---
title: Linear Momentum and Collisions
date: 2024-10-27
author: Mars Ultor
---

# Linear Momentum, Newton's second law rewritten

> [!NOTE]
> This note will only discuss *linear* momentum, impulse, and other collision parameters, not *angular* momentum.

Momentum is defined as  the product of mass times the velocity of an object. This is written mathematically as:
$$
\vec{p}=m\vec{v}
$$
Velocity is a vector, therefore momentum is also a vector. There is also no SI unit for momentum. In its base, it's written as kg$\cdot$m$/$s, although you may sometimes see it as N s, or Newton seconds. What this can be interpreted as is that the momentum of an object is a measure of its tendency to move. Furthermore, a force must be exerted  to change the momentum of an object, therefore **momentum is always conserved.** Consequently, Newton's second law of motion can be rewritten as:

>   The rate of change of momentum of an object is equal to the net force applied on it.

Put mathematically:
$$
\sum{\vec{F}}=\frac{\Delta \vec{p}}{\Delta t}=\frac{\mathrm{d}\vec{p}}{\mathrm{d}t}
$$
If you don't understand the calculus, come to understand it or, don't. In the above equation, the following observations can be made:

-   $\sum{\vec{F}}$ is the net force applied on the object, as a vector. 
-   $\Delta\vec{p}$ or $\mathrm{d}\vec{p}$ is the differential change of the momentum resulting so.

Keeping in line with the previous definition for force, it is possible to derive the old equation for it using this one. This particular note however, will leave it as an exercise for the reader to do so.

> [!NOTE] Example 1
>   A tennis ball leaves the racket with the speed of $55 \mathrm{m}/\mathrm{s}$. If the tennis ball has a mass of $0.060 \mathrm{kg}$, and remains in contact with the racket for $4\mathrm{ms}$, estimate the average for of the ball. Would this force be large enough to lift a $60$-kg person?

>   [!faq]- Answer
>   $||\vec{F_{\text{net}}}||\approx 800\mathrm{N}$. It does, if exerted for a longer time.

## Impulse 

Impulse is simply the differntial of momentum such that:
$$
J=\Delta\vec{p}=\mathrm{d}\vec{p}=\vec{F}\Delta t=\vec{F}\mathrm{d}\vec{t}
$$
Like many other things in physics, this is just another quantity to keep track of.

# Conservation of Energy in Collisions

When collisions are perfectly elastic, meaning all energy in motion continues to remain in the form of kinetic energy. Mathematically as long as the mass does not change:
$$
\tfrac{1}{2}m_Av^2_A+\tfrac{1}{2}m_Bv^2_B=\tfrac{1}{2}m_Av^{'2}_A+\tfrac{1}{2}m_Bv^{'2}_B
$$

If the collision is not perfectly elastic, it is called an **inelastic collision** meaning energy is allowed to change forms, to sound, heat, and other forms. In that case:
$$
\tfrac{1}{2}m_Av^2_A+\tfrac{1}{2}m_Bv^2_B=\tfrac{1}{2}m_Av^{'2}_A+\tfrac{1}{2}m_Bv^{'2}_B+\text{nonconservative energy losses}
$$
This does not mean, however, that the energy is lost. It continues to propagate and increase entropy in a closed system.

