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 one year ago
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 \({\bf \vec{F}} = m{\bf \vec{a}}\)
 one year ago
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 \({\bf \vec{F}} = m{\bf \vec{a}}\)

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DLS
 one year ago
Best ResponseYou've already chosen the best response.1\[\LARGE F=\frac{dp}{dt}\] Can we use this? :P

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0force is equal to change in momentum divided by change in time

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0@amistre64 Isn't that just what DLS said...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0i was just seeing if i remembered how to interpret the question correctly :)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0\[F = \dfrac{\Delta p}{\Delta t} = \dfrac{\Delta mv}{\Delta t}\]Is the above true? How did you derive it?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0\[\vec F\Delta t=\Delta \vec p = m(\vec v_f\vec v_i)\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0not sure what you mean by how you derive it.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Does not seem so convincing. :

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0do 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

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0No, not that one. lol, got it

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0changing an objects momentum requires applying a force over a given interval of time, right?\[\vec F\Delta t=\Delta \vec p\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0I'd rather say...\[\vec F \Delta t \propto \Delta \vec p\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0that style went out with Descartes :)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0I feel like a newbie here!

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Can 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
 one year ago
Best ResponseYou've already chosen the best response.1You are indirectly asking for the proof of F=ma which only Isaac Newton knows.

DLS
 one year ago
Best ResponseYou've already chosen the best response.1I think @shubhamsrg can derive F=ma. Its a piece of cake for him.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Of course, of course!

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0an 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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0over 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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