg \cdot m/s + (3 kg) v_{2 f}

10 = - 2 + 3 v_{2 f}

10 + 2 = 3 v_{2 f}

12 = 3 v_{2 f}

v_{2 f} = \frac{12}{3}

v_{2 f} = 4 m/s

Final Answer

v_{2 f} = 4 m/s

Formula Guide

Info

DifficultyIntermediate

Read Time6 min

PrerequisitesBasic Physics

Tools

Source & review

StatusNeeds review

Last reviewedPending

Source

FormuLab initial formula library

Initial content draft pending verification against authoritative course or textbook sources.

Assumptions

• Use SI units unless the formula or course context states otherwise.
• Check notation, sign conventions, and applicability against your course material.
• Confirm variables and assumptions before using the formula in graded or professional work.