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 = 2.6 g at a rate of R = 140 bullets/min. The speed of each bullet is v = 510 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?
To find the average force exerted by the stream of bullets on Superman's chest, we can use the concept of impulse and momentum.
First, let's calculate the total momentum change of the bullets.
The momentum of each bullet is given by:
p = mv
Where m is the mass of the bullet and v is the velocity of the bullet.
In this case, m = 2.6 g = 2.6 × 10^(-3) kg (converting grams to kilograms) and v = 510 m/s.
So, the momentum of each bullet is:
p = (2.6 × 10^(-3)) kg × 510 m/s
Next, let's calculate the rate at which bullets are fired, given as 140 bullets per minute. We can convert this to bullets per second for convenience:
R = 140 bullets/min = 140/60 bullets/s ≈ 2.33 bullets/s
Now, let's calculate the total momentum change per second (rate of change of momentum) caused by the bullets:
Δp/Δt = p × R
The average force exerted by the stream of bullets on Superman's chest is equal to the rate of change of momentum. Therefore:
F = Δp/Δt
Substituting the values, we have:
F = (2.6 × 10^(-3) kg × 510 m/s) × 2.33 bullets/s
Calculating this expression will give us the average force exerted by the stream of bullets on Superman's chest.