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 6.4 g bullets at the rate of 92 bullets/min, and the speed of each bullet is 400 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

Magnitude of the average force = 7.85 N

Explanation:

Data given:

Mass of bullets, m = 6.4 g

Rate of bullets/min, r = 92 bullets/min

Speed of each bullet, v = 400 m/s

Change in momentum here, Δ B = Bf - Bi

where f is the final and i is the initial

Note that change in momentum = force X time

So, Δ B = m(vf - vi)

              = 2mv

             = 2 X 0.0064 kg X 400 m/s        (convert g to kg)

            = 5.12 kg.m/s (for one bullet)

        so for the 92 bullets = 92 X 5.12 kg.m/s

           = 471.04 kg.m/s

         The force = Δ B ÷ Δt

where t = time measured in 60 seconds

        =  471.04 kg.m/s ÷ 60 seconds

       = 7.85 N

wvbbvvv