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

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The discussion centers on calculating the average force exerted by bullets bouncing off Superman's chest, using Newton's laws. Given the bullet mass of 1.8 g, a firing rate of 200 bullets per minute, and a speed of 460 m/s, participants explore the change in momentum and force. The change in momentum for one bullet is confirmed to be 1.7 kgm/s. By calculating the total number of bullets impacting per second, the average force can be derived. The conversation concludes with a successful resolution of the problem.
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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?


THANK YOU!
 
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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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