# Force due to Collision and Acceleration Problems

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)

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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 $$a_{y}=-g$$ which means that acceleration taking place during a free fall is equal to negative gravity or $$-9.8 m/s^{2}$$

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?

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 $$m/s$$ and $$m/s^{2}$$ 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.

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.... >.>

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

ok, I'm just gonna type in exactly what I wrote down on my scratch paper.

Given:
$$\Delta Y = 2$$ m
$$v_{i} = 0$$ m/s
$$95 kg/s = rate$$
$$g = -9.8 m/s^{2}$$

Find: $$v_{f}$$

Solution:
$$v_{fy}^2 - v_{iy}^2 = -2g\Delta Y$$
$$v_{fy}^2 = -2g\Delta Y + v_{iy}^2$$
$$v_{fy}^2 = -2(-9.8)(2) + (0)^2$$
$$v_{fy} = \sqrt{-2(-9.8)(2)}$$
$$v_{fy} = 6.26 m/s$$

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.....