Question

A 0.43 kg football traveling at a speed of 18.0 m/s is caught and brought to rest by the receiver in 0.052 s. What is the magnitude of the force that was applied to the football?0.402 N41.7 N149 N2680 N

Answer 1

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:

`I=p_2-p_1`

Where:

`\begin{gathered} I=\text{ impulse} \\ p_2,p_1=\text{ final and initial momentum} \end{gathered}`

Since the ball is brought to rest this means that the final momentum is 0:

`I=-p_1`

the formula for impulse is given by:

`I=Ft`

Subsituting in the formula:

`Ft=-p_2`

The momentum is given by:

`p_2=-mv_0`

Where:

`\begin{gathered} m=\text{ mass} \\ v_0=\text{ initial velocity} \end{gathered}`

Substituting:

`\begin{gathered} Ft=-(-mv_0) \\ Ft=mv_0 \end{gathered}`

Now, we divide both sides by the time:

`F=(mv_0)/(t)`

Now, we plug in the values:

`F=((0.43kg)(18(m)/(s)))/(0.052s)`

Solving the operations:

`F=149N`

Therefore, the force exerted was 149N.