# 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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