Question

Raindrops fall on brian's head at the rate of 4 drops per second. each raindrop has a mass of 1.6 mg and falls with a speed of 25 m/s. assuming that on making contact with brian's head the drops come to rest and do not rebound, calculate the force felt by brian.

Answer 1

This problem relates to impulse and momentum of an object, it is a property related to an object's motion and mass. The Given are Time t=1sec. Mass=4 x 1.6mg then convert to Kg m=6.4 x 10⁻³kg. Initial Velocity Vi=0 and Final Velocity Vf=25m/s. The equation to follow f=ma but a=Vf-Vi/t
F=m(Vf-Vi/t). Force should be in "Newton"

Answer 2

Because the raindrops do not rebound, the KE (kinetic energy) of each drop is dissipated upon impact.

Given:
m = 1.6 mg = 1.6 x 10⁻³ kg per drop
v = 25 m/s, the speed of each drop
4 drops fall per second.

By definition,
KE = (1/2)*m*v²

Therefore energy dissipated per second is

`E= (1)/(2)(1.6 * 10^(-3) \, (kg)/(drop) )*(4 \, (drops)/(s)) *(25 \, (m)/(s) )^(2) = 2 \, (J)/(s)`

Another way to look at the problem is in terms of momentum of impact.
Because there are 4 drops per second, the time per impact may be approximated as Δt = 1/4 = 0.25 s.
The momentum is
P = F*Δt = m*v
where F = the force of impact

Therefore
F = (m*v)/Δt
= [(1.6 x 10⁻³ kg)*(25 m/s)] / (0.25 s)
= 0.16 N

Answer: 0.16 N