Question

Figure 9-53 shows an approximate plot of force magnitude F versus time t during the collision of a 58 gSuperball with a wall. The initial velocity of the ball is 34 m/sperpendicular to the wall; the ball rebounds directly back with approximately the same speed, also perpendicular to the wall. What is${\mathit{F}}_{\mathbf{m}\mathbf{a}\mathbf{x}}$the maximum magnitude of the force on the ball from the wall during the collision?

Fig. 9-53

Short answer

The maximum magnitude of the force on the ball from the wall is $F_{max}\mathrm{is}9.9\times {10}^2\mathrm{N}$.

Step-by-step solution

Understanding the given information

1. The mass of the superball,$m\mathrm{is}58\times {10}^{-3}\mathrm{kg}$.
2. The initial speed of the ball= final speed of the ball= 34 m/s.

Concept and formula used in the given question

The graphical presentation of force and time indicates the way an impulse act on an object. Using the graph, we can determine the momentum change or force in a given situation.

$$\begin{gathered} \overrightarrow{\mathbf{J}}\mathbf{=}\mathbf{∆}\overrightarrow{\mathbf{p}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{m}\mathbf{∆}\overrightarrow{\mathbf{v}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}{\mathbf{F}}_{\mathbf{avg}}\mathbf{∆}\mathbf{t}\mathbf{} \end{gathered}$$

Calculation for the maximum magnitude of the force on the ball from the wall

The impulse – linear momentum theorem gives

$$\begin{gathered} J=m∆v \\ =F_{avg}∆t \end{gathered}$$

The impulse can be calculated as

$$\begin{gathered} J=m∆v \\ =58\times {10}^{-3}\times (34-(-34\left)\right) \\ =58\times {10}^{-3}\times 68 \\ =3.94N.s \end{gathered}$$

From the graph we can see that

role="math" localid="1661240669180" $$\begin{gathered} ∆p=areaunderthecurvefromt=0tot=6.0s \\ =areaoftrapezoid \\ =\left(\frac{1}{2}\times (a+b)\times h\right) \\ =\left(\frac{1}{2}\times (2+6)\times {10}^{-3}\times F_{max}\right) \end{gathered}$$

Using the two equations of the $∆p$, we get

$$\begin{gathered} 3.94=\left(\frac{1}{2}\times (2+6)\times {10}^{-3}\times F_{max}\right) \\ {F}_{max}=\frac{3.94}{4\times {10}^{-3}} \\ =9.9\times {10}^2\mathrm{N} \end{gathered}$$