Question

A 0.50 kg football is thrown with a velocity of 15 m/s to the right. A stationary receiver catches the ball and brings it to rest in 0.020 s. What is the force exerted on the receiver?

Answer 1

`F_"avg"=375"N"`

Explanation:

The average force exerted can be calculated using the following equation:

`F_"avg"=(Deltap)/(Deltat)`

• Where `Deltap` is the change in momentum and `Deltat` is the change in time.

We are given the following information:

• `|->m=0.50"kg"`
• `|->v_i=15" m"//"s to the right"`
• `|->v_f=0`
• `|->Deltat=0.020"s"`

Momentum is given by `vecp=mvecv`.

Therefore, the change in momentum `Deltap` is:

`Deltap=p_f-p_i=mv_f-mv_i`

We can calculate `Deltap`:

`Deltap=(0.50"kg")(0-15"m"//"s")`

`=>-7.5" kgm"//"s"`

• Note that the negative sign indicates a decrease in momentum.

`F_"avg"=(-7.5" kgm"//"s")/(0.020"s")`

`=color(darkblue)(375"N")`