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Find the average force exerted on Superman's chest

Superman leaps in front of Lois Lane to save her from a volley of bullets. In a 1 minute interval, an automatic weapon fires 149 bullets, each of mass 8.0 g, at 430 m/s. The bullets strike his mighty chest, which has an area of 0.74 m2. Find the average force exerted on Superman's chest if the bullets bounce back after an elastic, head-on collision.


i dont c what the area has to do with it i tried multiplyin

number of bullets x mass in kg x velocity / by the time in seconds 60
 

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D H
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Almost! That is the force if the collisions were purely inelastic (i.e., if the bullets stuck to Superman). These are elastic collisions.
 
then what would it be? the equation cant figure it out
 
D H
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What happens in an elastic collision?
 
well the v is halved right? so 1/2 mv2
 
D H
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No. What happens in an elastic collision?
 
momentum is conserved??? like KE0=KEf
 
D H
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You are having conceptual issues, so please take this one step at a time. What happens in an elastic collision?
 
im honestly clueless at this point... energy is conserved?
 
D H
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What happens to the bullets? What is their velocity (NOT SPEED) before and after the collision?
 
oh its in the opposite direction
 
D H
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OK. So how much has the momentum changed? Remember that momentum, like velocity, is a vector quantity.
 
twice as much?
 
D H
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That's right. I don't like the question mark, however. It means you are guessing.
 
no im not, coz it acts like distance vs displacement kinda, so i still dont get whats wrong with the equation
 
D H
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Spell things out then. Draw pictures. Momentum is conserved here, so the momentum changes incurred the bullets also happen to Superman (but in exactly the opposite sense, of course). So what exactly happens to each bullet? Be specific. Don't say "twice as much". (Twice as much as what?)

I have an early meeting tomorrow so I need to go to bed. Good luck!
 
never mind i got it!
 

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