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How to derive \({\bf \vec{F}} = \dfrac{m(v_f - v_i)}{t}\) or \({\rm \vec{F}} \propto \dfrac{m(v_f - v_i)}{t}\), for that matter? Note, you do not have to use \({\bf \vec{F}} = m{\bf \vec{a}}\)

Physics
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  • DLS
\[\LARGE F=\frac{dp}{dt}\] Can we use this? :P
Good idea! ^_^ No.

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Other answers:

force is equal to change in momentum divided by change in time
  • DLS
@amistre64 Isn't that just what DLS said...
i was just seeing if i remembered how to interpret the question correctly :)
\[F = \dfrac{\Delta p}{\Delta t} = \dfrac{\Delta mv}{\Delta t}\]Is the above true? How did you derive it?
Ft = p = mv2-mv1
\[\vec F\Delta t=\Delta \vec p = m(\vec v_f-\vec v_i)\]
not sure what you mean by how you derive it.
Does not seem so convincing. :-|
do we have to first prove that momentum is a product of mass and velocity? i never know when i have to reinvent the wheel on these things
No, not that one. lol, got it
changing an objects momentum requires applying a force over a given interval of time, right?\[\vec F\Delta t=\Delta \vec p\]
I'd rather say...\[\vec F \Delta t \propto \Delta \vec p\]
that style went out with Descartes :)
I feel like a newbie here!
Can you give me a link to the derivation? I got how you mean \(\Delta \vec p =m(\vec v _f - \vec v_i)\) and there's only a little work left... but I didn't get how \(\vec F \Delta t = \Delta \vec p \).
  • DLS
You are indirectly asking for the proof of F=ma which only Isaac Newton knows.
  • DLS
I think @shubhamsrg can derive F=ma. Its a piece of cake for him.
Of course, of course!
an object remains at a constant velocity (change in velocity is acceleration) unless acted upon by a force. therefore in order to change an objects momentum, you would need to apply a constant force
over a given time interval, the mass "acquires" a new velocity, and therefore a change in momentum. does that sound about right?

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