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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 :

Answer:

[tex]\Delta F = 13.52~N/m^2[/tex]

Explanation:

We will use the relationship between impulse and momentum.

[tex]\vec{J} = \Delta \vec{P}\\\vec{F}\Delta t = \vec{P}_2 - \vec{P}_1[/tex]

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

[tex]\Delta P_{Superman} = 0 - P_{bullets}[/tex]

[tex]P_{bullets} = mv = (160\times 8\times 10^{-3})450 = 576[/tex]

[tex]F*(60s) = 576\\F = 9.6 N[/tex]

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

[tex]\Delta F = F/Area = 9.6 / 0.71 = 13.52~N/m^2[/tex]

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