# Calculating impulse of a constant force

**Impulse** is the change in momentum of an object. Impulse is interchangeably symbolized as change in momentum, $\Delta p$, or simply $\vec{J}$. This falls out of the mathematical definition of impulse:

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vec{J}=\Delta \vec{p}.$

In investigations of the processes that cause an object's momentum to change, such as cars colliding, or a ball bouncing, impulse is the path to studying the forces involved in the change.

As it is an expression of momentum, an acceptable unit for impulse is the $\text{Kg m s}^{-1}$, however common units for impulse include the Newton-second ($\text{Ns}$).

## Impulse in terms of force

When a force is applied to an object, no matter what the state of that object's motion is, acceleration will occur. This is in fact Newton's second law, $F = ma$. The equation for impulse, $\Delta p,$ can in fact be derived from this law:

$F=ma, ~a=\frac{\Delta v}{\Delta t} \Rightarrow F=m \frac{\Delta v}{\Delta t}=\frac{m\Delta v}{\Delta t}.$

Since $m \Delta v = \Delta p$ and $F=\frac{\Delta p}{\Delta t}$ see below, we have the following theorem:

$\Delta p = F \Delta t,$

where $\Delta p$ represents impulse, $F$ represents applied force, and $\Delta t$ represents the time over which the force is applied.

## A force of $10N$ was applied to an object for $5s$. Calculate the impulse experienced by the object.

Let $\Delta p$ represent impulse, $F$ force, and $\Delta t$ the time that the force acted. Then

$\Delta p = F \Delta t=10\times 5= 50 \text{ (Ns)}.$

## Restatement of Newton's Second Law

Newton's second law states that the acceleration of an object due to a net force is in the same direction as that force and inversely proportional to the mass of the object, or, mathematically, $\vec{F}=m\vec{a}$.

Since an acceleration causes a change in velocity, a reformulation of Newton's second law is possible in terms of velocity.

$F=ma=m \frac{\Delta v}{\Delta t} = \frac{m\Delta v}{\Delta t} = \frac{\Delta p}{\Delta t}$

$F=\frac{\Delta p}{\Delta t}$

The equality of the two formulations of Newton's second law is best demonstrated through a simple example.

A 2 $\text{kg}$ ball is dropped near the surface of a planet. If $v(t=5\text{s})=10 \frac{\text(m)}{\text{s}},$ find the average force on the ball throughout its fall.

Method 1: $F=ma$The problem provides $m=2\text{kg}$. The acceleration is not given, but since the motion takes place "near the surface of a planet", it is assumed to be constant. Hence, find it using one dimensional kinematics:

$a=\dfrac{\Delta v}{\Delta t} = \dfrac{v_f-v_i}{\Delta t}=\dfrac{10\dfrac{\text{m}}{\text{s}} - 0\dfrac{\text{m}}{\text{s}}} {5 \text{s}}=2\dfrac{\text{m}}{\text{s}^2}.$

Finally, evaluate Newton's second law:

$F=ma=(2\text{kg})(2\dfrac{\text{m}}{\text{s}^2})=4\text{N}.$

Method 2:$F=\frac{\Delta p}{\Delta t}$$F=\frac{\Delta p}{\Delta t}=\dfrac{p_f-p_i}{\Delta t}.$

Since $p=mv$,

$F=\dfrac{mv_f-mv_i}{\Delta t}=m\dfrac{v_f-v_i}{\Delta t}=(2\text{kg})(\dfrac{10\frac{\text{m}}{\text{s}} - 0\frac{\text{m}}{\text{s}}} {5 \text{s}}) =4\text{N}.$

An object which was initially at rest has a momentum of $10\text{ Kg ms}^{-1}$ after being influenced by a force for $2s$. Calculate the magnitude of the force that acted upon the object.

The $\text{ Kg ms}^{-1}$ is equivalent to the $Ns$, so substitute that value, along with the provided time, into $F=\frac{\Delta p}{\Delta t}$ to obtain:

$F=\dfrac{\Delta p}{\Delta t}=\dfrac{p_f-p_i}{\Delta t}=\dfrac{10-0}{2}=5\text{N}.$

A 1 $\text{kg}$ basketball is dropped from a height of 5 $\text{m}$, bounces off the ground, and returns to a maximum height of 1.8 $\text{m}$. Find the magnitude of the average force (in $\text{N}$) between the ball and the ground if the ball is in contact with the ground for 0.1 s. (Use $g=10 \text{m}/\text{s}^2$ for the acceleration due to gravity.)

**Cite as:**Calculating impulse of a constant force.

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