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Wednesday, October 6, 2010
Ch7 2 #3
4 comments:
Erica
said...
I'm not understanding this. Part A numbers are, m=2kg, v=1.3m/s, and total time=1.5s (0.5s was labeled on the the middle interval so I tripled it.) F=p*t and p=m*v so I plugged in: F=(2kg)*(1.3m/s)/1.5s= 1.7 I put in 1.7 for the answer and it was right even though it asked for velocity and what I calculated was force. I'm not even sure what the graph of Fx has to do with anything. Can someone please explain how to do part A and C?
For the actual question though, I can't seem to get the equation right. I thought the answer would be (mb * vi^2)/(mb + mw), however this turns out to be wrong.
4 comments:
I'm not understanding this.
Part A numbers are, m=2kg, v=1.3m/s, and total time=1.5s (0.5s was labeled on the the middle interval so I tripled it.)
F=p*t and p=m*v so I plugged in:
F=(2kg)*(1.3m/s)/1.5s= 1.7
I put in 1.7 for the answer and it was right even though it asked for velocity and what I calculated was force. I'm not even sure what the graph of Fx has to do with anything. Can someone please explain how to do part A and C?
Hi Erica,
I'm confused by which problem you are asking about. Is it really Ch7_1 #3 (which has a graph)?
Think of the force in the graph as an impulse, and use it to calculate the change in momentum as described in class on Monday.
For the actual question though, I can't seem to get the equation right. I thought the answer would be (mb * vi^2)/(mb + mw), however this turns out to be wrong.
The expression mb*vi^2/(mb+mw) has units of velocity squared and not velocity. It looks like you're close, but maybe have an algebra error somewhere?
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