Law of conservation of momentum

[FarazAli]

I can't seem to figure out what the damn question is asking :D. Can anyone explain what the question is asking?

During a chicago storm, winds can whip horizontaly at speeds of 100 km/h. If the air strikes a person at the rate of 40kg/s per square meter and is brought to rest, calculate the force of the wind on a person. Assume the person's area to be 1.50m high and 0.50m wide. Compare to the typical maximum force of friction (μ = 1.0) between the person and the ground, if the person has a mass of 70kg.

What comes to rest here? is it the air or the person?

Also on a side note, can anyone explain whether the transfer of force during impulse instantaneous or not?

[Galileo]

If the air strikes a person at the rate of 40kg/s per square meter and is brought to rest, calculate the force of the wind on a person.

What comes to rest here? is it the air or the person?
The air is. Consider the person standing still. The rate of change in momentum of the air is equal and opposite to the force on the person.

[FarazAli]

Ok this is how I solved the problem. I doubt it's right.

I see that the force is in units of kg/sm^2 and I have an area in square meters, so I multiplied the two of them to get this
(\frac{40 kg}{m^2s})(1.0m)(0.5m) = 30 \frac{kg}{s}

I plugged that into \Delta{p} = F\Delta{t} = (\frac{m\Delta{v}}{\Delta{t}} \times \Delta{t}) = m\Delta{v} and got \frac{30kg}{s} \times \frac{27.8m}{s} = 8.3\times10^2N

[thermodynamicaldude]

looks good to me. :-D

[Galileo]

That's correct. In 1 second 30 kg of air having a speed of 27.8 m/s comes to rest.
So the change in momentum (which equals the force) is 30 \cdot 27.8 = 8.3 \cdot 10^2N