Question

Superman leaps in front of Lois Lane to save her from a volley of bullets. In a 1-minute interval, an automatic weapon fires 147 bullets, each of mass 8.0 g, at 350 m/s. The bullets strike his mighty chest, which has an area of 0.74 m2. Find the average force exerted on Superman's chest if the bullets bounce back after an elastic, head-on collision.

Answer 1

Answer:

F/Area = 9.27 N / m^2

Explanation:

Given:

- mass of 1 bullet m = 0.008 kg

- number of bullets fired n = 147

- time interval dt = 60 s

- Initial velocity v_i = 350 m/s

- Area of chest A = 0.74 m^2

Find:

Average force imparted on the chest F

Solution:

The number of bullets fired per second:

- 147 / 60 s = 2.45 bullets / s

Momentum of single bullet:

- mass of one bullet * speed of bullet = 0.008 * 350 = 2.8 kg m/s.  

Total momentum per second:

- number of bullets fired per second * Momentum of single bullet

 = 2.8 * 2.45 = 6.86 kg m/s

Momentum = Force * time

- f = 6.86*1 = 6.86 N

Superman experiences a constant average force of 6.86 N for 1 minute

Assuming the bullets are spread out evenly over the chest:

- (Force / Area) = 6.86 / 0.74 = 9.27 N / m^2

So the average force experienced by superman per unit area is 9.27 N/m^2