1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Raindrops problem

  1. Oct 23, 2012 #1
    The problem is:
    Raindrops keep falling on your head at a rate of 4 drops per second. The raindrops each have a mass of 1.6 x 10^-6 kg and fall with a velocity of 25 m/s. Assuming that on making contact the drops come to a rest and do not rebound, calculate the perceived force each second.

    There are two approaches and I don't know why one is more correct than the other. The first is the solution I was given in the text. The second is another solution.

    1st:
    a=Δv/Δt=0-25/1=25 m/s^2.
    F(one drop)=ma=1.6x10^-6*-25m/s^2=4x10^-5 N.
    F(4 drops)=4*F(one drop)=1.6x10^-4 N.<-------------
    (along with considerations for negative signs, not shown)

    2nd:
    Since there are 4 drops per second, there is .25 seconds per drop.
    a=Δv/Δt=25/.25=100m/s^2.
    F(one drop)=1.6x10^-6*100m/s^2=1.6x10^-6 N.
    F(4 drops)=4*F(one drop)=6.4x10^-4 N.<---------------
    (again with directional considerations not shown)

    Is this simply a poorly worded problem, or is there some subtlety I'm missing? Thanks.
     
  2. jcsd
  3. Oct 23, 2012 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I don't like the way the explanation is worded, but it gets the right answer. In the first part it computes the force if the rate is one drop per second, then multiplies by 4. In your attempt you the factor of in earlier, but applied it a second time.
    My objection to the wording is that you cannot compute a force per drop. It depends how long the raindrop takes to come to rest. What you can compute is the impulse (change in momentum). You can then average that over time by multiplying by the drop rate, and this gives you an average force.
     
  4. Oct 23, 2012 #3
    So,

    Impulse (one drop) = change in momentum
    = pf - pi
    = 0 - mvi
    = 1.6x10^-6kg*25m/s=4x10^-4 N*s (per drop)

    Then, 4x10^-5 N*s/drop(4 drop/s) = 1.6x10^-4 N.

    That makes perfect sense to me. The reason for the confusion is that impulse has not been introduced in the course/text yet.

    EDIT: and thanks for the quick response.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Raindrops problem
  1. Raindrop Problem (Replies: 2)

  2. Speed of raindrops (Replies: 5)

  3. Raindrops in tiles (Replies: 3)

Loading...