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Speed of catching a falling ruler

  1. Apr 15, 2010 #1
    1. The problem statement, all variables and given/known data

    Okay, I've been set a piece of coursework to compare the reaction times of catching a ruler, with, or without a certain variable. (I've chosen either eye.)
    I'm having trouble working out the exact amount of time it takes the person to catch the ruler. (Yeah, I have attempted it. :) )

    2. Relevant equations

    (Explained in section 3.)

    3. The attempt at a solution

    The ruler weighs 0.025kg.

    I hold the ruler at 1.04m. (An average of 10 times holding the ruler at a neutral position)

    To calculate the gravitational potential energy I do:

    0.025 x 1.04 x 10

    This gives me = 0.26 (Joules)

    The ruler falls 12 cm, (the person caught it at 18cm)

    So 0.025 x 0.12 x 10 = the kinetic energy = 0.03

    The formula for speed is :

    Speed squared = kinetic energy/(0.5 times the weight)

    0.03 divided by 0.0125 = 2.4

    The square root of 2.4 = 1.549193338

    To two decimal places = 1.55m/s

    And if the ruler travelled 12 cm at 1.55m/s

    0.12/1.55 = 0.0774 seconds……

    I must have gone wrong somewhere...because I swear its 'physically' (pun intended) impossible for them to catch it that quickly?
  2. jcsd
  3. Apr 15, 2010 #2
    If the ruler falls 12cm you find the time taken using
    S=ut + ½gt²
    s=12cm, that is 0.12m
    and u=0 if it started from rest.
    The reaction time will be between 1 and 2 tenths of a second, which seems about right.
  4. Apr 15, 2010 #3
    Sorry, would you mind using words to write it out? This is only GCSE so i'm not too good with all the proper letters and stuff :(
  5. Apr 15, 2010 #4
    so, you are using work- kinetic energy theorem. I'm trying to understand your problem, if anywhere I go wrong then let me know by posting here.

    The ruler weighs 0.025kg.

    so m = 0.025 kg

    I hold the ruler at 1.04m.

    so h = 1.04 m

    I'm unable to understand this sentence and your solution from here. And most probably you are wrong from here. Also remember that potential energy is calculated from the bottom. I mean height is calculated from the ground level. so if the body falls 12 cm then potential energy will be mg (h -12). Secondly, loss in potential energy is equal to gain in kinetic energy.
  6. Apr 15, 2010 #5
    Ahhh sorry. The way the test works is that I hold the '30cm' mark directly above their fingers, so its almost touching. So the person's fingers are on '18cm' on the ruler.
  7. Apr 15, 2010 #6
    And as you might have seen. I don't actually use the potential energy at all, I'm just showing that I know how to, to gain extra marks.
  8. Apr 15, 2010 #7
    Have you done the equations of uniform motion at GCSE yet? See below.
    If you are trying to use kinetic and potential energy considerations you will not get the time directly.

    There is a set of 4 equations you can use to solve problems of objects accelerating when you know some of these:
    s=distance travelled
    t=time taken
    u=initial velocity
    v=final velocity
    I mentioned one of them in my post. This was the one to use in this case. g is acceleration due to gravity, =9.8m/s/s
  9. Apr 15, 2010 #8
    okay, then potential energy at the point where you stop the scale is

    = 0.025 * 10 * 0.92

    = 0.23 Joule

    Now loss in energy = 0.26 - 0.23

    =0.03 Joule

    This much energy is converted into kinetic energy

    0.03 = (1/2) 0.025. v^2

    v = sqrt of (2 * 0.03)/0.023

    Now try to calculate. But remember don't trust blindly on me cause I'm noob like you in physics.
  10. Apr 15, 2010 #9
    on second line 0.92 is because the height of the scale from the ground is (1.04 - 0.12) m
  11. Apr 15, 2010 #10


    Staff: Mentor

    The weight or mass and energy don't enter into this problem at all. Basically all you're doing is measuring how far the ruler falls before the other person catches it. From this distance you can calculate the person's reaction time. The governing equation is s = (1/2)gt2, with g ~ 9.8 m/sec2.
  12. Apr 15, 2010 #11
    Guys I'm seriously confused now :(
    So is what I'm doing in my original post....correct? Or at least somewhat? I don't need to go into detail about drag or anything for GCSE do i...?
  13. Apr 15, 2010 #12
    What I'm trying to say is....have I just gone about it in a different way? Could someone use the g ~ 9.8m/sec^2 thing for my example in my first post and see if it gets the same outcome?
  14. Apr 15, 2010 #13
    so I should continue with my last post.

    after calculation, I've got v = 1.61 m/s

    Now v = u + a.t

    1.6 = 0 + 10 t

    t = 1.6/10

    t = 0.16 second.

    Or you can also get this result by using equation

    v^2 = u^2 + 2.a.s

    v^2 = 2*10*0.12

    then apply,

    v = u + a.t

    Or simply you can use as told by most of here,

    s = u*t + (1/2) a* t^2

    => s = (1/2)a * t^2

    => 0.12 = (1/2) 10 * t^2

    => t = 0.154 second
    Last edited: Apr 15, 2010
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