Force due to Collision and Acceleration Problems

  • #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?



Homework Equations


The only equation I can find in my notes that doesn't ask for the m of the glove is
F=ma=m[(v2-v1)/t]
Then for velocity its v=delta x/delta t

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)
 

Answers and Replies

  • #2
16
0
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]
 
  • #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?
 
  • #4
16
0
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.
 
  • #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.... >.>
 
  • #6
ooh, okay... mini brainwave...

F= ma= m(delta v/delta t) when delta t= delta x/v
So:
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
 
  • #7
16
0
ok, I'm just gonna type in exactly what I wrote down on my scratch paper.

Given:
[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]

Solution:
[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]
 
  • #8
given:
m=140gm=.14kg
u=32m/s
v=0m/s
a=?
s=25cm=.25m
f=m*a


we will get the value of acceleration by the formula
v^2=u^2+2as


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

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