Understanding the Relationship Between Force, Acceleration, and Momentum

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We know that
##F=ma## or ##F=ma=m \frac{v}{t}##, so the acceleration is rate of slowing or speeding velocity, then does force changes the momentum ##p=mv## or ##p=Ft##.

I'm waiting an answer, thanks!:)
 
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-Physician said:
We know that
##F=ma## or ##F=ma=m \frac{v}{t}##, so the acceleration is rate of slowing or speeding velocity, then does force changes the momentum ##p=mv## or ##p=Ft##.

I'm waiting an answer, thanks!:)
What exactly is your question? Yes, a net force on a body does change its momentum.
 
Uhm, the answer was the answer that i requested, thanks!
 
Doc Al said:
Yes, a net force on a body does change its momentum.
Doc Al, the correct expression of this would be FΔt = Δp, no? Ft = p strikes me as imprecise, confusing Δp with p, which is mv. In other words, we use FT or FΔt to find change in momentum, and mv to find momentum. Correct?
 
i have another question , why we couldn't gather first law from second law ? and does 2nd law have any prove ?
 
zoobyshoe said:
Doc Al, the correct expression of this would be FΔt = Δp, no? Ft = p strikes me as imprecise, confusing Δp with p, which is mv. In other words, we use FT or FΔt to find change in momentum, and mv to find momentum. Correct?
Yes, I already pointed that out yesterday:
https://www.physicsforums.com/showthread.php?t=576484
 
zoobyshoe said:
Doc Al, the correct expression of this would be FΔt = Δp, no? Ft = p strikes me as imprecise, confusing Δp with p, which is mv. In other words, we use FT or FΔt to find change in momentum, and mv to find momentum. Correct?
Right. Sure, his statements were inaccurate. But first I wanted to find out what his question was, as it wasn't clear to me just what he was trying to do.
 
Doc Al said:
Right. Sure, his statements were inaccurate. But first I wanted to find out what his question was, as it wasn't clear to me just what he was trying to do.
Me either. I just wanted to clarify because I'm talking to another guy in another thread who seemed to think Ft = p, as opposed to Δp. Seeing it twice made me wonder if it was somehow acceptable in some way shape or form I wasn't aware of. Thanks.