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 6.4 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

Answer 1

Answer:

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