Example on Analyzing a Force Graph

Analyse the Force vs. time graph provided and answer the following questions:

• What is the impulse for the interval \(\text{0}\) \(\text{s}\) to \(\text{3}\) \(\text{s}\)?
• What is the impulse for the interval \(\text{3}\) \(\text{s}\) to \(\text{6}\) \(\text{s}\)?
• What is the change in momentum for the interval \(\text{0}\) \(\text{s}\) to \(\text{6}\) \(\text{s}\)?
• What is the impulse for the interval \(\text{6}\) \(\text{s}\) to \(\text{20}\) \(\text{s}\)?
• What is the impulse for the interval \(\text{0}\) \(\text{s}\) to \(\text{20}\) \(\text{s}\)?

Step 1: Identify what information is given and what is being asked for

A graph of force versus time is provided. We are asked to determine both impulseand change in momentum from it.

We know that the area under the graph is the impulse and we can relate impulse tochange in momentum with the impulse-momentum theorem.

We need to calculate the area under the graph for the various intervals to determine impulseand then work from there.

Step 2: Impulse for interval \(\text{0}\) \(\text{s}\) to \(\text{3}\) \(\text{s}\)

We need to calculate the area of the shaded portion under the graph. This is a triangle witha base of \(\text{3}\) \(\text{s}\) and a height of\(\text{3}\) \(\text{N}\) therefore:

\begin{align*}\text{Impulse} &= \frac{1}{2}bh \\& = \frac{1}{2}(3)(3) \\& = \text{4.5}\text{ N·s}\end{align*}

The impulse is \(\text{4.5}\) \(\text{N·s}\) in the positive direction.

Step 2: Impulse for interval \(\text{3}\) \(\text{s}\) to \(\text{6}\) \(\text{s}\)

We need to calculate the area of the shaded portion under the graph. This is a triangle witha base of \(\text{3}\) \(\text{s}\) and a height of\(-\text{3}\) \(\text{N}\). Note that the force has anegative value so is pointing in the negative direction.

\begin{align*}\text{Impulse} &= \frac{1}{2}bh \\& = \frac{1}{2}(3)(-3) \\& = -\text{4.5}\text{ N·s}\end{align*}

The impulse is \(\text{4.5}\) \(\text{N·s}\) in the negative direction.

Step 3: What is the change in momentum for the interval \(\text{0}\) \(\text{s}\) to \(\text{6}\) \(\text{s}\)

From the impulse-momentum theorem we know that that impulse is equal to the change inmomentum. We have worked out the impulse for the two sub-intervals making up\(\text{0}\) \(\text{s}\) to\(\text{6}\) \(\text{s}\). We can sum them to find theimpulse for the total interval:

\begin{align*}\text{impulse}_{0-6} & = \text{impulse}_{0-3} + \text{impulse}_{3-6} \\&= (\text{4.5})+(-\text{4.5}) \\&= \text{0}\text{ N·s}\end{align*}

The positive impulse in the first 3 seconds is exactly opposite to the impulse in the second3 second interval making the total impulse for the first 6 seconds zero:

\[\text{impulse}_{0-6} = \text{0}\text{ N·s}\]

From the impulse-momentum theorem we know that:

\[\Delta \vec{p} = \text{impulse} = \text{0}\text{ N·s}\]

Step 4: What is the impulse for the interval \(\text{6}\) \(\text{s}\) to \(\text{20}\) \(\text{s}\)

We need to calculate the area of the shaded portion under the graph.This is divided into two areas, \(\text{6}\) \(\text{s}\) to \(\text{12}\) \(\text{s}\) and \(\text{12}\) \(\text{s}\) to \(\text{20}\) \(\text{s}\), which we need to sum to get thetotal impulse.

\begin{align*}\text{Impulse}_{6-12} &= (6)(-3) \\& = -\text{18}\text{ N·s}\end{align*}\begin{align*}\text{Impulse}_{12-20} &= (8)(2) \\& = +\text{16}\text{ N·s}\end{align*}

The total impulse is the sum of the two:

\begin{align*}\text{Impulse}_{6-20} &= \text{Impulse}_{6-12} + \text{Impulse}_{12-20} \\&= (-18) + (16) \\&= -\text{2}\text{ N·s}\end{align*}

The impulse is \(\text{2}\) \(\text{N·s}\) in the negative direction.

Step 5: What is the impulse of the entire period

The impulse is \(\text{2}\) \(\text{N·s}\) in the negative direction.