We are asked to determine the force exerted on a football when it is brought to rest in seconds. To do that we will use the formula for impulse:
[tex]I=p_2-p_1[/tex]Where:
[tex]\begin{gathered} I=\text{ impulse} \\ p_2,p_1=\text{ final and initial momentum} \end{gathered}[/tex]Since the ball is brought to rest this means that the final momentum is 0:
[tex]I=-p_1[/tex]the formula for impulse is given by:
[tex]I=Ft[/tex]Subsituting in the formula:
[tex]Ft=-p_2[/tex]The momentum is given by:
[tex]p_2=-mv_0[/tex]Where:
[tex]\begin{gathered} m=\text{ mass} \\ v_0=\text{ initial velocity} \end{gathered}[/tex]Substituting:
[tex]\begin{gathered} Ft=-(-mv_0) \\ Ft=mv_0 \end{gathered}[/tex]Now, we divide both sides by the time:
[tex]F=\frac{mv_0}{t}[/tex]Now, we plug in the values:
[tex]F=\frac{(0.43kg)(18\frac{m}{s})}{0.052s}[/tex]Solving the operations:
[tex]F=149N[/tex]Therefore, the force exerted was 149N.
