Figure 9-26 shows graphs of force magnitude versus time for a body involved in a collision. Rank the graphs according to the magnitude of the impulse on the body, greatest first.

1. For Case (a), the Nature of the graph is rectangular with Height = $2{\mathrm{F}}_0$and Base =$6{\mathrm{t}}_0$
2. For Case (b), the Nature of the graph is rectangular with Height =$4{\mathrm{F}}_0$ and Base =$3{\mathrm{t}}_0$
3. For Case (c), the Nature of the graph is triangular with Height = $2{\mathrm{F}}_0$and Base =$12{\mathrm{t}}_0$

The magnitude of impulse on an object which is involved in a collision is equal to the force acting on it multiplied by the small-time interval of action of the force. Thus, the graph of Force versus time is given. So, we calculate the area under the curve to get the magnitude of the impulse.

Formulae:

The area of the rectangular graph, $\mathrm{A}=\mathrm{Height}\times \mathrm{Base}$ (1)

The area of the triangular graph, $\mathrm{A}=\frac{1}{2}\times \mathrm{Height}\times \mathrm{Base}$ (2)

For Case (a)

The area under the curve will be given using equation (1) as follows:

$$\begin{gathered} {\mathrm{J}}_1=\left(2\mathrm{F}\right)\times \left(6{\mathrm{t}}_0\right) \\ =12{\mathrm{F}}_0{\mathrm{t}}_0 \end{gathered}$$

For Case (b)

The area under the curve will be given using equation (1) as follows:

$$\begin{gathered} {\mathrm{J}}_2=\left(4{\mathrm{F}}_0\right)\times \left(3{\mathrm{t}}_0\right) \\ =12{\mathrm{F}}_0{\mathrm{t}}_0 \end{gathered}$$

For Case (c)

The area under the curve will be given using equation (2) as follows:

$$\begin{gathered} {\mathrm{J}}_3=\frac{1}{2}\times \left(2{\mathrm{F}}_0\right)\times \left(12{\mathrm{t}}_0\right) \\ =12{\mathrm{F}}_0{\mathrm{t}}_0 \end{gathered}$$

As we can see that the magnitude for the impulse for all three cases is the same, so we can say that rank for all the graphs is the same.

The rank of the graphs according to the magnitude of the impulse on the body will be

(a) = (b) = (c) , greatest first