Question

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 = 1.7 g at a rate of R = 165 bullets/min. The speed of each bullet is v = 380 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?

Answer 1

Answer:

1.77 N

Explanation:

Mass of bullet m = 1.7 x 10⁻³ kg

velocity v = 380 m /s .

Momentum of one bullet

= 1.7 x 10⁻³ x 380

= 646 x 10⁻³ kg m/s

momentum of 165 bullets

= 165 x 646 x 10⁻³ kg m/s

= 106.59 kg m/s

Final momentum after bouncing

= 0

change in momentum

= 106.59  - 0

= 106.59 kg m/s

This change occurs in one minute

so rate of change in momentum

= 106.59 / 60 kg m/s per second

= 1.77 kg m/s per second

rate of change in momentum = force

This is force on superman's chest

Force required = 1.77 N .