Question

1. Anthrax spores can cause a very bad disease in humans and animals. The spores look like the inset micrograph. Although not quite spherical, when they are in the air they act very much like spheres with a diameter of about 0.9 µm (.9e-6 m) and density (when they are dried out) of 1.42 g/cubic centimeter (1,420 kg/m3), and a total mass of about 540 fg (femtograms) which in kilograms is about 5.42 e-16 kg. When the spores are falling in air, they reach a terminal velocity of about 4.05e-5 m/s at which time the air resistance (friction) on the spores equals the downward force of gravity. When someone infected sneezes, they may throw the spores straight out (in the x-direction) into the air with a good bit of speed, about 20 m/s. Assuming the air resistance is independent of velocity (i.e. the same for all speeds), and that its magnitude is equal to the gravitational force on the spore, (a) how much time will it take for the spores to lose all of their x-velocity? (b) How far will the spores travel during this time? (Note that the assumptions here imply that the acceleration due to air resistance is constant. This is not quite true but it makes the problem easier). (c) Assume now that there is a slight breeze blowing, about 2 m/s. If the spores fall with their terminal velocity from the height of the sneeze (about 2 m) during which time the air carries them along how long will it take them to reach the ground and how far will they travel?

Answer 1

Given:

acceleration due to air resistance, a = g, acceleration due to gravity ( because gravitational force and air resistance are equal here)

Initial velocity of the spores after sneeze, u = 20 m/s

Terminal velocity, V_t = 4.05e-5 m/s = 4.05×10^-5m/ss

Mass of the spore,m = 540 fg = 5.42×10^-16 kg

height of sneeze = 2m

a) In x-direction there is only one force acting i.e air resisitance. And it is constant. So we can use Newton's 1^st equation of motion

V= u + at

0 = 20+(-g)t

t = 20/9.8 g = 9.8 m/s²

t= 2.04 s

b) Here we can use both 2^nd or 3rd equation of motion. I am using 3rd .

V² = u² + 2as

0 = 20² + 2(-a)s

s = 400/(2×9.8)

= 20.40 m

c) Lets say that the breeze is blowing in x-direction as well. Its velocity is 2m/s.

The spore will come down with the terminal velocity V_t = 4.05×10^-5 m/s.

Time taken to reach groung, t = (height of sneeze) ÷ (terminal velocity)

t= 2/(4.05×10^-5)

= 49382.71 s

Initially the velocity of spore is greater than the velocity of breeze. So the breeze is not of any work for the spore till it matches with the velocity of breeze. But as we have seen in part (a) , the time to vanish its x- velocity is about 2s only, the time to reach the velocity of 2m/s will be less than 2 s.

So we can neglect this 2s motion of the spore, and can consider its x-velocity to be constant which is equal to the velocity of breeze.

Distance travelled S = (breeze velocity) × (time)

= 2× 49382 71

= 98765.43 m