Superman and the Kinetic Theory of Gases

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SUMMARY

The discussion revolves around calculating the average force exerted on Superman's chest when he intercepts bullets using the Kinetic Theory of Gases. The problem states that 150 bullets, each with a mass of 8.0g and traveling at 400 m/s, strike Superman's chest with an area of 0.75 m². After applying the Kinetic Theory of Gases equation, the calculated pressure is 8.53 x 107 Pa, but the textbook answer indicates a force of 16N. The user seeks clarification on the volume calculation and the relationship between work and pressure in this context.

PREREQUISITES
  • Understanding of Kinetic Theory of Gases, specifically the equation P = (2/3)(N/V)(0.5mv²)
  • Basic principles of momentum and elastic collisions
  • Knowledge of pressure calculations in physics
  • Familiarity with algebra-based physics concepts
NEXT STEPS
  • Review the derivation and application of the Kinetic Theory of Gases equations
  • Study momentum conservation in elastic collisions
  • Explore the relationship between work, pressure, and force in physics
  • Investigate alternative methods for solving problems involving kinetic energy and momentum
USEFUL FOR

This discussion is beneficial for physics students, particularly those studying mechanics and thermal physics, as well as educators seeking to clarify concepts related to momentum and pressure in elastic collisions.

science.girl
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Homework Statement



Superman leaps in front of Lois Lane to save her from a volley of bullets. In a 1-minute interval, and automatic weapon fires 150 bullets, each of mass 8.0g, at 400 m/s. The bullets strike his mighty chest, which has an area of 0.75 m2. Find the average force exerted on Superman's chest if the bullets bounce back after an elastic, head-on collision.


Homework Equations



Kinetic Theory of Gases: P = \frac{2}{3}(\frac{N}{V})(.5mv^2)

where .5mv^2 is the average kinetic energy per molecule.


The Attempt at a Solution


After substituting values into the Kinetic Theory of Gases:

P = \frac{2}{3}(\frac{N}{V})(.5mv^2)

P = (2/3)(150 bullets/V)(.598.0g)(400m/s)^2

If V = 0.75m^2 * 1m = .75m^3, then P = 8.53 * 10^7 Pa.

However, the answer in the textbook is 16N, I'm just not sure if:
1. V (described above) is accurate.
2. There is an equation to relate work and pressure. (Note: This is a algebra-based course.)
 
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This question can just as easily be solved by considering it kinematically as it can through a kinetic theory of gases perspective. Anyways, basically the jist of it is that, for each bullet, superman has to change its momentum by 2mv (mv to stop is and another mv to get it going back in the opposite direction at the same speed). Over the course of a minute he does this 150 times. so the total force divided by the total time
 
maverick_starstrider said:
This question can just as easily be solved by considering it kinematically as it can through a kinetic theory of gases perspective. Anyways, basically the jist of it is that, for each bullet, superman has to change its momentum by 2mv (mv to stop is and another mv to get it going back in the opposite direction at the same speed). Over the course of a minute he does this 150 times. so the total force divided by the total time

Well, I thought about solving it this way, but my professor wants me to relate these problems to thermal physics :(

Any help?
 

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