Question

Superman leaps in front of Lois Lane to save her from a

volleyof bullets. In a 1 minute interval, an automatic weapon
fires160 bullets, each of mass 8.0 g, at450 m/s. The bullets strike
his mightychest, which has an area of 0.71m2. Find the
average force exerted on Superman's chestif the bullets bounce back
after an elastic, head-on collision.Express your anwser in
N.

Answer 1

`\Delta F = 13.52~N/m^2`

Step-by-step explanation:

We will use the relationship between impulse and momentum.

`\vec{J} = \Delta \vec{P}\\\vec{F}\Delta t = \vec{P}_2 - \vec{P}_1`

We are considering the change of momentum of Superman at the moment of collision.

`\Delta P_(Superman) = 0 - P_(bullets)`

`P_(bullets) = mv = (160* 8* 10^(-3))450 = 576`

`F*(60s) = 576\\F = 9.6 N`

This force is distributed throughout Kal-El’s chest. So, the average force is

`\Delta F = F/Area = 9.6 / 0.71 = 13.52~N/m^2`