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ParthKohli
 3 years 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 \({\bf \vec{F}} = m{\bf \vec{a}}\)
ParthKohli
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.1\[\LARGE F=\frac{dp}{dt}\] Can we use this? :P

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

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

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

ParthKohli
 3 years 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
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0not sure what you mean by how you derive it.

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

amistre64
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0No, not that one. lol, got it

amistre64
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0I'd rather say...\[\vec F \Delta t \propto \Delta \vec p\]

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

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

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

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

amistre64
 3 years 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
 3 years 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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