Question

It is well known that bullets and other missiles fired at Superman simply bounce off his chest. Suppose that a gangster sprays Superman's chest with 6.5 g bullets at the rate of 170 bullets/min, and the speed of each bullet is 540 m/s. Suppose too that the bullets rebound straight back with no change in speed. What is the magnitude of the average force on Superman's chest from the stream of bullets

Answer 1

Answer:

$F = 76.05\,N$

Explanation:

The phenomenon can be described and analyzed by means of the Principle of Momentum Conservation and the Impact Theorem:

$(6.5\times 10^{-3}\,kg)\cdot (650\,\frac{bullets}{min} )\cdot (\frac{1\,min}{60\,s} )\cdot (540\,\frac{m}{s} ) - F = (6.5\times 10^{-3}\,kg)\cdot (650\,\frac{bullets}{min} )\cdot (\frac{1\,min}{60\,s} )\cdot (-540\,\frac{m}{s} )$

The magnitude of the average force on Superman's chest is:

$F = 76.05\,N$