The discussion centers on the relationship between force, acceleration, and momentum, specifically addressing the equations F=ma and p=mv. Participants confirm that a net force acting on an object changes its momentum, with the correct expression being FΔt = Δp, which distinguishes between momentum (p) and change in momentum (Δp). The conversation also touches on the clarity of these concepts and the confusion surrounding the use of Ft = p.
PREREQUISITESStudents of physics, educators teaching mechanics, and anyone interested in the fundamental principles of motion and forces.
What exactly is your question? Yes, a net force on a body does change its momentum.-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!:)
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?Doc Al said:Yes, a net force on a body does change its momentum.
Yes, I already pointed that out yesterday: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.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?
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.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.