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Force due to Collision and Acceleration Problems

  1. Oct 16, 2007 #1
    1. A baseball (m=140g) travelling 32 m/s moves a fielder's glove backward 25 cm when the ball is caught. What was the average force exerted by the ball on the glove?

    2. Relevant equations
    The only equation I can find in my notes that doesn't ask for the m of the glove is
    Then for velocity its v=delta x/delta t

    3. The attempt at a solution
    v=x/t -> t=x/v=.25m/32 m/s =0.0078s
    F=(140g)(0-32 m/s /0.0078s) = -574358.97
    which I'm pretty sure is wrong... yet I'm not sure where I went wrong, or if I was even on the right track in the first place...

    And random other question:
    If you drop an object, what's the formula to figure out its acceleration? I mean, it has something to do with the weight as well as g, but I don't think it's a=mg, is it? (considering W=mgh, it just doesn't seem right to me)
  2. jcsd
  3. Oct 16, 2007 #2
    I can answer your random question (i'll also take a look at the actual problem to see if I can contribute any insight) but [tex]a_{y}=-g[/tex] which means that acceleration taking place during a free fall is equal to negative gravity or [tex]-9.8 m/s^{2}[/tex]
  4. Oct 16, 2007 #3
    Or just 9.8 if I decide that down is my positive value and up is the negative... (my prof likes elevator questions for some reason, so we do a lot of positive downward directions)
    Anyway, then that means I did a couple of my other questions correctly (wonder of wonder)
    But... is it possible for velocity to equal acceleration?
  5. Oct 16, 2007 #4
    Now keep in mind that velocity and acceleration two different things. One measures the change in distance over a period of time and the other is the change in velocity over a period of time. Hence the units of measure being [tex]m/s[/tex] and [tex]m/s^{2}[/tex] respectively. So, to answer your question in a rather round about question, velocity and acceleration can have the same values but that does not mean they equal each other.
  6. Oct 16, 2007 #5
    Okay, I just have this question where:
    Water falls onto a water wheel from a height of 2.0 m at a rate of 95kg/s. (a) If
    this water wheel is set up to provide electricity output, what is its maximum
    power output? (b) What is the speed of the water as it hits the wheel?

    I got a fine, (P=W/t=mgh/t=(95kg)(9.8m/s^2)(2.0m)/1s= 1862 J/s) but its b that messes with me...

    I think I just have problems with speed and acceleration.... >.>
  7. Oct 16, 2007 #6
    ooh, okay... mini brainwave...

    F= ma= m(delta v/delta t) when delta t= delta x/v
    F= 0.14kg(0-(-32m/s)/0.0078s)= 574.36N

    Does that look better than my original -574358.97N? lol

    Still don't know how to do the water wheel part b question... :S
  8. Oct 16, 2007 #7
    ok, I'm just gonna type in exactly what I wrote down on my scratch paper.

    [tex]\Delta Y = 2[/tex] m
    [tex]v_{i} = 0[/tex] m/s
    [tex]95 kg/s = rate [/tex]
    [tex]g = -9.8 m/s^{2}[/tex]

    Find: [tex]v_{f}[/tex]

    [tex]v_{fy}^2 - v_{iy}^2 = -2g\Delta Y[/tex]
    [tex]v_{fy}^2 = -2g\Delta Y + v_{iy}^2[/tex]
    [tex]v_{fy}^2 = -2(-9.8)(2) + (0)^2[/tex]
    [tex]v_{fy} = \sqrt{-2(-9.8)(2)}[/tex]
    [tex]v_{fy} = 6.26 m/s [/tex]
  9. Aug 14, 2010 #8

    we will get the value of acceleration by the formula

    now calculate and simply put the values u get the answer.....
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