What is the Average Force of Bullets Bouncing off Superman's Chest?

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The average force exerted by bullets bouncing off Superman's chest can be calculated using Newton's 2nd and 3rd laws. Given that each bullet has a mass of 1.8 g (0.0018 kg) and travels at a speed of 460 m/s, with a rate of 200 bullets per minute (approximately 3.33 bullets per second), the total change in momentum per second is 1.7 kg·m/s for one bullet. Therefore, the average force exerted by the stream of bullets on Superman's chest is 5.67 N, calculated by multiplying the total change in momentum per second by the number of bullets hitting per second.

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I can't figure this one out. Please help.

It is well known that bullets fired at Superman simply bounce off his chest. Suppose that a gangster sprays Superman's chest with bullets of mass m = 1.8 g at a rate of R = 200 bullets/min. The speed of each bullet is v = 460 m/s. Suppose the bullets rebound with no change in speed. What is the average force exerted by the stream of bullets on Superman's chest?


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Use Newton's 2nd and 3rd laws. Force equals the rate of change of momentum.
 
Well I have tried using Impulse (change in p = F * time) and I am not getting it right. The change in momentum is 1.656, right? And then you should be able to just plug in...using 60 sec for time, right?
 
Correct, 1.7 kgm/s is the change in momentum of one bullet. Now just calculate how many bullets hit per second for the total change in momentum per second.
 
I got it! Thanks!
 

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