Linear momentum tells us both something about the object and something about its motion. The linear momentum of a particle of mass m moving with velocity v is defined to be:

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

**Components (rectangular coordinates):**

p_{x} = mv_{x}

p_{y} = mv_{y}

p_{z} = mv_{z}

The concept of momentum is very useful in treating problems involving collisions and for analyzing rocket propulsion.

Momentum is a vector, and its direction is along v.

Dimension: ML/T

Unit: kg.m/s

### Newton’s Second Law Restated

If $m$ is a cosntant,

$$\begin{aligned} \frac{d \vec{p}}{dt} &= \frac{d}{dt} \left( m \vec{v} \right) \\ &= m \frac{d \vec{v}}{dt} \\ &= m \vec{a} \end{aligned}$$

In terms of linear momentum, Newton’s second law can be written as:

$$\vec{F} = \frac{d \vec{p}}{dt}$$

The above equation states that the time rate of change of linear momentum of a particle is equal to the resultant force acting on the particle.

Knowing that the change in momentum caused by a force is useful in solving some problems.

Assume that a single force $\vec{F}$ acts on a particle and that this force varies with time.

$$\begin{aligned} \vec{F} &= \frac{d\vec{p}}{dt} \\ d \vec{p} &= \vec{F} \, dt \end{aligned}$$

Integrating the above, we have:

$$\begin{aligned} \int \, d\vec{p} &= \int \vec{F} \, dt \\ \Delta \vec{p} &= \int\limits_{t_{i}}^{t_{f}} \vec{F} \, dt \\ \vec{p}_{f}-\vec{p}_{i} &= \int\limits_{t_{i}}^{t_{f}} \vec{F} \, dt \end{aligned}$$

### Impulse

The impulse of a force $\vec{F}\left( t \right)$ acting on a particle from time $t_{i}$ to $t_{f}$ is:

$$\vec{J} = \vec{p}_{f}-\vec{p}_{i} = \int\limits_{t_{i}}^{t_{f}} \vec{F} \, dt$$

Impulse of a force $\vec{F}$ acting on a particle equals the change in momentum of the particle caused by that force.

Impulse is a vector and has the same dimensions as momentum $\left( \frac{\text{ML}}{\text{T}} \right)$.

### Impulsive Force

Consider a non-constant force acting on an object, we can find the impulse by find the average force, $\vec{F}_{av}$:

$$\vec{F}_{av} = \frac{1}{\Delta t} \int\limits_{t_{1}}^{t_{2}} \left( \sum \vec{F} \right) \, dt$$

Hence, we have:

$$\begin{aligned} \vec{J} &= \Delta \vec{p} \\ &= \left( \vec{F}_{av} \right) \left( t_{2}-t_{1} \right) \\ &= \vec{F}_{av} \Delta t \end{aligned}$$

We can use impulse approximation to solve problems. We assume that the impulsive force acts for a short time but is much larger than any other force present and very little motion takes place during this time. (Usually also neglect effects of external forces during this time)

Using $\vec{p}_{i}$ and $\vec{p}_{f}$ to represent the momenta immediately before and after the collision. The impulse, $\vec{p}_{f}-\vec{p}_{i}$ is the same if the area under the $F-t$ curves are the same. The impulsive force is bigger if it acts over a short time and is smaller if it acts over a longer time.

### Conservation Of Linear Momentum

If there are no external forces acting on an isolated system of particles:

$$\begin{aligned} \sum \vec{F} &= 0 \\ \sum \frac{d \vec{p}}{dt} &= 0 \\ \sum \vec{p} &= \text{constant} \\ \sum \vec{p}_{i} &= \sum \vec{p}_{f} \end{aligned}$$

The total momentum of an isolated system at all times equals its initial momentum.

The total momentum of an isolated system remains constant if there are no external forces present.